cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A005408 The odd numbers: a(n) = 2*n + 1.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131
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Comments

Leibniz's series: Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) (cf. A072172).
Beginning of the ordering of the natural numbers used in Sharkovski's theorem - see the Cielsielski-Pogoda paper.
The Sharkovski ordering begins with the odd numbers >= 3, then twice these numbers, then 4 times them, then 8 times them, etc., ending with the powers of 2 in decreasing order, ending with 2^0 = 1.
Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(6).
Also continued fraction for coth(1) (A073747 is decimal expansion). - Rick L. Shepherd, Aug 07 2002
a(1) = 1; a(n) is the smallest number such that a(n) + a(i) is composite for all i = 1 to n-1. - Amarnath Murthy, Jul 14 2003
Smallest number greater than n, not a multiple of n, but containing it in binary representation. - Reinhard Zumkeller, Oct 06 2003
Numbers n such that phi(2n) = phi(n), where phi is Euler's totient (A000010). - Lekraj Beedassy, Aug 27 2004
Pi*sqrt(2)/4 = Sum_{n>=0} (-1)^floor(n/2)/(2n+1) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 ... [since periodic f(x)=x over -Pi < x < Pi = 2(sin(x)/1 - sin(2x)/2 + sin(3x)/3 - ...) using x = Pi/4 (Maor)]. - Gerald McGarvey, Feb 04 2005
For n > 1, numbers having 2 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005
a(n) = shortest side a of all integer-sided triangles with sides a <= b <= c and inradius n >= 1.
First differences of squares (A000290). - Lekraj Beedassy, Jul 15 2006
The odd numbers are the solution to the simplest recursion arising when assuming that the algorithm "merge sort" could merge in constant unit time, i.e., T(1):= 1, T(n):= T(floor(n/2)) + T(ceiling(n/2)) + 1. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 14 2006
2n-5 counts the permutations in S_n which have zero occurrences of the pattern 312 and one occurrence of the pattern 123. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
For n > 0: number of divisors of (n-1)th power of any squarefree semiprime: a(n) = A000005(A001248(k)^(n-1)); a(n) = A000005(A000302(n-1)) = A000005(A001019(n-1)) = A000005(A009969(n-1)) = A000005(A087752(n-1)). - Reinhard Zumkeller, Mar 04 2007
For n > 2, a(n-1) is the least integer not the sum of < n n-gonal numbers (0 allowed). - Jonathan Sondow, Jul 01 2007
A134451(a(n)) = abs(A134452(a(n))) = 1; union of A134453 and A134454. - Reinhard Zumkeller, Oct 27 2007
Numbers n such that sigma(2n) = 3*sigma(n). - Farideh Firoozbakht, Feb 26 2008
a(n) = A139391(A016825(n)) = A006370(A016825(n)). - Reinhard Zumkeller, Apr 17 2008
Number of divisors of 4^(n-1) for n > 0. - J. Lowell, Aug 30 2008
Equals INVERT transform of A078050 (signed - cf. comments); and row sums of triangle A144106. - Gary W. Adamson, Sep 11 2008
Odd numbers(n) = 2*n+1 = square pyramidal number(3*n+1) / triangular number(3*n+1). - Pierre CAMI, Sep 27 2008
A000035(a(n))=1, A059841(a(n))=0. - Reinhard Zumkeller, Sep 29 2008
Multiplicative closure of A065091. - Reinhard Zumkeller, Oct 14 2008
a(n) is also the maximum number of triangles that n+2 points in the same plane can determine. 3 points determine max 1 triangle; 4 points can give 3 triangles; 5 points can give 5; 6 points can give 7 etc. - Carmine Suriano, Jun 08 2009
Binomial transform of A130706, inverse binomial transform of A001787(without the initial 0). - Philippe Deléham, Sep 17 2009
Also the 3-rough numbers: positive integers that have no prime factors less than 3. - Michael B. Porter, Oct 08 2009
Or n without 2 as prime factor. - Juri-Stepan Gerasimov, Nov 19 2009
Given an L(2,1) labeling l of a graph G, let k be the maximum label assigned by l. The minimum k possible over all L(2,1) labelings of G is denoted by lambda(G). For n > 0, this sequence gives lambda(K_{n+1}) where K_{n+1} is the complete graph on n+1 vertices. - K.V.Iyer, Dec 19 2009
A176271 = odd numbers seen as a triangle read by rows: a(n) = A176271(A002024(n+1), A002260(n+1)). - Reinhard Zumkeller, Apr 13 2010
For n >= 1, a(n-1) = numbers k such that arithmetic mean of the first k positive integers is an integer. A040001(a(n-1)) = 1. See A145051 and A040001. - Jaroslav Krizek, May 28 2010
Union of A179084 and A179085. - Reinhard Zumkeller, Jun 28 2010
For n>0, continued fraction [1,1,n] = (n+1)/a(n); e.g., [1,1,7] = 8/15. - Gary W. Adamson, Jul 15 2010
Numbers that are the sum of two sequential integers. - Dominick Cancilla, Aug 09 2010
Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n - h)/4 (h and n in A000027), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 4). Also a(n)^2 - 1 == 0 (mod 8). - Bruno Berselli, Nov 17 2010
A004767 = a(a(n)). - Reinhard Zumkeller, Jun 27 2011
A001227(a(n)) = A000005(a(n)); A048272(a(n)) < 0. - Reinhard Zumkeller, Jan 21 2012
a(n) is the minimum number of tosses of a fair coin needed so that the probability of more than n heads is at least 1/2. In fact, Sum_{k=n+1..2n+1} Pr(k heads|2n+1 tosses) = 1/2. - Dennis P. Walsh, Apr 04 2012
A007814(a(n)) = 0; A037227(a(n)) = 1. - Reinhard Zumkeller, Jun 30 2012
1/N (i.e., 1/1, 1/2, 1/3, ...) = Sum_{j=1,3,5,...,infinity} k^j, where k is the infinite set of constants 1/exp.ArcSinh(N/2) = convergents to barover(N). The convergent to barover(1) or [1,1,1,...] = 1/phi = 0.6180339..., whereas c.f. barover(2) converges to 0.414213..., and so on. Thus, with k = 1/phi we obtain 1 = k^1 + k^3 + k^5 + ..., and with k = 0.414213... = (sqrt(2) - 1) we get 1/2 = k^1 + k^3 + k^5 + .... Likewise, with the convergent to barover(3) = 0.302775... = k, we get 1/3 = k^1 + k^3 + k^5 + ..., etc. - Gary W. Adamson, Jul 01 2012
Conjecture on primes with one coach (A216371) relating to the odd integers: iff an integer is in A216371 (primes with one coach either of the form 4q-1 or 4q+1, (q > 0)); the top row of its coach is composed of a permutation of the first q odd integers. Example: prime 19 (q = 5), has 5 terms in each row of its coach: 19: [1, 9, 5, 7, 3] ... [1, 1, 1, 2, 4]. This is interpreted: (19 - 1) = (2^1 * 9), (19 - 9) = (2^1 * 5), (19 - 5) = (2^1 - 7), (19 - 7) = (2^2 * 3), (19 - 3) = (2^4 * 1). - Gary W. Adamson, Sep 09 2012
A005408 is the numerator 2n-1 of the term (1/m^2 - 1/n^2) = (2n-1)/(mn)^2, n = m+1, m > 0 in the Rydberg formula, while A035287 is the denominator (mn)^2. So the quotient a(A005408)/a(A035287) simulates the Hydrogen spectral series of all hydrogen-like elements. - Freimut Marschner, Aug 10 2013
This sequence has unique factorization. The primitive elements are the odd primes (A065091). (Each term of the sequence can be expressed as a product of terms of the sequence. Primitive elements have only the trivial factorization. If the products of terms of the sequence are always in the sequence, and there is a unique factorization of each element into primitive elements, we say that the sequence has unique factorization. So, e.g., the composite numbers do not have unique factorization, because for example 36 = 4*9 = 6*6 has two distinct factorizations.) - Franklin T. Adams-Watters, Sep 28 2013
These are also numbers k such that (k^k+1)/(k+1) is an integer. - Derek Orr, May 22 2014
a(n-1) gives the number of distinct sums in the direct sum {1,2,3,..,n} + {1,2,3,..,n}. For example, {1} + {1} has only one possible sum so a(0) = 1. {1,2} + {1,2} has three distinct possible sums {2,3,4} so a(1) = 3. {1,2,3} + {1,2,3} has 5 distinct possible sums {2,3,4,5,6} so a(2) = 5. - Derek Orr, Nov 22 2014
The number of partitions of 4*n into at most 2 parts. - Colin Barker, Mar 31 2015
a(n) is representable as a sum of two but no fewer consecutive nonnegative integers, e.g., 1 = 0 + 1, 3 = 1 + 2, 5 = 2 + 3, etc. (see A138591). - Martin Renner, Mar 14 2016
Unique solution a( ) of the complementary equation a(n) = a(n-1)^2 - a(n-2)*b(n-1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences. - Clark Kimberling, Nov 21 2017
Also the number of maximal and maximum cliques in the n-centipede graph. - Eric W. Weisstein, Dec 01 2017
Lexicographically earliest sequence of distinct positive integers such that the average of any number of consecutive terms is always an integer. (For opposite property see A042963.) - Ivan Neretin, Dec 21 2017
Maximum number of non-intersecting line segments between vertices of a convex (n+2)-gon. - Christoph B. Kassir, Oct 21 2022
a(n) is the number of parking functions of size n+1 avoiding the patterns 123, 132, and 231. - Lara Pudwell, Apr 10 2023

Examples

			G.f. = q + 3*q^3 + 5*q^5 + 7*q^7 + 9*q^9 + 11*q^11 + 13*q^13 + 15*q^15 + ...
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 28.
  • T. Dantzig, The Language of Science, 4th Edition (1954) page 276.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 73.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.1 Terminology, p. 264.
  • D. Hök, Parvisa mönster i permutationer [Swedish], (2007).
  • E. Maor, Trigonometric Delights, Princeton University Press, NJ, 1998, pp. 203-205.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

See A120062 for sequences related to integer-sided triangles with integer inradius n.
Cf. A001651 (n=1 or 2 mod 3), A047209 (n=1 or 4 mod 5).
Cf. A003558, A216371, A179480 (relating to the Coach theorem).
Cf. A000754 (boustrophedon transform).

Programs

Formula

a(n) = 2*n + 1. a(-1 - n) = -a(n). a(n+1) = a(n) + 2.
G.f.: (1 + x) / (1 - x)^2.
E.g.f.: (1 + 2*x) * exp(x).
G.f. with interpolated zeros: (x^3+x)/((1-x)^2 * (1+x)^2); e.g.f. with interpolated zeros: x*(exp(x)+exp(-x))/2. - Geoffrey Critzer, Aug 25 2012
a(n) = L(n,-2)*(-1)^n, where L is defined as in A108299. - Reinhard Zumkeller, Jun 01 2005
Euler transform of length 2 sequence [3, -1]. - Michael Somos, Mar 30 2007
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = v * (1 + 2*u) * (1 - 2*u + 16*v) - (u - 4*v)^2 * (1 + 2*u + 2*u^2). - Michael Somos, Mar 30 2007
a(n) = b(2*n + 1) where b(n) = n if n is odd is multiplicative. [This seems to say that A000027 is multiplicative? - R. J. Mathar, Sep 23 2011]
From Hieronymus Fischer, May 25 2007: (Start)
a(n) = (n+1)^2 - n^2.
G.f. g(x) = Sum_{k>=0} x^floor(sqrt(k)) = Sum_{k>=0} x^A000196(k). (End)
a(0) = 1, a(1) = 3, a(n) = 2*a(n-1) - a(n-2). - Jaume Oliver Lafont, May 07 2008
a(n) = A000330(A016777(n))/A000217(A016777(n)). - Pierre CAMI, Sep 27 2008
a(n) = A034856(n+1) - A000217(n) = A005843(n) + A000124(n) - A000217(n) = A005843(n) + 1. - Jaroslav Krizek, Sep 05 2009
a(n) = (n - 1) + n (sum of two sequential integers). - Dominick Cancilla, Aug 09 2010
a(n) = 4*A000217(n)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. - Bruno Berselli, Nov 17 2010
n*a(2n+1)^2+1 = (n+1)*a(2n)^2; e.g., 3*15^2+1 = 4*13^2. - Charlie Marion, Dec 31 2010
arctanh(x) = Sum_{n>=0} x^(2n+1)/a(n). - R. J. Mathar, Sep 23 2011
a(n) = det(f(i-j+1))A113311(n);%20for%20n%20%3C%200%20we%20have%20f(n)=0.%20-%20_Mircea%20Merca">{1<=i,j<=n}, where f(n) = A113311(n); for n < 0 we have f(n)=0. - _Mircea Merca, Jun 23 2012
G.f.: Q(0), where Q(k) = 1 + 2*(k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
a(n) = floor(sqrt(2*A000384(n+1))). - Ivan N. Ianakiev, Jun 17 2013
a(n) = 3*A000330(n)/A000217(n), n > 0. - Ivan N. Ianakiev, Jul 12 2013
a(n) = Product_{k=1..2*n} 2*sin(Pi*k/(2*n+1)) = Product_{k=1..n} (2*sin(Pi*k/(2*n+1)))^2, n >= 0 (undefined product = 1). See an Oct 09 2013 formula contribution in A000027 with a reference. - Wolfdieter Lang, Oct 10 2013
Noting that as n -> infinity, sqrt(n^2 + n) -> n + 1/2, let f(n) = n + 1/2 - sqrt(n^2 + n). Then for n > 0, a(n) = round(1/f(n))/4. - Richard R. Forberg, Feb 16 2014
a(n) = Sum_{k=0..n+1} binomial(2*n+1,2*k)*4^(k)*bernoulli(2*k). - Vladimir Kruchinin, Feb 24 2015
a(n) = Sum_{k=0..n} binomial(6*n+3, 6*k)*Bernoulli(6*k). - Michel Marcus, Jan 11 2016
a(n) = A000225(n+1) - A005803(n+1). - Miquel Cerda, Nov 25 2016
O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient function A000010. - Peter Bala, Mar 22 2019
Sum_{n>=0} 1/a(n)^2 = Pi^2/8 = A111003. - Bernard Schott, Dec 10 2020
Sum_{n >= 1} (-1)^n/(a(n)*a(n+1)) = Pi/4 - 1/2 = 1/(3 + (1*3)/(4 + (3*5)/(4 + ... + (4*n^2 - 1)/(4 + ... )))). Cf. A016754. - Peter Bala, Mar 28 2024
a(n) = A055112(n)/oblong(n) = A193218(n+1)/Hex number(n). Compare to the Sep 27 2008 comment by Pierre CAMI. - Klaus Purath, Apr 23 2024
a(k*m) = k*a(m) - (k-1). - Ya-Ping Lu, Jun 25 2024
a(n) = A000217(a(n))/n for n > 0. - Stefano Spezia, Feb 15 2025

Extensions

Incorrect comment and example removed by Joerg Arndt, Mar 11 2010
Peripheral comments deleted by N. J. A. Sloane, May 09 2022

A000578 The cubes: a(n) = n^3.

Original entry on oeis.org

0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507
Offset: 0

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Comments

a(n) is the sum of the next n odd numbers; i.e., group the odd numbers so that the n-th group contains n elements like this: (1), (3, 5), (7, 9, 11), (13, 15, 17, 19), (21, 23, 25, 27, 29), ...; then each group sum = n^3 = a(n). Also the median of each group = n^2 = mean. As the sum of first n odd numbers is n^2 this gives another proof of the fact that the n-th partial sum = (n(n + 1)/2)^2. - Amarnath Murthy, Sep 14 2002
Total number of triangles resulting from criss-crossing cevians within a triangle so that two of its sides are each n-partitioned. - Lekraj Beedassy, Jun 02 2004. See Propp and Propp-Gubin for a proof.
Also structured triakis tetrahedral numbers (vertex structure 7) (cf. A100175 = alternate vertex); structured tetragonal prism numbers (vertex structure 7) (cf. A100177 = structured prisms); structured hexagonal diamond numbers (vertex structure 7) (cf. A100178 = alternate vertex; A000447 = structured diamonds); and structured trigonal anti-diamond numbers (vertex structure 7) (cf. A100188 = structured anti-diamonds). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {4, 3}.
Least multiple of n such that every partial sum is a square. - Amarnath Murthy, Sep 09 2005
Draw a regular hexagon. Construct points on each side of the hexagon such that these points divide each side into equally sized segments (i.e., a midpoint on each side or two points on each side placed to divide each side into three equally sized segments or so on), do the same construction for every side of the hexagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least one side of the polygon. The result is a triangular tiling of the hexagon and the creation of a number of smaller regular hexagons. The equation gives the total number of regular hexagons found where n = the number of points drawn + 1. For example, if 1 point is drawn on each side then n = 1 + 1 = 2 and a(n) = 2^3 = 8 so there are 8 regular hexagons in total. If 2 points are drawn on each side then n = 2 + 1 = 3 and a(n) = 3^3 = 27 so there are 27 regular hexagons in total. - Noah Priluck (npriluck(AT)gmail.com), May 02 2007
The solutions of the Diophantine equation: (X/Y)^2 - X*Y = 0 are of the form: (n^3, n) with n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^2k - XY = 0 are of the form: (m*n^(2k + 1), m*n^(2k - 1)) with m >= 1, k >= 1 and n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^(2k + 1) - XY = 0 are of the form: (m*n^(k + 1), m*n^k) with m >= 1, k >= 1 and n >= 1. - Mohamed Bouhamida, Oct 04 2007
Except for the first two terms, the sequence corresponds to the Wiener indices of C_{2n} i.e., the cycle on 2n vertices (n > 1). - K.V.Iyer, Mar 16 2009
Totally multiplicative sequence with a(p) = p^3 for prime p. - Jaroslav Krizek, Nov 01 2009
Sums of rows of the triangle in A176271, n > 0. - Reinhard Zumkeller, Apr 13 2010
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Numbers n for which order of torsion subgroup t of the elliptic curve y^2 = x^3 - n is t = 2. - Artur Jasinski, Jun 30 2010
The sequence with the lengths of the Pisano periods mod k is 1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, ... for k >= 1, apparently multiplicative and derived from A000027 by dividing every ninth term through 3. Cubic variant of A186646. - R. J. Mathar, Mar 10 2011
The number of atoms in a bcc (body-centered cubic) rhombic hexahedron with n atoms along one edge is n^3 (T. P. Martin, Shells of atoms, eq. (8)). - Brigitte Stepanov, Jul 02 2011
The inverse binomial transform yields the (finite) 0, 1, 6, 6 (third row in A019538 and A131689). - R. J. Mathar, Jan 16 2013
Twice the area of a triangle with vertices at (0, 0), (t(n - 1), t(n)), and (t(n), t(n - 1)), where t = A000217 are triangular numbers. - J. M. Bergot, Jun 25 2013
If n > 0 is not congruent to 5 (mod 6) then A010888(a(n)) divides a(n). - Ivan N. Ianakiev, Oct 16 2013
For n > 2, a(n) = twice the area of a triangle with vertices at points (binomial(n,3),binomial(n+2,3)), (binomial(n+1,3),binomial(n+1,3)), and (binomial(n+2,3),binomial(n,3)). - J. M. Bergot, Jun 14 2014
Determinants of the spiral knots S(4,k,(1,1,-1)). a(k) = det(S(4,k,(1,1,-1))). - Ryan Stees, Dec 14 2014
One of the oldest-known examples of this sequence is shown in the Senkereh tablet, BM 92698, which displays the first 32 terms in cuneiform. - Charles R Greathouse IV, Jan 21 2015
From Bui Quang Tuan, Mar 31 2015: (Start)
We construct a number triangle from the integers 1, 2, 3, ... 2*n-1 as follows. The first column contains all the integers 1, 2, 3, ... 2*n-1. Each succeeding column is the same as the previous column but without the first and last items. The last column contains only n. The sum of all the numbers in the triangle is n^3.
Here is the example for n = 4, where 1 + 2*2 + 3*3 + 4*4 + 3*5 + 2*6 + 7 = 64 = a(4):
1
2 2
3 3 3
4 4 4 4
5 5 5
6 6
7
(End)
For n > 0, a(n) is the number of compositions of n+11 into n parts avoiding parts 2 and 3. - Milan Janjic, Jan 07 2016
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
Number of inequivalent face colorings of the cube using at most n colors such that each color appears at least twice. - David Nacin, Feb 22 2017
Consider A = {a,b,c} a set with three distinct members. The number of subsets of A is 8, including {a,b,c} and the empty set. The number of subsets from each of those 8 subsets is 27. If the number of such iterations is n, then the total number of subsets is a(n-1). - Gregory L. Simay, Jul 27 2018
By Fermat's Last Theorem, these are the integers of the form x^k with the least possible value of k such that x^k = y^k + z^k never has a solution in positive integers x, y, z for that k. - Felix Fröhlich, Jul 27 2018

Examples

			For k=3, b(3) = 2 b(2) - b(1) = 4-1 = 3, so det(S(4,3,(1,1,-1))) = 3*3^2 = 27.
For n=3, a(3) = 3 + (3*0^2 + 3*0 + 3*1^2 + 3*1 + 3*2^2 + 3*2) = 27. - _Patrick J. McNab_, Mar 28 2016
		

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 191.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 43, 64, 81.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 292.
  • T. Aaron Gulliver, "Sequences from cubes of integers", International Mathematical Journal, 4 (2003), no. 5, 439 - 445. See http://www.m-hikari.com/z2003.html for information about this journal. [I expanded the reference to make this easier to find. - N. J. A. Sloane, Feb 18 2019]
  • J. Propp and A. Propp-Gubin, "Counting Triangles in Triangles", Pi Mu Epsilon Journal (to appear).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 6-7.
  • D. Wells, You Are A Mathematician, pp. 238-241, Penguin Books 1995.

Crossrefs

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
For sums of cubes, cf. A000537 (partial sums), A003072, A003325, A024166, A024670, A101102 (fifth partial sums).
Cf. A001158 (inverse Möbius transform), A007412 (complement), A030078(n) (cubes of primes), A048766, A058645 (binomial transform), A065876, A101094, A101097.
Subsequence of A145784.
Cf. A260260 (comment). - Bruno Berselli, Jul 22 2015
Cf. A000292 (tetrahedral numbers), A005900 (octahedral numbers), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Cf. A098737 (main diagonal).

Programs

  • Haskell
    a000578 = (^ 3)
    a000578_list = 0 : 1 : 8 : zipWith (+)
       (map (+ 6) a000578_list)
       (map (* 3) $ tail $ zipWith (-) (tail a000578_list) a000578_list)
    -- Reinhard Zumkeller, Sep 05 2015, May 24 2012, Oct 22 2011
    
  • Magma
    [ n^3 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 14 2014
    
  • Magma
    I:=[0,1,8,27]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jul 05 2014
    
  • Maple
    A000578 := n->n^3;
    seq(A000578(n), n=0..50);
    isA000578 := proc(r)
        local p;
        if r = 0 or r =1 then
            true;
        else
            for p in ifactors(r)[2] do
                if op(2, p) mod 3 <> 0 then
                    return false;
                end if;
            end do:
            true ;
        end if;
    end proc: # R. J. Mathar, Oct 08 2013
  • Mathematica
    Table[n^3, {n, 0, 30}] (* Stefan Steinerberger, Apr 01 2006 *)
    CoefficientList[Series[x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jul 05 2014 *)
    Accumulate[Table[3n^2+3n+1,{n,0,20}]] (* or *) LinearRecurrence[{4,-6,4,-1},{1,8,27,64},20](* Harvey P. Dale, Aug 18 2018 *)
  • Maxima
    A000578(n):=n^3$
    makelist(A000578(n),n,0,30); /* Martin Ettl, Nov 03 2012 */
    
  • PARI
    A000578(n)=n^3 \\ M. F. Hasler, Apr 12 2008
    
  • PARI
    is(n)=ispower(n,3) \\ Charles R Greathouse IV, Feb 20 2012
    
  • Python
    A000578_list, m = [], [6, -6, 1, 0]
    for _ in range(10**2):
        A000578_list.append(m[-1])
        for i in range(3):
            m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015
    
  • Scheme
    (define (A000578 n) (* n n n)) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = Sum_{i=0..n-1} A003215(i).
Multiplicative with a(p^e) = p^(3e). - David W. Wilson, Aug 01 2001
G.f.: x*(1+4*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
Dirichlet generating function: zeta(s-3). - Franklin T. Adams-Watters, Sep 11 2005, Amarnath Murthy, Sep 09 2005
E.g.f.: (1+3*x+x^2)*x*exp(x). - Franklin T. Adams-Watters, Sep 11 2005 - Amarnath Murthy, Sep 09 2005
a(n) = Sum_{i=1..n} (Sum_{j=i..n+i-1} A002024(j,i)). - Reinhard Zumkeller, Jun 24 2007
a(n) = lcm(n, (n - 1)^2) - (n - 1)^2. E.g.: lcm(1, (1 - 1)^2) - (1 - 1)^2 = 0, lcm(2, (2 - 1)^2) - (2 - 1)^2 = 1, lcm(3, (3 - 1)^2) - (3 - 1)^2 = 8, ... - Mats Granvik, Sep 24 2007
Starting (1, 8, 27, 64, 125, ...), = binomial transform of [1, 7, 12, 6, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
a(n) = A007531(n) + A000567(n). - Reinhard Zumkeller, Sep 18 2009
a(n) = binomial(n+2,3) + 4*binomial(n+1,3) + binomial(n,3). [Worpitzky's identity for cubes. See. e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019]
a(n) = n + 6*binomial(n+1,3) = binomial(n,1)+6*binomial(n+1,3). - Ron Knott, Jun 10 2019
A010057(a(n)) = 1. - Reinhard Zumkeller, Oct 22 2011
a(n) = A000537(n) - A000537(n-1), difference between 2 squares of consecutive triangular numbers. - Pierre CAMI, Feb 20 2012
a(n) = A048395(n) - 2*A006002(n). - J. M. Bergot, Nov 25 2012
a(n) = 1 + 7*(n-1) + 6*(n-1)*(n-2) + (n-1)*(n-2)*(n-3). - Antonio Alberto Olivares, Apr 03 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6. - Ant King Apr 29 2013
a(n) = A000330(n) + Sum_{i=1..n-1} A014105(i), n >= 1. - Ivan N. Ianakiev, Sep 20 2013
a(k) = det(S(4,k,(1,1,-1))) = k*b(k)^2, where b(1)=1, b(2)=2, b(k) = 2*b(k-1) - b(k-2) = b(2)*b(k-1) - b(k-2). - Ryan Stees, Dec 14 2014
For n >= 1, a(n) = A152618(n-1) + A033996(n-1). - Bui Quang Tuan, Apr 01 2015
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Jon Tavasanis, Feb 21 2016
a(n) = n + Sum_{j=0..n-1} Sum_{k=1..2} binomial(3,k)*j^(3-k). - Patrick J. McNab, Mar 28 2016
a(n) = A000292(n-1) * 6 + n. - Zhandos Mambetaliyev, Nov 24 2016
a(n) = n*binomial(n+1, 2) + 2*binomial(n+1, 3) + binomial(n,3). - Tony Foster III, Nov 14 2017
From Amiram Eldar, Jul 02 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(3) (A002117).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*zeta(3)/4 (A197070). (End)
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi.
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)/(3*Pi). (End)
a(n) = Sum_{d|n} sigma_3(d)*mu(n/d) = Sum_{d|n} A001158(d)*A008683(n/d). Moebius transform of sigma_3(n). - Ridouane Oudra, Apr 15 2021

A000244 Powers of 3: a(n) = 3^n.

Original entry on oeis.org

1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987
Offset: 0

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Author

Keywords

Comments

Same as Pisot sequences E(1, 3), L(1, 3), P(1, 3), T(1, 3). Essentially same as Pisot sequences E(3, 9), L(3, 9), P(3, 9), T(3, 9). See A008776 for definitions of Pisot sequences.
Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2n + 2, s(0) = 1, s(2n+2) = 3. - Herbert Kociemba, Jun 10 2004
a(1) = 1, a(n+1) is the least number such that there are a(n) even numbers between a(n) and a(n+1). Generalization for the sequence of powers of k: 1, k, k^2, k^3, k^4, ... There are a(n) multiples of k-1 between a(n) and a(n+1). - Amarnath Murthy, Nov 28 2004
a(n) = sum of (n+1)-th row in Triangle A105728. - Reinhard Zumkeller, Apr 18 2005
With p(n) being the number of integer partitions of n, p(i) being the number of parts of the i-th partition of n, d(i) being the number of different parts of the i-th partition of n, m(i, j) being the multiplicity of the j-th part of the i-th partition of n, Sum_{i = 1..p(n)} being the sum over i and Product_{j = 1..d(i)} being the product over j, one has: a(n) = Sum_{i = 1..p(n)} (p(i)!/(Product_{j = 1..d(i)} m(i, j)!))*2^(p(i) - 1). - Thomas Wieder, May 18 2005
For any k > 1 in the sequence, k is the first prime power appearing in the prime decomposition of repunit R_k, i.e., of A002275(k). - Lekraj Beedassy, Apr 24 2006
a(n-1) is the number of compositions of compositions. In general, (k+1)^(n-1) is the number of k-levels nested compositions (e.g., 4^(n-1) is the number of compositions of compositions of compositions, etc.). Each of the n - 1 spaces between elements can be a break for one of the k levels, or not a break at all. - Franklin T. Adams-Watters, Dec 06 2006
Let S be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xSy if x is a subset of y. Then a(n) = |S|. - Ross La Haye, Dec 22 2006
From Manfred Boergens, Mar 28 2023: (Start)
With regard to the comment by Ross La Haye:
Cf. A001047 if either nonempty subsets are considered or x is a proper subset of y.
Cf. a(n+1) in A028243 if nonempty subsets are considered and x is a proper subset of y. (End)
If X_1, X_2, ..., X_n is a partition of the set {1, 2, ..., 2*n} into blocks of size 2 then, for n >= 1, a(n) is equal to the number of functions f : {1, 2, ..., 2*n} -> {1, 2} such that for fixed y_1, y_2, ..., y_n in {1, 2} we have f(X_i) <> {y_i}, (i = 1, 2, ..., n). - Milan Janjic, May 24 2007
This is a general comment on all sequences of the form a(n) = [(2^k)-1]^n for all positive integers k. Example 1.1.16 of Stanley's "Enumerative Combinatorics" offers a slightly different version. a(n) in the number of functions f:[n] into P([k]) - {}. a(n) is also the number of functions f:[k] into P([n]) such that the generalized intersection of f(i) for all i in [k] is the empty set. Where [n] = {1, 2, ..., n}, P([n]) is the power set of [n] and {} is the empty set. - Geoffrey Critzer, Feb 28 2009
a(n) = A064614(A000079(n)) and A064614(m)A000079(n). - Reinhard Zumkeller, Feb 08 2010
3^(n+1) = (1, 2, 2, 2, ...) dot (1, 1, 3, 9, ..., 3^n); e.g., 3^3 = 27 = (1, 2, 2, 2) dot (1, 1, 3, 9) = (1 + 2 + 6 + 18). - Gary W. Adamson, May 17 2010
a(n) is the number of generalized compositions of n when there are 3*2^i different types of i, (i = 1, 2, ...). - Milan Janjic, Sep 24 2010
For n >= 1, a(n-1) is the number of generalized compositions of n when there are 2^(i-1) different types of i, (i = 1, 2, ...). - Milan Janjic, Sep 24 2010
The sequence in question ("Powers of 3") also describes the number of moves of the k-th disk solving the [RED ; BLUE ; BLUE] or [RED ; RED ; BLUE] pre-colored Magnetic Tower of Hanoi puzzle (cf. A183111 - A183125).
a(n) is the number of Stern polynomials of degree n. See A057526. - T. D. Noe, Mar 01 2011
Positions of records in the number of odd prime factors, A087436. - Juri-Stepan Gerasimov, Mar 17 2011
Sum of coefficients of the expansion of (1+x+x^2)^n. - Adi Dani, Jun 21 2011
a(n) is the number of compositions of n elements among {0, 1, 2}; e.g., a(2) = 9 since there are the 9 compositions 0 + 0, 0 + 1, 1 + 0, 0 + 2, 1 + 1, 2 + 0, 1 + 2, 2 + 1, and 2 + 2. [From Adi Dani, Jun 21 2011; modified by editors.]
Except the first two terms, these are odd numbers n such that no x with 2 <= x <= n - 2 satisfy x^(n-1) == 1 (mod n). - Arkadiusz Wesolowski, Jul 03 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 3-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Explanation from David Applegate, Feb 20 2017: (Start)
Since the preceding comment appears in a large number of sequences, it might be worth adding a proof.
The number of compositions of n into exactly k parts is binomial(n-1,k-1).
For a p-colored composition of n such that no adjacent parts have the same color, there are exactly p choices for the color of the first part, and p-1 choices for the color of each additional part (any color other than the color of the previous one). So, for a partition into k parts, there are p (p-1)^(k-1) valid colorings.
Thus the number of p-colored compositions of n into exactly k parts such that no adjacent parts have the same color is binomial(n-1,k-1) p (p-1)^(k-1).
The total number of p-colored compositions of n such that no adjacent parts have the same color is then
Sum_{k=1..n} binomial(n-1,k-1) * p * (p-1)^(k-1) = p^n.
To see this, note that the binomial expansion of ((p - 1) + 1)^(n - 1) = Sum_{k = 0..n - 1} binomial(n - 1, k) (p - 1)^k 1^(n - 1 - k) = Sum_{k = 1..n} binomial(n - 1, k - 1) (p - 1)^(k - 1).
(End)
Also, first and least element of the matrix [1, sqrt(2); sqrt(2), 2]^(n+1). - M. F. Hasler, Nov 25 2011
One-half of the row sums of the triangular version of A035002. - J. M. Bergot, Jun 10 2013
Form an array with m(0,n) = m(n,0) = 2^n; m(i,j) equals the sum of the terms to the left of m(i,j) and the sum of the terms above m(i,j), which is m(i,j) = Sum_{k=0..j-1} m(i,k) + Sum_{k=0..i-1} m(k,j). The sum of the terms in antidiagonal(n+1) = 4*a(n). - J. M. Bergot, Jul 10 2013
a(n) = A007051(n+1) - A007051(n), and A007051 are the antidiagonal sums of an array defined by m(0,k) = 1 and m(n,k) = Sum_{c = 0..k - 1} m(n, c) + Sum_{r = 0..n - 1} m(r, k), which is the sum of the terms to left of m(n, k) plus those above m(n, k). m(1, k) = A000079(k); m(2, k) = A045623(k + 1); m(k + 1, k) = A084771(k). - J. M. Bergot, Jul 16 2013
Define an array to have m(0,k) = 2^k and m(n,k) = Sum_{c = 0..k - 1} m(n, c) + Sum_{r = 0..n - 1} m(r, k), which is the sum of the terms to the left of m(n, k) plus those above m(n, k). Row n = 0 of the array comprises A000079, column k = 0 comprises A011782, row n = 1 comprises A001792. Antidiagonal sums of the array are a(n): 1 = 3^0, 1 + 2 = 3^1, 2 + 3 + 4 = 3^2, 4 + 7 + 8 + 8 = 3^3. - J. M. Bergot, Aug 02 2013
The sequence with interspersed zeros and o.g.f. x/(1 - 3*x^2), A(2*k) = 0, A(2*k + 1) = 3^k = a(k), k >= 0, can be called hexagon numbers. This is because the algebraic number rho(6) = 2*cos(Pi/6) = sqrt(3) of degree 2, with minimal polynomial C(6, x) = x^2 - 3 (see A187360, n = 6), is the length ratio of the smaller diagonal and the side in the hexagon. Hence rho(6)^n = A(n-1)*1 + A(n)*rho(6), in the power basis of the quadratic number field Q(rho(6)). One needs also A(-1) = 1. See also a Dec 02 2010 comment and the P. Steinbach reference given in A049310. - Wolfdieter Lang, Oct 02 2013
Numbers k such that sigma(3k) = 3k + sigma(k). - Jahangeer Kholdi, Nov 23 2013
All powers of 3 are perfect totient numbers (A082897), since phi(3^n) = 2 * 3^(n - 1) for n > 0, and thus Sum_{i = 0..n} phi(3^i) = 3^n. - Alonso del Arte, Apr 20 2014
The least number k > 0 such that 3^k ends in n consecutive decreasing digits is a 3-term sequence given by {1, 13, 93}. The consecutive increasing digits are {3, 23, 123}. There are 100 different 3-digit endings for 3^k. There are no k-values such that 3^k ends in '012', '234', '345', '456', '567', '678', or '789'. The k-values for which 3^k ends in '123' are given by 93 mod 100. For k = 93 + 100*x, the digit immediately before the run of '123' is {9, 5, 1, 7, 3, 9, 5, 1, 3, 7, ...} for x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...}, respectively. Thus we see the digit before '123' will never be a 0. So there are no further terms. - Derek Orr, Jul 03 2014
All elements of A^n where A = (1, 1, 1; 1, 1, 1; 1, 1, 1). - David Neil McGrath, Jul 23 2014
Counts all walks of length n (open or closed) on the vertices of a triangle containing a loop at each vertex starting from any given vertex. - David Neil McGrath, Oct 03 2014
a(n) counts walks (closed) on the graph G(1-vertex;1-loop,1-loop,1-loop). - David Neil McGrath, Dec 11 2014
2*a(n-2) counts all permutations of a solitary closed walk of length (n) from the vertex of a triangle that contains 2 loops on each of the remaining vertices. In addition, C(m,k)=2*(2^m)*B(m+k-2,m) counts permutations of walks that contain (m) loops and (k) arcs. - David Neil McGrath, Dec 11 2014
a(n) is the sum of the coefficients of the n-th layer of Pascal's pyramid (a.k.a., Pascal's tetrahedron - see A046816). - Bob Selcoe, Apr 02 2016
Numbers n such that the trinomial x^(2*n) + x^n + 1 is irreducible over GF(2). Of these only the trinomial for n=1 is primitive. - Joerg Arndt, May 16 2016
Satisfies Benford's law [Berger-Hill, 2011]. - N. J. A. Sloane, Feb 08 2017
a(n-1) is also the number of compositions of n if the parts can be runs of any length from 1 to n, and can contain any integers from 1 to n. - Gregory L. Simay, May 26 2017
Also the number of independent vertex sets and vertex covers in the n-ladder rung graph n P_2. - Eric W. Weisstein, Sep 21 2017
Also the number of (not necessarily maximal) cliques in the n-cocktail party graph. - Eric W. Weisstein, Nov 29 2017
a(n-1) is the number of 2-compositions of n; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020
a(n) is the number of faces of any dimension (vertices, edges, square faces, etc.) of the n-dimensional hypercube. For example, the 0-dimensional hypercube is a point, and its only face is itself. The 1-dimensional hypercube is a line, which has two vertices and an edge. The 2-dimensional hypercube is a square, which has four vertices, four edges, and a square face. - Kevin Long, Mar 14 2023
Number of pairs (A,B) of subsets of M={1,2,...,n} with union(A,B)=M. For nonempty subsets cf. A058481. - Manfred Boergens, Mar 28 2023
From Jianing Song, Sep 27 2023: (Start)
a(n) is the number of disjunctive clauses of n variables up to equivalence. A disjunctive clause is a propositional formula of the form l_1 OR ... OR l_m, where l_1, ..., l_m are distinct elements in {x_1, ..., x_n, NOT x_1, ..., NOT x_n} for n variables x_1, ... x_n, and no x_i and NOT x_i appear at the same time. For each 1 <= i <= n, we can have neither of x_i or NOT x_i, only x_i or only NOT x_i appearing in a disjunctive clause, so the number of such clauses is 3^n. Viewing the propositional formulas of n variables as functions {0,1}^n -> {0,1}, a disjunctive clause corresponds to a function f such that the inverse image of 0 is of the form A_1 X ... X A_n, where A_i is nonempty for all 1 <= i <= n. Since each A_i has 3 choices ({0}, {1} or {0,1}), we also find that the number of disjunctive clauses of n variables is 3^n.
Equivalently, a(n) is the number of conjunctive clauses of n variables. (End)
The finite subsequence a(2), a(3), a(4), a(5) = 9, 27, 81, 243 is one of only two geometric sequences that can be formed with all interior angles (all integer, in degrees) of a simple polygon. The other sequence is a subsequence of A007283 (see comment there). - Felix Huber, Feb 15 2024

Examples

			G.f. = 1 + 3*x + 9*x^2 + 27*x^3 + 81*x^4 + 243*x^5 + 729*x^6 + 2187*x^7 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A008776 (2*a(n), and first differences).
a(n) = A092477(n, 2) for n > 0.
a(n) = A159991(n) / A009964(n).
Cf. A100772, A035002. Row sums of A125076 and A153279.
a(n) = A217764(0, n).
Cf. A046816, A006521, A014945, A275414 (multisets).
The following are parallel families: A000079 (2^n), A004094 (2^n reversed), A028909 (2^n sorted up), A028910 (2^n sorted down), A036447 (double and reverse), A057615 (double and sort up), A263451 (double and sort down); A000244 (3^n), A004167 (3^n reversed), A321540 (3^n sorted up), A321539 (3^n sorted down), A163632 (triple and reverse), A321542 (triple and sort up), A321541 (triple and sort down).

Programs

Formula

a(n) = 3^n.
a(0) = 1; a(n) = 3*a(n-1).
G.f.: 1/(1-3*x).
E.g.f.: exp(3*x).
a(n) = n!*Sum_{i + j + k = n, i, j, k >= 0} 1/(i!*j!*k!). - Benoit Cloitre, Nov 01 2002
a(n) = Sum_{k = 0..n} 2^k*binomial(n, k), binomial transform of A000079.
a(n) = A090888(n, 2). - Ross La Haye, Sep 21 2004
a(n) = 2^(2n) - A005061(n). - Ross La Haye, Sep 10 2005
a(n) = A112626(n, 0). - Ross La Haye, Jan 11 2006
Hankel transform of A007854. - Philippe Deléham, Nov 26 2006
a(n) = 2*StirlingS2(n+1,3) + StirlingS2(n+2,2) = 2*(StirlingS2(n+1,3) + StirlingS2(n+1,2)) + 1. - Ross La Haye, Jun 26 2008
a(n) = 2*StirlingS2(n+1, 3) + StirlingS2(n+2, 2) = 2*(StirlingS2(n+1, 3) + StirlingS2(n+1, 2)) + 1. - Ross La Haye, Jun 09 2008
Sum_{n >= 0} 1/a(n) = 3/2. - Gary W. Adamson, Aug 29 2008
If p(i) = Fibonacci(2i-2) and if A is the Hessenberg matrix of order n defined by A(i, j) = p(j-i+1), (i <= j), A(i, j) = -1, (i = j+1), and A(i, j) = 0 otherwise, then, for n >= 1, a(n-1) = det A. - Milan Janjic, May 08 2010
G.f. A(x) = M(x)/(1-M(x))^2, M(x) - o.g.f for Motzkin numbers (A001006). - Vladimir Kruchinin, Aug 18 2010
a(n) = A133494(n+1). - Arkadiusz Wesolowski, Jul 27 2011
2/3 + 3/3^2 + 2/3^3 + 3/3^4 + 2/3^5 + ... = 9/8. [Jolley, Summation of Series, Dover, 1961]
a(n) = Sum_{k=0..n} A207543(n,k)*4^(n-k). - Philippe Deléham, Feb 25 2012
a(n) = Sum_{k=0..n} A125185(n,k). - Philippe Deléham, Feb 26 2012
Sum_{n > 0} Mobius(n)/a(n) = 0.181995386702633887827... (see A238271). - Alonso del Arte, Aug 09 2012. See also the sodium 3s orbital energy in table V of J. Chem. Phys. 53 (1970) 348.
a(n) = (tan(Pi/3))^(2*n). - Bernard Schott, May 06 2022
a(n-1) = binomial(2*n-1, n) + Sum_{k >= 1} binomial(2*n, n+3*k)*(-1)^k. - Greg Dresden, Oct 14 2022
G.f.: Sum_{k >= 0} x^k/(1-2*x)^(k+1). - Kevin Long, Mar 14 2023

A002113 Palindromes in base 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454, 464, 474, 484, 494, 505, 515
Offset: 1

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Comments

n is a palindrome (i.e., a(k) = n for some k) if and only if n = A004086(n). - Reinhard Zumkeller, Mar 10 2002
It seems that if n*reversal(n) is in the sequence then n = 3 or all digits of n are less than 3. - Farideh Firoozbakht, Nov 02 2014
The position of a palindrome within the sequence can be determined almost without calculation: If the palindrome has an even number of digits, prepend a 1 to the front half of the palindrome's digits. If the number of digits is odd, prepend the value of front digit + 1 to the digits from position 2 ... central digit. Examples: 98766789 = a(19876), 515 = a(61), 8206028 = a(9206), 9230329 = a(10230). - Hugo Pfoertner, Aug 14 2015
This sequence is an additive basis of order at most 49, see Banks link. - Charles R Greathouse IV, Aug 23 2015
The order has been reduced from 49 to 3; see the Cilleruelo-Luca and Cilleruelo-Luca-Baxter links. - Jonathan Sondow, Nov 27 2017
See A262038 for the "next palindrome" and A261423 for the "preceding palindrome" functions. - M. F. Hasler, Sep 09 2015
The number of palindromes with d digits is 10 if d = 1, and otherwise it is 9 * 10^(floor((d - 1)/2)). - N. J. A. Sloane, Dec 06 2015
Sequence A033665 tells how many iterations of the Reverse-then-add function A056964 are needed to reach a palindrome; numbers for which this will never happen are Lychrel numbers (A088753) or rather Kin numbers (A023108). - M. F. Hasler, Apr 13 2019
This sequence is an additive basis of order 3, see Cilleruelo, Luca, & Baxter and Sigg. - Charles R Greathouse IV, Apr 08 2025

References

  • Karl G. Kröber, "Palindrome, Perioden und Chaoten: 66 Streifzüge durch die palindromischen Gefilde" (1997, Deutsch-Taschenbücher; Bd. 99) ISBN 3-8171-1522-9.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 50-52.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 120.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Subsequence of A061917 and A221221.
A110745 is a subsequence.
Union of A056524 and A056525.
Palindromes in bases 2 through 11: A006995 and A057148, A014190 and A118594, A014192 and A118595, A029952 and A118596, A029953 and A118597, A029954 and A118598, A029803 and A118599, A029955 and A118600, this sequence, A029956. Also A262065 (base 60), A262069 (subsequence).
Palindromic primes: A002385. Palindromic nonprimes: A032350.
Palindromic-pi: A136687.
Cf. A029742 (complement), A086862 (first differences).
Palindromic floor function: A261423, also A261424. Palindromic ceiling: A262038.
Cf. A004086 (read n backwards), A064834, A118031, A136522 (characteristic function), A178788.
Ways to write n as a sum of three palindromes: A261132, A261422.
Minimal number of palindromes that add to n using greedy algorithm: A088601.
Minimal number of palindromes that add to n: A261675.

Programs

  • GAP
    Filtered([0..550],n->ListOfDigits(n)=Reversed(ListOfDigits(n))); # Muniru A Asiru, Mar 08 2019
    
  • Haskell
    a002113 n = a002113_list !! (n-1)
      a002113_list = filter ((== 1) . a136522) [1..] -- Reinhard Zumkeller, Oct 09 2011
    
  • Haskell
    import Data.List.Ordered (union)
      a002113_list = union a056524_list a056525_list -- Reinhard Zumkeller, Jul 29 2015, Dec 28 2011
    
  • Magma
    [n: n in [0..600] | Intseq(n, 10) eq Reverse(Intseq(n, 10))]; // Vincenzo Librandi, Nov 03 2014
    
  • Maple
    read transforms; t0:=[]; for n from 0 to 2000 do if digrev(n) = n then t0:=[op(t0),n]; fi; od: t0;
    # Alternatively, to get all palindromes with <= N digits in the list "Res":
    N:=5;
    Res:= $0..9:
    for d from 2 to N do
      if d::even then
        m:= d/2;
        Res:= Res, seq(n*10^m + digrev(n),n=10^(m-1)..10^m-1);
      else
        m:= (d-1)/2;
        Res:= Res, seq(seq(n*10^(m+1)+y*10^m+digrev(n),y=0..9),n=10^(m-1)..10^m-1);
      fi
    od: Res:=[Res]: # Robert Israel, Aug 10 2014
    # A variant: Gets all base-10 palindromes with exactly d digits, in the list "Res"
    d:=4:
    if d=1 then Res:= [$0..9]:
    elif d::even then
        m:= d/2:
        Res:= [seq(n*10^m + digrev(n), n=10^(m-1)..10^m-1)]:
    else
        m:= (d-1)/2:
        Res:= [seq(seq(n*10^(m+1)+y*10^m+digrev(n), y=0..9), n=10^(m-1)..10^m-1)]:
    fi:
    Res; # N. J. A. Sloane, Oct 18 2015
    isA002113 := proc(n)
        simplify(digrev(n) = n) ;
    end proc: # R. J. Mathar, Sep 09 2015
  • Mathematica
    palQ[n_Integer, base_Integer] := Module[{idn = IntegerDigits[n, base]}, idn == Reverse[idn]]; (* then to generate any base-b sequence for 1 < b < 37, replace the 10 in the following instruction with b: *) Select[Range[0, 1000], palQ[#, 10] &]
    base10Pals = {0}; r = 2; Do[Do[AppendTo[base10Pals, n * 10^(IntegerLength[n] - 1) + FromDigits@Rest@Reverse@IntegerDigits[n]], {n, 10^(e - 1), 10^e - 1}]; Do[AppendTo[base10Pals, n * 10^IntegerLength[n] + FromDigits@Reverse@IntegerDigits[n]], {n, 10^(e - 1), 10^e - 1}], {e, r}]; base10Pals (* Arkadiusz Wesolowski, May 04 2012 *)
    nthPalindromeBase[n_, b_] := Block[{q = n + 1 - b^Floor[Log[b, n + 1 - b^Floor[Log[b, n/b]]]], c = Sum[Floor[Floor[n/((b + 1) b^(k - 1) - 1)]/(Floor[n/((b + 1) b^(k - 1) - 1)] - 1/b)] - Floor[Floor[n/(2 b^k - 1)]/(Floor[n/(2 b^k - 1)] - 1/ b)], {k, Floor[Log[b, n]]}]}, Mod[q, b] (b + 1)^c * b^Floor[Log[b, q]] + Sum[Floor[Mod[q, b^(k + 1)]/b^k] b^(Floor[Log[b, q]] - k) (b^(2 k + c) + 1), {k, Floor[Log[b, q]]}]] (* after the work of Eric A. Schmidt, works for all integer bases b > 2 *)
    Array[nthPalindromeBase[#, 10] &, 61, 0] (* please note that Schmidt uses a different, a more natural and intuitive offset, that of a(1) = 1. - Robert G. Wilson v, Sep 22 2014 and modified Nov 28 2014 *)
    Select[Range[10^3], PalindromeQ] (* Michael De Vlieger, Nov 27 2017 *)
    nLP[cn_Integer]:=Module[{s,len,half,left,pal,fdpal},s=IntegerDigits[cn]; len=Length[s]; half=Ceiling[len/2]; left=Take[s,half]; pal=Join[left,Reverse[ Take[left,Floor[len/2]]]]; fdpal=FromDigits[pal]; Which[cn==9,11,fdpal>cn,fdpal,True,left=IntegerDigits[ FromDigits[left]+1]; pal=Join[left,Reverse[Take[left,Floor[len/2]]]]; FromDigits[pal]]]; NestList[nLP,0,100] (* Harvey P. Dale, Dec 10 2024 *)
  • PARI
    is_A002113(n)=Vecrev(n=digits(n))==n \\ M. F. Hasler, Nov 17 2008, updated Apr 26 2014, Jun 19 2018
    
  • PARI
    is(n)=n=digits(n);for(i=1,#n\2,if(n[i]!=n[#n+1-i],return(0))); 1 \\ Charles R Greathouse IV, Jan 04 2013
    
  • PARI
    a(n)={my(d,i,r);r=vector(#digits(n-10^(#digits(n\11)))+#digits(n\11));n=n-10^(#digits(n\11));d=digits(n);for(i=1,#d,r[i]=d[i];r[#r+1-i]=d[i]);sum(i=1,#r,10^(#r-i)*r[i])} \\ David A. Corneth, Jun 06 2014
    
  • PARI
    \\ recursive--feed an element a(n) and it gives a(n+1)
    nxt(n)=my(d=digits(n));i=(#d+1)\2;while(i&&d[i]==9,d[i]=0;d[#d+1-i]=0;i--);if(i,d[i]++;d[#d+1-i]=d[i],d=vector(#d+1);d[1]=d[#d]=1);sum(i=1,#d,10^(#d-i)*d[i]) \\ David A. Corneth, Jun 06 2014
    
  • PARI
    \\ feed a(n), returns n.
    inv(n)={my(d=digits(n));q=ceil(#d/2);sum(i=1,q,10^(q-i)*d[i])+10^floor(#d/2)} \\ David A. Corneth, Jun 18 2014
    
  • PARI
    inv_A002113(P)={P\(P=10^(logint(P+!P,10)\/2))+P} \\ index n of palindrome P = a(n), much faster than above: no sum is needed. - M. F. Hasler, Sep 09 2018
    
  • PARI
    A002113(n,L=logint(n,10))=(n-=L=10^max(L-(n<11*10^(L-1)),0))*L+fromdigits(Vecrev(digits(if(nM. F. Hasler, Sep 11 2018
    
  • Python
    # edited by M. F. Hasler, Jun 19 2018
    def A002113_list(nMax):
      mlist=[]
      for n in range(nMax+1):
         mstr=str(n)
         if mstr==mstr[::-1]:
            mlist.append(n)
      return mlist # Bill McEachen, Dec 17 2010
    
  • Python
    from itertools import chain
    A002113 = sorted(chain(map(lambda x:int(str(x)+str(x)[::-1]),range(1,10**3)),map(lambda x:int(str(x)+str(x)[-2::-1]), range(10**3)))) # Chai Wah Wu, Aug 09 2014
    
  • Python
    from itertools import chain, count
    A002113 = chain(k for k in count(0) if str(k) == str(k)[::-1])
    print([next(A002113) for k in range(60)]) # Jan P. Hartkopf, Apr 10 2021
    
  • Python
    is_A002113 = lambda n: (s:=str(n))[::-1]==s # M. F. Hasler, May 23 2024
    
  • Python
    from math import log10, floor
    def A002113(n):
      if n < 2: return 0
      P = 10**floor(log10(n//2)); M = 11*P
      s = str(n - (P if n < M else M-P))
      return int(s + s[-2 if n < M else -1::-1]) # M. F. Hasler, Jun 06 2024
    
  • SageMath
    [n for n in (0..515) if Word(n.digits()).is_palindrome()] # Peter Luschny, Sep 13 2018
    
  • Scala
    def palQ(n: Int, b: Int = 10): Boolean = n - Integer.parseInt(n.toString.reverse) == 0
    (0 to 999).filter(palQ()) // _Alonso del Arte, Nov 10 2019

Formula

A136522(a(n)) = 1.
A178788(a(n)) = 0 for n > 9. - Reinhard Zumkeller, Jun 30 2010
A064834(a(n)) = 0. - Reinhard Zumkeller, Sep 18 2013
a(n+1) = A262038(a(n)+1). - M. F. Hasler, Sep 09 2015
Sum_{n>=2} 1/a(n) = A118031. - Amiram Eldar, Oct 17 2020
a(n) = (floor(d(n)/(c(n)*9 + 1)))*10^A055642(d(n)) + A004086(d(n)) where b(n, k) = ceiling(log((n + 1)/k)/log(10)), c(n) = b(n, 2) - b(n, 11) and d(n) = (n - A086573(b(n*(2 - c(n)), 2) - 1)/2 - 1). - Alan Michael Gómez Calderón, Mar 11 2025

A002378 Oblong (or promic, pronic, or heteromecic) numbers: a(n) = n*(n+1).

Original entry on oeis.org

0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462, 506, 552, 600, 650, 702, 756, 812, 870, 930, 992, 1056, 1122, 1190, 1260, 1332, 1406, 1482, 1560, 1640, 1722, 1806, 1892, 1980, 2070, 2162, 2256, 2352, 2450, 2550
Offset: 0

Views

Author

Keywords

Comments

4*a(n) + 1 are the odd squares A016754(n).
The word "pronic" (used by Dickson) is incorrect. - Michael Somos
According to the 2nd edition of Webster, the correct word is "promic". - R. K. Guy
a(n) is the number of minimal vectors in the root lattice A_n (see Conway and Sloane, p. 109).
Let M_n denote the n X n matrix M_n(i, j) = (i + j); then the characteristic polynomial of M_n is x^(n-2) * (x^2 - a(n)*x - A002415(n)). - Benoit Cloitre, Nov 09 2002
The greatest LCM of all pairs (j, k) for j < k <= n for n > 1. - Robert G. Wilson v, Jun 19 2004
First differences are a(n+1) - a(n) = 2*n + 2 = 2, 4, 6, ... (while first differences of the squares are (n+1)^2 - n^2 = 2*n + 1 = 1, 3, 5, ...). - Alexandre Wajnberg, Dec 29 2005
25 appended to these numbers corresponds to squares of numbers ending in 5 (i.e., to squares of A017329). - Lekraj Beedassy, Mar 24 2006
A rapid (mental) multiplication/factorization technique -- a generalization of Lekraj Beedassy's comment: For all bases b >= 2 and positive integers n, c, d, k with c + d = b^k, we have (n*b^k + c)*(n*b^k + d) = a(n)*b^(2*k) + c*d. Thus the last 2*k base-b digits of the product are exactly those of c*d -- including leading 0(s) as necessary -- with the preceding base-b digit(s) the same as a(n)'s. Examples: In decimal, 113*117 = 13221 (as n = 11, b = 10 = 3 + 7, k = 1, 3*7 = 21, and a(11) = 132); in octal, 61*67 = 5207 (52 is a(6) in octal). In particular, for even b = 2*m (m > 0) and c = d = m, such a product is a square of this type. Decimal factoring: 5609 is immediately seen to be 71*79. Likewise, 120099 = 301*399 (k = 2 here) and 99990000001996 = 9999002*9999998 (k = 3). - Rick L. Shepherd, Jul 24 2021
Number of circular binary words of length n + 1 having exactly one occurrence of 01. Example: a(2) = 6 because we have 001, 010, 011, 100, 101 and 110. Column 1 of A119462. - Emeric Deutsch, May 21 2006
The sequence of iterated square roots sqrt(N + sqrt(N + ...)) has for N = 1, 2, ... the limit (1 + sqrt(1 + 4*N))/2. For N = a(n) this limit is n + 1, n = 1, 2, .... For all other numbers N, N >= 1, this limit is not a natural number. Examples: n = 1, a(1) = 2: sqrt(2 + sqrt(2 + ...)) = 1 + 1 = 2; n = 2, a(2) = 6: sqrt(6 + sqrt(6 + ...)) = 1 + 2 = 3. - Wolfdieter Lang, May 05 2006
Nonsquare integers m divisible by ceiling(sqrt(m)), except for m = 0. - Max Alekseyev, Nov 27 2006
The number of off-diagonal elements of an (n + 1) X (n + 1) matrix. - Artur Jasinski, Jan 11 2007
a(n) is equal to the number of functions f:{1, 2} -> {1, 2, ..., n + 1} such that for a fixed x in {1, 2} and a fixed y in {1, 2, ..., n + 1} we have f(x) <> y. - Aleksandar M. Janjic and Milan Janjic, Mar 13 2007
Numbers m >= 0 such that round(sqrt(m+1)) - round(sqrt(m)) = 1. - Hieronymus Fischer, Aug 06 2007
Numbers m >= 0 such that ceiling(2*sqrt(m+1)) - 1 = 1 + floor(2*sqrt(m)). - Hieronymus Fischer, Aug 06 2007
Numbers m >= 0 such that fract(sqrt(m+1)) > 1/2 and fract(sqrt(m)) < 1/2 where fract(x) is the fractional part (fract(x) = x - floor(x), x >= 0). - Hieronymus Fischer, Aug 06 2007
X values of solutions to the equation 4*X^3 + X^2 = Y^2. To find Y values: b(n) = n(n+1)(2n+1). - Mohamed Bouhamida, Nov 06 2007
Nonvanishing diagonal of A132792, the infinitesimal Lah matrix, so "generalized factorials" composed of a(n) are given by the elements of the Lah matrix, unsigned A111596, e.g., a(1)*a(2)*a(3) / 3! = -A111596(4,1) = 24. - Tom Copeland, Nov 20 2007
If Y is a 2-subset of an n-set X then, for n >= 2, a(n-2) is the number of 2-subsets and 3-subsets of X having exactly one element in common with Y. - Milan Janjic, Dec 28 2007
a(n) coincides with the vertex of a parabola of even width in the Redheffer matrix, directed toward zero. An integer p is prime if and only if for all integer k, the parabola y = kx - x^2 has no integer solution with 1 < x < k when y = p; a(n) corresponds to odd k. - Reikku Kulon, Nov 30 2008
The third differences of certain values of the hypergeometric function 3F2 lead to the squares of the oblong numbers i.e., 3F2([1, n + 1, n + 1], [n + 2, n + 2], z = 1) - 3*3F2([1, n + 2, n + 2], [n + 3, n + 3], z = 1) + 3*3F2([1, n + 3, n + 3], [n + 4, n + 4], z = 1) - 3F2([1, n + 4, n + 4], [n + 5, n + 5], z = 1) = (1/((n+2)*(n+3)))^2 for n = -1, 0, 1, 2, ... . See also A162990. - Johannes W. Meijer, Jul 21 2009
Generalized factorials, [a.(n!)] = a(n)*a(n-1)*...*a(0) = A010790(n), with a(0) = 1 are related to A001263. - Tom Copeland, Sep 21 2011
For n > 1, a(n) is the number of functions f:{1, 2} -> {1, ..., n + 2} where f(1) > 1 and f(2) > 2. Note that there are n + 1 possible values for f(1) and n possible values for f(2). For example, a(3) = 12 since there are 12 functions f from {1, 2} to {1, 2, 3, 4, 5} with f(1) > 1 and f(2) > 2. - Dennis P. Walsh, Dec 24 2011
a(n) gives the number of (n + 1) X (n + 1) symmetric (0, 1)-matrices containing two ones (see [Cameron]). - L. Edson Jeffery, Feb 18 2012
a(n) is the number of positions of a domino in a rectangled triangular board with both legs equal to n + 1. - César Eliud Lozada, Sep 26 2012
a(n) is the number of ordered pairs (x, y) in [n+2] X [n+2] with |x-y| > 1. - Dennis P. Walsh, Nov 27 2012
a(n) is the number of injective functions from {1, 2} into {1, 2, ..., n + 1}. - Dennis P. Walsh, Nov 27 2012
a(n) is the sum of the positive differences of the partition parts of 2n + 2 into exactly two parts (see example). - Wesley Ivan Hurt, Jun 02 2013
a(n)/a(n-1) is asymptotic to e^(2/n). - Richard R. Forberg, Jun 22 2013
Number of positive roots in the root system of type D_{n + 1} (for n > 2). - Tom Edgar, Nov 05 2013
Number of roots in the root system of type A_n (for n > 0). - Tom Edgar, Nov 05 2013
From Felix P. Muga II, Mar 18 2014: (Start)
a(m), for m >= 1, are the only positive integer values t for which the Binet-de Moivre formula for the recurrence b(n) = b(n-1) + t*b(n-2) with b(0) = 0 and b(1) = 1 has a root of a square. PROOF (as suggested by Wolfdieter Lang, Mar 26 2014): The sqrt(1 + 4t) appearing in the zeros r1 and r2 of the characteristic equation is (a positive) integer for positive integer t precisely if 4t + 1 = (2m + 1)^2, that is t = a(m), m >= 1. Thus, the characteristic roots are integers: r1 = m + 1 and r2 = -m.
Let m > 1 be an integer. If b(n) = b(n-1) + a(m)*b(n-2), n >= 2, b(0) = 0, b(1) = 1, then lim_{n->oo} b(n+1)/b(n) = m + 1. (End)
Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices (chromatic polynomial) of the complete graphs (here simply K_2). - Tom Copeland, Apr 05 2014
The set of integers k for which k + sqrt(k + sqrt(k + sqrt(k + sqrt(k + ...) ... is an integer. - Leslie Koller, Apr 11 2014
a(n-1) is the largest number k such that (n*k)/(n+k) is an integer. - Derek Orr, May 22 2014
Number of ways to place a domino and a singleton on a strip of length n - 2. - Ralf Stephan, Jun 09 2014
With offset 1, this appears to give the maximal number of crossings between n nonconcentric circles of equal radius. - Felix Fröhlich, Jul 14 2014
For n > 1, the harmonic mean of the n values a(1) to a(n) is n + 1. The lowest infinite sequence of increasing positive integers whose cumulative harmonic mean is integral. - Ian Duff, Feb 01 2015
a(n) is the maximum number of queens of one color that can coexist without attacking one queen of the opponent's color on an (n+2) X (n+2) chessboard. The lone queen can be placed in any position on the perimeter of the board. - Bob Selcoe, Feb 07 2015
With a(0) = 1, a(n-1) is the smallest positive number not in the sequence such that Sum_{i = 1..n} 1/a(i-1) has a denominator equal to n. - Derek Orr, Jun 17 2015
The positive members of this sequence are a proper subsequence of the so-called 1-happy couple products A007969. See the W. Lang link there, eq. (4), with Y_0 = 1, with a table at the end. - Wolfdieter Lang, Sep 19 2015
For n > 0, a(n) is the reciprocal of the area bounded above by y = x^(n-1) and below by y = x^n for x in the interval [0, 1]. Summing all such areas visually demonstrates the formula below giving Sum_{n >= 1} 1/a(n) = 1. - Rick L. Shepherd, Oct 26 2015
It appears that, except for a(0) = 0, this is the set of positive integers n such that x*floor(x) = n has no solution. (For example, to get 3, take x = -3/2.) - Melvin Peralta, Apr 14 2016
If two independent real random variables, x and y, are distributed according to the same exponential distribution: pdf(x) = lambda * exp(-lambda * x), lambda > 0, then the probability that n - 1 <= x/y < n is given by 1/a(n). - Andres Cicuttin, Dec 03 2016
a(n) is equal to the sum of all possible differences between n different pairs of consecutive odd numbers (see example). - Miquel Cerda, Dec 04 2016
a(n+1) is the dimension of the space of vector fields in the plane with polynomial coefficients up to order n. - Martin Licht, Dec 04 2016
It appears that a(n) + 3 is the area of the largest possible pond in a square (A268311). - Craig Knecht, May 04 2017
Also the number of 3-cycles in the (n+3)-triangular honeycomb acute knight graph. - Eric W. Weisstein, Jul 27 2017
Also the Wiener index of the (n+2)-wheel graph. - Eric W. Weisstein, Sep 08 2017
The left edge of a Floyd's triangle that consists of even numbers: 0; 2, 4; 6, 8, 10; 12, 14, 16, 18; 20, 22, 24, 26, 28; ... giving 0, 2, 6, 12, 20, ... The right edge generates A028552. - Waldemar Puszkarz, Feb 02 2018
a(n+1) is the order of rowmotion on a poset obtained by adjoining a unique minimal (or maximal) element to a disjoint union of at least two chains of n elements. - Nick Mayers, Jun 01 2018
From Juhani Heino, Feb 05 2019: (Start)
For n > 0, 1/a(n) = n/(n+1) - (n-1)/n.
For example, 1/6 = 2/3 - 1/2; 1/12 = 3/4 - 2/3.
Corollary of this:
Take 1/2 pill.
Next day, take 1/6 pill. 1/2 + 1/6 = 2/3, so your daily average is 1/3.
Next day, take 1/12 pill. 2/3 + 1/12 = 3/4, so your daily average is 1/4.
And so on. (End)
From Bernard Schott, May 22 2020: (Start)
For an oblong number m >= 6 there exists a Euclidean division m = d*q + r with q < r < d which are in geometric progression, in this order, with a common integer ratio b. For b >= 2 and q >= 1, the Euclidean division is m = qb*(qb+1) = qb^2 * q + qb where (q, qb, qb^2) are in geometric progression.
Some examples with distinct ratios and quotients:
6 | 4 30 | 25 42 | 18
----- ----- -----
2 | 1 , 5 | 1 , 6 | 2 ,
and also:
42 | 12 420 | 100
----- -----
6 | 3 , 20 | 4 .
Some oblong numbers also satisfy a Euclidean division m = d*q + r with q < r < d that are in geometric progression in this order but with a common noninteger ratio b > 1 (see A335064). (End)
For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {2, 2n}]. For n=1, this collapses to [1; {2}]. - Magus K. Chu, Sep 09 2022
a(n-2) is the maximum irregularity over all trees with n vertices. The extremal graphs are stars. (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, May 29 2023
For n > 0, number of diagonals in a regular 2*(n+1)-gon that are not parallel to any edge (cf. A367204). - Paolo Xausa, Mar 30 2024
a(n-1) is the maximum Zagreb index over all trees with n vertices. The extremal graphs are stars. (The Zagreb index of a graph is the sum of the squares of the degrees over all vertices of the graph.) - Allan Bickle, Apr 11 2024
For n >= 1, a(n) is the determinant of the distance matrix of a cycle graph on 2*n + 1 vertices (if the length of the cycle is even such a determinant is zero). - Miquel A. Fiol, Aug 20 2024
For n > 1, the continued fraction expansion of sqrt(16*a(n)) is [2n+1; {1, 2n-1, 1, 8n+2}]. - Magus K. Chu, Nov 20 2024
For n>=2, a(n) is the number of faces on a n+1-zone rhombic zonohedron. Each pair of a collection of great circles on a sphere intersects at two points, so there are 2*binomial(n+1,2) intersections. The dual of the implied polyhedron is a rhombic zonohedron, its faces corresponding to the intersections. - Shel Kaphan, Aug 12 2025

Examples

			a(3) = 12, since 2(3)+2 = 8 has 4 partitions with exactly two parts: (7,1), (6,2), (5,3), (4,4). Taking the positive differences of the parts in each partition and adding, we get: 6 + 4 + 2 + 0 = 12. - _Wesley Ivan Hurt_, Jun 02 2013
G.f. = 2*x + 6*x^2 + 12*x^3 + 20*x^4 + 30*x^5 + 42*x^6 + 56*x^7 + ... - _Michael Somos_, May 22 2014
From _Miquel Cerda_, Dec 04 2016: (Start)
a(1) = 2, since 45-43 = 2;
a(2) = 6, since 47-45 = 2 and 47-43 = 4, then 2+4 = 6;
a(3) = 12, since 49-47 = 2, 49-45 = 4, and 49-43 = 6, then 2+4+6 = 12. (End)
		

References

  • W. W. Berman and D. E. Smith, A Brief History of Mathematics, 1910, Open Court, page 67.
  • J. H. Conway and R. K. Guy, The Book of Numbers, 1996, p. 34.
  • J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag.
  • L. E. Dickson, History of the Theory of Numbers, Vol. 1: Divisibility and Primality. New York: Chelsea, p. 357, 1952.
  • L. E. Dickson, History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Chelsea, pp. 6, 232-233, 350 and 407, 1952.
  • H. Eves, An Introduction to the History of Mathematics, revised, Holt, Rinehart and Winston, 1964, page 72.
  • Nicomachus of Gerasa, Introduction to Arithmetic, translation by Martin Luther D'Ooge, Ann Arbor, University of Michigan Press, 1938, p. 254.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 291.
  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), pp. 980-981.
  • C. S. Ogilvy and J. T. Anderson, Excursions in Number Theory, Oxford University Press, 1966, pp. 61-62.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 54-55.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • F. J. Swetz, From Five Fingers to Infinity, Open Court, 1994, p. 219.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 2-6.

Crossrefs

Partial sums of A005843 (even numbers). Twice triangular numbers (A000217).
1/beta(n, 2) in A061928.
A036689 and A036690 are subsequences. Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488. - Bruno Berselli, Jun 10 2013
Row n=2 of A185651.
Cf. A007745, A169810, A213541, A005369 (characteristic function).
Cf. A281026. - Bruno Berselli, Jan 16 2017
Cf. A045943 (4-cycles in triangular honeycomb acute knight graph), A028896 (5-cycles), A152773 (6-cycles).
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
A335064 is a subsequence.
Second column of A003506.
Cf. A002378, A046092, A028896 (irregularities of maximal k-degenerate graphs).
Cf. A347213 (Dgf at s=4).
Cf. A002378, A152811, A371912 (Zagreb indices of maximal k-degenerate graphs).

Programs

Formula

G.f.: 2*x/(1-x)^3. - Simon Plouffe in his 1992 dissertation.
a(n) = a(n-1) + 2*n, a(0) = 0.
Sum_{n >= 1} a(n) = n*(n+1)*(n+2)/3 (cf. A007290, partial sums).
Sum_{n >= 1} 1/a(n) = 1. (Cf. Tijdeman)
Sum_{n >= 1} (-1)^(n+1)/a(n) = log(4) - 1 = A016627 - 1 [Jolley eq (235)].
1 = 1/2 + Sum_{n >= 1} 1/(2*a(n)) = 1/2 + 1/4 + 1/12 + 1/24 + 1/40 + 1/60 + ... with partial sums: 1/2, 3/4, 5/6, 7/8, 9/10, 11/12, 13/14, ... - Gary W. Adamson, Jun 16 2003
a(n)*a(n+1) = a(n*(n+2)); e.g., a(3)*a(4) = 12*20 = 240 = a(3*5). - Charlie Marion, Dec 29 2003
Sum_{k = 1..n} 1/a(k) = n/(n+1). - Robert G. Wilson v, Feb 04 2005
a(n) = A046092(n)/2. - Zerinvary Lajos, Jan 08 2006
Log 2 = Sum_{n >= 0} 1/a(2n+1) = 1/2 + 1/12 + 1/30 + 1/56 + 1/90 + ... = (1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + ... = Sum_{n >= 0} (-1)^n/(n+1) = A002162. - Gary W. Adamson, Jun 22 2003
a(n) = A110660(2*n). - N. J. A. Sloane, Sep 21 2005
a(n-1) = n^2 - n = A000290(n) - A000027(n) for n >= 1. a(n) is the inverse (frequency distribution) sequence of A000194(n). - Mohammad K. Azarian, Jul 26 2007
(2, 6, 12, 20, 30, ...) = binomial transform of (2, 4, 2). - Gary W. Adamson, Nov 28 2007
a(n) = 2*Sum_{i=0..n} i = 2*A000217(n). - Artur Jasinski, Jan 09 2007, and Omar E. Pol, May 14 2008
a(n) = A006503(n) - A000292(n). - Reinhard Zumkeller, Sep 24 2008
a(n) = A061037(4*n) = (n+1/2)^2 - 1/4 = ((2n+1)^2 - 1)/4 = (A005408(n)^2 - 1)/4. - Paul Curtz, Oct 03 2008 and Klaus Purath, Jan 13 2022
a(0) = 0, a(n) = a(n-1) + 1 + floor(x), where x is the minimal positive solution to fract(sqrt(a(n-1) + 1 + x)) = 1/2. - Hieronymus Fischer, Dec 31 2008
E.g.f.: (x+2)*x*exp(x). - Geoffrey Critzer, Feb 06 2009
Product_{i >= 2} (1-1/a(i)) = -2*sin(Pi*A001622)/Pi = -2*sin(A094886)/A000796 = 2*A146481. - R. J. Mathar, Mar 12 2009, Mar 15 2009
E.g.f.: ((-x+1)*log(-x+1)+x)/x^2 also Integral_{x = 0..1} ((-x+1)*log(-x+1) + x)/x^2 = zeta(2) - 1. - Stephen Crowley, Jul 11 2009
a(A007018(n)) = A007018(n+1), i.e., A007018(n+1) = A007018(n)-th oblong numbers. - Jaroslav Krizek, Sep 13 2009
a(n) = floor((n + 1/2)^2). a(n) = A035608(n) + A004526(n+1). - Reinhard Zumkeller, Jan 27 2010
a(n) = 2*(2*A006578(n) - A035608(n)). - Reinhard Zumkeller, Feb 07 2010
a(n-1) = floor(n^5/(n^3 + n^2 + 1)). - Gary Detlefs, Feb 11 2010
For n > 1: a(n) = A173333(n+1, n-1). - Reinhard Zumkeller, Feb 19 2010
a(n) = A004202(A000217(n)). - Reinhard Zumkeller, Feb 12 2011
a(n) = A188652(2*n+1) + 1. - Reinhard Zumkeller, Apr 13 2011
For n > 0 a(n) = 1/(Integral_{x=0..Pi/2} 2*(sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011
a(n) = A002061(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(0) = 0, a(n) = A005408(A034856(n)) - A005408(n-1). - Ivan N. Ianakiev, Dec 06 2012
a(n) = A005408(A000096(n)) - A005408(n). - Ivan N. Ianakiev, Dec 07 2012
a(n) = A001318(n) + A085787(n). - Omar E. Pol, Jan 11 2013
Sum_{n >= 1} 1/(a(n))^(2s) = Sum_{t = 1..2*s} binomial(4*s - t - 1, 2*s - 1) * ( (1 + (-1)^t)*zeta(t) - 1). See Arxiv:1301.6293. - R. J. Mathar, Feb 03 2013
a(n)^2 + a(n+1)^2 = 2 * a((n+1)^2), for n > 0. - Ivan N. Ianakiev, Apr 08 2013
a(n) = floor(n^2 * e^(1/n)) and a(n-1) = floor(n^2 / e^(1/n)). - Richard R. Forberg, Jun 22 2013
a(n) = 2*C(n+1, 2), for n >= 0. - Felix P. Muga II, Mar 11 2014
A005369(a(n)) = 1. - Reinhard Zumkeller, Jul 05 2014
Binomial transform of [0, 2, 2, 0, 0, 0, ...]. - Alois P. Heinz, Mar 10 2015
a(2n) = A002943(n) for n >= 0, a(2n-1) = A002939(n) for n >= 1. - M. F. Hasler, Oct 11 2015
For n > 0, a(n) = 1/(Integral_{x=0..1} (x^(n-1) - x^n) dx). - Rick L. Shepherd, Oct 26 2015
a(n) = A005902(n) - A007588(n). - Peter M. Chema, Jan 09 2016
For n > 0, a(n) = lim_{m -> oo} (1/m)*1/(Sum_{i=m*n..m*(n+1)} 1/i^2), with error of ~1/m. - Richard R. Forberg, Jul 27 2016
From Ilya Gutkovskiy, Jul 28 2016: (Start)
Dirichlet g.f.: zeta(s-2) + zeta(s-1).
Convolution of nonnegative integers (A001477) and constant sequence (A007395).
Sum_{n >= 0} a(n)/n! = 3*exp(1). (End)
From Charlie Marion, Mar 06 2020: (Start)
a(n)*a(n+2k-1) + (n+k)^2 = ((2n+1)*k + n^2)^2.
a(n)*a(n+2k) + k^2 = ((2n+1)*k + a(n))^2. (End)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi. - Amiram Eldar, Jan 20 2021
A generalization of the Dec 29 2003 formula, a(n)*a(n+1) = a(n*(n+2)), follows. a(n)*a(n+k) = a(n*(n+k+1)) + (k-1)*n*(n+k+1). - Charlie Marion, Jan 02 2023
a(n) = A016742(n) - A049450(n). - Leo Tavares, Mar 15 2025

Extensions

Additional comments from Michael Somos
Comment and cross-reference added by Christopher Hunt Gribble, Oct 13 2009

A000129 Pell numbers: a(0) = 0, a(1) = 1; for n > 1, a(n) = 2*a(n-1) + a(n-2).

Original entry on oeis.org

0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428, 38613965, 93222358, 225058681, 543339720, 1311738121, 3166815962, 7645370045, 18457556052, 44560482149, 107578520350, 259717522849
Offset: 0

Views

Author

Keywords

Comments

Sometimes also called lambda numbers.
Also denominators of continued fraction convergents to sqrt(2): 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129.
Number of lattice paths from (0,0) to the line x=n-1 consisting of U=(1,1), D=(1,-1) and H=(2,0) steps (i.e., left factors of Grand Schroeder paths); for example, a(3)=5, counting the paths H, UD, UU, DU and DD. - Emeric Deutsch, Oct 27 2002
a(2*n) with b(2*n) := A001333(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = +1 (see Emerson reference). a(2*n+1) with b(2*n+1) := A001333(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = -1.
Bisection: a(2*n+1) = T(2*n+1, sqrt(2))/sqrt(2) = A001653(n), n >= 0 and a(2*n) = 2*S(n-1,6) = 2*A001109(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003
Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the denominators. - Amarnath Murthy, Mar 22 2003
This is also the Horadam sequence (0,1,1,2). Limit_{n->oo} a(n)/a(n-1) = sqrt(2) + 1 = A014176. - Ross La Haye, Aug 18 2003
Number of 132-avoiding two-stack sortable permutations.
From Herbert Kociemba, Jun 02 2004: (Start)
For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 3.
Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 2. (End)
Counts walks of length n from a vertex of a triangle to another vertex to which a loop has been added. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
Apart from initial terms, Pisot sequence P(2,5). See A008776 for definition of Pisot sequences. - David W. Wilson
Sums of antidiagonals of A038207 [Pascal's triangle squared]. - Ross La Haye, Oct 28 2004
The Pell primality test is "If N is an odd prime, then P(N)-Kronecker(2,N) is divisible by N". "Most" composite numbers fail this test, so it makes a useful pseudoprimality test. The odd composite numbers which are Pell pseudoprimes (i.e., that pass the above test) are in A099011. - Jack Brennen, Nov 13 2004
a(n) = sum of n-th row of triangle in A008288 = A094706(n) + A000079(n). - Reinhard Zumkeller, Dec 03 2004
Pell trapezoids (cf. A084158); for n > 0, A001109(n) = (a(n-1) + a(n+1))*a(n)/2; e.g., 1189 = (12+70)*29/2. - Charlie Marion, Apr 01 2006
(0!a(1), 1!a(2), 2!a(3), 3!a(4), ...) and (1,-2,-2,0,0,0,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Oct 29 2007
Let C = (sqrt(2)+1) = 2.414213562..., then for n > 1, C^n = a(n)*(1/C) + a(n+1). Example: C^3 = 14.0710678... = 5*(0.414213562...) + 12. Let X = the 2 X 2 matrix [0, 1; 1, 2]; then X^n * [1, 0] = [a(n-1), a(n); a(n), a(n+1)]. a(n) = numerator of n-th convergent to (sqrt(2)-1) = 0.414213562... = [2, 2, 2, ...], the convergents being [1/2, 2/5, 5/12, ...]. - Gary W. Adamson, Dec 21 2007
A = sqrt(2) = 2/2 + 2/5 + 2/(5*29) + 2/(29*169) + 2/(169*985) + ...; B = ((5/2) - sqrt(2)) = 2/2 + 2/(2*12) + 2/(12*70) + 2/(70*408) + 2/(408*2378) + ...; A+B = 5/2. C = 1/2 = 2/(1*5) + 2/(2*12) + 2/(5*29) + 2/(12*70) + 2/(29*169) + ... - Gary W. Adamson, Mar 16 2008
From Clark Kimberling, Aug 27 2008: (Start)
Related convergents (numerator/denominator):
lower principal convergents: A002315/A001653
upper principal convergents: A001541/A001542
intermediate convergents: A052542/A001333
lower intermediate convergents: A005319/A001541
upper intermediate convergents: A075870/A002315
principal and intermediate convergents: A143607/A002965
lower principal and intermediate convergents: A143608/A079496
upper principal and intermediate convergents: A143609/A084068. (End)
Equals row sums of triangle A143808 starting with offset 1. - Gary W. Adamson, Sep 01 2008
Binomial transform of the sequence:= 0,1,0,2,0,4,0,8,0,16,..., powers of 2 alternating with zeros. - Philippe Deléham, Oct 28 2008
a(n) is also the sum of the n-th row of the triangle formed by starting with the top two rows of Pascal's triangle and then each next row has a 1 at both ends and the interior values are the sum of the three numbers in the triangle above that position. - Patrick Costello (pat.costello(AT)eku.edu), Dec 07 2008
Starting with offset 1 = eigensequence of triangle A135387 (an infinite lower triangular matrix with (2,2,2,...) in the main diagonal and (1,1,1,...) in the subdiagonal). - Gary W. Adamson, Dec 29 2008
Starting with offset 1 = row sums of triangle A153345. - Gary W. Adamson, Dec 24 2008
From Charlie Marion, Jan 07 2009: (Start)
In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
let a(k,0) = 1, a(k,1) = 2k; for n > 0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2)
and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
let b(k,0) = 1, b(k,1) = 2k+1; for n > 0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2)
and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
for example, if k=1 and n=3, then a(1,n) = a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
(End)
Starting with offset 1 = row sums of triangle A155002, equivalent to the statement that the Fibonacci sequence convolved with the Pell sequence prefaced with a "1": (1, 1, 2, 5, 12, 29, ...) = (1, 2, 5, 12, 29, ...). - Gary W. Adamson, Jan 18 2009
It appears that P(p) == 8^((p-1)/2) (mod p), p = prime; analogous to [Schroeder, p. 90]: Fp == 5^((p-1)/2) (mod p). Example: Given P(11) = 5741, == 8^5 (mod 11). Given P(17) = 11336689, == 8^8 (mod 17) since 17 divides (8^8 - P(17)). - Gary W. Adamson, Feb 21 2009
Equals eigensequence of triangle A154325. - Gary W. Adamson, Feb 12 2009
Another combinatorial interpretation of a(n-1) arises from a simple tiling scenario. Namely, a(n-1) gives the number of ways of tiling a 1 X n rectangle with indistinguishable 1 X 2 rectangles and 1 X 1 squares that come in two varieties, say, A and B. For example, with C representing the 1 X 2 rectangle, we obtain a(4)=12 from AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB, AC, BC, CA and CB. - Martin Griffiths, Apr 25 2009
a(n+1) = 2*a(n) + a(n-1), a(1)=1, a(2)=2 was used by Theon from Smyrna. - Sture Sjöstedt, May 29 2009
The n-th Pell number counts the perfect matchings of the edge-labeled graph C_2 x P_(n-1), or equivalently, the number of domino tilings of a 2 X (n-1) cylindrical grid. - Sarah-Marie Belcastro, Jul 04 2009
As a fraction: 1/79 = 0.0126582278481... or 1/9799 = 0.000102051229...(1/119 and 1/10199 for sequence in reverse). - Mark Dols, May 18 2010
Limit_{n->oo} (a(n)/a(n-1) - a(n-1)/a(n)) tends to 2.0. Example: a(7)/a(6) - a(6)/a(7) = 169/70 - 70/169 = 2.0000845... - Gary W. Adamson, Jul 16 2010
Numbers k such that 2*k^2 +- 1 is a square. - Vincenzo Librandi, Jul 18 2010
Starting (1, 2, 5, ...) = INVERTi transform of A006190: (1, 3, 10, 33, 109, ...). - Gary W. Adamson, Aug 06 2010
[u,v] = [a(n), a(n-1)] generates all Pythagorean triples [u^2-v^2, 2uv, u^2+v^2] whose legs differ by 1. - James R. Buddenhagen, Aug 14 2010
An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 21 and 336, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A078057. - Johannes W. Meijer, Aug 15 2010
Let the 2 X 2 square matrix A=[2, 1; 1, 0] then a(n) = the (1,1) element of A^(n-1). - Carmine Suriano, Jan 14 2011
Define a t-circle to be a first-quadrant circle tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the t-circle with radius 1. Then for n > 0, define C(n) to be the next larger t-circle which is tangent to C(n - 1). C(n) has radius A001333(2n) + a(2n)*sqrt(2) and each of the coordinates of its point of intersection with C(n + 1) is a(2n + 1) + (A001333(2n + 1)*sqrt(2))/2. See similar Comments for A001109 and A001653, Sep 14 2005. - Charlie Marion, Jan 18 2012
A001333 and A000129 give the diagonal numbers described by Theon from Smyrna. - Sture Sjöstedt, Oct 20 2012
Pell numbers could also be called "silver Fibonacci numbers", since, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio, while a(n+1) = ceiling(delta*a(n)), if n is even and a(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1+sqrt(2) is the silver ratio. - Vladimir Shevelev, Feb 22 2013
a(n) is the number of compositions (ordered partitions) of n-1 into two sorts of 1's and one sort of 2's. Example: the a(3)=5 compositions of 3-1=2 are 1+1, 1+1', 1'+1, 1'+1', and 2. - Bob Selcoe, Jun 21 2013
Between every two consecutive squares of a 1 X n array there is a flap that can be folded over one of the two squares. Two flaps can be lowered over the same square in 2 ways, depending on which one is on top. The n-th Pell number counts the ways n-1 flaps can be lowered. For example, a sideway representation for the case n = 3 squares and 2 flaps is \\., .//, \./, ./., .\., where . is an empty square. - Jean M. Morales, Sep 18 2013
Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A005319(k)*(a(n-2k+1) - a(n-2k)) + a(n-4k) = A075870(k)*(a(n-2k+2) - a(n-2k+1)) - a(n-4k+2). - Charlie Marion, Nov 26 2013
An alternative formulation of the combinatorial tiling interpretation listed above: Except for n=0, a(n-1) is the number of ways of partial tiling a 1 X n board with 1 X 1 squares and 1 X 2 dominoes. - Matthew Lehman, Dec 25 2013
Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A077444(k)*a(n-2k+1) + a(n-4k+2). This formula generalizes the formula used to define this sequence. - Charlie Marion, Jan 30 2014
a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 1, 1; 1, 1, 1; 0, 1, 1], [0, 1, 1; 0, 1, 1; 1, 1, 1], [0, 1, 0; 1, 1, 1; 1, 1, 1] or [0, 0, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
a(n+1) counts closed walks on K2 containing two loops on the other vertex. Equivalently the (1,1) entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1;1,2). - David Neil McGrath, Oct 28 2014
For n >= 1, a(n) equals the number of ternary words of length n-1 avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)). - Tom Edgar, Jan 28 2015
A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Jan 03 2017
a(n) is the number of compositions (ordered partitions) of n-1 into two kinds of parts, n and n', when the order of the 1 does not matter, or equivalently, when the order of the 1' does not matter. Example: When the order of the 1 does not matter, the a(3)=5 compositions of 3-1=2 are 1+1, 1+1'=1+1, 1'+1', 2 and 2'. (Contrast with entry from Bob Selcoe dated Jun 21 2013.) - Gregory L. Simay, Sep 07 2017
Number of weak orderings R on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n and for which {1,...,n} has exactly one minimal element for the weak ordering R. - J. Devillet, Sep 28 2017
Also the number of matchings in the (n-1)-centipede graph. - Eric W. Weisstein, Sep 30 2017
Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0)=1. Let A_1(r,n) = Sum_{j=0..n} A(r,j) and let A_s(r,n) = Sum_{j=0..n} A_(s-1)(r,j). Then A_0(1,n) + A_2(3,n-4) + A_4(5,n-8) + ... + A_(2j) (2j+1, n-4j) = a(n) without the initial 0. - Gregory L. Simay, May 25 2018
(1, 2, 5, 12, 29, ...) is the fourth INVERT transform of (1, -2, 5, -12, 29, ...), as shown in A073133. - Gary W. Adamson, Jul 17 2019
Number of 2-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
Also called the 2-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence. - Michael A. Allen, Jan 23 2023
Named by Lucas (1878) after the English mathematician John Pell (1611-1685). - Amiram Eldar, Oct 02 2023
a(n) is the number of compositions of n when there are F(i) parts of size i, with i,n >= 1, F(n) the Fibonacci numbers, A000045(n) (see example below). - Enrique Navarrete, Dec 15 2023

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 12*x^4 + 29*x^5 + 70*x^6 + 169*x^7 + 408*x^8 + 985*x^9 + ...
From _Enrique Navarrete_, Dec 15 2023: (Start)
From the comment on compositions with Fibonacci number of parts, F(n), there are F(1)=1 type of 1, F(2)=1 type of 2, F(3)=2 types of 3, F(4)=3 types of 4, F(5)=5 types of 5 and F(6)=8 types of 6.
The following table gives the number of compositions of n=6 with Fibonacci number of parts:
Composition, number of such compositions, number of compositions of this type:
6,           1,     8;
5+1,         2,    10;
4+2,         2,     6;
3+3,         1,     4;
4+1+1,       3,     9;
3+2+1,       6,    12;
2+2+2,       1,     1;
3+1+1+1,     4,     8;
2+2+1+1,     6,     6;
2+1+1+1+1,   5,     5;
1+1+1+1+1+1, 1,     1;
for a total of a(6)=70 compositions of n=6. (End).
		

References

  • J. Austin and L. Schneider, Generalized Fibonacci sequences in Pythagorean triple preserving sequences, Fib. Q., 58:1 (2020), 340-350.
  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 76.
  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 941.
  • J. M. Borwein, D. H. Bailey, and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 53.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 204.
  • John Derbyshire, Prime Obsession, Joseph Henry Press, 2004, see p. 16.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.1.
  • Shaun Giberson and Thomas J. Osler, Extending Theon's Ladder to Any Square Root, Problem 3858, Elementa, No. 4 1996.
  • R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 288.
  • Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • Paulo Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 43.
  • Paulo Ribenboim, My Numbers, My Friends: Popular Lectures on Number Theory, Springer-Verlag, NY, 2000, p. 3.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 46, 61.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
  • Manfred R. Schroeder, "Number Theory in Science and Communication", 5th ed., Springer-Verlag, 2009, p. 90.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 34.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 62.

Crossrefs

Partial sums of A001333.
2nd row of A172236.
a(n) = A054456(n-1, 0), n>=1 (first column of triangle).
Cf. A175181 (Pisano periods), A214028 (Entry points), A214027 (number of zeros in a fundamental period).
A077985 is a signed version.
INVERT transform of Fibonacci numbers (A000045).
Cf. A038207.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A048739.
Cf. A073133.
Cf. A041085.
Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), this sequence (k=2), A006190 (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), A154597 (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).

Programs

  • GAP
    a := [0,1];; for n in [3..10^3] do a[n] := 2 * a[n-1] + a[n-2]; od; A000129 := a; # Muniru A Asiru, Oct 16 2017
    
  • Haskell
    a000129 n = a000129_list !! n
    a000129_list = 0 : 1 : zipWith (+) a000129_list (map (2 *) $ tail a000129_list)
    -- Reinhard Zumkeller, Jan 05 2012, Feb 05 2011
    
  • Magma
    [0] cat [n le 2 select n else 2*Self(n-1) + Self(n-2): n in [1..35]]; // Vincenzo Librandi, Aug 08 2015
    
  • Maple
    A000129 := proc(n) option remember; if n <=1 then n; else 2*procname(n-1)+procname(n-2); fi; end;
    a:= n-> (<<2|1>, <1|0>>^n)[1, 2]: seq(a(n), n=0..40); # Alois P. Heinz, Aug 01 2008
    A000129 := n -> `if`(n<2, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1)):
    seq(simplify(A000129(n)), n=0..31); # Peter Luschny, Dec 17 2015
  • Mathematica
    CoefficientList[Series[x/(1 - 2*x - x^2), {x, 0, 60}], x] (* Stefan Steinerberger, Apr 08 2006 *)
    Expand[Table[((1 + Sqrt[2])^n - (1 - Sqrt[2])^n)/(2Sqrt[2]), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
    LinearRecurrence[{2, 1}, {0, 1}, 60] (* Harvey P. Dale, Jan 04 2012 *)
    a[ n_] := With[ {s = Sqrt@2}, ((1 + s)^n - (1 - s)^n) / (2 s)] // Simplify; (* Michael Somos, Jun 01 2013 *)
    Table[Fibonacci[n, 2], {n, 0, 20}] (* Vladimir Reshetnikov, May 08 2016 *)
    Fibonacci[Range[0, 20], 2] (* Eric W. Weisstein, Sep 30 2017 *)
    a[ n_] := ChebyshevU[n - 1, I] / I^(n - 1); (* Michael Somos, Oct 30 2021 *)
  • Maxima
    a[0]:0$
    a[1]:1$
    a[n]:=2*a[n-1]+a[n-2]$
    A000129(n):=a[n]$
    makelist(A000129(n),n,0,30); /* Martin Ettl, Nov 03 2012 */
    
  • Maxima
    makelist((%i)^(n-1)*ultraspherical(n-1,1,-%i),n,0,24),expand; /* Emanuele Munarini, Mar 07 2018 */
    
  • PARI
    for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[2, 1]; if (a > 10^(10^3 - 6), break); write("b000129.txt", n, " ", a)); \\ Harry J. Smith, Jun 12 2009
    
  • PARI
    {a(n) = imag( (1 + quadgen( 8))^n )}; /* Michael Somos, Jun 01 2013 */
    
  • PARI
    {a(n) = if( n<0, -(-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [2, 1]}; /* Michael Somos, Jun 01 2013 */
    
  • PARI
    a(n)=([2, 1; 1, 0]^n)[2,1] \\ Charles R Greathouse IV, Mar 04 2014
    
  • PARI
    {a(n) = polchebyshev(n-1, 2, I) / I^(n-1)}; /* Michael Somos, Oct 30 2021 */
    
  • Python
    from itertools import islice
    def A000129_gen(): # generator of terms
        a, b = 0, 1
        yield from [a,b]
        while True:
            a, b = b, a+2*b
            yield b
    A000129_list = list(islice(A000129_gen(),20)) # Chai Wah Wu, Jan 11 2022
  • Sage
    [lucas_number1(n, 2, -1) for n in range(30)]  # Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: x/(1 - 2*x - x^2). - Simon Plouffe in his 1992 dissertation.
a(2n+1)=A001653(n). a(2n)=A001542(n). - Ira Gessel, Sep 27 2002
G.f.: Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (2*k + x)/(1 + 2*k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 1 + k)/(1 + k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 3 - k)/(1 - k*x) ) may all be proved using telescoping series. - Peter Bala, Jan 04 2015
a(n) = 2*a(n-1) + a(n-2), a(0)=0, a(1)=1.
a(n) = ((1 + sqrt(2))^n - (1 - sqrt(2))^n)/(2*sqrt(2)).
For initial values a(0) and a(1), a(n) = ((a(0)*sqrt(2)+a(1)-a(0))*(1+sqrt(2))^n + (a(0)*sqrt(2)-a(1)+a(0))*(1-sqrt(2))^n)/(2*sqrt(2)). - Shahreer Al Hossain, Aug 18 2019
a(n) = integer nearest a(n-1)/(sqrt(2) - 1), where a(0) = 1. - Clark Kimberling
a(n) = Sum_{i, j, k >= 0: i+j+2k = n} (i+j+k)!/(i!*j!*k!).
a(n)^2 + a(n+1)^2 = a(2n+1) (1999 Putnam examination).
a(2n) = 2*a(n)*A001333(n). - John McNamara, Oct 30 2002
a(n) = ((-i)^(n-1))*S(n-1, 2*i), with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x)=0, S(-2, x)= -1.
Binomial transform of expansion of sinh(sqrt(2)x)/sqrt(2). E.g.f.: exp(x)sinh(sqrt(2)x)/sqrt(2). - Paul Barry, May 09 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k+1)*2^k. - Paul Barry, May 13 2003
a(n-2) + a(n) = (1 + sqrt(2))^(n-1) + (1 - sqrt(2))^(n-1) = A002203(n-1). (A002203(n))^2 - 8(a(n))^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003
Unreduced g.f.: x(1+x)/(1 - x - 3x^2 - x^3); a(n) = a(n-1) + 3*a(n-2) + a(n-2). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2k). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
Apart from initial terms, inverse binomial transform of A052955. - Paul Barry, May 23 2004
a(n)^2 + a(n+2k+1)^2 = A001653(k)*A001653(n+k); e.g., 5^2 + 70^2 = 5*985. - Charlie Marion Aug 03 2005
a(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)*(1+(-1)^(n-k))*2^k/2. - Paul Barry, Aug 28 2005
a(n) = a(n-1) + A001333(n-1) = A001333(n) - a(n-1) = A001109(n)/A001333(n) = sqrt(A001110(n)/A001333(n)^2) = ceiling(sqrt(A001108(n)/2)). - Henry Bottomley, Apr 18 2000
a(n) = F(n, 2), the n-th Fibonacci polynomial evaluated at x=2. - T. D. Noe, Jan 19 2006
Define c(2n) = -A001108(n), c(2n+1) = -A001108(n+1) and d(2n) = d(2n+1) = A001652(n); then ((-1)^n)*(c(n) + d(n)) = a(n). [Proof given by Max Alekseyev.] - Creighton Dement, Jul 21 2005
a(r+s) = a(r)*a(s+1) + a(r-1)*a(s). - Lekraj Beedassy, Sep 03 2006
a(n) = (b(n+1) + b(n-1))/n where {b(n)} is the sequence A006645. - Sergio Falcon, Nov 22 2006
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = a(n) = Pell numbers, L(n) = A002203 = companion Pell numbers (A002203):
For a >= b and odd b, F(a+b) + F(a-b) = L(a)*F(b).
For a >= b and even b, F(a+b) + F(a-b) = F(a)*L(b).
For a >= b and odd b, F(a+b) - F(a-b) = F(a)*L(b).
For a >= b and even b, F(a+b) - F(a-b) = L(a)*F(b).
F(n+m) + (-1)^m*F(n-m) = F(n)*L(m).
F(n+m) - (-1)^m*F(n-m) = L(n)*F(m).
F(n+m+k) + (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = F(n)*L(m)*L(k).
F(n+m+k) - (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = L(n)*L(m)*F(k).
F(n+m+k) + (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = L(n)*F(m)*L(k).
F(n+m+k) - (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = 8*F(n)*F(m)*F(k). (End)
a(n+1)*a(n) = 2*Sum_{k=0..n} a(k)^2 (a similar relation holds for A001333). - Creighton Dement, Aug 28 2007
a(n+1) = Sum_{k=0..n} binomial(n+1,2k+1) * 2^k = Sum_{k=0..n} A034867(n,k) * 2^k = (1/n!) * Sum_{k=0..n} A131980(n,k) * 2^k. - Tom Copeland, Nov 30 2007
Equals row sums of unsigned triangle A133156. - Gary W. Adamson, Apr 21 2008
a(n) (n >= 3) is the determinant of the (n-1) X (n-1) tridiagonal matrix with diagonal entries 2, superdiagonal entries 1 and subdiagonal entries -1. - Emeric Deutsch, Aug 29 2008
a(n) = A000045(n) + Sum_{k=1..n-1} A000045(k)*a(n-k). - Roger L. Bagula and Gary W. Adamson, Sep 07 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
fract((1+sqrt(2))^n) = (1/2)*(1 + (-1)^n) - (-1)^n*(1+sqrt(2))^(-n) = (1/2)*(1 + (-1)^n) - (1-sqrt(2))^n.
See A001622 for a general formula concerning the fractional parts of powers of numbers x > 1, which satisfy x - x^(-1) = floor(x).
a(n) = round((1+sqrt(2))^n/(2*sqrt(2))) for n > 0. (End) [last formula corrected by Josh Inman, Mar 05 2024]
a(n) = ((4+sqrt(18))*(1+sqrt(2))^n + (4-sqrt(18))*(1-sqrt(2))^n)/4 offset 0. - Al Hakanson (hawkuu(AT)gmail.com), Aug 08 2009
If p[i] = Fibonacci(i) and if A is the Hessenberg matrix of order n defined by A[i,j] = p[j-i+1] when i<=j, A[i,j]=-1 when i=j+1, and A[i,j]=0 otherwise, then, for n >= 1, a(n) = det A. - Milan Janjic, May 08 2010
a(n) = 3*a(n-1) - a(n-2) - a(n-3), n > 2. - Gary Detlefs, Sep 09 2010
From Charlie Marion, Apr 13 2011: (Start)
a(n) = 2*(a(2k-1) + a(2k))*a(n-2k) - a(n-4k).
a(n) = 2*(a(2k) + a(2k+1))*a(n-2k-1) + a(n-4k-2). (End)
G.f.: x/(1 - 2*x - x^2) = sqrt(2)*G(0)/4; G(k) = ((-1)^k) - 1/(((sqrt(2) + 1)^(2*k)) - x*((sqrt(2) + 1)^(2*k))/(x + ((sqrt(2) - 1)^(2*k + 1))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 02 2011
In general, for n > k, a(n) = a(k+1)*a(n-k) + a(k)*a(n-k-1). See definition of Pell numbers and the formula for Sep 04 2008. - Charlie Marion, Jan 17 2012
Sum{n>=1} (-1)^(n-1)/(a(n)*a(n+1)) = sqrt(2) - 1. - Vladimir Shevelev, Feb 22 2013
From Vladimir Shevelev, Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = a(n) + sqrt(2*a^2(n) + (-1)^n);
(2) a(n+1)^2 - a(n)*a(n+2) = (-1)^n;
(3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(4) a(n)/a(n+1) = sqrt(2) - 1 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
a(-n) = -(-1)^n * a(n). - Michael Somos, Jun 01 2013
G.f.: G(0)/(2+2*x) - 1/(1+x), where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 10 2013
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 + x)/( x*(4*k+4 + x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013
a(n) = Sum_{r=0..n-1} Sum_{k=0..n-r-1} binomial(r+k,k)*binomial(k,n-k-r-1). - Peter Luschny, Nov 16 2013
a(n) = Sum_{k=1,3,5,...<=n} C(n,k)*2^((k-1)/2). - Vladimir Shevelev, Feb 06 2014
a(2n) = 2*a(n)*(a(n-1) + a(n)). - John Blythe Dobson, Mar 08 2014
a(k*n) = a(k)*a(k*n-k+1) + a(k-1)*a(k*n-k). - Charlie Marion, Mar 27 2014
a(k*n) = 2*a(k)*(a(k*n-k)+a(k*n-k-1)) + (-1)^k*a(k*n-2k). - Charlie Marion, Mar 30 2014
a(n+1) = (1+sqrt(2))*a(n) + (1-sqrt(2))^n. - Art DuPre, Apr 04 2014
a(n+1) = (1-sqrt(2))*a(n) + (1+sqrt(2))^n. - Art DuPre, Apr 04 2014
a(n) = F(n) + Sum_{k=1..n} F(k)*a(n-k), n >= 0 where F(n) the Fibonacci numbers A000045. - Ralf Stephan, May 23 2014
a(n) = round(sqrt(a(2n) + a(2n-1)))/2. - Richard R. Forberg, Jun 22 2014
a(n) = Product_{k divides n} A008555(k). - Tom Edgar, Jan 28 2015
a(n+k)^2 - A002203(k)*a(n)*a(n+k) + (-1)^k*a(n)^2 = (-1)^n*a(k)^2. - Alexander Samokrutov, Aug 06 2015
a(n) = 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1) for n >= 2. - Peter Luschny, Dec 17 2015
a(n+1) = Sum_{k=0..n} binomial(n,k)*2^floor(k/2). - Tony Foster III, May 07 2017
a(n) = exp((i*Pi*n)/2)*sinh(n*arccosh(-i))/sqrt(2). - Peter Luschny, Mar 07 2018
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)^2 - a(n-2)^2 = 2*a(n-1)*(a(n) + a(n-2)) (see A005319);
(2) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(3) Sum_{k=2..n+1} a(k)*a(k-1) = a(n+1)^2 if n is odd, else a(n+1)^2 - 1 if n is even;
(4) a(n) - a(n-2*k+1) = (A077444(k) - 1)*a(n-2*k+1) + a(n-4*k+2);
(5) Sum_{k=n..n+9} a(k) = 41*A001333(n+5). (End)
From Kai Wang, Dec 30 2019: (Start)
a(m+r)*a(n+s) - a(m+s)*a(n+r) = -(-1)^(n+s)*a(m-n)*a(r-s).
a(m+r)*a(n+s) + a(m+s)*a(n+r) = (2*A002203(m+n+r+s) - (-1)^(n+s)*A002203(m-n)*A002203(r-s))/8.
A002203(m+r)*A002203(n+s) - A002203(m+s)*A002203(n+r) = (-1)^(n+s)*8*a(m-n)*a(r-s).
A002203(m+r)*A002203(n+s) - 8*a(m+s)*a(n+r) = (-1)^(n+s)*A002203(m-n)*A002203(r-s).
A002203(m+r)*A002203(n+s) + 8*a(m+s)*a(n+r) = 2*A002203(m+n+r+s)+ (-1)^(n+s)*8*a(m-n)*a(r-s). (End)
From Kai Wang, Jan 12 2020: (Start)
a(n)^2 - a(n+1)*a(n-1) = (-1)^(n-1).
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = (-1)^n*a(m-n).
a(m-n) = (-1)^n (a(m)*A002203(n) - A002203(m)*a(n))/2.
a(m+n) = (a(m)*A002203(n) + A002203(m)*a(n))/2.
A002203(n)^2 - A002203(n+r)*A002203(n-r) = (-1)^(n-r-1)*8*a(r)^2.
A002203(m)*A002203(n+1) - A002203(m+1)*A002203(n) = (-1)^(n-1)*8*a(m-n).
A002203(m-n) = (-1)^(n)*(A002203(m)*A002203(n) - 8*a(m)*a(n) )/2.
A002203(m+n) = (A002203(m)*A002203(n) + 8*a(m)*a(n) )/2. (End)
From Kai Wang, Mar 03 2020: (Start)
Sum_{m>=1} arctan(2/a(2*m+1)) = arctan(1/2).
Sum_{m>=2} arctan(2/a(2*m+1)) = arctan(1/12).
In general, for n > 0,
Sum_{m>=n} arctan(2/a(2*m+1)) = arctan(1/a(2*n)). (End)
a(n) = (A001333(n+3*k) + (-1)^(k-1)*A001333(n-3*k)) / (20*A041085(k-1)) for any k>=1. - Paul Curtz, Jun 23 2021
Sum_{i=0..n} a(i)*J(n-i) = (a(n+1) + a(n) - J(n+2))/2 for J(n) = A001045(n). - Greg Dresden, Jan 05 2022
From Peter Bala, Aug 20 2022: (Start)
Sum_{n >= 1} 1/(a(2*n) + 1/a(2*n)) = 1/2.
Sum_{n >= 1} 1/(a(2*n+1) - 1/a(2*n+1)) = 1/4. Both series telescope - see A075870 and A005319.
Product_{n >= 1} ( 1 + 2/a(2*n) ) = 1 + sqrt(2).
Product_{n >= 2} ( 1 - 2/a(2*n) ) = (1/3)*(1 + sqrt(2)). (End)
G.f. = 1/(1 - Sum_{k>=1} Fibonacci(k)*x^k). - Enrique Navarrete, Dec 17 2023
Sum_{n >=1} 1/a(n) = 1.84220304982752858079237158327980838... - R. J. Mathar, Feb 05 2024
a(n) = ((3^(n+1) + 1)^(n-1) mod (9^(n+1) - 2)) mod (3^(n+1) - 1). - Joseph M. Shunia, Jun 06 2024

A005843 The nonnegative even numbers: a(n) = 2n.

Original entry on oeis.org

0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120
Offset: 0

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-2, -4, -6, -8, -10, -12, -14, ... are the trivial zeros of the Riemann zeta function. - Vivek Suri (vsuri(AT)jhu.edu), Jan 24 2008
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-2) is the number of 2-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
A134452(a(n)) = 0; A134451(a(n)) = 2 for n > 0. - Reinhard Zumkeller, Oct 27 2007
Omitting the initial zero gives the number of prime divisors with multiplicity of product of terms of n-th row of A077553. - Ray Chandler, Aug 21 2003
A059841(a(n))=1, A000035(a(n))=0. - Reinhard Zumkeller, Sep 29 2008
(APSO) Alternating partial sums of (a-b+c-d+e-f+g...) = (a+b+c+d+e+f+g...) - 2*(b+d+f...), it appears that APSO(A005843) = A052928 = A002378 - 2*(A116471), with A116471=2*A008794. - Eric Desbiaux, Oct 28 2008
A056753(a(n)) = 1. - Reinhard Zumkeller, Aug 23 2009
Twice the nonnegative numbers. - Juri-Stepan Gerasimov, Dec 12 2009
The number of hydrogen atoms in straight-chain (C(n)H(2n+2)), branched (C(n)H(2n+2), n > 3), and cyclic, n-carbon alkanes (C(n)H(2n), n > 2). - Paul Muljadi, Feb 18 2010
For n >= 1; a(n) = the smallest numbers m with the number of steps n of iterations of {r - (smallest prime divisor of r)} needed to reach 0 starting at r = m. See A175126 and A175127. A175126(a(n)) = A175126(A175127(n)) = n. Example (a(4)=8): 8-2=6, 6-2=4, 4-2=2, 2-2=0; iterations has 4 steps and number 8 is the smallest number with such result. - Jaroslav Krizek, Feb 15 2010
For n >= 1, a(n) = numbers k such that arithmetic mean of the first k positive integers is not integer. A040001(a(n)) > 1. See A145051 and A040001. - Jaroslav Krizek, May 28 2010
Union of A179082 and A179083. - Reinhard Zumkeller, Jun 28 2010
a(k) is the (Moore lower bound on and the) order of the (k,4)-cage: the smallest k-regular graph having girth four: the complete bipartite graph with k vertices in each part. - Jason Kimberley, Oct 30 2011
For n > 0: A048272(a(n)) <= 0. - Reinhard Zumkeller, Jan 21 2012
Let n be the number of pancakes that have to be divided equally between n+1 children. a(n) is the minimal number of radial cuts needed to accomplish the task. - Ivan N. Ianakiev, Sep 18 2013
For n > 0, a(n) is the largest number k such that (k!-n)/(k-n) is an integer. - Derek Orr, Jul 02 2014
a(n) when n > 2 is also the number of permutations simultaneously avoiding 213, 231 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl Aug 07 2014
It appears that for n > 2, a(n) = A020482(n) + A002373(n), where all sequences are infinite. This is consistent with Goldbach's conjecture, which states that every even number > 2 can be expressed as the sum of two prime numbers. - Bob Selcoe, Mar 08 2015
Number of partitions of 4n into exactly 2 parts. - Colin Barker, Mar 23 2015
Number of neighbors in von Neumann neighborhood. - Dmitry Zaitsev, Nov 30 2015
Unique solution b( ) of the complementary equation a(n) = a(n-1)^2 - a(n-2)*b(n-1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences. - Clark Kimberling, Nov 21 2017
Also the maximum number of non-attacking bishops on an (n+1) X (n+1) board (n>0). (Cf. A000027 for rooks and queens (n>3), A008794 for kings or A030978 for knights.) - Martin Renner, Jan 26 2020
Integer k is even positive iff phi(2k) > phi(k), where phi is Euler's totient (A000010) [see reference De Koninck & Mercier]. - Bernard Schott, Dec 10 2020
Number of 3-permutations of n elements avoiding the patterns 132, 213, 312 and also number of 3-permutations avoiding the patterns 213, 231, 321. See Bonichon and Sun. - Michel Marcus, Aug 20 2022
a(n) gives the y-value of the integral solution (x,y) of the Pellian equation x^2 - (n^2 + 1)*y^2 = 1. The x-value is given by 2*n^2 + 1 (see Tattersall). - Stefano Spezia, Jul 24 2025

Examples

			G.f. = 2*x + 4*x^2 + 6*x^3 + 8*x^4 + 10*x^5 + 12*x^6 + 14*x^7 + 16*x^8 + ...
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 28.
  • J.-M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 529a pp. 71 and 257, Ellipses, 2004, Paris.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 256.

Crossrefs

a(n)=2*A001477(n). - Juri-Stepan Gerasimov, Dec 12 2009
Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), this sequence (g=4), A002522 (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Oct 30 2011
Cf. A231200 (boustrophedon transform).

Programs

Formula

G.f.: 2*x/(1-x)^2.
E.g.f.: 2*x*exp(x). - Geoffrey Critzer, Aug 25 2012
G.f. with interpolated zeros: 2x^2/((1-x)^2 * (1+x)^2); e.g.f. with interpolated zeros: x*sinh(x). - Geoffrey Critzer, Aug 25 2012
Inverse binomial transform of A036289, n*2^n. - Joshua Zucker, Jan 13 2006
a(0) = 0, a(1) = 2, a(n) = 2a(n-1) - a(n-2). - Jaume Oliver Lafont, May 07 2008
a(n) = Sum_{k=1..n} floor(6n/4^k + 1/2). - Vladimir Shevelev, Jun 04 2009
a(n) = A034856(n+1) - A000124(n) = A000217(n) + A005408(n) - A000124(n) = A005408(n) - 1. - Jaroslav Krizek, Sep 05 2009
a(n) = Sum_{k>=0} A030308(n,k)*A000079(k+1). - Philippe Deléham, Oct 17 2011
Digit sequence 22 read in base n-1. - Jason Kimberley, Oct 30 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Dec 23 2011
a(n) = 2*n = Product_{k=1..2*n-1} 2*sin(Pi*k/(2*n)), n >= 0 (undefined product := 1). See an Oct 09 2013 formula contribution in A000027 with a reference. - Wolfdieter Lang, Oct 10 2013
From Ilya Gutkovskiy, Aug 19 2016: (Start)
Convolution of A007395 and A057427.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/2 = (1/2)*A002162 = (1/10)*A016655. (End)
From Bernard Schott, Dec 10 2020: (Start)
Sum_{n>=1} 1/a(n)^2 = Pi^2/24 = A222171.
Sum_{n>=1} (-1)^(n+1)/a(n)^2 = Pi^2/48 = A245058. (End)

A001045 Jacobsthal sequence (or Jacobsthal numbers): a(n) = a(n-1) + 2*a(n-2), with a(0) = 0, a(1) = 1; also a(n) = nearest integer to 2^n/3.

Original entry on oeis.org

0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381, 174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243, 89478485, 178956971, 357913941, 715827883, 1431655765, 2863311531, 5726623061, 11453246123
Offset: 0

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Don Knuth points out (personal communication) that Jacobsthal may never have seen the actual values of this sequence. However, Horadam uses the name "Jacobsthal sequence", such an important sequence needs a name, and there is a law that says the name for something should never be that of its discoverer. - N. J. A. Sloane, Dec 26 2020
Number of ways to tile a 3 X (n-1) rectangle with 1 X 1 and 2 X 2 square tiles.
Also, number of ways to tile a 2 X (n-1) rectangle with 1 X 2 dominoes and 2 X 2 squares. - Toby Gottfried, Nov 02 2008
Also a(n) counts each of the following four things: n-ary quasigroups of order 3 with automorphism group of order 3, n-ary quasigroups of order 3 with automorphism group of order 6, (n-1)-ary quasigroups of order 3 with automorphism group of order 2 and (n-2)-ary quasigroups of order 3. See the McKay-Wanless (2008) paper. - Ian Wanless, Apr 28 2008
Also the number of ways to tie a necktie using n + 2 turns. So three turns make an "oriental", four make a "four in hand" and for 5 turns there are 3 methods: "Kelvin", "Nicky" and "Pratt". The formula also arises from a special random walk on a triangular grid with side conditions (see Fink and Mao, 1999). - arne.ring(AT)epost.de, Mar 18 2001
Also the number of compositions of n + 1 ending with an odd part (a(2) = 3 because 3, 21, 111 are the only compositions of 3 ending with an odd part). Also the number of compositions of n + 2 ending with an even part (a(2) = 3 because 4, 22, 112 are the only compositions of 4 ending with an even part). - Emeric Deutsch, May 08 2001
Arises in study of sorting by merge insertions and in analysis of a method for computing GCDs - see Knuth reference.
Number of perfect matchings of a 2 X n grid upon replacing unit squares with tetrahedra (C_4 to K_4):
o----o----o----o...
| \/ | \/ | \/ |
| /\ | /\ | /\ |
o----o----o----o... - Roberto E. Martinez II, Jan 07 2002
Also the numerators of the reduced fractions in the alternating sum 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 + ... - Joshua Zucker, Feb 07 2002
Also, if A(n), B(n), C(n) are the angles of the n-orthic triangle of ABC then A(1) = Pi - 2*A, A(n) = s(n)*Pi + (-2)^n*A where s(n) = (-1)^(n-1) * a(n) [1-orthic triangle = the orthic triangle of ABC, n-orthic triangle = the orthic triangle of the (n-1)-orthic triangle]. - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), Jun 05 2002
Also the number of words of length n+1 in the two letters s and t that reduce to the identity 1 by using the relations sss = 1, tt = 1 and stst = 1. The generators s and t and the three stated relations generate the group S3. - John W. Layman, Jun 14 2002
Sums of pairs of consecutive terms give all powers of 2 in increasing order. - Amarnath Murthy, Aug 15 2002
Excess clockwise moves (over counterclockwise) needed to move a tower of size n to the clockwise peg is -(-1)^n*(2^n - (-1)^n)/3; a(n) is its unsigned version. - Wouter Meeussen, Sep 01 2002
Also the absolute value of the number represented in base -2 by the string of n 1's, the negabinary repunit. The Mersenne numbers (A000225 and its subsequences) are the binary repunits. - Rick L. Shepherd, Sep 16 2002
Note that 3*a(n) + (-1)^n = 2^n is significant for Pascal's triangle A007318. It arises from a Jacobsthal decomposition of Pascal's triangle illustrated by 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = (7 + 35 + 1) + (1 + 35 + 7) + (21 + 21) = 43 + 43 + 42 = 3a(7) - 1; 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = (1 + 56 + 28) + (28 + 56 + 1) + (8 + 70 + 8) = 85 + 85 + 86 = 3a(8)+1. - Paul Barry, Feb 20 2003
Number of positive integers requiring exactly n signed bits in the nonadjacent form representation.
Equivalently, number of length-(n-1) words with letters {0, 1, 2} where no two consecutive letters are nonzero, see example and fxtbook link. - Joerg Arndt, Nov 10 2012
Counts walks between adjacent vertices of a triangle. - Paul Barry, Nov 17 2003
Every amphichiral rational knot written in Conway notation is a palindromic sequence of numbers, not beginning or ending with 1. For example, for 4 <= n <= 12, the amphichiral rational knots are: 2 2, 2 1 1 2, 4 4, 3 1 1 3, 2 2 2 2, 4 1 1 4, 3 1 1 1 1 3, 2 3 3 2, 2 1 2 2 1 2, 2 1 1 1 1 1 1 2, 6 6, 5 1 1 5, 4 2 2 4, 3 3 3 3, 2 4 4 2, 3 2 1 1 2 3, 3 1 2 2 1 3, 2 2 2 2 2 2, 2 2 1 1 1 1 2 2, 2 1 2 1 1 2 1 2, 2 1 1 1 1 1 1 1 1 2. For the number of amphichiral rational knots for n=2*k (k=1, 2, 3, ...), we obtain the sequence 0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, ... - Slavik Jablan, Dec 26 2003
a(n+2) counts the binary sequences of total length n made up of codewords from C = {0, 10, 11}. - Paul Barry, Jan 23 2004
Number of permutations with no fixed points avoiding 231 and 132.
The n-th entry (n > 1) of the sequence is equal to the 2,2-entry of the n-th power of the unnormalized 4 X 4 Haar matrix: [1 1 1 0 / 1 1 -1 0 / 1 1 0 1 / 1 1 0 -1]. - Simone Severini, Oct 27 2004
a(n) is the number of Motzkin (n+1)-sequences whose flatsteps all occur at level 1 and whose height is less than or equal to 2. For example, a(4) = 5 counts UDUFD, UFDUD, UFFFD, UFUDD, UUDFD. - David Callan, Dec 09 2004
a(n+1) gives row sums of A059260. - Paul Barry, Jan 26 2005
If (m + n) is odd, then 3*(a(m) + a(n)) is always of the form a^2 + 2*b^2, where a and b both equal powers of 2; consequently every factor of (a(m) + a(n)) is always of the form a^2 + 2*b^2. - Matthew Vandermast, Jul 12 2003
Number of "0,0" in f_{n+1}, where f_0 = "1" and f_{n+1} = a sequence formed by changing all "1"s in f_n to "1,0" and all "0"s in f_n to "0,1". - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Sep 22 2006
All prime Jacobsthal numbers A049883[n] = {3, 5, 11, 43, 683, 2731, 43691, ...} have prime indices except for a(4) = 5. All prime Jacobsthal numbers with prime indices (all but a(4) = 5) are of the form (2^p + 1)/3 - the Wagstaff primes A000979[n]. Indices of prime Jacobsthal numbers are listed in A107036[n] = {3, 4, 5, 7, 11, 13, 17, 19, 23, 31, 43, 61, ...}. For n>1 A107036[n] = A000978[n] Numbers n such that (2^n + 1)/3 is prime. - Alexander Adamchuk, Oct 03 2006
Correspondence: a(n) = b(n)*2^(n-1), where b(n) is the sequence of the arithmetic means of previous two terms defined by b(n) = 1/2*(b(n-1) + b(n-2)) with initial values b(0) = 0, b(1) = 1; the g.f. for b(n) is B(x) := x/(1-(x^1+x^2)/2), so the g.f. A(x) for a(n) satisfies A(x) = B(2*x)/2. Because b(n) converges to the limit lim (1-x)*B(x) = 1/3*(b(0) + 2*b(1)) = 2/3 (for x --> 1), it follows that a(n)/2^(n-1) also converges to 2/3 (see also A103770). - Hieronymus Fischer, Feb 04 2006
Inverse: floor(log_2(a(n))) = n - 2 for n >= 2. Also: log_2(a(n) + a(n-1)) = n - 1 for n >= 1 (see also A130249). Characterization: x is a Jacobsthal number if and only if there is a power of 4 (= c) such that x is a root of p(x) = 9*x*(x-c) + (c-1)*(2*c+1) (see also the indicator sequence A105348). - Hieronymus Fischer, May 17 2007
This sequence counts the odd coefficients in the expansion of (1 + x + x^2)^(2^n - 1), n >= 0. - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Oct 18 2007, Jan 08 2008
2^(n+1) = 2*A005578(n) + 2*a(n) + 2*A000975(n-1). Let A005578(n), a(n), A000975(n-1) = triangle (a, b, c). Then ((S-c), (S-b), (S-a)) = (A005578(n-1), a(n-1), A000975(n-2)). Example: (a, b, c) = (11, 11, 10) = (A005578(5), a(5), A000975(4)). Then ((S-c), (S-b), (S-a)) = (6, 5, 5) = (A005578(4), a(4), A000975(3)). - Gary W. Adamson, Dec 24 2007
Sequence is identical to the absolute values of its inverse binomial transform. A similar result holds for [0,A001045*2^n]. - Paul Curtz, Jan 17 2008
From a(2) on (i.e., 1, 3, 5, 11, 21, ...) also: least odd number such that the subsets of {a(2), ..., a(n)} sum to 2^(n-1) different values, cf. A138000 and A064934. It is interesting to note the pattern of numbers occurring (or not occurring) as such a sum (A003158). - M. F. Hasler, Apr 09 2008
a(n) is the term (5, 1) of n-th power of the 5 X 5 matrix shown in A121231. - Gary W. Adamson, Oct 03 2008
A147612(a(n)) = 1. - Reinhard Zumkeller, Nov 08 2008
a(n+1) = Sum(A153778(i): 2^n <= i < 2^(n+1)). - Reinhard Zumkeller, Jan 01 2009
It appears that a(n) is also the number of integers between 2^n and 2^(n+1) that are divisible by 3 with no remainder. - John Fossaceca (john(AT)fossaceca.net), Jan 31 2009
Number of pairs of consecutive odious (or evil) numbers between 2^(n+1) and 2^(n+2), inclusive. - T. D. Noe, Feb 05 2009
Equals eigensequence of triangle A156319. - Gary W. Adamson, Feb 07 2009
A three-dimensional interpretation of a(n+1) is that it gives the number of ways of filling a 2 X 2 X n hole with 1 X 2 X 2 bricks. - Martin Griffiths, Mar 28 2009
Starting with offset 1 = INVERTi transform of A002605: (1, 2, 6, 16, 44, ...). - Gary W. Adamson, May 12 2009
Convolved with (1, 2, 2, 2, ...) = A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 23 2009
The product of a pair of successive terms is always a triangular number. - Giuseppe Ottonello, Jun 14 2009
Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := -2, A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n) = (-1)^(n-1)*det(A). - Milan Janjic, Jan 26 2010
Let R denote the irreducible representation of the symmetric group S_3 of dimension 2, and let s and t denote respectively the sign and trivial irreducible representations of dimension 1. The decomposition of R^n into irreducible representations consists of a(n) copies of R and a(n-1) copies of each of s and t. - Andrew Rupinski, Mar 12 2010
As a fraction: 1/88 = 0.0113636363... or 1/9898 = 0.00010103051121... - Mark Dols, May 18 2010
Starting with "1" = the INVERT transform of (1, 0, 2, 0, 4, 0, 8, ...); e.g., a(7) = 43 = (1, 1, 1, 3, 5, 11, 21) dot (8, 0, 4, 0, 2, 0, 1) = (8 + 4 + 10 + 21) = 43. - Gary W. Adamson, Oct 28 2010
Rule 28 elementary cellular automaton (A266508) generates this sequence. - Paul Muljadi, Jan 27 2011
This is a divisibility sequence. - Michael Somos, Feb 06 2011
From L. Edson Jeffery, Apr 04 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(6,2) =
(0 0 1)
(0 2 0)
(2 0 1).
Then a(n+1) = (Trace(U^n))/3, a(n+1) = ((U^n){3, 3})/3, a(n) = ((U^n){1, 3})/3 and a(n) = ((U^(n+1))_{1, 1})/2. (End)
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 3*a(n-1) equals the number of 3-colored compositions of n with all parts greater than or equal to 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
This sequence is connected with the Collatz problem. We consider the array T(i, j) where the i-th row gives the parity trajectory of i, for example for i = 6, the infinite trajectory is 6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1... and T(6, j) = [0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, ..., 1, 0, 0, 1, ...]. Now, we consider the sum of the digits "1" of each column. We obtain the sequence a(n) = Sum_{k = 1..2^n} T(k, n) = Sum {k = 1..2^n} digits "1" of the n-th column. Because a(n) + a(n+1) = 2^n, then a(n+1) = Number of digits "0" among the 2^n elements of the n-th column. - _Michel Lagneau, Jan 11 2012
3!*a(n-1) is apparently the trace of the n-th power of the adjacency matrix of the complete 3-graph, a 3 X 3 matrix with diagonal elements all zero and off-diagonal all ones. The off-diagonal elements for the n-th power are all equal to a(n) while each diagonal element seems to be a(n) + 1 for an even power and a(n) - 1 for an odd. These are related to the lengths of closed paths on the graph (see Delfino and Viti's paper). - Tom Copeland, Nov 06 2012
From Paul Curtz, Dec 11 2012: (Start)
2^n * a(-n) = (-1)^(n-1) * a(n), which extends the sequence to negative indices: ..., -5/16, 3/8, -1/4, 1/2, 0, 1, 1, 3, 5, ...
The "autosequence" property with respect to the binomial transform mentioned in my comment of Jan 17 2008 is still valid if the term a(-1) is added to the array of the sequence and its iterated higher-order differences in subsequent rows:
0 1/2 1/2 3/2 5/2 11/2 ...
1/2 0 1 1 3 5 ...
-1/2 1 0 2 2 6 ...
3/2 -1 2 0 4 4 ...
-5/2 3 -2 4 0 8 ...
11/2 -5 6 -4 8 0 ...
The main diagonal in this array contains 0's. (End)
Assign to a triangle T(n, 0) = 1 and T(n+1, 1) = n; T(r, c) = T(r-1, c-1) + T(r-1, c-2) + T(r-2, c-2). Then T(n+1, n) - T(n, n) = a(n). - J. M. Bergot, May 02 2013
a(n+1) counts clockwise walks on n points on a circle that take steps of length 1 and 2, return to the starting point after two full circuits, and do not duplicate any steps (USAMO 2013, problem 5). - Kiran S. Kedlaya, May 11 2013
Define an infinite square array m by m(n, 0) = m(0, n) = a(n) in top row and left column and m(i, j) = m(i, j-1) + m(i-1, j-1) otherwise, then m(n+1, n+1) = 3^(n-1). - J. M. Bergot, May 10 2013
a(n) is the number of compositions (ordered partitions) of n - 1 into one sort of 1's and two sorts of 2's. Example: the a(4) = 5 compositions of 3 are 1 + 1 + 1, 1 + 2, 1 + 2', 2 + 1 and 2' + 1. - Bob Selcoe, Jun 24 2013
Without 0, a(n)/2^n equals the probability that n will occur as a partial sum in a randomly-generated infinite sequence of 1's and 2's. The limiting ratio is 2/3. - Bob Selcoe, Jul 04 2013
Number of conjugacy classes of Z/2Z X Z/2Z in GL(2,2^(n+1)). - Jared Warner, Aug 18 2013
a(n) is the top left entry of the (n-1)-st power of the 3 X 3 matrix [1, 1, 1, 1, 0, 0, 1, 0, 0]. a(n) is the top left entry of the (n+1)-st power of any of the six 3 X 3 matrices [0, 1, 0; 1, 1, 1; 0, 1, 0], [0, 1, 1; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 1, 1; 1, 1, 0], [0, 1, 1; 1, 0, 1; 0, 1, 1], [0, 0, 1; 0, 0, 1; 1, 1, 1] or [0, 1, 0; 1, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
This is the only integer sequence from the family of homogeneous linear recurrence of order 2 given by a(n) = k*a(n-1) + t*a(n-2) with positive integer coefficients k and t and initial values a(0) = 0 and a(1) = 1 whose ratio a(n+1)/a(n) converges to 2 as n approaches infinity. - Felix P. Muga II, Mar 14 2014
This is the Lucas sequence U(1, -2). - Felix P. Muga II, Mar 21 2014
sqrt(a(n+1) * a(n-1)) -> a(n) + 3/4 if n is even, and -> a(n) - 3/4 if n is odd, for n >= 2. - Richard R. Forberg, Jun 24 2014
a(n+1) counts closed walks on the end vertices of P_3 containing one loop at the middle vertex. a(n-1) counts closed walks on the middle vertex of P_3 containing one loop at that vertex. - David Neil McGrath, Nov 07 2014
From César Eliud Lozada, Jan 21 2015: (Start)
Let P be a point in the plane of a triangle ABC (with sides a, b, c) and barycentric coordinates P = [x:y:z]. The Complement of P with respect to ABC is defined to be Complement(P) = [b*y + c*z : c*z + a*x : a*x + b*y].
Then, for n >= 1, Complement(Complement(...(Complement(P))..)) = (n times) =
[2*a(n-1)*a*x + (2*a(n-1) - (-1)^n)*(b*y + c*z):
2*a(n-1)*b*y + (2*a(n-1) - (-1)^n)*(c*z + a*x):
2*a(n-1)*c*z + (2*a(n-1) - (-1)^n)*(a*x + b*y)]. (End)
a(n) (n >= 2) is the number of induced hypercubes of the Fibonacci cube Gamma(n-2). See p. 513 of the Klavzar reference. Example: a(5) = 11. Indeed, the Fibonacci cube Gamma(3) is <>- (cycle C(4) with a pendant edge) and the hypercubes are: 5 vertices, 5 edges, and 1 square. - Emeric Deutsch, Apr 07 2016
If the sequence of points {P_i(x_i, y_i)} on the cubic y = a*x^3 + b*x^2 + c*x + d has the property that the segment P_i(x_i, y_i) P_i+1(x_i+1, y_i+1) is always tangent to the cubic at P_i+1(x_i+1, y_i+1) then a(n) = -2^n*a/b*(x_(n+1)-(-1/2)^n*x_1). - Michael Brozinsky, Aug 01 2016
With the quantum integers defined by [n+1]A000225%20are%20given%20by%20q%20=%20sqrt(2).%20Cf.%20A239473.%20-%20_Tom%20Copeland">q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)), the Jacobsthal numbers are a(n+1) = (-1)^n*q^n [n+1]_q with q = i * sqrt(2) for i^2 = -1, whereas the signed Mersenne numbers A000225 are given by q = sqrt(2). Cf. A239473. - _Tom Copeland, Sep 05 2016
Every positive integer has a unique expression as a sum of Jacobsthal numbers in which the index of the smallest summand is odd, with a(1) and a(2) both allowed. See the L. Carlitz, R. Scoville, and V. E. Hoggatt, Jr. reference. - Ira M. Gessel, Dec 31 2016. See A280049 for these expansions. - N. J. A. Sloane, Dec 31 2016
For n > 0, a(n) equals the number of ternary words of length n-1 in which 0 and 1 avoid runs of odd lengths. - Milan Janjic, Jan 08 2017
For n > 0, a(n) equals the number of orbits of the finite group PSL(2,2^n) acting on subsets of size 4 for the 2^n+1 points of the projective line. - Paul M. Bradley, Jan 31 2017
For n > 1, number of words of length n-2 over alphabet {1,2,3} such that no odd letter is followed by an odd letter. - Armend Shabani, Feb 17 2017
Also, the decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 678", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. See A283641. - Robert Price, Mar 12 2017
Also the number of independent vertex sets and vertex covers in the 2 X (n-2) king graph. - Eric W. Weisstein, Sep 21 2017
From César Eliud Lozada, Dec 14 2017: (Start)
Let T(0) be a triangle and let T(1) be the medial triangle of T(0), T(2) the medial triangle of T(1) and, in general, T(n) the medial triangle of T(n-1). The barycentric coordinates of the first vertex of T(n) are [2*a(n-1)/a(n), 1, 1], for n > 0.
Let S(0) be a triangle and let S(1) be the antimedial triangle of S(0), S(2) the antimedial triangle of S(1) and, in general, S(n) the antimedial triangle of S(n-1). The barycentric coordinates of the first vertex of S(n) are [-a(n+1)/a(n), 1, 1], for n > 0. (End)
a(n) is also the number of derangements in S_{n+1} with empty peak set. - Isabella Huang, Apr 01 2018
For n > 0, gcd(a(n), a(n+1)) = 1. - Kengbo Lu, Jul 27 2020
Number of 2-compositions of n+1 with 1 not allowed as a part; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
The number of Hamiltonian paths of the flower snark graph of even order 2n > 2 is 12*a(n-1). - Don Knuth, Dec 25 2020
When set S = {1, 2, ..., 2^n}, n>=0, then the largest subset T of S with the property that if x is in T, then 2*x is not in T, has a(n+1) elements. For example, for n = 4, #S = 16, a(5) = 11 with T = {1, 3, 4, 5, 7, 9, 11, 12, 13, 15, 16} (see Hassan Tarfaoui link, Concours Général 1991). - Bernard Schott, Feb 14 2022
a(n) is the number of words of length n over a binary alphabet whose position in the lexicographic order is one more than a multiple of three. a(3) = 3: aaa, abb, bba. - Alois P. Heinz, Apr 13 2022
Named by Horadam (1988) after the German mathematician Ernst Jacobsthal (1882-1965). - Amiram Eldar, Oct 02 2023
Define the sequence u(n) = (u(n-1) + u(n-2))/u(n-3) with u(0) = 0, u(1) = 1, u(2) = u(3) = -1. Then u(4*n) = -1 + (-1)^n/a(n+1), u(4*n+1) = 2 - (-1)^n/a(n+1), u(4*n+2) = u(4*n+3) = -1. For example, a(3) = 3 and u(8) = -2/3, u(9) = 5/3, u(10) = u(11) = -1. - Michael Somos, Oct 24 2023
From Miquel A. Fiol, May 25 2024: (Start)
Also, a(n) is the number of (3-color) states of a cycle (n+1)-pole C_{n+1} with n+1 terminals (or semiedges).
For instance, for n=3, the a(3)=3 states (3-coloring of the terminals) of C_4 are
a a a a a b
a a b b a b (End)
Also, with offset 1, the cogrowth sequence of the 6-element dihedral group D3. - Sean A. Irvine, Nov 04 2024

Examples

			a(2) = 3 because the tiling of the 3 X 2 rectangle has either only 1 X 1 tiles, or one 2 X 2 tile in one of two positions (together with two 1 X 1 tiles).
From _Joerg Arndt_, Nov 10 2012: (Start)
The a(6)=21 length-5 ternary words with no two consecutive letters nonzero are (dots for 0's)
[ 1]   [ . . . . ]
[ 2]   [ . . . 1 ]
[ 3]   [ . . . 2 ]
[ 4]   [ . . 1 . ]
[ 5]   [ . . 2 . ]
[ 6]   [ . 1 . . ]
[ 7]   [ . 1 . 1 ]
[ 8]   [ . 1 . 2 ]
[ 9]   [ . 2 . . ]
[10]   [ . 2 . 1 ]
[11]   [ . 2 . 2 ]
[12]   [ 1 . . . ]
[13]   [ 1 . . 1 ]
[14]   [ 1 . . 2 ]
[15]   [ 1 . 1 . ]
[16]   [ 1 . 2 . ]
[17]   [ 2 . . . ]
[18]   [ 2 . . 1 ]
[19]   [ 2 . . 2 ]
[20]   [ 2 . 1 . ]
[21]   [ 2 . 2 . ]
(End)
G.f. = x + x^2 + 3*x^3 + 5*x^4 + 11*x^5 + 21*x^6 + 43*x^7 + 85*x^8 + 171*x^9 + ...
		

References

  • Jathan Austin and Lisa Schneider, Generalized Fibonacci sequences in Pythagorean triple preserving sequences, Fib. Q., 58:1 (2020), 340-350.
  • Thomas Fink and Yong Mao, The 85 ways to tie a tie, Fourth Estate, London, 1999; Die 85 Methoden eine Krawatte zu binden. Hoffmann und Kampe, Hamburg, 1999.
  • International Mathematical Olympiad 2001, Hong Kong Preliminary Selection Contest Problem #16.
  • Jablan S. and Sazdanovic R., LinKnot: Knot Theory by Computer, World Scientific Press, 2007. See p. 80.
  • Ernst Erich Jacobsthal, Fibonaccische Polynome und Kreisteilungsgleichungen, Sitzungsber. Berliner Math. Gesell. 17 (1919-1920), 43-57.
  • Tanya Khovanova, "Coins and Logic", Chapter 6, The Mathematics of Various Entertaining Subjects: Volume 3 (2019), Jennifer Beineke & Jason Rosenhouse, eds. Princeton University Press, Princeton and Oxford, p. 73. ISBN: 0691182582, 978-0691182582.
  • Donald E. Knuth, Art of Computer Programming, Vol. 3, Sect. 5.3.1, Eq. 13.
  • Thomas Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.
  • Steven Roman, Introduction to Coding and Information Theory, Springer Verlag, 1996, 41-42.
  • P. D. Seymour and D. J. A. Welsh, Combinatorial applications of an inequality form statistical mechanics, Math. Proc. Camb. Phil. Soc. 77 (1975), 485-495. [Although Daykin et al. (1979) claim that the present sequence is studied in this article, it does not seem to be explicitly mentioned. Note that definition of log-convex in (3.1) is wrong. - N. J. A. Sloane, Dec 26 2020]
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Robert M. Young, Excursions in Calculus, MAA, 1992, p. 239

Crossrefs

Partial sums of this sequence give A000975, where there are additional comments from B. E. Williams and Bill Blewett on the tie problem.
A002487(a(n)) = A000045(n).
Row sums of A059260, A156667 and A134317. Equals A026644(n-2)+1 for n > 1.
a(n) = A073370(n-1, 0), n >= 1 (first column of triangle).
Cf. A266508 (binary), A081857 (base 4), A147612 (characteristic function).
Cf. A049883 = primes in this sequence, A107036 = indices of primes, A129738.
Cf. A091084 (mod 10), A239473, A280049.
Bisections: A002450, A007583.
Cf. A077925 (signed version).

Programs

  • Haskell
    a001045 = (`div` 3) . (+ 1) . a000079
    a001045_list = 0 : 1 :
       zipWith (+) (map (2 *) a001045_list) (tail a001045_list)
    -- Reinhard Zumkeller, Mar 24 2013, Jan 05 2012, Feb 05 2011
    
  • Magma
    [n le 2 select n-1 else Self(n-1)+2*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jun 27 2016
    
  • Maple
    A001045 := proc(n)
      (2^n-(-1)^n)/3 ;
    end proc: # R. J. Mathar, Dec 18 2012
  • Mathematica
    Jacob0[n_] := (2^n - (-1)^n)/3; Table[Jacob0[n], {n, 0, 33}] (* Robert G. Wilson v, Dec 05 2005 *)
    Array[(2^# - (-1)^#)/3 &, 33, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
    LinearRecurrence[{1, 2}, {0, 1}, 40] (* Harvey P. Dale, Nov 30 2011 *)
    CoefficientList[Series[x/(1 - x - 2 x^2), {x, 0, 34}], x] (* Robert G. Wilson v, Jul 21 2015 *)
    Table[(2^n - (-1)^n)/3, {n, 0, 20}] (* Eric W. Weisstein, Sep 21 2017 *)
    Table[Abs[QBinomial[n, 1, -2]], {n, 0, 35}] (* John Keith, Jan 29 2022 *)
  • Maxima
    a[0]:0$
    a[n]:=2^(n-1)-a[n-1]$
    A001045(n):=a[n]$
    makelist(A001045(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n) = (2^n - (-1)^n) / 3
    
  • PARI
    M=[1,1,0;1,0,1;0,1,1];for(i=0,34,print1((M^i)[2,1],",")) \\ Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005
    
  • PARI
    a=0; for(n=0,34,print1(a,", "); a=2*(a-n%2)+1) \\ K. Spage, Aug 22 2014
    
  • Python
    # A001045.py
    def A001045():
        a, b = 0, 1
        while True:
            yield a
            a, b = b, b+2*a
    sequence = A001045()
    [next(sequence) for i in range(20)] # David Radcliffe, Jun 26 2016
    
  • Python
    [(2**n-(-1)**n)//3 for n in range(40)] # Gennady Eremin, Mar 03 2022
  • Sage
    [lucas_number1(n, 1, -2) for n in range(34)]  # Zerinvary Lajos, Apr 22 2009
    # Alternatively:
    a = BinaryRecurrenceSequence(1,2)
    [a(n) for n in (0..34)] # Peter Luschny, Aug 29 2016
    

Formula

a(n) = 2^(n-1) - a(n-1). a(n) = 2*a(n-1) - (-1)^n = (2^n - (-1)^n)/3.
G.f.: x/(1 - x - 2*x^2) = x/((x+1)*(1-2*x)). Simon Plouffe in his 1992 dissertation
E.g.f.: (exp(2*x) - exp(-x))/3.
a(2*n) = 2*a(2*n-1)-1 for n >= 1, a(2*n+1) = 2*a(2*n)+1 for n >= 0. - Lee Hae-hwang, Oct 11 2002; corrected by Mario Catalani (mario.catalani(AT)unito.it), Dec 04 2002
Also a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x, y) = x*F(n-1)(x, y) + y*F(n-2)(x, y), with y=2*x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 04 2002
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*3^(k-1). - Paul Barry, Apr 02 2003
The ratios a(n)/2^(n-1) converge to 2/3 and every fraction after 1/2 is the arithmetic mean of the two preceding fractions. - Gary W. Adamson, Jul 05 2003
a(n) = U(n-1, i/(2*sqrt(2)))*(-i*sqrt(2))^(n-1) with i^2=-1. - Paul Barry, Nov 17 2003
a(n+1) = Sum_{k=0..ceiling(n/2)} 2^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(2*n) = A002450(n) = (4^n - 1)/3; a(2*n+1) = A007583(n) = (2^(2*n+1) + 1)/3. - Philippe Deléham, Mar 27 2004
a(n) = round(2^n/3) = (2^n + (-1)^(n-1))/3 so lim_{n->infinity} 2^n/a(n) = 3. - Gerald McGarvey, Jul 21 2004
a(n) = Sum_{k=0..n-1} (-1)^k*2^(n-k-1) = Sum_{k=0..n-1}, 2^k*(-1)^(n-k-1). - Paul Barry, Jul 30 2004
a(n+1) = Sum_{k=0..n} binomial(k, n-k)*2^(n-k). - Paul Barry, Oct 07 2004
a(n) = Sum_{k=0..n-1} W(n-k, k)*(-1)^(n-k)*binomial(2*k,k), W(n, k) as in A004070. - Paul Barry, Dec 17 2004
From Paul Barry, Jan 17 2005: (Start)
a(n) = Sum_{k=0..n} k*binomial(n-1, (n-k)/2)*(1+(-1)^(n+k))*floor((2*k+1)/3).
a(n+1) = Sum_{k=0..n} k*binomial(n-1, (n-k)/2)*(1+(-1)^(n+k))*(A042965(k)+0^k). (End)
From Paul Barry, Jan 17 2005: (Start)
a(n+1) = ceiling(2^n/3) + floor(2^n/3) = (ceiling(2^n/3))^2 - (floor(2^n/3))^2.
a(n+1) = A005578(n) + A000975(n-1) = A005578(n)^2 - A000975(n-1)^2. (End)
a(n+1) = Sum_{k=0..n} Sum_{j=0..n} (-1)^(n-j)*binomial(j, k). - Paul Barry, Jan 26 2005
Let M = [1, 1, 0; 1, 0, 1; 0, 1, 1], then a(n) = (M^n)[2, 1], also matrix characteristic polynomial x^3 - 2*x^2 - x + 2 defines the three-step recursion a(0)=0, a(1)=1, a(2)=1, a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n > 2. - Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005
a(n) = ceiling(2^(n+1)/3) - ceiling(2^n/3) = A005578(n+1) - A005578(n). - Paul Barry, Oct 08 2005
a(n) = floor(2^(n+1)/3) - floor(2^n/3) = A000975(n) - A000975(n-1). - Paul Barry, Oct 08 2005
From Paul Barry, Feb 20 2003: (Start)
a(n) = Sum_{k=0..floor(n/3)} binomial(n, f(n-1)+3*k);
a(n) = Sum_{k=0..floor(n/3)} binomial(n, f(n-2)+3*k), where f(n)=A080425(n). (End)
From Miklos Kristof, Mar 07 2007: (Start)
a(2*n) = (1/3)*Product_{d|n} cyclotomic(d,4).
a(2*n+1) = (1/3)*Product_{d|2*n+1} cyclotomic(2*d,2). (End)
From Hieronymus Fischer, Apr 23 2007: (Start)
The a(n) are closely related to nested square roots; this is 2*sin(2^(-n)*Pi/2*a(n)) = sqrt(2-sqrt(2-sqrt(2-sqrt(...sqrt(2)))...) {with the '2' n times, n >= 0}.
Also 2*cos(2^(-n)*Pi*a(n)) = sqrt(2-sqrt(2-sqrt(2-sqrt(...sqrt(2)))...) {with the '2' n-1 times, n >= 1} as well as
2*sin(2^(-n)*3/2*Pi*a(n)) = sqrt(2+sqrt(2+sqrt(2+sqrt(...sqrt(2)))...) {with the '2' n times, n >= 0} and
2*cos(2^(-n)*3*Pi*a(n)) = -sqrt(2+sqrt(2+sqrt(2+sqrt(...sqrt(2)))...) {with the '2' n-1 times, n >= 1}.
a(n) = 2^(n+1)/Pi*arcsin(b(n+1)/2) where b(n) is defined recursively by b(0)=2, b(n)=sqrt(2-b(n-1)).
There is a similar formula regarding the arccos function, this is a(n) = 2^n/Pi*arccos(b(n)/2).
With respect to the sequence c(n) defined recursively by c(0)=-2, c(n)=sqrt(2+c(n-1)), the following formulas hold true: a(n) = 2^n/3*(1-(-1)^n*(1-2/Pi*arcsin(c(n+1)/2))); a(n) = 2^n/3*(1-(-1)^n*(1-1/Pi*arccos(-c(n)/2))).
(End)
Sum_{k=0..n} A039599(n,k)*a(k) = A049027(n), for n >= 1. - Philippe Deléham, Jun 10 2007
Sum_{k=0..n} A039599(n,k)*a(k+1) = A067336(n). - Philippe Deléham, Jun 10 2007
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0,] = [A005578(n), a(n), A000975(n-1)]. - Gary W. Adamson, Dec 24 2007
a(n) + a(n+5) = 11*2^n. - Paul Curtz, Jan 17 2008
a(n) = Sum_{k=1..n} K(2, k)*a(n - k), where K(n,k) = k if 0 <= k <= n and K(n,k)=0 otherwise. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, the Fibonacci sequence can be generated in several ways using the K-coefficient.) - Thomas Wieder, Jan 13 2008
a(n) + a(n+2*k+1) = a(2*k+1)*2^n. - Paul Curtz, Feb 12 2008
a(n) = lower left term in the 2 X 2 matrix [0,2; 1,1]^n. - Gary W. Adamson, Mar 02 2008
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-2)^(n-k). -Philippe Deléham, Oct 26 2008
a(n) = sqrt(8*a(n-1)*a(n-2) + 1). E.g., sqrt(3*5*8+1) = 11, sqrt(5*11*8+1) = 21. - Giuseppe Ottonello, Jun 14 2009
Let p[i] = Fibonacci(i-1) and let A be the Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, May 08 2010
a(p-1) = p*A007663(n)/3 if n > 1, and a(p-1) = p*A096060(n) if n > 2, with p=prime(n). - Jonathan Sondow, Jul 19 2010
Algebraically equivalent to replacing the 5's with 9's in the explicit (Binet) formula for the n-th term in the Fibonacci sequence: The formula for the n-th term in the Fibonacci sequence is F(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)). Replacing the 5's with 9's gives ((1+sqrt(9))^n - (1-sqrt(9))^n)/(2^n*sqrt(9)) = (2^n+(-1)^(n+1))/3 = (2^n-(-1)^(n))/3 = a(n). - Jeffrey R. Goodwin, May 27 2011
For n > 1, a(n) = A000975(n-1) + (1 + (-1)^(n-1))/2. - Vladimir Shevelev, Feb 27 2012
From Sergei N. Gladkovskii, Jun 12 2012: (Start)
G.f.: x/(1-x-2*x^2) = G(0)/3; G(k) = 1 - ((-1)^k)/(2^k - 2*x*4^k/(2*x*2^k - ((-1)^k)/G(k+1))); (continued fraction 3 kind, 3-step).
E.g.f.: G(0)/3; G(k) = 1 - ((-1)^k)/(2^k - 2*x*4^k/(2*x*2^k - ((-1)^k)*(k+1)/G(k+1))); (continued fraction 3rd kind, 3-step). (End)
a(n) = 2^k * a(n-k) + (-1)^(n+k)*a(k). - Paul Curtz, Jean-François Alcover, Dec 11 2012
a(n) = sqrt((A014551(n))^2 + (-1)^(n-1)*2^(n+2))/3. - Vladimir Shevelev, Mar 13 2013
G.f.: Q(0)/3, where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 21 2013
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(2*k+1 + 2*x)/( x*(2*k+2 + 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 29 2013
G.f.: Q(0) -1, where Q(k) = 1 + 2*x^2 + (k+2)*x - x*(k+1 + 2*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n+2) = Sum_{k=0..n} A108561(n,k)*(-2)^k. - Philippe Deléham, Nov 17 2013
a(n) = (Sum_{k=1..n, k odd} C(n,k)*3^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(-n) = -(-1)^n * a(n) / 2^n for all n in Z. - Michael Somos, Mar 18 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-3)^k = (2^n - (-1)^n)/3 = (-1)^(n-1)*Sum_{k=0..n-1} (-2)^k. Equals (-1)^(n-1)*Phi(n,-2), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
From Peter Bala, Apr 06 2015: (Start)
a(2*n)/a(n) = A014551(n) for n >= 1; a(3*n)/a(n) = 3*A245489(n) for n >= 1.
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n.
exp( Sum_{n >= 1} a(3*n)/a(n)*x^n/n ) = Sum_{n >= 0} A084175(n+1)*x^n.
exp( Sum_{n >= 1} a(4*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015266(n+3)*(-x)^n.
exp( Sum_{n >= 1} a(5*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015287(n+4)*x^n.
exp( Sum_{n >= 1} a(6*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015305(n+5)*(-x)^n.
exp( Sum_{n >= 1} a(7*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015323(n+6)*x^n.
exp( Sum_{n >= 1} a(8*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015338(n+7)*(-x)^n.
exp( Sum_{n >= 1} a(9*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015356(n+8)*x^n.
exp( Sum_{n >= 1} a(10*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015371(n+9)*(-x)^n. (End)
a(n) = (1-(-1)^n)/2 + floor((2^n)/3). - Reiner Moewald, Jun 05 2015
a(n+k)^2 - A014551(k)*a(n)*a(n+k) + (-2)^k*a(n)^2 = (-2)^n*a(k)^2, for n >= 0 and k >= 0. - Alexander Samokrutov, Jul 21 2015
Dirichlet g.f.: (PolyLog(s,2) + (1 - 2^(1-s))*zeta(s))/3. - Ilya Gutkovskiy, Jun 27 2016
From Yuchun Ji, Apr 08 2018: (Start)
a(m)*a(n) + a(m-1)*a(n-1) - 2*a(m-2)*a(n-2) = 2^(m+n-3).
a(m+n-1) = a(m)*a(n) + 2*a(m-1)*a(n-1); a(m+n) = a(m+1)*a(n+1) - 4*a(m-1)*a(n-1).
a(2*n-1) = a(n)^2 + 2*a(n-1)^2; a(2*n) = a(n+1)^2 - 4*a(n-1)^2. (End)
a(n+4) = a(n) + 5*2^n, a(0) = 0, a(1..4) = [1,1,3,5]. That is to say, for n > 0, the ones digits of Jacobsthal numbers follow the pattern 1,1,3,5,1,1,3,5,1,1,3,5,.... - Yuchun Ji, Apr 25 2019
a(n) mod 10 = A091084(n). - Alois P. Heinz, Apr 25 2019
The sequence starting with "1" is the second INVERT transform of (1, -1, 3, -5, 11, -21, 43, ...). - Gary W. Adamson, Jul 08 2019
From Kai Wang, Jan 14 2020: (Start)
a(n)^2 - a(n+1)*a(n-1) = (-2)^(n-1).
a(n)^2 - a(n+r)*a(n-r) = (-2)^(n-r)*a(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = (-2)^n*a(m-n).
a(m-n) = (-1)^n*(a(m)*A014551(n) - A014551(m)*a(n))/(2^(n+1)).
a(m+n) = (a(m)*A014551(n) + A014551(m)*a(n))/2.
A014551(n)^2 - A014551(n+r)*A014551(n-r) = 9*(-1)^(n-r-1)*2^(n-r)*a(r)^2 .
A014551(m)*A014551(n+1) - A014551(m+1)*A014551(n) = 9*(-1)^(n-1)*2^(n)*a(m-n).
A014551(m-n) = (-1)^(n)*(A014551(m)*A014551(n) - 9*a(m)*a(n))/2^(n+1).
A014551(m+n) = (A014551(m)*A014551(n) + 9*a(m)*a(n))/2.
a(n) = Sum_{i=0..n-1; j=0..n-1; i+2*j=n-1} 2^j*((i+j)!/(i!*j!)). (End)
For n > 0, 1/(2*a(n+1)) = Sum_{m>=n} a(m)/(a(m+1)*a(m+2)). - Kai Wang, Mar 03 2020
For 4 > h >= 0, k >= 0, a(4*k+h) mod 5 = a(h) mod 5. - Kai Wang, May 07 2020
From Kengbo Lu, Jul 27 2020: (Start)
a(n) = 1 + Sum_{k=0..n-1} a(k) if n odd; a(n) = Sum_{k=0..n-1} a(k) if n even.
a(n) = F(n) + Sum_{k=0..n-2} a(k)*F(n-k-1), where F denotes the Fibonacci numbers.
a(n) = b(n) + Sum_{k=0..n-1} a(k)*b(n-k), where b(n) is defined through b(0) = 0, b(1) = 1, b(n) = 2*b(n-2).
a(n) = 1 + 2*Sum_{k=0..n-2} a(k).
a(m+n) = a(m)*a(n+1) + 2*a(m-1)*a(n).
a(2*n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)*2^(i+j). (End)
G.f.: x/(1 - x - 2*x^2) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 2*x)/(1 + k*x) (a telescoping series). - Peter Bala, May 08 2024
a(n) = Sum_{r>=0} F(n-2r, r), where F(n, 0) is the n-th Fibonacci number and F(n,r) = Sum_{j=1..n} F(n+1-j, r-1) F(j, r-1). - Gregory L. Simay, Aug 31 2024
From Peter Bala, Jun 27 2025: (Start)
The following are all examples of telescoping infinite products:
Product_{n >= 1} (1 + 2^n/a(2*n+2)) = 2, since 1 + 2^n/a(2*n+2) = b(n+1)/b(n), where b(n) = 2 - 3/(2^n + 1).
Product_{n >= 1} (1 - 2^n/a(2*n+2)) = 2/5, since 1 - 2^n/a(2*n+2) = c(n+1)/c(n), where c(n) = 2 + 3/(2^n - 1).
Product_{n >= 1} (1 + (-2)^n/a(2*n+2)) = 2/3, since 1 + (-2)^n/a(2*n+2) = d(n+1)/d(n), where d(n) = 2 - 1/(1 + (-2)^n).
Product_{n >= 1} (1 - (-2)^n/a(2*n+2)) = 6/5, since 1 - (-2)^n/a(2*n+2) = e(n+1)/e(n), where e(n) = 2 - 1/(1 - (-2)^n). (End)

Extensions

Thanks to Don Knuth, who pointed out several missing references, including Brocard (1880), which although it was mentioned in the 1973 Handbook of Integer Sequences, was omitted from the 1995 "Encyclopedia". - N. J. A. Sloane, Dec 26 2020

A000035 Period 2: repeat [0, 1]; a(n) = n mod 2; parity of n.

Original entry on oeis.org

0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0
Offset: 0

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Keywords

Comments

Least significant bit of n, lsb(n).
Also decimal expansion of 1/99.
Also the binary expansion of 1/3. - Robert G. Wilson v, Sep 01 2015
a(n) = A134451(n) mod 2. - Reinhard Zumkeller, Oct 27 2007 [Corrected by Jianing Song, Nov 22 2019]
Characteristic function of odd numbers: a(A005408(n)) = 1, a(A005843(n)) = 0. - Reinhard Zumkeller, Sep 29 2008
A102370(n) modulo 2. - Philippe Deléham, Apr 04 2009
Base b expansion of 1/(b^2-1) for any b >= 2 is 0.0101... (A005563 has b^2-1). - Rick L. Shepherd, Sep 27 2009
Let A be the Hessenberg n X n matrix defined by: A[1,j] = j mod 2, A[i,i] := 1, A[i,i-1] = -1, and A[i,j] = 0 otherwise. Then, for n >= 1, a(n) = (-1)^n*charpoly(A,1). - Milan Janjic, Jan 24 2010
From R. J. Mathar, Jul 15 2010: (Start)
The sequence is the principal Dirichlet character of the reduced residue system mod 2 or mod 4 or mod 8 or mod 16 ...
Associated Dirichlet L-functions are for example L(2,chi) = Sum_{n>=1} a(n)/n^2 == A111003,
or L(3,chi) = Sum_{n>=1} a(n)/n^3 = 1.05179979... = 7*A002117/8,
or L(4,chi) = Sum_{n>=1} a(n)/n^4 = 1.014678... = A092425/96. (End)
Also parity of the nonnegative integers A001477. - Omar E. Pol, Jan 17 2012
a(n) = (4/n), where (k/n) is the Kronecker symbol. See the Eric Weisstein link. - Wolfdieter Lang, May 28 2013
Also the inverse binomial transform of A131577. - Paul Curtz, Nov 16 2016 [an observation forwarded by Jean-François Alcover]
The emanation sequence for the globe category. That is take the globe category, take the corresponding polynomial comonad, consider its carrier polynomial as a generating function, and take the corresponding sequence. - David Spivak, Sep 25 2020
For n > 0, a(n) is the alternating sum of the product of n increasing and n decreasing odd factors. For example, a(4) = 1*7 - 3*5 + 5*3 - 7*1 and a(5) = 1*9 - 3*7 + 5*5 - 7*3 + 9*1. - Charlie Marion, Mar 24 2022

Examples

			G.f. = x + x^3 + x^5 + x^7 + x^9 + x^11 + x^13 + x^15 + ... - _Michael Somos_, Feb 20 2024
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Ones complement of A059841.
Cf. A053644 for most significant bit.
This is Guy Steele's sequence GS(1, 2) (see A135416).
Period k zigzag sequences: this sequence (k=2), A007877 (k=4), A260686 (k=6), A266313 (k=8), A271751 (k=10), A271832 (k=12), A279313 (k=14), A279319 (k=16), A158289 (k=18).
Cf. A154955 (Mobius transform), A131577 (binomial transform).
Cf. A111003 (Dgf at s=2), A233091 (Dgf at s=3), A300707 (Dgf at s=4).
Parity of A005811.

Programs

Formula

a(n) = (1 - (-1)^n)/2.
a(n) = n mod 2.
a(n) = 1 - a(n-1).
Multiplicative with a(p^e) = p mod 2. - David W. Wilson, Aug 01 2001
G.f.: x/(1-x^2). E.g.f.: sinh(x). - Paul Barry, Mar 11 2003
a(n) = (A000051(n) - A014551(n))/2. - Mario Catalani (mario.catalani(AT)unito.it), Aug 30 2003
a(n) = ceiling((-2)^(-n-1)). - Reinhard Zumkeller, Apr 19 2005
Dirichlet g.f.: (1-1/2^s)*zeta(s). - R. J. Mathar, Mar 04 2011
a(n) = ceiling(n/2) - floor(n/2). - Arkadiusz Wesolowski, Sep 16 2012
a(n) = ceiling( cos(Pi*(n-1))/2 ). - Wesley Ivan Hurt, Jun 16 2013
a(n) = floor((n-1)/2) - floor((n-2)/2). - Mikael Aaltonen, Feb 26 2015
Dirichlet g.f.: L(chi(2),s) with chi(2) the principal Dirichlet character modulo 2. - Ralf Stephan, Mar 27 2015
a(n) = 0^^n = 0^(0^(0...)) (n times), where we take 0^0 to be 1. - Natan Arie Consigli, May 02 2015
Euler transform and inverse Moebius transform of length 2 sequence [0, 1]. - Michael Somos, Feb 20 2024

A000302 Powers of 4: a(n) = 4^n.

Original entry on oeis.org

1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736, 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656
Offset: 0

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Author

Keywords

Comments

Same as Pisot sequences E(1, 4), L(1, 4), P(1, 4), T(1, 4). Essentially same as Pisot sequences E(4, 16), L(4, 16), P(4, 16), T(4, 16). See A008776 for definitions of Pisot sequences.
The convolution square root of this sequence is A000984, the central binomial coefficients: C(2n,n). - T. D. Noe, Jun 11 2002
With P(n) being the number of integer partitions of n, p(i) as the number of parts of the i-th partition of n, d(i) as the number of different parts of the i-th partition of n, m(i, j) the multiplicity of the j-th part of the i-th partition of n, one has a(n) = Sum_{i = 1..P(n)} p(i)!/(Product_{j = 1..d(i)} m(i, j)!) * 2^(n-1). - Thomas Wieder, May 18 2005
Sums of rows of the triangle in A122366. - Reinhard Zumkeller, Aug 30 2006
Hankel transform of A076035. - Philippe Deléham, Feb 28 2009
Equals the Catalan sequence: (1, 1, 2, 5, 14, ...), convolved with A032443: (1, 3, 11, 42, ...). - Gary W. Adamson, May 15 2009
Sum of coefficients of expansion of (1 + x + x^2 + x^3)^n.
a(n) is number of compositions of natural numbers into n parts less than 4. For example, a(2) = 16 since there are 16 compositions of natural numbers into 2 parts less than 4.
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 4-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Squares in A002984. - Reinhard Zumkeller, Dec 28 2011
Row sums of Pascal's triangle using the rule that going left increases the value by a factor of k = 3. For example, the first three rows are {1}, {3, 1}, and {9, 6, 1}. Using this rule gives row sums as (k+1)^n. - Jon Perry, Oct 11 2012
First differences of A002450. - Omar E. Pol, Feb 20 2013
Sum of all peak heights in Dyck paths of semilength n+1. - David Scambler, Apr 22 2013
Powers of 4 exceed powers of 2 by A020522 which is the m-th oblong number A002378(m), m being the n-th Mersenne number A000225(n); hence, we may write, a(n) = A000079(n) + A002378(A000225(n)). - Lekraj Beedassy, Jan 17 2014
a(n) is equal to 1 plus the sum for 0 < k < 2^n of the numerators and denominators of the reduced fractions k/2^n. - J. M. Bergot, Jul 13 2015
Binomial transform of A000244. - Tony Foster III, Oct 01 2016
From Ilya Gutkovskiy, Oct 01 2016: (Start)
Number of nodes at level n regular 4-ary tree.
Partial sums of A002001. (End)
Satisfies Benford's law [Berger-Hill, 2011]. - N. J. A. Sloane, Feb 08 2017
Also the number of connected dominating sets in the (n+1)-barbell graph. - Eric W. Weisstein, Jun 29 2017
Side length of the cells at level n in a pyramid scheme where a square grid is decomposed into overlapping 2 X 2 blocks (cf. Kropatsch, 1985). - Felix Fröhlich, Jul 04 2019
a(n-1) is the number of 3-compositions of n; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020

References

  • H. W. Gould, Combinatorial Identities, 1972, eq. (1.93), p. 12.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, eq. (5.39), p. 187.
  • D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Crossrefs

Cf. A024036, A052539, A032443, A000351 (Binomial transform).
Cf. A249307.
Cf. A083420.

Programs

Formula

a(n) = 4^n.
a(0) = 1; a(n) = 4*a(n-1).
G.f.: 1/(1-4*x).
E.g.f.: exp(4*x).
a(n) = Sum_{k = 0..n} binomial(2k, k) * binomial(2(n - k), n - k). - Benoit Cloitre, Jan 26 2003 [See Graham et al., eq. (5.39), p. 187. - Wolfdieter Lang, Aug 16 2019]
1 = Sum_{n >= 1} 3/a(n) = 3/4 + 3/16 + 3/64 + 3/256 + 3/1024, ...; with partial sums: 3/4, 15/16, 63/64, 255/256, 1023/1024, ... - Gary W. Adamson, Jun 16 2003
a(n) = A001045(2*n) + A001045(2*n+1). - Paul Barry, Apr 27 2004
A000005(a(n)) = A005408(n+1). - Reinhard Zumkeller, Mar 04 2007
a(n) = Sum_{j = 0..n} 2^(n - j)*binomial(n + j, j). - Peter C. Heinig (algorithms(AT)gmx.de), Apr 06 2007
Hankel transform of A115967. - Philippe Deléham, Jun 22 2007
a(n) = 6*Stirling2(n+1, 4) + 6*Stirling2(n+1, 3) + 3*Stirling2(n+1, 2) + 1 = 2*Stirling2(2^n, 2^n - 1) + Stirling2(n+1, 2) + 1. - Ross La Haye, Jun 26 2008
a(n) = A159991(n)/A001024(n) = A047653(n) + A181765(n). A160700(a(n)) = A010685(n). - Reinhard Zumkeller, May 02 2009
a(n) = A188915(A006127(n)). - Reinhard Zumkeller, Apr 14 2011
a(n) = Sum_{k = 0..n} binomial(2*n+1, k). - Mircea Merca, Jun 25 2011
Sum_{n >= 1} Mobius(n)/a(n) = 0.1710822479183... - R. J. Mathar, Aug 12 2012
a(n) = Sum_{k = 0..n} binomial(2*k + x, k)*binomial(2*(n - k) - x, n - k) for every real number x. - Rui Duarte and António Guedes de Oliveira, Feb 16 2013
a(n) = 5*a(n - 1) - 4*a(n - 2). - Jean-Bernard François, Sep 12 2013
a(n) = (2*n+1) * binomial(2*n,n) * Sum_{j=0..n} (-1)^j/(2*j+1)*binomial(n,j). - Vaclav Kotesovec, Sep 15 2013
a(n) = A000217(2^n - 1) + A000217(2^n). - J. M. Bergot, Dec 28 2014
a(n) = (2^n)^2 = A000079(n)^2. - Doug Bell, Jun 23 2015
a(n) = A002063(n)/3 - A004171(n). - Zhandos Mambetaliyev, Nov 19 2016
a(n) = (1/2) * Product_{k = 0..n} (1 + (2*n + 1)/(2*k + 1)). - Peter Bala, Mar 06 2018
a(n) = A001045(n+1)*A001045(n+2) + A001045(n)^2. - Ezhilarasu Velayutham, Aug 30 2019
a(n) = 1 + 3*Sum_{k=0..n} binomial(2*n, n+k)*(k|9), where (k|9) is the Jacobi symbol. - Greg Dresden, Oct 11 2022
a(n) = Sum_{k = 0..n} binomial(2*n+1, 2*k) = Sum_{k = 0..n} binomial(2*n+1, 2*k+1). - Sela Fried, Mar 23 2023

Extensions

Partially edited by Joerg Arndt, Mar 11 2010
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