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
Offset: 0

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Author

Keywords

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

A007310 Numbers congruent to 1 or 5 mod 6.

Original entry on oeis.org

1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 163, 167, 169, 173, 175
Offset: 1

Views

Author

C. Christofferson (Magpie56(AT)aol.com)

Keywords

Comments

Numbers n such that phi(4n) = phi(3n). - Benoit Cloitre, Aug 06 2003
Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5-rough numbers. (Edited by M. F. Hasler, Nov 01 2014: merged with comments from Zak Seidov, Apr 26 2007 and Michael B. Porter, Oct 09 2009)
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 38 ).
Numbers k such that k mod 2 = 1 and (k+1) mod 3 <> 1. - Klaus Brockhaus, Jun 15 2004
Also numbers n such that the sum of the squares of the first n integers is divisible by n, or A000330(n) = n*(n+1)*(2*n+1)/6 is divisible by n. - Alexander Adamchuk, Jan 04 2007
Numbers n such that the sum of squares of n consecutive integers is divisible by n, because A000330(m+n) - A000330(m) = n*(n+1)*(2*n+1)/6 + n*(m^2+n*m+m) is divisible by n independent of m. - Kaupo Palo, Dec 10 2016
A126759(a(n)) = n + 1. - Reinhard Zumkeller, Jun 16 2008
Terms of this sequence (starting from the second term) are equal to the result of the expression sqrt(4!*(k+1) + 1) - but only when this expression yields integral values (that is when the parameter k takes values, which are terms of A144065). - Alexander R. Povolotsky, Sep 09 2008
For n > 1: a(n) is prime if and only if A075743(n-2) = 1; a(2*n-1) = A016969(n-1), a(2*n) = A016921(n-1). - Reinhard Zumkeller, Oct 02 2008
A156543 is a subsequence. - Reinhard Zumkeller, Feb 10 2009
Numbers n such that ChebyshevT(x, x/2) is not an integer (is integer/2). - Artur Jasinski, Feb 13 2010
If 12*k + 1 is a perfect square (k = 0, 2, 4, 10, 14, 24, 30, 44, ... = A152749) then the square root of 12*k + 1 = a(n). - Gary Detlefs, Feb 22 2010
A089128(a(n)) = 1. Complement of A047229(n+1) for n >= 1. See A164576 for corresponding values A175485(a(n)). - Jaroslav Krizek, May 28 2010
Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (with h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 6). Also a(n)^2 - 1 == 0 (mod 12). - Bruno Berselli, Nov 05 2010 - Nov 17 2010
Numbers n such that ( Sum_{k = 1..n} k^14 ) mod n = 0. (Conjectured) - Gary Detlefs, Dec 27 2011
From Peter Bala, May 02 2018: (Start)
The above conjecture is true. Apply Ireland and Rosen, Proposition 15.2.2. with m = 14 to obtain the congruence 6*( Sum_{k = 1..n} k^14 )/n = 7 (mod n), true for all n >= 1. Suppose n is coprime to 6, then 6 is a unit in Z/nZ, and it follows from the congruence that ( Sum_{k = 1..n} k^14 )/n is an integer. On the other hand, if either 2 divides n or 3 divides n then the congruence shows that ( Sum_{k = 1..n} k^14 )/n cannot be integral. (End)
A126759(a(n)) = n and A126759(m) < n for m < a(n). - Reinhard Zumkeller, May 23 2013
(a(n-1)^2 - 1)/24 = A001318(n), the generalized pentagonal numbers. - Richard R. Forberg, May 30 2013
Numbers k for which A001580(k) is divisible by 3. - Bruno Berselli, Jun 18 2014
Numbers n such that sigma(n) + sigma(2n) = sigma(3n). - Jahangeer Kholdi and Farideh Firoozbakht, Aug 15 2014
a(n) are values of k such that Sum_{m = 1..k-1} m*(k-m)/k is an integer. Sums for those k are given by A062717. Also see Detlefs formula below based on A062717. - Richard R. Forberg, Feb 16 2015
a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)/6 is integral, and also such that the sequence c(n) = n*(n + m)*(n + 2*m)*(n + 3*m)/24 is integral. Cf. A007775. - Peter Bala, Nov 13 2015
Along with 2, these are the numbers k such that the k-th Fibonacci number is coprime to every Lucas number. - Clark Kimberling, Jun 21 2016
This sequence is the Engel expansion of 1F2(1; 5/6, 7/6; 1/36) + 1F2(1; 7/6, 11/6; 1/36)/5. - Benedict W. J. Irwin, Dec 16 2016
The sequence a(n), n >= 4 is generated by the successor of the pair of polygonal numbers {P_s(4) + 1, P_(2*s - 1)(3) + 1}, s >= 3. - Ralf Steiner, May 25 2018
The asymptotic density of this sequence is 1/3. - Amiram Eldar, Oct 18 2020
Also, the only vertices in the odd Collatz tree A088975 that are branch values to other odd nodes t == 1 (mod 2) of A005408. - Heinz Ebert, Apr 14 2021
From Flávio V. Fernandes, Aug 01 2021: (Start)
For any two terms j and k, the product j*k is also a term (the same property as p^n and smooth numbers).
From a(2) to a(phi(A033845(n))), or a((A033845(n))/3), the terms are the totatives of the A033845(n) itself. (End)
Also orders n for which cyclic and semicyclic diagonal Latin squares exist (see A123565 and A342990). - Eduard I. Vatutin, Jul 11 2023
If k is in the sequence, then k*2^m + 3 is also in the sequence, for all m > 0. - Jules Beauchamp, Aug 29 2024

Examples

			G.f. = x + 5*x^2 + 7*x^3 + 11*x^4 + 13*x^5 + 17*x^6 + 19*x^7 + 23*x^8 + ...
		

References

  • K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1980.

Crossrefs

A005408 \ A016945. Union of A016921 and A016969; union of A038509 and A140475. Essentially the same as A038179. Complement of A047229. Subsequence of A186422.
Cf. A000330, A001580, A002194, A019670, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885-A175887.
For k-rough numbers with other values of k, see A000027, A005408, A007775, A008364-A008366, A166061, A166063.
Cf. A126760 (a left inverse).
Row 3 of A260717 (without the initial 1).
Cf. A105397 (first differences).

Programs

Formula

a(n) = (6*n + (-1)^n - 3)/2. - Antonio Esposito, Jan 18 2002
a(n) = a(n-1) + a(n-2) - a(n-3), n >= 4. - Roger L. Bagula
a(n) = 3*n - 1 - (n mod 2). - Zak Seidov, Jan 18 2006
a(1) = 1 then alternatively add 4 and 2. a(1) = 1, a(n) = a(n-1) + 3 + (-1)^n. - Zak Seidov, Mar 25 2006
1 + 1/5^2 + 1/7^2 + 1/11^2 + ... = Pi^2/9 [Jolley]. - Gary W. Adamson, Dec 20 2006
For n >= 3 a(n) = a(n-2) + 6. - Zak Seidov, Apr 18 2007
From R. J. Mathar, May 23 2008: (Start)
Expand (x+x^5)/(1-x^6) = x + x^5 + x^7 + x^11 + x^13 + ...
O.g.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^2). (End)
a(n) = 6*floor(n/2) - 1 + 2*(n mod 2). - Reinhard Zumkeller, Oct 02 2008
1 + 1/5 - 1/7 - 1/11 + + - - ... = Pi/3 = A019670 [Jolley eq (315)]. - Jaume Oliver Lafont, Oct 23 2009
a(n) = ( 6*A062717(n)+1 )^(1/2). - Gary Detlefs, Feb 22 2010
a(n) = 6*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), with n > 1. - Bruno Berselli, Nov 05 2010
a(n) = 6*n - a(n-1) - 6 for n>1, a(1) = 1. - Vincenzo Librandi, Nov 18 2010
Sum_{n >= 1} (-1)^(n+1)/a(n) = A093766 [Jolley eq (84)]. - R. J. Mathar, Mar 24 2011
a(n) = 6*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
a(n) = 3*n + ((n+1) mod 2) - 2. - Gary Detlefs, Jan 08 2012
a(n) = 2*n + 1 + 2*floor((n-2)/2) = 2*n - 1 + 2*floor(n/2), leading to the o.g.f. given by R. J. Mathar above. - Wolfdieter Lang, Jan 20 2012
1 - 1/5 + 1/7 - 1/11 + - ... = Pi*sqrt(3)/6 = A093766 (L. Euler). - Philippe Deléham, Mar 09 2013
1 - 1/5^3 + 1/7^3 - 1/11^3 + - ... = Pi^3*sqrt(3)/54 (L. Euler). - Philippe Deléham, Mar 09 2013
gcd(a(n), 6) = 1. - Reinhard Zumkeller, Nov 14 2013
a(n) = sqrt(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5)/sqrt(2). - Alexander R. Povolotsky, May 16 2014
a(n) = 3*n + 6/(9*n mod 6 - 6). - Mikk Heidemaa, Feb 05 2016
From Mikk Heidemaa, Feb 11 2016: (Start)
a(n) = 2*floor(3*n/2) - 1.
a(n) = A047238(n+1) - 1. (suggested by Michel Marcus) (End)
E.g.f.: (2 + (6*x - 3)*exp(x) + exp(-x))/2. - Ilya Gutkovskiy, Jun 18 2016
From Bruno Berselli, Apr 27 2017: (Start)
a(k*n) = k*a(n) + (4*k + (-1)^k - 3)/2 for k>0 and odd n, a(k*n) = k*a(n) + k - 1 for even n. Some special cases:
k=2: a(2*n) = 2*a(n) + 3 for odd n, a(2*n) = 2*a(n) + 1 for even n;
k=3: a(3*n) = 3*a(n) + 4 for odd n, a(3*n) = 3*a(n) + 2 for even n;
k=4: a(4*n) = 4*a(n) + 7 for odd n, a(4*n) = 4*a(n) + 3 for even n;
k=5: a(5*n) = 5*a(n) + 8 for odd n, a(5*n) = 5*a(n) + 4 for even n, etc. (End)
From Antti Karttunen, May 20 2017: (Start)
a(A273669(n)) = 5*a(n) = A084967(n).
a((5*n)-3) = A255413(n).
A126760(a(n)) = n. (End)
a(2*m) = 6*m - 1, m >= 1; a(2*m + 1) = 6*m + 1, m >= 0. - Ralf Steiner, May 17 2018
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(3) (A002194).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi/3 (A019670). (End)

A001651 Numbers not divisible by 3.

Original entry on oeis.org

1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29, 31, 32, 34, 35, 37, 38, 40, 41, 43, 44, 46, 47, 49, 50, 52, 53, 55, 56, 58, 59, 61, 62, 64, 65, 67, 68, 70, 71, 73, 74, 76, 77, 79, 80, 82, 83, 85, 86, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 103, 104
Offset: 1

Views

Author

Keywords

Comments

Inverse binomial transform of A084858. - Benoit Cloitre, Jun 12 2003
Earliest monotonic sequence starting with (1,2) and satisfying the condition: "a(n)+a(n-1) is not in the sequence." - Benoit Cloitre, Mar 25 2004. [The numbers of the form a(n)+a(n-1) form precisely the complement with respect to the positive integers. - David W. Wilson, Feb 18 2012]
a(1) = 1; a(n) is least number which is relatively prime to the sum of all the previous terms. - Amarnath Murthy, Jun 18 2001
For n > 3, numbers having 3 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005
Also numbers n such that (n+1)*(n+2)/6 = A000292(n)/n is an integer. - Ctibor O. Zizka, Oct 15 2010
Notice the 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, n natural numbers), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 3). - Bruno Berselli, Nov 17 2010
A001651 mod 9 gives A141425. - Paul Curtz, Dec 31 2010. (Correct for the modified offset 1. - M. F. Hasler, Apr 07 2015)
The set of natural numbers (1, 2, 3, ...), sequence A000027; represents the numbers of ordered compositions of n using terms in the signed set: (1, 2, -4, -5, 7, 8, -10, -11, 13, 14, ...). This follows from (1, 2, 3, ...) being the INVERT transform of A011655, signed and beginning: (1, 1, 0, -1, -1, 0, 1, 1, 0, ...). - Gary W. Adamson, Apr 28 2013
Union of A047239 and A047257. - Wesley Ivan Hurt, Dec 19 2013
Numbers whose sum of digits (and digital root) is != 0 (mod 3). - Joerg Arndt, Aug 29 2014
The number of partitions of 3*(n-1) into at most 2 parts. - Colin Barker, Apr 22 2015
a(n) is the number of partitions of 3*n into two distinct parts. - L. Edson Jeffery, Jan 14 2017
Conjectured (and like even easily proved) to be the graph bandwidth of the complete bipartite graph K_{n,n}. - Eric W. Weisstein, Apr 24 2017
Numbers k such that Fibonacci(k) mod 4 = 1 or 3. Equivalently, sequence lists the indices of the odd Fibonacci numbers (see A014437). - Bruno Berselli, Oct 17 2017
Minimum value of n_3 such that the "rectangular spiral pattern" is the optimal solution for Ripà's n_1 X n_2 x n_3 Dots Problem, for any n_1 = n_2. For example, if n_1 = n_2 = 5, n_3 = floor((3/2)*(n_1 - 1)) + 1 = a(5). - Marco Ripà, Jul 23 2018
For n >= 54, a(n) = sat(n, P_n), the minimum number of edges in a P_n-saturated graph on n vertices, where P_n is the n-vertex path (see Dudek, Katona, and Wojda, 2003; Frick and Singleton, 2005). - Danny Rorabaugh, Nov 07 2017
From Roger Ford, May 09 2021: (Start)
a(n) is the smallest sum of arch lengths for the top arches of a semi-meander with n arches. An arch length is the number of arches covered + 1.
/\ The top arch has a length of 3. /\ The top arch has a length of 3.
/ \ Both bottom arches have a //\\ The middle arch has a length of 2.
//\/\\ length of 1. ///\\\ The bottom arch has a length of 1.
Example: a(6) = 8 /\ /\
//\\ /\ //\\ /\ 2 + 1 + 1 + 2 + 1 + 1 = 8. (End)
This is the lexicographically earliest increasing sequence of positive integers such that no polynomial of degree d can be fitted to d+2 consecutive terms (equivalently, such that no iterated difference is zero). - Pontus von Brömssen, Dec 26 2021

Examples

			G.f.: x + 2*x^2 + 4*x^3 + 5*x^4 + 7*x^5 + 8*x^6 + 10*x^7 + 11*x^8 + 13*x^9 + ...
		

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

Programs

  • GAP
    Filtered([0..110],n->n mod 3<>0); # Muniru A Asiru, Jul 24 2018
    
  • Haskell
    a001651 = (`div` 2) . (subtract 1) . (* 3)
    a001651_list = filter ((/= 0) . (`mod` 3)) [1..]
    -- Reinhard Zumkeller, Jul 07 2012, Aug 23 2011
    
  • Magma
    [3*(2*n-1)/4-(-1)^n/4: n in [1..80]]; // Vincenzo Librandi, Jun 07 2011
    
  • Maple
    A001651 := n -> 3*floor(n/2) - (-1)^n; # Corrected by M. F. Hasler, Apr 07 2015
    A001651:=(1+z+z**2)/(z+1)/(z-1)**2; # Simon Plouffe in his 1992 dissertation
    a[1]:=1:a[2]:=2:for n from 3 to 100 do a[n]:=a[n-2]+3 od: seq(a[n], n=1..69); # Zerinvary Lajos, Mar 16 2008, offset corrected by M. F. Hasler, Apr 07 2015
  • Mathematica
    Select[Table[n,{n,200}],Mod[#,3]!=0&] (* Vladimir Joseph Stephan Orlovsky, Feb 18 2011 *)
    Drop[Range[200 + 1], {1, -1, 3}] - 1 (* József Konczer, May 24 2016 *)
    Floor[(3 Range[70] - 1)/2] (* Eric W. Weisstein, Apr 24 2017 *)
    CoefficientList[Series[(x^2 + x + 1)/((x - 1)^2 (x + 1)), {x, 0, 70}],
      x] (* or *)
    LinearRecurrence[{1, 1, -1}, {1, 2, 4}, 70] (* Robert G. Wilson v, Jul 25 2018 *)
  • PARI
    {a(n) = n + (n-1)\2}; /* Michael Somos, Jan 15 2011 */
    
  • PARI
    x='x+O('x^100); Vec(x*(1+x+x^2)/((1-x)*(1-x^2))) \\ Altug Alkan, Oct 22 2015
    
  • Python
    print([k for k in range(1, 105) if k%3]) # Michael S. Branicky, Sep 06 2021
    
  • Python
    def A001651(n): return (n<<1)-(n>>1)-1 # Chai Wah Wu, Mar 05 2024

Formula

a(n) = 3 + a(n-2) for n > 2.
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 3.
a(2*n+1) = 3*n+1, a(2*n) = 3*n-1.
G.f.: x * (1 + x + x^2) / ((1 - x) * (1 - x^2)). - Michael Somos, Jun 08 2000
a(n) = (4-n)*a(n-1) + 2*a(n-2) + (n-3)*a(n-3) (from the Carlitz et al. article).
a(n) = floor((3*n-1)/2). [Corrected by Gary Detlefs]
a(1) = 1, a(n) = 2*a(n-1) - 3*floor(a(n-1)/3). - Benoit Cloitre, Aug 17 2002
a(n+1) = 1 + n - n mod 2 + (n + n mod 2)/2. - Reinhard Zumkeller, Dec 17 2002
a(1) = 1, a(n+1) = a(n) + (a(n) mod 3). - Reinhard Zumkeller, Mar 23 2003
a(1) = 1, a(n) = 3*(n-1) - a(n-1). - Benoit Cloitre, Apr 12 2003
a(n) = 3*(2*n-1)/4 - (-1)^n/4. - Benoit Cloitre, Jun 12 2003
Nearest integer to (Sum_{k>=n} 1/k^3)/(Sum_{k>=n} 1/k^4). - Benoit Cloitre, Jun 12 2003
Partial sums of A040001. a(n) = A032766(n-1)+1. - Paul Barry, Sep 02 2003
a(n) = T(n, 1) = T(n, n-1), where T is the array in A026386. - Emeric Deutsch, Feb 18 2004
a(n) = sqrt(3*A001082(n)+1). - Zak Seidov, Dec 12 2007
a(n) = A077043(n) - A077043(n-1). - Reinhard Zumkeller, Dec 28 2007
a(n) = A001477(n-1) + A008619(n-1). - Yosu Yurramendi, Aug 10 2008
Euler transform of length 3 sequence [2, 1, -1]. - Michael Somos, Sep 06 2008
A011655(a(n)) = 1. - Reinhard Zumkeller, Nov 30 2009
a(n) = n - 1 + ceiling(n/2). - Michael Somos, Jan 15 2011
a(n) = 3*A000217(n)+1 - 2*Sum_{i=1..n-1} a(i), for n>1. - Bruno Berselli, Nov 17 2010
a(n) = 3*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
A215879(a(n)) > 0. - Reinhard Zumkeller, Dec 28 2012 [More precisely, A215879 is the characteristic function of A001651. - M. F. Hasler, Apr 07 2015]
a(n) = 2n - 1 - floor(n/2). - Wesley Ivan Hurt, Oct 25 2013
a(n) = (3n - 2 + (n mod 2)) / 2. - Wesley Ivan Hurt, Mar 31 2014
a(n) = A000217(n) - A000982(n-1). - Bui Quang Tuan, Mar 28 2015
1/1^3 - 1/2^3 + 1/4^3 - 1/5^3 + 1/7^3 - 1/8^3 + ... = 4 Pi^3/(3 sqrt(3)). - M. F. Hasler, Mar 29 2015
E.g.f.: (4 + sinh(x) - cosh(x) + 3*(2*x - 1)*exp(x))/4. - Ilya Gutkovskiy, May 24 2016
a(n) = a(n+k-1) + a(n-k) - a(n-1) for n > k >= 0. - Bob Selcoe, Feb 03 2017
a(n) = -a(1-n) for all n in Z. - Michael Somos, Jul 31 2018
a(n) = n + A004526(n-1). - David James Sycamore, Sep 06 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(3*sqrt(3)) (A073010). - Amiram Eldar, Dec 04 2021
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 1.
Product_{n>=2} (1 + (-1)^n/a(n)) = 2*Pi/(3*sqrt(3)) (A248897). (End)

Extensions

This is a list, so the offset should be 1. I corrected this and adjusted some of the comments and formulas. Other lines probably also need to be adjusted. - N. J. A. Sloane, Jan 01 2011
Offset of pre-2011 formulas verified or corrected by M. F. Hasler, Apr 07-18 2015 and by Danny Rorabaugh, Oct 23 2015

A047209 Numbers that are congruent to {1, 4} mod 5.

Original entry on oeis.org

1, 4, 6, 9, 11, 14, 16, 19, 21, 24, 26, 29, 31, 34, 36, 39, 41, 44, 46, 49, 51, 54, 56, 59, 61, 64, 66, 69, 71, 74, 76, 79, 81, 84, 86, 89, 91, 94, 96, 99, 101, 104, 106, 109, 111, 114, 116, 119, 121, 124, 126, 129, 131, 134, 136, 139, 141, 144, 146, 149, 151, 154
Offset: 1

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Author

Keywords

Comments

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 72 ).
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, n natural numbers), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in our case, a(n)^2 - 1 == 0 (mod 5). - Bruno Berselli, Nov 17 2010
The sum of the alternating series (-1)^(n+1)/a(n) from n=1 to infinity is (Pi/5)*cot(Pi/5), that is (1/5)*sqrt(1 + 2/sqrt(5))*Pi. - Jean-François Alcover, May 03 2013
These numbers appear in the product of a Rogers-Ramanujan identity. See A003114 also for references. - Wolfdieter Lang, Oct 29 2016
Let m be a product of any number of terms of this sequence. Then m - 1 or m + 1 is divisible by 5. Closed under multiplication. - David A. Corneth, May 11 2018

Crossrefs

Cf. A005408 (n=1 or 3 mod 4), A007310 (n=1 or 5 mod 6).
Cf. A045468 (primes), A032527 (partial sums).

Programs

Formula

G.f.: (1+3x+x^2)/((1-x)(1-x^2)).
a(n) = floor((5n-2)/2). [corrected by Reinhard Zumkeller, Jul 19 2013]
a(1) = 1, a(n) = 5(n-1) - a(n-1). - Benoit Cloitre, Apr 12 2003
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = (10*n + (-1)^n - 5)/4.
a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.
a(n) = a(n-2) + 5 for n > 2.
a(n) = 5*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1.
a(n)^2 = 5*A036666(n) + 1 (cf. also Comments). (End)
a(n) = 5*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
E.g.f.: 1 + ((10*x - 5)*exp(x) + exp(-x))/4. - David Lovler, Aug 23 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = phi (A001622).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/5) * cosec(Pi/5) (A352324). (End)

Extensions

Edited by Michael Somos, Sep 22 2002

A047522 Numbers that are congruent to {1, 7} mod 8.

Original entry on oeis.org

1, 7, 9, 15, 17, 23, 25, 31, 33, 39, 41, 47, 49, 55, 57, 63, 65, 71, 73, 79, 81, 87, 89, 95, 97, 103, 105, 111, 113, 119, 121, 127, 129, 135, 137, 143, 145, 151, 153, 159, 161, 167, 169, 175, 177, 183, 185, 191, 193, 199, 201, 207, 209, 215, 217, 223, 225, 231, 233
Offset: 1

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Author

Keywords

Comments

Also n such that Kronecker(2,n) = mu(gcd(2,n)). - Jon Perry and T. D. Noe, Jun 13 2003
Also n such that x^2 == 2 (mod n) has a solution. The primes are given in sequence A001132. - T. D. Noe, Jun 13 2003
As indicated in the formula, a(n) is related to the even triangular numbers. - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
Cf. property described by Gary Detlefs in A113801: more generally, these a(n) are of the form (2*h*n + (h-4)*(-1)^n-h)/4 (h,n natural numbers). Therefore a(n)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 8). Also a(n)^2 - 1 == 0 (mod 16). - Bruno Berselli, Nov 17 2010
A089911(3*a(n)) = 2. - Reinhard Zumkeller, Jul 05 2013
S(a(n+1)/2, 0) = (1/2)*(S(a(n+1), sqrt(2)) - S(a(n+1) - 2, sqrt(2))) = T(a(n+1), sqrt(2)/2) = cos(a(n+1)*Pi/4) = sqrt(2)/2 = A010503, identically for n >= 0, where S is the Chebyshev polynomial (A049310) here extended to fractional n, evaluated at x = 0. (For T see A053120.) - Wolfdieter Lang, Jun 04 2023

References

  • L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 16.

Crossrefs

Programs

  • Haskell
    a047522 n = a047522_list !! (n-1)
    a047522_list = 1 : 7 : map (+ 8) a047522_list
    -- Reinhard Zumkeller, Jan 07 2012
    
  • Mathematica
    Select[Range[1, 191, 2], JacobiSymbol[2, # ]==1&]
  • PARI
    a(n)=4*n-2+(-1)^n \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = sqrt(8*A014494(n)+1) = sqrt(16*ceiling(n/2)*(2*n+1)+1) = sqrt(8*A056575(n)-8*(2n+1)*(-1)^n+1). - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
1 - 1/7 + 1/9 - 1/15 + 1/17 - ... = (Pi/8)*(1 + sqrt(2)). [Jolley] - Gary W. Adamson, Dec 16 2006
From R. J. Mathar, Feb 19 2009: (Start)
a(n) = 4n - 2 + (-1)^n = a(n-2) + 8.
G.f.: x(1+6x+x^2)/((1+x)(1-x)^2). (End)
a(n) = 8*n - a(n-1) - 8. - Vincenzo Librandi, Aug 06 2010
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 8*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
E.g.f.: 1 + (4*x - 1)*cosh(x) + (4*x - 3)*sinh(x). - Stefano Spezia, May 13 2021
E.g.f.: 1 + (4*x - 3)*exp(x) + 2*cosh(x). - David Lovler, Jul 16 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(2+sqrt(2)) (A179260).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/8)*cosec(Pi/8) (A352125). (End)

A056020 Numbers that are congruent to +-1 mod 9.

Original entry on oeis.org

1, 8, 10, 17, 19, 26, 28, 35, 37, 44, 46, 53, 55, 62, 64, 71, 73, 80, 82, 89, 91, 98, 100, 107, 109, 116, 118, 125, 127, 134, 136, 143, 145, 152, 154, 161, 163, 170, 172, 179, 181, 188, 190, 197, 199, 206, 208, 215, 217, 224, 226, 233, 235, 242, 244, 251, 253
Offset: 1

Views

Author

Robert G. Wilson v, Jun 08 2000

Keywords

Comments

Or, numbers k such that k^2 == 1 (mod 9).
Or, numbers k such that the iterative cycle j -> sum of digits of j^2 when started at k contains a 1. E.g., 8 -> 6+4 = 10 -> 1+0+0 = 1 and 17 -> 2+8+9 = 19 -> 3+6+1 = 10 -> 1+0+0 = 1. - Asher Auel, May 17 2001

Crossrefs

Cf. A007953, A047522 (n=1 or 7 mod 8), A090771 (n=1 or 9 mod 10).
Cf. A129805 (primes), A195042 (partial sums).
Cf. A381319 (general case mod n^2).

Programs

  • Haskell
    a056020 n = a056020_list !! (n-1)
    a05602_list = 1 : 8 : map (+ 9) a056020_list
    -- Reinhard Zumkeller, Jan 07 2012
  • Mathematica
    Select[ Range[ 300 ], PowerMod[ #, 2, 3^2 ]==1& ]
    (* or *)
    LinearRecurrence[{1, 1, -1}, {1, 8, 10}, 67] (* Mike Sheppard, Feb 18 2025 *)
  • PARI
    a(n)=9*(n>>1)+if(n%2,1,-1) \\ Charles R Greathouse IV, Jun 29 2011
    
  • PARI
    for(n=1,40,print1(9*n-8,", ",9*n-1,", ")) \\ Charles R Greathouse IV, Jun 29 2011
    

Formula

a(1) = 1; a(n) = 9(n-1) - a(n-1). - Rolf Pleisch, Jan 31 2008 [Offset corrected by Jon E. Schoenfield, Dec 22 2008]
From R. J. Mathar, Feb 10 2008: (Start)
O.g.f.: 1 + 5/(4(x+1)) + 27/(4(-1+x)) + 9/(2(-1+x)^2).
a(n+1) - a(n) = A010697(n). (End)
a(n) = (9*A132355(n) + 1)^(1/2). - Gary Detlefs, Feb 22 2010
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = a(n-2) + 9, for n > 2.
a(n) = 9*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), n > 1. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/9)*cot(Pi/9) = A019676 * A019968. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((18*x - 9)*exp(x) + 5*exp(-x))/4. - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/9) (A332437).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/9)*cosec(Pi/9). (End)
From Mike Sheppard, Feb 18 2025: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3).
a(n) ~ (3^2/2)*n. (End)

A090771 Numbers that are congruent to {1, 9} mod 10.

Original entry on oeis.org

1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69, 71, 79, 81, 89, 91, 99, 101, 109, 111, 119, 121, 129, 131, 139, 141, 149, 151, 159, 161, 169, 171, 179, 181, 189, 191, 199, 201, 209, 211, 219, 221, 229, 231, 239, 241, 249, 251, 259, 261, 269, 271, 279, 281
Offset: 1

Views

Author

Giovanni Teofilatto, Feb 07 2004

Keywords

Comments

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, n natural numbers), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 10). - Bruno Berselli, Nov 17 2010

Crossrefs

Cf. A056020 (n = 1 or 8 mod 9), A175885 (n = 1 or 10 mod 11).
Cf. A045468 (primes), A195142 (partial sums).

Programs

Formula

a(n) = sqrt(40*A057569(n) + 1). - Gary Detlefs, Feb 22 2010
From Bruno Berselli, Sep 16 2010 - Nov 17 2010: (Start)
G.f.: x*(1 + 8*x + x^2)/((1 + x)*(1 - x)^2).
a(n) = (10*n + 3*(-1)^n - 5)/2.
a(n) = -a(-n + 1) = a(n-1) + a(n-2) - a(n-3) = a(n-2) + 10.
a(n) = 10*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
a(n) = 10*n - a(n-1) - 10 (with a(1) = 1). - Vincenzo Librandi, Nov 16 2010
a(n) = sqrt(10*A132356(n-1) + 1). - Ivan N. Ianakiev, Nov 09 2012
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/10)*cot(Pi/10) = A000796 * A019970 / 10 = sqrt(5 + 2*sqrt(5))*Pi/10. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((10*x - 5)*exp(x) + 3*exp(-x))/2. - David Lovler, Sep 03 2022
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(phi+2) (A188593).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi*phi/5 = A094888/10. (End)

Extensions

Edited and extended by Ray Chandler, Feb 10 2004

A047336 Numbers that are congruent to {1, 6} mod 7.

Original entry on oeis.org

1, 6, 8, 13, 15, 20, 22, 27, 29, 34, 36, 41, 43, 48, 50, 55, 57, 62, 64, 69, 71, 76, 78, 83, 85, 90, 92, 97, 99, 104, 106, 111, 113, 118, 120, 125, 127, 132, 134, 139, 141, 146, 148, 153, 155, 160, 162, 167, 169, 174, 176, 181, 183, 188, 190, 195, 197, 202, 204, 209
Offset: 1

Views

Author

Keywords

Comments

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, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2-1 == 0 (mod 7). - Bruno Berselli, Nov 17 2010

Crossrefs

Programs

  • Haskell
    a047336 n = a047336_list !! (n-1)
    a047336_list = 1 : 6 : map (+ 7) a047336_list
    -- Reinhard Zumkeller, Jan 07 2012
    
  • Magma
    [n: n in [1..210]| n mod 7 in {1,6}]; // Bruno Berselli, Feb 22 2011
    
  • Mathematica
    Rest[Flatten[Table[{7i-1,7i+1},{i,0,40}]]] (* Harvey P. Dale, Nov 20 2010 *)
  • PARI
    a(n)=n\2*7-(-1)^n \\ Charles R Greathouse IV, May 02 2016

Formula

a(1) = 1; a(n) = 7(n-1) - a(n-1). - Rolf Pleisch, Jan 31 2008 (corrected by Jon E. Schoenfield, Dec 22 2008)
a(n) = (7/2)*(n-(1-(-1)^n)/2) - (-1)^n. - Rolf Pleisch, Nov 02 2010
From Bruno Berselli, Nov 17 2010: (Start)
G.f.: x*(1+5*x+x^2)/((1+x)*(1-x)^2).
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = a(n-2)+7.
a(n) = 7*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
a(n) = 7*floor(n/2)+(-1)^(n+1). - Gary Detlefs, Dec 29 2011
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/7)*cot(Pi/7) = A019674 * A178818. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((14*x - 7)*exp(x) + 3*exp(-x))/4. - David Lovler, Sep 01 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/7) (A160389).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/7) * cosec(Pi/7) (A371858). (End)

Extensions

More terms from Jon E. Schoenfield, Jan 18 2009

A091998 Numbers that are congruent to {1, 11} mod 12.

Original entry on oeis.org

1, 11, 13, 23, 25, 35, 37, 47, 49, 59, 61, 71, 73, 83, 85, 95, 97, 107, 109, 119, 121, 131, 133, 143, 145, 155, 157, 167, 169, 179, 181, 191, 193, 203, 205, 215, 217, 227, 229, 239, 241, 251, 253, 263, 265, 275, 277, 287, 289, 299, 301, 311, 313, 323, 325, 335
Offset: 1

Views

Author

Ray Chandler, Feb 21 2004

Keywords

Comments

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), then ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in our case, a(n)^2 - 1 == 0 (mod 12). Also a(n)^2 - 1 == 0 (mod 24).

Crossrefs

First row of A092260.
Cf. A175885 (n == 1 or 10 (mod 11)), A175886 (n == 1 or 12 (mod 13)).
Cf. A097933 (primes), A195143 (partial sums).

Programs

  • Haskell
    a091998 n = a091998_list !! (n-1)
    a091998_list = 1 : 11 : map (+ 12) a091998_list
    -- Reinhard Zumkeller, Jan 07 2012
    
  • Magma
    [ n: n in [1..350] | n mod 12 eq 1 or n mod 12 eq 11 ];
    
  • Mathematica
    LinearRecurrence[{1,1,-1},{1,11,13},100] (* Harvey P. Dale, Jul 26 2017 *)
  • PARI
    is(n)=n=n%12;n==11 || n==1 \\ Charles R Greathouse IV, Jul 02 2013

Formula

a(n) = 12*n - a(n-1) - 12 (with a(1)=1). - Vincenzo Librandi, Nov 16 2010
a(n) = 6*n + 2*(-1)^n - 3.
G.f.: x*(1+10*x+x^2)/((1+x)*(1-x)^2).
a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.
a(n) = a(n-2) + 12 for n > 2.
a(n) = 12*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2 + sqrt(3))*Pi/12. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + (6*x - 3)*exp(x) + 2*exp(-x). - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(2 + sqrt(3)) = 2*cos(Pi/12) (A188887).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/3)*cos(Pi/12). (End)

Extensions

Formulae and comment added by Bruno Berselli, Nov 17 2010 - Nov 18 2010

A175885 Numbers that are congruent to {1, 10} mod 11.

Original entry on oeis.org

1, 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98, 100, 109, 111, 120, 122, 131, 133, 142, 144, 153, 155, 164, 166, 175, 177, 186, 188, 197, 199, 208, 210, 219, 221, 230, 232, 241, 243, 252, 254, 263, 265, 274, 276, 285, 287, 296, 298
Offset: 1

Views

Author

Bruno Berselli, Oct 08 2010 - Nov 17 2010

Keywords

Comments

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, n natural numbers), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 11).

Crossrefs

Cf. A090771 (n==1 or 9 mod 10), A091998 (n==1 or 11 mod 12).
Cf. A195043 (partial sums).

Programs

Formula

G.f.: x*(1+9*x+x^2)/((1+x)*(1-x)^2).
a(n) = (22*n + 7*(-1)^n - 11)/4.
a(n) = -a(-n+1) = a(n-2) + 11 = a(n-1) + a(n-2) - a(n-3).
a(n) = 11*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1.
a(n) = A195312(n) + A195312(n-1) = A195313(n) - A195313(n-2). - Bruno Berselli, Sep 18 2011
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/11)*cot(Pi/11). - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((22*x - 11)*exp(x) + 7*exp(-x))/4. - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/11).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/11)*cosec(Pi/11). (End)
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