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.

Previous Showing 41-50 of 852 results. Next

A000071 a(n) = Fibonacci(n) - 1.

Original entry on oeis.org

0, 0, 1, 2, 4, 7, 12, 20, 33, 54, 88, 143, 232, 376, 609, 986, 1596, 2583, 4180, 6764, 10945, 17710, 28656, 46367, 75024, 121392, 196417, 317810, 514228, 832039, 1346268, 2178308, 3524577, 5702886, 9227464, 14930351, 24157816, 39088168, 63245985, 102334154
Offset: 1

Views

Author

Keywords

Comments

a(n) is the number of allowable transition rules for passing from one change to the next (on n-1 bells) in the English art of bell-ringing. This is also the number of involutions in the symmetric group S_{n-1} which can be represented as a product of transpositions of consecutive numbers from {1, 2, ..., n-1}. Thus for n = 6 we have a(6) from (12), (12)(34), (12)(45), (23), (23)(45), (34), (45), for instance. See my 1983 Math. Proc. Camb. Phil. Soc. paper. - Arthur T. White, letter to N. J. A. Sloane, Dec 18 1986
Number of permutations p of {1, 2, ..., n-1} such that max|p(i) - i| = 1. Example: a(4) = 2 since only the permutations 132 and 213 of {1, 2, 3} satisfy the given condition. - Emeric Deutsch, Jun 04 2003 [For a(5) = 4 we have 2143, 1324, 2134 and 1243. - Jon Perry, Sep 14 2013]
Number of 001-avoiding binary words of length n-3. a(n) is the number of partitions of {1, ..., n-1} into two blocks in which only 1- or 2-strings of consecutive integers can appear in a block and there is at least one 2-string. E.g., a(6) = 7 because the enumerated partitions of {1, 2, 3, 4, 5} are 124/35, 134/25, 14/235, 13/245, 1245/3, 145/23, 125/34. - Augustine O. Munagi, Apr 11 2005
Numbers for which only one Fibonacci bit-representation is possible and for which the maximal and minimal Fibonacci bit-representations (A104326 and A014417) are equal. For example, a(12) = 10101 because 8 + 3 + 1 = 12. - Casey Mongoven, Mar 19 2006
Beginning with a(2), the "Recamán transform" (see A005132) of the Fibonacci numbers (A000045). - Nick Hobson, Mar 01 2007
Starting with nonzero terms, a(n) gives the row sums of triangle A158950. - Gary W. Adamson, Mar 31 2009
a(n+2) is the minimum number of elements in an AVL tree of height n. - Lennert Buytenhek (buytenh(AT)wantstofly.org), May 31 2010
a(n) is the number of branch nodes in the Fibonacci tree of order n-1. A Fibonacci tree of order n (n >= 2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node (see the Knuth reference, p. 417). - Emeric Deutsch, Jun 14 2010
a(n+3) is the number of distinct three-strand positive braids of length n (cf. Burckel). - Maxime Bourrigan, Apr 04 2011
a(n+1) is the number of compositions of n with maximal part 2. - Joerg Arndt, May 21 2013
a(n+2) is the number of leafs of great-grandparent DAG (directed acyclic graph) of height n. A great-grandparent DAG of height n is a single node for n = 1; for n > 1 each leaf of ggpDAG(n-1) has two child nodes where pairs of adjacent new nodes are merged into single node if and only if they have disjoint grandparents and same great-grandparent. Consequence: a(n) = 2*a(n-1) - a(n-3). - Hermann Stamm-Wilbrandt, Jul 06 2014
2 and 7 are the only prime numbers in this sequence. - Emmanuel Vantieghem, Oct 01 2014
From Russell Jay Hendel, Mar 15 2015: (Start)
We can establish Gerald McGarvey's conjecture mentioned in the Formula section, however we require n > 4. We need the following 4 prerequisites.
(1) a(n) = F(n) - 1, with {F(n)}A000045.%20(2)%20(Binet%20form)%20F(n)%20=%20(d%5En%20-%20e%5En)/sqrt(5)%20with%20d%20=%20phi%20and%20e%20=%201%20-%20phi,%20de%20=%20-1%20and%20d%20+%20e%20=%201.%20It%20follows%20that%20a(n)%20=%20(d(n)%20-%20e(n))/sqrt(5)%20-%201.%20(3)%20To%20prove%20floor(x)%20=%20y%20is%20equivalent%20to%20proving%20that%20x%20-%20y%20lies%20in%20the%20half-open%20interval%20%5B0,%201).%20(4)%20The%20series%20%7Bs(n)%20=%20c1%20x%5En%20+%20c2%7D">{n >= 1} the Fibonacci numbers A000045. (2) (Binet form) F(n) = (d^n - e^n)/sqrt(5) with d = phi and e = 1 - phi, de = -1 and d + e = 1. It follows that a(n) = (d(n) - e(n))/sqrt(5) - 1. (3) To prove floor(x) = y is equivalent to proving that x - y lies in the half-open interval [0, 1). (4) The series {s(n) = c1 x^n + c2}{n >= 1}, with -1 < x < 0, and c1 and c2 positive constants, converges by oscillation with s(1) < s(3) < s(5) < ... < s(6) < s(4) < s(2). If follows that for any odd n, the open interval (s(n), s(n+1)) contains the subsequence {s(t)}_{t >= n + 2}. Using these prerequisites we can analyze the conjecture.
Using prerequisites (2) and (3) we see we must prove, for all n > 4, that d((d^(n-1) - e^(n-1))/sqrt(5) - 1) - (d^n - e^n)/sqrt(5) + 1 + c lies in the interval [0, 1). But de = -1, implying de^(n-1) = -e^(n-2). It follows that we must equivalently prove (for all n > 4) that E(n, c) = (e^(n-2) + e^n)/sqrt(5) + 1 - d + c = e^(n-2) (e^2 + 1)/sqrt(5) + e + c lies in [0, 1). Clearly, for any particular n, E(n, c) has extrema (maxima, minima) when c = 2*(1-d) and c = (1+d)*(1-d). Therefore, the proof is completed by using prerequisite (4). It suffices to verify E(5, 2*(1-d)) = 0, E(6, 2*(1-d)) = 0.236068, E(5, (1-d)*(1+d)) = 0.618034, E(6, (1-d)*(1+d)) = 0.854102, all lie in [0, 1).
(End)
a(n) can be shown to be the number of distinct nonempty matchings on a path with n vertices. (A matching is a collection of disjoint edges.) - Andrew Penland, Feb 14 2017
Also, for n > 3, the lexicographically earliest sequence of positive integers such that {phi*a(n)} is located strictly between {phi*a(n-1)} and {phi*a(n-2)}. - Ivan Neretin, Mar 23 2017
From Eric M. Schmidt, Jul 17 2017: (Start)
Number of sequences (e(1), ..., e(n-2)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) <= e(k). [Martinez and Savage, 2.5]
Number of sequences (e(1), ..., e(n-2)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) >= e(j) <= e(k) and e(i) != e(k). [Martinez and Savage, 2.5]
(End)
Numbers whose Zeckendorf (A014417) and dual Zeckendorf (A104326) representations are the same: alternating digits of 1 and 0. - Amiram Eldar, Nov 01 2019
a(n+2) is the length of the longest array whose local maximum element can be found in at most n reveals. See link to the puzzle by Alexander S. Kulikov. - Dmitry Kamenetsky, Aug 08 2020
a(n+2) is the number of nonempty subsets of {1,2,...,n} that contain no consecutive elements. For example, the a(6)=7 subsets of {1,2,3,4} are {1}, {2}, {3}, {4}, {1,3}, {1,4} and {2,4}. - Muge Olucoglu, Mar 21 2021
a(n+3) is the number of allowed patterns of length n in the even shift (that is, a(n+3) is the number of binary words of length n in which there are an even number of 0s between any two occurrences of 1). For example, a(7)=12 and the 12 allowed patterns of length 4 in the even shift are 0000, 0001, 0010, 0011, 0100, 0110, 0111, 1000, 1001, 1100, 1110, 1111. - Zoran Sunic, Apr 06 2022
Conjecture: for k a positive odd integer, the sequence {a(k^n): n >= 1} is a strong divisibility sequence; that is, for n, m >= 1, gcd(a(k^n), a(k^m)) = a(k^gcd(n,m)). - Peter Bala, Dec 05 2022
In general, the sum of a second-order linear recurrence having signature (c,d) will be a third-order recurrence having a signature (c+1,d-c,-d). - Gary Detlefs, Jan 05 2023
a(n) is the number of binary strings of length n-2 whose longest run of 1's is of length 1, for n >= 3. - Félix Balado, Apr 03 2025

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 1.
  • GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See page 28.
  • M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 64.
  • D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 155.
  • 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).
  • J. L. Yucas, Counting special sets of binary Lyndon words, Ars Combin., 31 (1991), 21-29.

Crossrefs

Antidiagonal sums of array A004070.
Right-hand column 2 of triangle A011794.
Related to sum of Fibonacci(kn) over n. Cf. A099919, A058038, A138134, A053606.
Subsequence of A226538. Also a subsequence of A061489.

Programs

  • Haskell
    a000071 n = a000071_list !! n
    a000071_list = map (subtract 1) $ tail a000045_list
    -- Reinhard Zumkeller, May 23 2013
    
  • Magma
    [Fibonacci(n)-1: n in [1..60]]; // Vincenzo Librandi, Apr 04 2011
    
  • Maple
    A000071 := proc(n) combinat[fibonacci](n)-1 ; end proc; # R. J. Mathar, Apr 07 2011
    a:= n-> (Matrix([[1, 1, 0], [1, 0, 0], [1, 0, 1]])^(n-1))[3, 2]; seq(a(n), n=1..50); # Alois P. Heinz, Jul 24 2008
  • Mathematica
    Fibonacci[Range[40]] - 1 (* or *) LinearRecurrence[{2, 0, -1}, {0, 0, 1}, 40] (* Harvey P. Dale, Aug 23 2013 *)
    Join[{0}, Accumulate[Fibonacci[Range[0, 39]]]] (* Alonso del Arte, Oct 22 2017, based on Giorgi Dalakishvili's formula *)
  • PARI
    {a(n) = if( n<1, 0, fibonacci(n)-1)};
    
  • SageMath
    [fibonacci(n)-1 for n in range(1,60)] # G. C. Greubel, Oct 21 2024

Formula

a(n) = A000045(n) - 1.
a(0) = -1, a(1) = 0; thereafter a(n) = a(n-1) + a(n-2) + 1.
a(n) = A101220(1, 1, n-2), for n > 1.
G.f.: x^3/((1-x-x^2)*(1-x)). - Simon Plouffe in his 1992 dissertation, dropping initial 0's
a(n) = 2*a(n-1) - a(n-3). - R. H. Hardin, Apr 02 2011
Partial sums of Fibonacci numbers. - Wolfdieter Lang
a(n) = -1 + (A*B^n + C*D^n)/10, with A, C = 5 +- 3*sqrt(5), B, D = (1 +- sqrt(5))/2. - Ralf Stephan, Mar 02 2003
a(1) = 0, a(2) = 0, a(3) = 1, then a(n) = ceiling(phi*a(n-1)) where phi is the golden ratio (1 + sqrt(5))/2. - Benoit Cloitre, May 06 2003
Conjecture: for all c such that 2*(2 - Phi) <= c < (2 + Phi)*(2 - Phi) we have a(n) = floor(Phi*a(n-1) + c) for n > 4. - Gerald McGarvey, Jul 22 2004. This is true provided n > 3 is changed to n > 4, see proof in Comments section. - Russell Jay Hendel, Mar 15 2015
a(n) = Sum_{k = 0..floor((n-2)/2)} binomial(n-k-2, k+1). - Paul Barry, Sep 23 2004
a(n+3) = Sum_{k = 0..floor(n/3)} binomial(n-2*k, k)*(-1)^k*2^(n-3*k). - Paul Barry, Oct 20 2004
a(n+1) = Sum(binomial(n-r, r)), r = 1, 2, ... which is the case t = 2 and k = 2 in the general case of t-strings and k blocks: a(n+1, k, t) = Sum(binomial(n-r*(t-1), r)*S2(n-r*(t-1)-1, k-1)), r = 1, 2, ... - Augustine O. Munagi, Apr 11 2005
a(n) = Sum_{k = 0..n-2} k*Fibonacci(n - k - 3). - Ross La Haye, May 31 2006
a(n) = term (3, 2) in the 3 X 3 matrix [1, 1, 0; 1, 0, 0; 1, 0, 1]^(n-1). - Alois P. Heinz, Jul 24 2008
For n >= 4, a(n) = ceiling(phi*a(n-1)), where phi is the golden ratio. - Vladimir Shevelev, Jul 04 2010
Closed-form without two leading zeros g.f.: 1/(1 - 2*x - x^3); ((5 + 2*sqrt(5))*((1 + sqrt(5))/2)^n + (5 - 2*sqrt(5))*((1 - sqrt(5))/2)^n - 5)/5; closed-form with two leading 0's g.f.: x^2/(1 - 2*x - x^3); ((5 + sqrt(5))*((1 + sqrt(5))/2)^n + (5 - sqrt(5))*((1 - sqrt(5))/2)^n - 10)/10. - Tim Monahan, Jul 10 2011
A000119(a(n)) = 1. - Reinhard Zumkeller, Dec 28 2012
a(n) = A228074(n - 1, 2) for n > 2. - Reinhard Zumkeller, Aug 15 2013
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(4*k + 2 - x^2)/( x*(4*k + 4 - x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013
A083368(a(n+3)) = n. - Reinhard Zumkeller, Aug 10 2014
E.g.f.: 1 - exp(x) + 2*exp(x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Ilya Gutkovskiy, Jun 15 2016
a(n) = A000032(3+n) - 1 mod A000045(3+n). - Mario C. Enriquez, Apr 01 2017
a(n) = Sum_{i=0..n-2} Fibonacci(i). - Giorgi Dalakishvili (mcnamara_gio(AT)yahoo.com), Apr 02 2005 [corrected by Doug Bell, Jun 01 2017]
a(n+2) = Sum_{j = 0..floor(n/2)} Sum_{k = 0..j} binomial(n - 2*j, k+1)*binomial(j, k). - Tony Foster III, Sep 08 2017
From Peter Bala, Nov 12 2021: (Start)
a(4*n) = Fibonacci(2*n+1)*Lucas(2*n-1) = A081006(n);
a(4*n+1) = Fibonacci(2*n)*Lucas(2*n+1) = A081007(n);
a(4*n+2) = Fibonacci(2*n)*Lucas(2*n+2) = A081008(n);
a(4*n+3) = Fibonacci(2*n+2)*Lucas(2*n+1) = A081009(n). (End)
G.f.: x^3/((1 - x - x^2)*(1 - x)) = Sum_{n >= 0} (-1)^n * x^(n+3) *( Product_{k = 1..n} (k - x)/Product_{k = 1..n+2} (1 - k*x) ) (a telescoping series). - Peter Bala, May 08 2024
Product_{n>=4} (1 + (-1)^n/a(n)) = 3*phi/4, where phi is the golden ratio (A001622). - Amiram Eldar, Nov 28 2024

Extensions

Edited by N. J. A. Sloane, Apr 04 2011

A003462 a(n) = (3^n - 1)/2.

Original entry on oeis.org

0, 1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573, 265720, 797161, 2391484, 7174453, 21523360, 64570081, 193710244, 581130733, 1743392200, 5230176601, 15690529804, 47071589413, 141214768240, 423644304721, 1270932914164
Offset: 0

Views

Author

Keywords

Comments

Partial sums of A000244. Values of base 3 strings of 1's.
a(n) = (3^n-1)/2 is also the number of different nonparallel lines determined by pair of vertices in the n dimensional hypercube. Example: when n = 2 the square has 4 vertices and then the relevant lines are: x = 0, y = 0, x = 1, y = 1, y = x, y = 1-x and when we identify parallel lines only 4 remain: x = 0, y = 0, y = x, y = 1 - x so a(2) = 4. - Noam Katz (noamkj(AT)hotmail.com), Feb 11 2001
Also number of 3-block bicoverings of an n-set (if offset is 1, cf. A059443). - Vladeta Jovovic, Feb 14 2001
3^a(n) is the highest power of 3 dividing (3^n)!. - Benoit Cloitre, Feb 04 2002
Apart from the a(0) and a(1) terms, maximum number of coins among which a lighter or heavier counterfeit coin can be identified (but not necessarily labeled as heavier or lighter) by n weighings. - Tom Verhoeff, Jun 22 2002, updated Mar 23 2017
n such that A001764(n) is not divisible by 3. - Benoit Cloitre, Jan 14 2003
Consider the mapping f(a/b) = (a + 2b)/(2a + b). Taking a = 1, b = 2 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the sequence 1/2, 4/5, 13/14, 40/41, ... converging to 1. Sequence contains the numerators = (3^n-1)/2. The same mapping for N, i.e., f(a/b) = (a + Nb)/(a+b) gives fractions converging to N^(1/2). - Amarnath Murthy, Mar 22 2003
Binomial transform of A000079 (with leading zero). - Paul Barry, Apr 11 2003
With leading zero, inverse binomial transform of A006095. - Paul Barry, Aug 19 2003
Number of walks of length 2*n + 2 in the path graph P_5 from one end to the other one. Example: a(2) = 4 because in the path ABCDE we have ABABCDE, ABCBCDE, ABCDCDE and ABCDEDE. - Emeric Deutsch, Apr 02 2004
The number of triangles of all sizes (not counting holes) in Sierpiński's triangle after n inscriptions. - Lee Reeves (leereeves(AT)fastmail.fm), May 10 2004
Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2*n + 1, s(0) = 1, s(2n+1) = 4. - Herbert Kociemba, Jun 10 2004
Number of non-degenerate right-angled incongruent integer-edged Heron triangles whose circumdiameter is the product of n distinct primes of shape 4k + 1. - Alex Fink and R. K. Guy, Aug 18 2005
Also numerator of the sum of the reciprocals of the first n powers of 3, with A000244 being the sequence of denominators. With the exception of n < 2, the base 10 digital root of a(n) is always 4. In base 3 the digital root of a(n) is the same as the digital root of n. - Alonso del Arte, Jan 24 2006
The sequence 3*a(n), n >= 1, gives the number of edges of the Hanoi graph H_3^{n} with 3 pegs and n >= 1 discs. - Daniele Parisse, Jul 28 2006
Numbers n such that a(n) is prime are listed in A028491 = {3, 7, 13, 71, 103, 541, 1091, ...}. 2^(m+1) divides a(2^m*k) for m > 0. 5 divides a(4k). 5^2 divides a(20k). 7 divides a(6k). 7^2 divides a(42k). 11^2 divides a(5k). 13 divides a(3k). 17 divides a(16k). 19 divides a(18k). 1093 divides a(7k). 41 divides a(8k). p divides a((p-1)/5) for prime p = {41, 431, 491, 661, 761, 1021, 1051, 1091, 1171, ...}. p divides a((p-1)/4) for prime p = {13, 109, 181, 193, 229, 277, 313, 421, 433, 541, ...}. p divides a((p-1)/3) for prime p = {61, 67, 73, 103, 151, 193, 271, 307, 367, ...} = A014753, 3 and -3 are both cubes (one implies other) mod these primes p = 1 mod 6. p divides a((p-1)/2) for prime p = {11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, ...} = A097933(n). p divides a(p-1) for prime p > 7. p^2 divides a(p*(p-1)k) for all prime p except p = 3. p^3 divides a(p*(p-1)*(p-2)k) for prime p = 11. - Alexander Adamchuk, Jan 22 2007
Let P(A) be the power set of an n-element set A. Then a(n) = the number of [unordered] pairs of elements {x,y} of P(A) for which x and y are disjoint [and both nonempty]. Wieder calls these "disjoint usual 2-combinations". - Ross La Haye, Jan 10 2008 [This is because each of the elements of {1, 2, ..., n} can be in the first subset, in the second or in neither. Because there are three options for each, the total number of options is 3^n. However, since the sets being empty is not an option we subtract 1 and since the subsets are unordered we then divide by 2! (The number of ways two objects can be arranged.) Thus we obtain (3^n-1)/2 = a(n). - Chayim Lowen, Mar 03 2015]
Also, still with P(A) being the power set of a n-element set A, a(n) is the number of 2-element subsets {x,y} of P(A) such that the union of x and y is equal to A. Cf. A341590. - Fabio Visonà, Feb 20 2021
Starting with offset 1 = binomial transform of A003945: (1, 3, 6, 12, 24, ...) and double bt of (1, 2, 1, 2, 1, 2, ...); equals polcoeff inverse of (1, -4, 3, 0, 0, 0, ...). - Gary W. Adamson, May 28 2009
Also the constant of the polynomials C(x) = 3x + 1 that form a sequence by performing this operation repeatedly and taking the result at each step as the input at the next. - Nishant Shukla (n.shukla722(AT)gmail.com), Jul 11 2009
It appears that this is A120444(3^n-1) = A004125(3^n) - A004125(3^n-1), where A004125 is the sum of remainders of n mod k for k = 1, 2, 3, ..., n. - John W. Layman, Jul 29 2009
Subsequence of A134025; A171960(a(n)) = a(n). - Reinhard Zumkeller, Jan 20 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j] = 1, A[i, i] := 3, (i > 1), A[i, i-1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n) = det(A). - Milan Janjic, Jan 27 2010
This is the sequence A(0, 1; 2, 3; 2) = A(0, 1; 4, -3; 0) of the family of sequences [a, b:c, d:k] considered by Gary Detlefs, and treated as A(a, b; c, d; k) in the Wolfdieter Lang link given below. - Wolfdieter Lang, Oct 18 2010
It appears that if s(n) is a first order rational sequence of the form s(0) = 0, s(n) = (2*s(n-1)+1)/(s(n-1)+2), n > 0, then s(n)= a(n)/(a(n)+1). - Gary Detlefs, Nov 16 2010
This sequence also describes the total number of moves to solve the [RED ; BLUE ; BLUE] or [RED ; RED ; BLUE] pre-colored Magnetic Towers of Hanoi puzzle (cf. A183111 - A183125).
From Adi Dani, Jun 08 2011: (Start)
a(n) is number of compositions of odd numbers into n parts less than 3. For example, a(3) = 13 and there are 13 compositions odd numbers into 3 parts < 3:
1: (0, 0, 1), (0, 1, 0), (1, 0, 0);
3: (0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0), (1, 1, 1);
5: (1, 2, 2), (2, 1, 2), (2, 2, 1).
(End)
Pisano period lengths: 1, 2, 1, 2, 4, 2, 6, 4, 1, 4, 5, 2, 3, 6, 4, 8, 16, 2, 18, 4, ... . - R. J. Mathar, Aug 10 2012
a(n) is the total number of holes (triangles removed) after the n-th step of a Sierpiński triangle production. - Ivan N. Ianakiev, Oct 29 2013
a(n) solves Sum_{j = a(n) + 1 .. a(n+1)} j = k^2 for some integer k, given a(0) = 0 and requiring smallest a(n+1) > a(n). Corresponding k = 3^n. - Richard R. Forberg, Mar 11 2015
a(n+1) equals the number of words of length n over {0, 1, 2, 3} avoiding 01, 02 and 03. - Milan Janjic, Dec 17 2015
For n >= 1, a(n) is also the total number of words of length n, over an alphabet of three letters, such that one of the letters appears an odd number of times (See A006516 for 4 letter words, and the Balakrishnan reference there). - Wolfdieter Lang, Jul 16 2017
Also, the number of maximal cliques, maximum cliques, and cliques of size 4 in the n-Apollonian network. - Andrew Howroyd, Sep 02 2017
For n > 1, the number of triangles (cliques of size 3) in the (n-1)-Apollonian network. - Andrew Howroyd, Sep 02 2017
a(n) is the largest number that can be represented with n trits in balanced ternary. Correspondingly, -a(n) is the smallest number that can be represented with n trits in balanced ternary. - Thomas König, Apr 26 2020
These form Sierpinski nesting-stars, which alternate pattern on 3^n+1/2 star numbers A003154, based on the square configurations of 9^n. The partial sums of 3^n are delineated according to the geometry of a hexagram, see illustrations in links. (3*a(n-1) + 1) create Sierpinski-anti-triangles, representing the number of holes in a (n+1) Sierpinski triangle (see illustrations). - John Elias, Oct 18 2021
For n > 1, a(n) is the number of iterations necessary to calculate the hyperbolic functions with CORDIC. - Mathias Zechmeister, Jul 26 2022
a(n) is the least number k such that A065363(k) = n. - Amiram Eldar, Sep 03 2022
For all n >= 0, Sum_{k=a(n)+1..a(n+1)} 1/k < Sum_{j=a(n+1)+1..a(n+2)} 1/j. These are the minimal points which partition the infinite harmonic series into a monotonically increasing sequence. Each partition approximates log(3) from below as n tends to infinity. - Joseph Wheat, Apr 15 2023
a(n) is also the number of 3-cycles in the n-Dorogovtsev-Goltsev-Mendes graph (using the convention the 0-Dorogovtsev-Goltsev-Mendes graph is P_2). - Eric W. Weisstein, Dec 06 2023

Examples

			There are 4 3-block bicoverings of a 3-set: {{1, 2, 3}, {1, 2}, {3}}, {{1, 2, 3}, {1, 3}, {2}}, {{1, 2, 3}, {1}, {2, 3}} and {{1, 2}, {1, 3}, {2, 3}}.
Ternary........Decimal
0.................0
1.................1
11................4
111..............13
1111.............40 etc. - _Zerinvary Lajos_, Jan 14 2007
There are altogether a(3) = 13 three letter words over {A,B,C} with say, A, appearing an odd number of times: AAA; ABC, ACB, ABB, ACC; BAC, CAB, BAB, CAC; BCA, CBA, BBA, CCA. - _Wolfdieter Lang_, Jul 16 2017
		

References

  • J. G. Mauldon, Strong solutions for the counterfeit coin problem, IBM Research Report RC 7476 (#31437) 9/15/78, IBM Thomas J. Watson Research Center, P. O. Box 218, Yorktown Heights, N. Y. 10598.
  • Paulo Ribenboim, The Book of Prime Number Records, Springer-Verlag, NY, 2nd ed., 1989, p. 60.
  • Paulo Ribenboim, The Little Book of Big Primes, Springer-Verlag, NY, 1991, p. 53.
  • Amir Sapir, The Tower of Hanoi with Forbidden Moves, The Computer J. 47 (1) (2004) 20, case three-in-a row, sequence a(n).
  • Robert Sedgewick, Algorithms, 1992, pp. 109.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Sequences used for Shell sort: A033622, A003462, A036562, A036564, A036569, A055875.
Cf. A179526 (repeats), A113047 (characteristic function).
Cf. A000225, A000392, A004125, A014753, A028491 (indices of primes), A059443 (column k = 3), A065363, A097933, A120444, A321872 (sum reciprocals).
Cf. A064099 (minimal number of weightings to detect lighter or heavier coin among n coins).
Cf. A039755 (column k = 1).
Cf. A006516 (binomial transform, and special 4 letter words).
Cf. A341590.
Cf. A003462(n) (3-cycles), A367967(n) (5-cycles), A367968(n) (6-cycles).

Programs

Formula

G.f.: x/((1-x)*(1-3*x)).
a(n) = 4*a(n-1) - 3*a(n-2), n > 1. a(0) = 0, a(1) = 1.
a(n) = 3*a(n-1) + 1, a(0) = 0.
E.g.f.: (exp(3*x) - exp(x))/2. - Paul Barry, Apr 11 2003
a(n+1) = Sum_{k = 0..n} binomial(n+1, k+1)*2^k. - Paul Barry, Aug 20 2004
a(n) = Sum_{i = 0..n-1} 3^i, for n > 0; a(0) = 0.
a(n) = A125118(n, 2) for n > 1. - Reinhard Zumkeller, Nov 21 2006
a(n) = StirlingS2(n+1, 3) + StirlingS2(n+1, 2). - Ross La Haye, Jan 10 2008
a(n) = Sum_{k = 0..n} A106566(n, k)*A106233(k). - Philippe Deléham, Oct 30 2008
a(n) = 2*a(n-1) + 3*a(n-2) + 2, n > 1. - Gary Detlefs, Jun 21 2010
a(n) = 3*a(n-1) + a(n-2) - 3*a(n-3) = 5*a(n-1) - 7*a(n-2) + 3*a(n-3), a(0) = 0, a(1) = 1, a(2) = 4. Observation by G. Detlefs. See the W. Lang comment and link. - Wolfdieter Lang, Oct 18 2010
A008344(a(n)) = 0, for n > 1. - Reinhard Zumkeller, May 09 2012
A085059(a(n)) = 1 for n > 0. - Reinhard Zumkeller, Jan 31 2013
G.f.: Q(0)/2 where Q(k) = 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - 1/(1 - 1/(3*9^k - 27*x*81^k/(9*x*9^k - 1/Q(k+1)))))); (continued fraction ). - Sergei N. Gladkovskii, Apr 12 2013
a(n) = A001065(3^n) where A001065(m) is the sum of the proper divisors of m for positive integer m. - Chayim Lowen, Mar 03 2015
a(n) = A000244(n) - A007051(n) = A007051(n)-1. - Yuchun Ji, Oct 23 2018
Sum_{n>=1} 1/a(n) = A321872. - Amiram Eldar, Nov 18 2020

Extensions

More terms from Michael Somos
Corrected my comment of Jan 10 2008. - Ross La Haye, Oct 29 2008
Removed comment that duplicated a formula. - Joerg Arndt, Mar 11 2010

A003215 Hex (or centered hexagonal) numbers: 3*n*(n+1)+1 (crystal ball sequence for hexagonal lattice).

Original entry on oeis.org

1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141, 1261, 1387, 1519, 1657, 1801, 1951, 2107, 2269, 2437, 2611, 2791, 2977, 3169, 3367, 3571, 3781, 3997, 4219, 4447, 4681, 4921, 5167, 5419, 5677, 5941, 6211, 6487, 6769
Offset: 0

Views

Author

Keywords

Comments

The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice.
Crystal ball sequence for A_2 lattice. - Michael Somos, Jun 03 2012
Sixth spoke of hexagonal spiral (cf. A056105-A056109).
Number of ordered integer triples (a,b,c), -n <= a,b,c <= n, such that a+b+c=0. - Benoit Cloitre, Jun 14 2003
Also the number of partitions of 6n into at most 3 parts, A001399(6n). - R. K. Guy, Oct 20 2003
Also, a(n) is the number of partitions of 6(n+1) into exactly 3 distinct parts. - William J. Keith, Jul 01 2004
Number of dots in a centered hexagonal figure with n+1 dots on each side.
Values of second Bessel polynomial y_2(n) (see A001498).
First differences of cubes (A000578). - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004
Final digits of Hex numbers (hex(n) mod 10) are periodic with palindromic period of length 5 {1, 7, 9, 7, 1}. Last two digits of Hex numbers (hex(n) mod 100) are periodic with palindromic period of length 100. - Alexander Adamchuk, Aug 11 2006
All divisors of a(n) are congruent to 1, modulo 6. Proof: If p is an odd prime different from 3 then 3n^2 + 3n + 1 = 0 (mod p) implies 9(2n + 1)^2 = -3 (mod p), whence p = 1 (mod 6). - Nick Hobson, Nov 13 2006
For n>=1, a(n) is the side of Outer Napoleon Triangle whose reference triangle is a right triangle with legs (3a(n))^(1/2) and 3n(a(n))^(1/2). - Tom Schicker (tschicke(AT)email.smith.edu), Apr 25 2007
Number of triples (a,b,c) where 0<=(a,b)<=n and c=n (at least once the term n). E.g., for n = 1: (0,0,1), (0,1,0), (1,0,0), (0,1,1), (1,0,1), (1,1,0), (1,1,1), so a(1)=7. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007
Equals the triangular numbers convolved with [1, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson and Alexander R. Povolotsky, May 29 2009
From Terry Stickels, Dec 07 2009: (Start)
Also the maximum number of viewable cubes from any one static point while viewing a cube stack of identical cubes of varying magnitude.
For example, viewing a 2 X 2 X 2 stack will yield 7 maximum viewable cubes.
If the stack is 3 X 3 X 3, the maximum number of viewable cubes from any one static position is 19, and so on.
The number of cubes in the stack must always be the same number for width, length, height (at true regular cubic stack) and the maximum number of visible cubes can always be found by taking any cubic number and subtracting the number of the cube that is one less.
Examples: 125 - 64 = 61, 64 - 27 = 37, 27 - 8 = 19. (End)
The sequence of digital roots of the a(n) is period 3: repeat [1,7,1]. - Ant King, Jun 17 2012
The average of the first n (n>0) centered hexagonal numbers is the n-th square. - Philippe Deléham, Feb 04 2013
A002024 is the following array A read along antidiagonals:
1, 2, 3, 4, 5, 6, ...
2, 3, 4, 5, 6, 7, ...
3, 4, 5, 6, 7, 8, ...
4, 5, 6, 7, 8, 9, ...
5, 6, 7, 8, 9, 10, ...
6, 7, 8, 9, 10, 11, ...
and a(n) is the hook sum Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013
a(n) is the sum of the terms in the n+1 X n+1 matrices minus those in n X n matrices in an array formed by considering A158405 an array (the beginning terms in each row are 1,3,5,7,9,11,...). - J. M. Bergot, Jul 05 2013
The formula also equals the product of the three distinct combinations of two consecutive numbers: n^2, (n+1)^2, and n*(n+1). - J. M. Bergot, Mar 28 2014
The sides of any triangle ABC are divided into 2n + 1 equal segments by 2n points: A_1, A_2, ..., A_2n in side a, and also on the sides b and c cyclically. If A'B'C' is the triangle delimited by AA_n, BB_n and CC_n cevians, we have (ABC)/(A'B'C') = a(n) (see Java applet link). - Ignacio Larrosa Cañestro, Jan 02 2015
a(n) is the maximal number of parts into which (n+1) triangles can intersect one another. - Ivan N. Ianakiev, Feb 18 2015
((2^m-1)n)^t mod a(n) = ((2^m-1)(n+1))^t mod a(n) = ((2^m-1)(2n+1))^t mod a(n), where m any positive integer, and t = 0(mod 6). - Alzhekeyev Ascar M, Oct 07 2016
((2^m-1)n)^t mod a(n) = ((2^m-1)(n+1))^t mod a(n) = a(n) - (((2^m-1)(2n+1))^t mod a(n)), where m any positive integer, and t = 3(mod 6). - Alzhekeyev Ascar M, Oct 07 2016
(3n+1)^(a(n)-1) mod a(n) = (3n+2)^(a(n)-1) mod a(n) = 1. If a(n) not prime, then always strong pseudoprime. - Alzhekeyev Ascar M, Oct 07 2016
Every positive integer is the sum of 8 hex numbers (zero included), at most 3 of which are greater than 1. - Mauro Fiorentini, Jan 01 2018
Area enclosed by the segment of Archimedean spiral between n*Pi/2 and (n+1)*Pi/2 in Pi^3/48 units. - Carmine Suriano, Apr 10 2018
This sequence contains all numbers k such that 12*k - 3 is a square. - Klaus Purath, Oct 19 2021
The continued fraction expansion of sqrt(3*a(n)) is [3n+1; {1, 1, 2n, 1, 1, 6n+2}]. For n = 0, this collapses to [1; {1, 2}]. - Magus K. Chu, Sep 12 2022

Examples

			G.f. = 1 + 7*x + 19*x^2 + 37*x^3 + 61*x^4 + 91*x^5 + 127*x^6 + 169*x^7 + 217*x^8 + ...
From _Omar E. Pol_, Aug 21 2011: (Start)
Illustration of initial terms:
.
.                                 o o o o
.                   o o o        o o o o o
.         o o      o o o o      o o o o o o
.   o    o o o    o o o o o    o o o o o o o
.         o o      o o o o      o o o o o o
.                   o o o        o o o o o
.                                 o o o o
.
.   1      7          19             37
.
(End)
From _Klaus Purath_, Dec 03 2021: (Start)
(1) a(19) is not a prime number, because besides a(19) = a(9) + P(29), a(19) = a(15) + P(20) = a(2) + P(33) is also true.
(2) a(25) is prime, because except for a(25) = a(12) + P(38) there is no other equation of this pattern. (End)
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 81.
  • M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 18.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=3 of A080853, and column k=2 of A047969.
See also A220083 for a list of numbers of the form n*P(s,n)-(n-1)*P(s,n-1), where P(s,n) is the n-th polygonal number with s sides.
Cf. A287326(A000124(n), 1).
Cf. A008292.
Cf. A154105.

Programs

Formula

a(n) = 3*n*(n+1) + 1, n >= 0 (see the name).
a(n) = (n+1)^3 - n^3 = a(-1-n).
G.f.: (1 + 4*x + x^2) / (1 - x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = 6*A000217(n) + 1.
a(n) = a(n-1) + 6*n = 2a(n-1) - a(n-2) + 6 = 3*a(n-1) - 3*a(n-2) + a(n-3) = A056105(n) + 5n = A056106(n) + 4*n = A056107(n) + 3*n = A056108(n) + 2*n = A056108(n) + n.
n-th partial arithmetic mean is n^2. - Amarnath Murthy, May 27 2003
a(n) = 1 + Sum_{j=0..n} (6*j). E.g., a(2)=19 because 1+ 6*0 + 6*1 + 6*2 = 19. - Xavier Acloque, Oct 06 2003
The sum of the first n hexagonal numbers is n^3. That is, Sum_{n>=1} (3*n*(n-1) + 1) = n^3. - Edward Weed (eweed(AT)gdrs.com), Oct 23 2003
a(n) = right term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 0 0 / 2 1 0 / 3 3 1]. M^n * [1 1 1] = [1 2n+1 a(n)]. E.g., a(4) = 61, right term in M^4 * [1 1 1], since M^4 * [1 1 1] = [1 9 61] = [1 2n+1 a(4)]. - Gary W. Adamson, Dec 22 2004
Row sums of triangle A130298. - Gary W. Adamson, Jun 07 2007
a(n) = 3*n^2 + 3*n + 1. Proof: 1) If n occurs once, it may be in 3 positions; for the two other ones, n terms are independently possible, then we have 3*n^2 different triples. 2) If the term n occurs twice, the third one may be placed in 3 positions and have n possible values, then we have 3*n more different triples. 3) The term n may occurs 3 times in one way only that gives the formula. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007
Binomial transform of [1, 6, 6, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 6, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
a(n) = (n-1)*A000166(n) + (n-2)*A000166(n-1) = (n-1)floor(n!*e^(-1)+1) + (n-2)*floor((n-1)!*e^(-1)+1) (with offset 0). - Gary Detlefs, Dec 06 2009
a(n) = A028896(n) + 1. - Omar E. Pol, Oct 03 2011
a(n) = integral( (sin((n+1/2)x)/sin(x/2))^3, x=0..Pi)/Pi. - Yalcin Aktar, Dec 03 2011
Sum_{n>=0} 1/a(n) = Pi/sqrt(3)*tanh(Pi/(2*sqrt(3))) = 1.305284153013581... - Ant King, Jun 17 2012
a(n) = A000290(n) + A000217(2n+1). - Ivan N. Ianakiev, Sep 24 2013
a(n) = A002378(n+1) + A056220(n) = A005408(n) + 2*A005449(n) = 6*A000217(n) + 1. - Ivan N. Ianakiev, Sep 26 2013
a(n) = 6*A000124(n) - 5. - Ivan N. Ianakiev, Oct 13 2013
a(n) = A239426(n+1) / A239449(n+1) = A215630(2*n+1,n+1). - Reinhard Zumkeller, Mar 19 2014
a(n) = A243201(n) / A002061(n + 1). - Mathew Englander, Jun 03 2014
a(n) = A101321(6,n). - R. J. Mathar, Jul 28 2016
E.g.f.: (1 + 6*x + 3*x^2)*exp(x). - Ilya Gutkovskiy, Jul 28 2016
a(n) = (A001844(n) + A016754(n))/2. - Bruce J. Nicholson, Aug 06 2017
a(n) = A045943(2n+1). - Miquel Cerda, Jan 22 2018
a(n) = 3*Integral_{x=n..n+1} x^2 dx. - Carmine Suriano, Apr 10 2018
a(n) = A287326(A000124(n), 1). - Kolosov Petro, Oct 22 2018
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=0} a(n)/n! = 10*e.
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 2/e. (End)
G.f.: polylog(-3, x)*(1-x)/x. See the Simon Plouffe formula above, and the g.f. of the rows of A008292 by Vladeta Jovovic, Sep 02 2002. - Wolfdieter Lang, May 08 2021
a(n) = T(n-1)^2 - 2*T(n)^2 + T(n+1)^2, n >= 1, T = triangular number A000217. - Klaus Purath, Oct 11 2021
a(n) = 1 + 2*Sum_{j=n..2n} j. - Klaus Purath, Oct 19 2021
a(n) = A069099(n+1) - A000217(n). - Klaus Purath, Nov 03 2021
From Leo Tavares, Dec 03 2021: (Start)
a(n) = A005448(n) + A140091(n);
a(n) = A001844(n) + A002378(n);
a(n) = A005891(n) + A000217(n);
a(n) = A000290(n) + A000384(n+1);
a(n) = A060544(n-1) + 3*A000217(n);
a(n) = A060544(n-1) + A045943(n).
a(2*n+1) = A154105(n).
(End)

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A001318 Generalized pentagonal numbers: m*(3*m - 1)/2, m = 0, +-1, +-2, +-3, ....

Original entry on oeis.org

0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117, 126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345, 376, 392, 425, 442, 477, 495, 532, 551, 590, 610, 651, 672, 715, 737, 782, 805, 852, 876, 925, 950, 1001, 1027, 1080, 1107, 1162, 1190, 1247, 1276, 1335
Offset: 0

Views

Author

Keywords

Comments

Partial sums of A026741. - Jud McCranie; corrected by Omar E. Pol, Jul 05 2012
From R. K. Guy, Dec 28 2005: (Start)
"Conway's relation twixt the triangular and pentagonal numbers: Divide the triangular numbers by 3 (when you can exactly):
0 1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 ...
0 - 1 2 .- .5 .7 .- 12 15 .- 22 26 .- .35 .40 .- ..51 ...
.....-.-.....+..+.....-..-.....+..+......-...-.......+....
"and you get the pentagonal numbers in pairs, one of positive rank and the other negative.
"Append signs according as the pair have the same (+) or opposite (-) parity.
"Then Euler's pentagonal number theorem is easy to remember:
"p(n-0) - p(n-1) - p(n-2) + p(n-5) + p(n-7) - p(n-12) - p(n-15) ++-- = 0^n
where p(n) is the partition function, the left side terminates before the argument becomes negative and 0^n = 1 if n = 0 and = 0 if n > 0.
"E.g. p(0) = 1, p(7) = p(7-1) + p(7-2) - p(7-5) - p(7-7) + 0^7 = 11 + 7 - 2 - 1 + 0 = 15."
(End)
The sequence may be used in order to compute sigma(n), as described in Euler's article. - Thomas Baruchel, Nov 19 2003
Number of levels in the partitions of n + 1 with parts in {1,2}.
a(n) is the number of 3 X 3 matrices (symmetrical about each diagonal) M = {{a, b, c}, {b, d, b}, {c, b, a}} such that a + b + c = b + d + b = n + 2, a,b,c,d natural numbers; example: a(3) = 5 because (a,b,c,d) = (2,2,1,1), (1,2,2,1), (1,1,3,3), (3,1,1,3), (2,1,2,3). - Philippe Deléham, Apr 11 2007
Also numbers a(n) such that 24*a(n) + 1 = (6*m - 1)^2 are odd squares: 1, 25, 49, 121, 169, 289, 361, ..., m = 0, +-1, +-2, ... . - Zak Seidov, Mar 08 2008
From Matthew Vandermast, Oct 28 2008: (Start)
Numbers n for which A000326(n) is a member of A000332. Cf. A145920.
This sequence contains all members of A000332 and all nonnegative members of A145919. For values of n such that n*(3*n - 1)/2 belongs to A000332, see A145919. (End)
Starting with offset 1 = row sums of triangle A168258. - Gary W. Adamson, Nov 21 2009
Starting with offset 1 = Triangle A101688 * [1, 2, 3, ...]. - Gary W. Adamson, Nov 27 2009
Starting with offset 1 can be considered the first in an infinite set generated from A026741. Refer to the array in A175005. - Gary W. Adamson, Apr 03 2010
Vertex number of a square spiral whose edges have length A026741. The two axes of the spiral forming an "X" are A000326 and A005449. The four semi-axes forming an "X" are A049452, A049453, A033570 and the numbers >= 2 of A033568. - Omar E. Pol, Sep 08 2011
A general formula for the generalized k-gonal numbers is given by n*((k - 2)*n - k + 4)/2, n=0, +-1, +-2, ..., k >= 5. - Omar E. Pol, Sep 15 2011
a(n) is the number of 3-tuples (w,x,y) having all terms in {0,...,n} and 2*w = 2*x + y. - Clark Kimberling, Jun 04 2012
Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. - Omar E. Pol, Aug 04 2012
a(n) is the sum of the largest parts of the partitions of n+1 into exactly 2 parts. - Wesley Ivan Hurt, Jan 26 2013
Conway's relation mentioned by R. K. Guy is a relation between triangular numbers and generalized pentagonal numbers, two sequences from different families, but as triangular numbers are also generalized hexagonal numbers in this case we have a relation between two sequences from the same family. - Omar E. Pol, Feb 01 2013
Start with the sequence of all 0's. Add n to each value of a(n) and the next n - 1 terms. The result is the generalized pentagonal numbers. - Wesley Ivan Hurt, Nov 03 2014
(6k + 1) | a(4k). (3k + 1) | a(4k+1). (3k + 2) | a(4k+2). (6k + 5) | a(4k+3). - Jon Perry, Nov 04 2014
Enge, Hart and Johansson proved: "Every generalised pentagonal number c >= 5 is the sum of a smaller one and twice a smaller one, that is, there are generalised pentagonal numbers a, b < c such that c = 2a + b." (see link theorem 5). - Peter Luschny, Aug 26 2016
The Enge, et al. result for c >= 5 also holds for c >= 2 if 0 is included as a generalized pentagonal number. That is, 2 = 2*1 + 0. - Michael Somos, Jun 02 2018
Suggestion for title, where n actually matches the list and b-file: "Generalized pentagonal numbers: k(n)*(3*k(n) - 1)/2, where k(n) = A001057(n) = [0, 1, -1, 2, -2, 3, -3, ...], n >= 0" - Daniel Forgues, Jun 09 2018 & Jun 12 2018
Generalized k-gonal numbers are the partial sums of the sequence formed by the multiples of (k - 4) and the odd numbers (A005408) interleaved, with k >= 5. - Omar E. Pol, Jul 25 2018
The last digits form a symmetric cycle of length 40 [0, 1, 2, 5, ..., 5, 2, 1, 0], i.e., a(n) == a(n + 40) (mod 10) and a(n) == a(40*k - n - 1) (mod 10), 40*k > n. - Alejandro J. Becerra Jr., Aug 14 2018
Only 2, 5, and 7 are prime. All terms are of the form k*(k+1)/6, where 3 | k or 3 | k+1. For k > 6, the value divisible by 3 must have another factor d > 2, which will remain after the division by 6. - Eric Snyder, Jun 03 2022
8*a(n) is the product of two even numbers one of which is n + n mod 2. - Peter Luschny, Jul 15 2022
a(n) is the dot product of [1, 2, 3, ..., n] and repeat[1, 1/2]. a(5) = 12 = [1, 2, 3, 4, 5] dot [1, 1/2, 1, 1/2, 1] = [1 + 1 + 3 + 2 + 5]. - Gary W. Adamson, Dec 10 2022
Every nonnegative number is the sum of four terms of this sequence [S. Realis]. - N. J. A. Sloane, May 07 2023
From Peter Bala, Jan 06 2025: (Start)
The sequence terms are the exponents in the expansions of the following infinite products:
1) Product_{n >= 1} (1 - s(n)*q^n) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ..., where s(n) = (-1)^(1 + mod(n+1,3)).
2) Product_{n >= 1} (1 - q^(2*n))*(1 - q^(3*n))^2/((1 - q^n)*(1 - q^(6*n))) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ....
3) Product_{n >= 1} (1 - q^n)*(1 - q^(4*n))*(1 - q^(6*n))^5/((1 - q^(2*n))*(1 - q^(3*n))*(1 - q^(12*n)))^2 = 1 - q + q^2 - q^5 - q^7 + q^12 - q^15 + q^22 + q^26 - q^35 + ....
4) Product_{n >= 1} (1 - q^(2*n))^13/((1 - (-1)^n*q^n)*(1 - q^(4*n)))^5 = 1 - 5*q + 7*q^2 - 11*q^5 + 13*q^7 - 17*q^12 + 19*q^15 - + .... See Oliver, Theorem 1.1. (End)

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 7*x^4 + 12*x^5 + 15*x^6 + 22*x^7 + 26*x^8 + 35*x^9 + ...
		

References

  • Enoch Haga, A strange sequence and a brilliant discovery, chapter 5 of Exploring prime numbers on your PC and the Internet, first revised ed., 2007 (and earlier ed.), pp. 53-70.
  • Ross Honsberger, Ingenuity in Mathematics, Random House, 1970, p. 117.
  • Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, (to appear), section 7.2.1.4, equation (18).
  • Ivan Niven and Herbert S. Zuckerman, An Introduction to the Theory of Numbers, 2nd ed., Wiley, NY, 1966, p. 231.
  • 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. A080995 (characteristic function), A026741 (first differences), A034828 (partial sums), A165211 (mod 2).
Cf. A000326 (pentagonal numbers), A005449 (second pentagonal numbers), A000217 (triangular numbers).
Indices of nonzero terms of A010815, i.e., the (zero-based) indices of 1-bits of the infinite binary word to which the terms of A068052 converge.
Union of A036498 and A036499.
Sequences of generalized k-gonal numbers: this sequence (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), A195160 (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), A303305 (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), A303814 (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).
Column 1 of A195152.
Squares in APs: A221671, A221672.
Quadrisection: A049453(k), A033570(k), A033568(k+1), A049452(k+1), k >= 0.
Cf. A002620.

Programs

  • GAP
    a:=[0,1,2,5];; for n in [5..60] do a[n]:=2*a[n-2]-a[n-4]+3; od; a; # Muniru A Asiru, Aug 16 2018
    
  • Haskell
    a001318 n = a001318_list !! n
    a001318_list = scanl1 (+) a026741_list -- Reinhard Zumkeller, Nov 15 2015
    
  • Magma
    [(6*n^2 + 6*n + 1 - (2*n + 1)*(-1)^n)/16 : n in [0..50]]; // Wesley Ivan Hurt, Nov 03 2014
    
  • Magma
    [(3*n^2 + 2*n + (n mod 2) * (2*n + 1)) div 8: n in [0..70]]; // Vincenzo Librandi, Nov 04 2014
    
  • Maple
    A001318 := -(1+z+z**2)/(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation; gives sequence without initial zero
    A001318 := proc(n) (6*n^2+6*n+1)/16-(2*n+1)*(-1)^n/16 ; end proc: # R. J. Mathar, Mar 27 2011
  • Mathematica
    Table[n*(n+1)/6, {n, Select[Range[0, 100], Mod[#, 3] != 1 &]}]
    Select[Accumulate[Range[0,200]]/3,IntegerQ] (* Harvey P. Dale, Oct 12 2014 *)
    CoefficientList[Series[x (1 + x + x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 70}], x] (* Vincenzo Librandi, Nov 04 2014 *)
    LinearRecurrence[{1,2,-2,-1,1},{0,1,2,5,7},70] (* Harvey P. Dale, Jun 05 2017 *)
    a[ n_] := With[{m = Quotient[n + 1, 2]}, m (3 m + (-1)^n) / 2]; (* Michael Somos, Jun 02 2018 *)
  • PARI
    {a(n) = (3*n^2 + 2*n + (n%2) * (2*n + 1)) / 8}; /* Michael Somos, Mar 24 2011 */
    
  • PARI
    {a(n) = if( n<0, n = -1-n); polcoeff( x * (1 - x^3) / ((1 - x) * (1-x^2))^2 + x * O(x^n), n)}; /* Michael Somos, Mar 24 2011 */
    
  • PARI
    {a(n) = my(m = (n+1) \ 2); m * (3*m + (-1)^n) / 2}; /* Michael Somos, Jun 02 2018 */
    
  • Python
    def a(n):
        p = n % 2
        return (n + p)*(3*n + 2 - p) >> 3
    print([a(n) for n in range(60)])  # Peter Luschny, Jul 15 2022
    
  • Python
    def A001318(n): return n*(n+1)-(m:=n>>1)*(m+1)>>1 # Chai Wah Wu, Nov 23 2024
  • Sage
    @CachedFunction
    def A001318(n):
        if n == 0 : return 0
        inc = n//2 if is_even(n) else n
        return inc + A001318(n-1)
    [A001318(n) for n in (0..59)] # Peter Luschny, Oct 13 2012
    

Formula

Euler: Product_{n>=1} (1 - x^n) = Sum_{n=-oo..oo} (-1)^n*x^(n*(3*n - 1)/2).
A080995(a(n)) = 1: complement of A090864; A000009(a(n)) = A051044(n). - Reinhard Zumkeller, Apr 22 2006
Euler transform of length-3 sequence [2, 2, -1]. - Michael Somos, Mar 24 2011
a(-1 - n) = a(n) for all n in Z. a(2*n) = A005449(n). a(2*n - 1) = A000326(n). - Michael Somos, Mar 24 2011. [The extension of the recurrence to negative indices satisfies the signature (1,2,-2,-1,1), but not the definition of the sequence m*(3*m -1)/2, because there is no m such that a(-1) = 0. - Klaus Purath, Jul 07 2021]
a(n) = 3 + 2*a(n-2) - a(n-4). - Ant King, Aug 23 2011
Product_{k>0} (1 - x^k) = Sum_{k>=0} (-1)^k * x^a(k). - Michael Somos, Mar 24 2011
G.f.: x*(1 + x + x^2)/((1 + x)^2*(1 - x)^3).
a(n) = n*(n + 1)/6 when n runs through numbers == 0 or 2 mod 3. - Barry E. Williams
a(n) = A008805(n-1) + A008805(n-2) + A008805(n-3), n > 2. - Ralf Stephan, Apr 26 2003
Sequence consists of the pentagonal numbers (A000326), followed by A000326(n) + n and then the next pentagonal number. - Jon Perry, Sep 11 2003
a(n) = (6*n^2 + 6*n + 1)/16 - (2*n + 1)*(-1)^n/16; a(n) = A034828(n+1) - A034828(n). - Paul Barry, May 13 2005
a(n) = Sum_{k=1..floor((n+1)/2)} (n - k + 1). - Paul Barry, Sep 07 2005
a(n) = A000217(n) - A000217(floor(n/2)). - Pierre CAMI, Dec 09 2007
If n even a(n) = a(n-1) + n/2 and if n odd a(n) = a(n-1) + n, n >= 2. - Pierre CAMI, Dec 09 2007
a(n)-a(n-1) = A026741(n) and it follows that the difference between consecutive terms is equal to n if n is odd and to n/2 if n is even. Hence this is a self-generating sequence that can be simply constructed from knowledge of the first term alone. - Ant King, Sep 26 2011
a(n) = (1/2)*ceiling(n/2)*ceiling((3*n + 1)/2). - Mircea Merca, Jul 13 2012
a(n) = (A008794(n+1) + A000217(n))/2 = A002378(n) - A085787(n). - Omar E. Pol, Jan 12 2013
a(n) = floor((n + 1)/2)*((n + 1) - (1/2)*floor((n + 1)/2) - 1/2). - Wesley Ivan Hurt, Jan 26 2013
From Oskar Wieland, Apr 10 2013: (Start)
a(n) = a(n+1) - A026741(n),
a(n) = a(n+2) - A001651(n),
a(n) = a(n+3) - A184418(n),
a(n) = a(n+4) - A007310(n),
a(n) = a(n+6) - A001651(n)*3 = a(n+6) - A016051(n),
a(n) = a(n+8) - A007310(n)*2 = a(n+8) - A091999(n),
a(n) = a(n+10)- A001651(n)*5 = a(n+10)- A072703(n),
a(n) = a(n+12)- A007310(n)*3,
a(n) = a(n+14)- A001651(n)*7. (End)
a(n) = (A007310(n+1)^2 - 1)/24. - Richard R. Forberg, May 27 2013; corrected by Zak Seidov, Mar 14 2015; further corrected by Jianing Song, Oct 24 2018
a(n) = Sum_{i = ceiling((n+1)/2)..n} i. - Wesley Ivan Hurt, Jun 08 2013
G.f.: x*G(0), where G(k) = 1 + x*(3*k + 4)/(3*k + 2 - x*(3*k + 2)*(3*k^2 + 11*k + 10)/(x*(3*k^2 + 11*k + 10) + (k + 1)*(3*k + 4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013
Sum_{n>=1} 1/a(n) = 6 - 2*Pi/sqrt(3). - Vaclav Kotesovec, Oct 05 2016
a(n) = Sum_{i=1..n} numerator(i/2) = Sum_{i=1..n} denominator(2/i). - Wesley Ivan Hurt, Feb 26 2017
a(n) = A000292(A001651(n))/A001651(n), for n>0. - Ivan N. Ianakiev, May 08 2018
a(n) = ((-5 + (-1)^n - 6n)*(-1 + (-1)^n - 6n))/96. - José de Jesús Camacho Medina, Jun 12 2018
a(n) = Sum_{k=1..n} k/gcd(k,2). - Pedro Caceres, Apr 23 2019
Quadrisection. For r = 0,1,2,3: a(r + 4*k) = 6*k^2 + sqrt(24*a(r) + 1)*k + a(r), for k >= 1, with inputs (k = 0) {0,1,2,5}. These are the sequences A049453(k), A033570(k), A033568(k+1), A049452(k+1), for k >= 0, respectively. - Wolfdieter Lang, Feb 12 2021
a(n) = a(n-4) + sqrt(24*a(n-2) + 1), n >= 4. - Klaus Purath, Jul 07 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 6*(log(3)-1). - Amiram Eldar, Feb 28 2022
a(n) = A002620(n) + A008805(n-1). Gary W. Adamson, Dec 10 2022
E.g.f.: (x*(7 + 3*x)*cosh(x) + (1 + 5*x + 3*x^2)*sinh(x))/8. - Stefano Spezia, Aug 01 2024

A000096 a(n) = n*(n+3)/2.

Original entry on oeis.org

0, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377, 1430, 1484, 1539, 1595, 1652, 1710, 1769
Offset: 0

Views

Author

Keywords

Comments

For n >= 1, a(n) is the maximal number of pieces that can be obtained by cutting an annulus with n cuts. See illustration. - Robert G. Wilson v
n(n-3)/2 (n >= 3) is the number of diagonals of an n-gon. - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr)
n(n-3)/2 (n >= 4) is the degree of the third-smallest irreducible presentation of the symmetric group S_n (cf. James and Kerber, Appendix 1).
a(n) is also the multiplicity of the eigenvalue (-2) of the triangle graph Delta(n+1). (See p. 19 in Biggs.) - Felix Goldberg (felixg(AT)tx.technion.ac.il), Nov 25 2001
For n > 3, a(n-3) = dimension of the traveling salesman polytope T(n). - Benoit Cloitre, Aug 18 2002
Also counts quasi-dominoes (quasi-2-ominoes) on an n X n board. Cf. A094170-A094172. - Jon Wild, May 07 2004
Coefficient of x^2 in (1 + x + 2*x^2)^n. - Michael Somos, May 26 2004
a(n) is the number of "prime" n-dimensional polyominoes. A "prime" n-polyomino cannot be formed by connecting any other n-polyominoes except for the n-monomino and the n-monomino is not prime. E.g., for n=1, the 1-monomino is the line of length 1 and the only "prime" 1-polyominoes are the lines of length 2 and 3. This refers to "free" n-dimensional polyominoes, i.e., that can be rotated along any axis. - Bryan Jacobs (bryanjj(AT)gmail.com), Apr 01 2005
Solutions to the quadratic equation q(m, r) = (-3 +- sqrt(9 + 8(m - r))) / 2, where m - r is included in a(n). Let t(m) = the triangular number (A000217) less than some number k and r = k - t(m). If k is neither prime nor a power of two and m - r is included in A000096, then m - q(m, r) will produce a value that shares a divisor with k. - Andrew S. Plewe, Jun 18 2005
Sum_{k=2..n+1} 4/(k*(k+1)*(k-1)) = ((n+3)*n)/((n+2)*(n+1)). Numerator(Sum_{k=2..n+1} 4/(k*(k+1)*(k-1))) = (n+3)*n/2. - Alexander Adamchuk, Apr 11 2006
Number of rooted trees with n+3 nodes of valence 1, no nodes of valence 2 and exactly two other nodes. I.e., number of planted trees with n+2 leaves and exactly two branch points. - Theo Johnson-Freyd (theojf(AT)berkeley.edu), Jun 10 2007
If X is an n-set and Y a fixed 2-subset of X then a(n-2) is equal to the number of (n-2)-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
For n >= 1, a(n) is the number of distinct shuffles of the identity permutation on n+1 letters with the identity permutation on 2 letters (12). - Camillia Smith Barnes, Oct 04 2008
If s(n) is a sequence defined as s(1) = x, s(n) = kn + s(n-1) + p for n > 1, then s(n) = a(n-1)*k + (n-1)*p + x. - Gary Detlefs, Mar 04 2010
The only primes are a(1) = 2 and a(2) = 5. - Reinhard Zumkeller, Jul 18 2011
a(n) = m such that the (m+1)-th triangular number minus the m-th triangular number is the (n+1)-th triangular number: (m+1)(m+2)/2 - m(m+1)/2 = (n+1)(n+2)/2. - Zak Seidov, Jan 22 2012
For n >= 1, number of different values that Sum_{k=1..n} c(k)*k can take where the c(k) are 0 or 1. - Joerg Arndt, Jun 24 2012
On an n X n chessboard (n >= 2), the number of possible checkmate positions in the case of king and rook versus a lone king is 0, 16, 40, 72, 112, 160, 216, 280, 352, ..., which is 8*a(n-2). For a 4 X 4 board the number is 40. The number of positions possible was counted including all mirror images and rotations for all four sides of the board. - Jose Abutal, Nov 19 2013
If k = a(i-1) or k = a(i+1) and n = k + a(i), then C(n, k-1), C(n, k), C(n, k+1) are three consecutive binomial coefficients in arithmetic progression and these are all the solutions. There are no four consecutive binomial coefficients in arithmetic progression. - Michael Somos, Nov 11 2015
a(n-1) is also the number of independent components of a symmetric traceless tensor of rank 2 and dimension n >= 1. - Wolfdieter Lang, Dec 10 2015
Numbers k such that 8k + 9 is a square. - Juri-Stepan Gerasimov, Apr 05 2016
Let phi_(D,rho) be the average value of a generic degree D monic polynomial f when evaluated at the roots of the rho-th derivative of f, expressed as a polynomial in the averaged symmetric polynomials in the roots of f. [See the Wojnar et al. link] The "last" term of phi_(D,rho) is a multiple of the product of all roots of f; the coefficient is expressible as a polynomial h_D(N) in N:=D-rho. These polynomials are of the form h_D(N)= ((-1)^D/(D-1)!)*(D-N)*N^chi*g_D(N) where chi = (1 if D is odd, 0 if D is even) and g_D(N) is a monic polynomial of degree (D-2-chi). Then a(n) are the negated coefficients of the next to the highest order term in the polynomials N^chi*g_D(N), starting at D=3. - Gregory Gerard Wojnar, Jul 19 2017
For n >= 2, a(n) is the number of summations required to solve the linear regression of n variables (n-1 independent variables and 1 dependent variable). - Felipe Pedraza-Oropeza, Dec 07 2017
For n >= 2, a(n) is the number of sums required to solve the linear regression of n variables: 5 for two variables (sums of X, Y, X^2, Y^2, X*Y), 9 for 3 variables (sums of X1, X2, Y1, X1^2, X1*X2, X1*Y, X2^2, X2*Y, Y^2), and so on. - Felipe Pedraza-Oropeza, Jan 11 2018
a(n) is the area of a triangle with vertices at (n, n+1), ((n+1)*(n+2)/2, (n+2)*(n+3)/2), ((n+2)^2, (n+3)^2). - J. M. Bergot, Jan 25 2018
Number of terms less than 10^k: 1, 4, 13, 44, 140, 446, 1413, 4471, 14141, 44720, 141420, 447213, ... - Muniru A Asiru, Jan 25 2018
a(n) is also the number of irredundant sets in the (n+1)-path complement graph for n > 2. - Eric W. Weisstein, Apr 11 2018
a(n) is also the largest number k such that the largest Dyck path of the symmetric representation of sigma(k) has exactly n peaks, n >= 1. (Cf. A237593.) - Omar E. Pol, Sep 04 2018
For n > 0, a(n) is the number of facets of associahedra. Cf. A033282 and A126216 and their refinements A111785 and A133437 for related combinatorial and analytic constructs. See p. 40 of Hanson and Sha for a relation to projective spaces and string theory. - Tom Copeland, Jan 03 2021
For n > 0, a(n) is the number of bipartite graphs with 2n or 2n+1 edges, no isolated vertices, and a stable set of cardinality 2. - Christian Barrientos, Jun 13 2022
For n >= 2, a(n-2) is the number of permutations in S_n which are the product of two different transpositions of adjacent points. - Zbigniew Wojciechowski, Mar 31 2023
a(n) represents the optimal stop-number to achieve the highest running score for the Greedy Pig game with an (n-1)-sided die with a loss on a 1. The total at which one should stop is a(s-1), e.g. for a 6-sided die, one should pass the die at 20. See Sparks and Haran. - Nicholas Stefan Georgescu, Jun 09 2024

Examples

			G.f. = 2*x + 5*x^2 + 9*x^3 + 14*x^4 + 20*x^5 + 27*x^6 + 35*x^7 + 44*x^8 + 54*x^9 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), Table 22.7, p. 797.
  • Norman Biggs, Algebraic Graph Theory, 2nd ed. Cambridge University Press, 1993.
  • G. James and A. Kerber, The Representation Theory of the Symmetric Group, Encyclopedia of Maths. and its Appls., Vol. 16, Addison-Wesley, 1981, Reading, MA, U.S.A.
  • D. G. Kendall et al., Shape and Shape Theory, Wiley, 1999; see p. 4.
  • 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

Complement of A007401. Column 2 of A145324. Column of triangle A014473, first skew subdiagonal of A033282, a diagonal of A079508.
Occurs as a diagonal in A074079/A074080, i.e., A074079(n+3, n) = A000096(n-1) for all n >= 2. Also A074092(n) = 2^n * A000096(n-1) after n >= 2.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488.
Similar sequences are listed in A316466.

Programs

Formula

G.f.: A(x) = x*(2-x)/(1-x)^3. a(n) = binomial(n+1, n-1) + binomial(n, n-1).
Connection with triangular numbers: a(n) = A000217(n+1) - 1.
a(n) = a(n-1) + n + 1. - Bryan Jacobs (bryanjj(AT)gmail.com), Apr 01 2005
a(n) = 2*t(n) - t(n-1) where t() are the triangular numbers, e.g., a(5) = 2*t(5) - t(4) = 2*15 - 10 = 20. - Jon Perry, Jul 23 2003
a(-3-n) = a(n). - Michael Somos, May 26 2004
2*a(n) = A008778(n) - A105163(n). - Creighton Dement, Apr 15 2005
a(n) = C(3+n, 2) - C(3+n, 1). - Zerinvary Lajos, Dec 09 2005
a(n) = A067550(n+1) / A067550(n). - Alexander Adamchuk, May 20 2006
a(n) = A126890(n,1) for n > 0. - Reinhard Zumkeller, Dec 30 2006
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Paul Curtz, Jan 02 2008
Starting (2, 5, 9, 14, ...) = binomial transform of (2, 3, 1, 0, 0, 0, ...). - Gary W. Adamson, Jul 03 2008
For n >= 0, a(n+2) = b(n+1) - b(n), where b(n) is the sequence A005586. - K.V.Iyer, Apr 27 2009
A002262(a(n)) = n. - Reinhard Zumkeller, May 20 2009
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n>=1, a(n-1)=coeff(charpoly(A,x),x^(n-2)). - Milan Janjic, Jul 08 2010
a(n) = Sum_{k=1..n} (k+1)!/k!. - Gary Detlefs, Aug 03 2010
a(n) = n(n+1)/2 + n = A000217(n) + n. - Zak Seidov, Jan 22 2012
E.g.f.: F(x) = 1/2*x*exp(x)*(x+4) satisfies the differential equation F''(x) - 2*F'(x) + F(x) = exp(x). - Peter Bala, Mar 14 2012
a(n) = binomial(n+3, 2) - (n+3). - Robert G. Wilson v, Mar 15 2012
a(n) = A181971(n+1, 2) for n > 0. - Reinhard Zumkeller, Jul 09 2012
a(n) = A214292(n+2, 1). - Reinhard Zumkeller, Jul 12 2012
G.f.: -U(0) where U(k) = 1 - 1/((1-x)^2 - x*(1-x)^4/(x*(1-x)^2 - 1/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 27 2012
A023532(a(n)) = 0. - Reinhard Zumkeller, Dec 04 2012
a(n) = A014132(n,n) for n > 0. - Reinhard Zumkeller, Dec 12 2012
a(n-1) = (1/n!)*Sum_{j=0..n} binomial(n,j)*(-1)^(n-j)*j^n*(j-1). - Vladimir Kruchinin, Jun 06 2013
a(n) = 2n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
a(n) = Sum_{i=2..n+1} i. - Wesley Ivan Hurt, Jun 28 2013
Sum_{n>0} 1/a(n) = 11/9. - Enrique Pérez Herrero, Nov 26 2013
a(n) = Sum_{i=1..n} (n - i + 2). - Wesley Ivan Hurt, Mar 31 2014
A023531(a(n)) = 1. - Reinhard Zumkeller, Feb 14 2015
For n > 0: a(n) = A101881(2*n-1). - Reinhard Zumkeller, Feb 20 2015
a(n) + a(n-1) = A008865(n+1) for all n in Z. - Michael Somos, Nov 11 2015
a(n+1) = A127672(4+n, n), n >= 0, where A127672 gives the coefficients of the Chebyshev C polynomials. See the Abramowitz-Stegun reference. - Wolfdieter Lang, Dec 10 2015
a(n) = (n+1)^2 - A000124(n). - Anton Zakharov, Jun 29 2016
Dirichlet g.f.: (zeta(s-2) + 3*zeta(s-1))/2. - Ilya Gutkovskiy, Jun 30 2016
a(n) = 2*A000290(n+3) - 3*A000217(n+3). - J. M. Bergot, Apr 04 2018
a(n) = Stirling2(n+2, n+1) - 1. - Peter Luschny, Jan 05 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/3 - 5/9. - Amiram Eldar, Jan 10 2021
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = 3.
Product_{n>=1} (1 - 1/a(n)) = 3*cos(sqrt(17)*Pi/2)/(4*Pi). (End)
Product_{n>=0} a(4*n+1)*a(4*n+4)/(a(4*n+2)*a(4*n+3)) = Pi/6. - Michael Jodl, Apr 05 2025

A000567 Octagonal numbers: n*(3*n-2). Also called star numbers.

Original entry on oeis.org

0, 1, 8, 21, 40, 65, 96, 133, 176, 225, 280, 341, 408, 481, 560, 645, 736, 833, 936, 1045, 1160, 1281, 1408, 1541, 1680, 1825, 1976, 2133, 2296, 2465, 2640, 2821, 3008, 3201, 3400, 3605, 3816, 4033, 4256, 4485, 4720, 4961, 5208, 5461
Offset: 0

Views

Author

Keywords

Comments

From Floor van Lamoen, Jul 21 2001: (Start)
Write 1,2,3,4,... in a hexagonal spiral around 0; then a(n) is the sequence found by reading the line from 0 in the direction 0,1,....
The spiral begins:
.
85--84--83--82--81--80
/ \
86 56--55--54--53--52 79
/ / \ \
87 57 33--32--31--30 51 78
/ / / \ \ \
88 58 34 16--15--14 29 50 77
/ / / / \ \ \ \
89 59 35 17 5---4 13 28 49 76
/ / / / / \ \ \ \ \
90 60 36 18 6 0 3 12 27 48 75
/ / / / / / / / / / /
91 61 37 19 7 1---2 11 26 47 74
\ \ \ \ \ . / / / /
92 62 38 20 8---9--10 25 46 73
\ \ \ \ . / / /
93 63 39 21--22--23--24 45 72
\ \ \ . / /
94 64 40--41--42--43--44 71
\ \ . /
95 65--66--67--68--69--70
\ .
96
.
(End)
From Lekraj Beedassy, Oct 02 2003: (Start)
Also the number of distinct three-cell blocks that may be removed out of A000217(n+1) square cells arranged in a stepping triangular array of side (n+1). A 5-layer triangular array of square cells, for instance, has vertices outlined thus:
x x
x x x
x x x x
x x x x x
x x x x x x
x x x x x x (End)
First derivative at n of A045991. - Ross La Haye, Oct 23 2004
Starting from n=1, the sequence corresponds to the Wiener index of K_{n,n} (the complete bipartite graph wherein each independent set has n vertices). - Kailasam Viswanathan Iyer, Mar 11 2009
Number of divisors of 24^(n-1) for n > 0 (cf A009968). - J. Lowell, Aug 30 2008
a(n) = A001399(6n-5), number of partitions of 6*n - 5 into parts < 4. For example a(2)=8 and partitions of 6*2 - 5 = 7 into parts < 4 are: [1,1,1,1,1,1,1], [1,1,1,1,1,2],[1,1,1,1,3], [1,1,1,2,2], [1,1,2,3], [1,2,2,2], [1,3,3], [2,2,3]. - Adi Dani, Jun 07 2011
Also, sequence found by reading the line from 0 in the direction 0, 8, ..., and the parallel line from 1 in the direction 1, 21, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Sep 10 2011
Partial sums give A002414. - Omar E. Pol, Jan 12 2013
Generate a Pythagorean triple using Euclid's formula with (n, n-1) to give A,B,C. a(n) = B + (A + C)/2. - J. M. Bergot, Jul 13 2013
The number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 773", based on the 5-celled von Neumann neighborhood. - Robert Price, May 23 2016
For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n-4; {1, 2n-2, 3, 2n-2, 1, 18n-8}]. For n=1, this collapses to [5; {5, 10}]. - Magus K. Chu, Oct 10 2022
a(n)*a(n+1) + 1 = (3n^2 + n - 1)^2. In general, a(n)*a(n+k) + k^2 = (3n^2 + (3k-2)n - k)^2. - Charlie Marion, May 23 2023

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 38.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 1.
  • 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 19-20.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 123.

Crossrefs

Cf. A014641, A014642, A014793, A014794, A001835, A016777, A045944, A093563 ((6, 1) Pascal, column m=2). A016921 (differences).
Cf. A005408 (the odd numbers).

Programs

  • GAP
    List([0..50], n -> n*(3*n-2)); # G. C. Greubel, Nov 15 2018
    
  • Haskell
    a000567 n = n * (3 * n - 2)  -- Reinhard Zumkeller, Dec 20 2012
    
  • Magma
    [n*(3*n-2) : n in [0..50]]; // Wesley Ivan Hurt, Oct 10 2021
  • Maple
    A000567 := proc(n)
        n*(3*n-2) ;
    end proc:
    seq(A000567(n), n=1..50) ;
  • Mathematica
    Table[n (3 n - 2), {n, 0, 50}] (* Harvey P. Dale, May 06 2012 *)
    Table[PolygonalNumber[RegularPolygon[8], n], {n, 0, 43}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
    PolygonalNumber[8, Range[0, 20]] (* Eric W. Weisstein, Sep 07 2017 *)
    LinearRecurrence[{3, -3, 1}, {1, 8, 21}, {0, 20}] (* Eric W. Weisstein, Sep 07 2017 *)
  • PARI
    a(n)=n*(3*n-2) \\ Charles R Greathouse IV, Jun 10 2011
    
  • PARI
    vector(50, n, n--; n*(3*n-2)) \\ G. C. Greubel, Nov 15 2018
    
  • Python
    # Intended to compute the initial segment of the sequence, not isolated terms.
    def aList():
         x, y = 1, 1
         yield 0
         while True:
             yield x
             x, y = x + y + 6, y + 6
    A000567 = aList()
    print([next(A000567) for i in range(49)]) # Peter Luschny, Aug 04 2019
    
  • Python
    [n*(3*n-2) for n in range(50)] # Gennady Eremin, Mar 10 2022
    
  • Sage
    [n*(3*n-2) for n in range(50)] # G. C. Greubel, Nov 15 2018
    

Formula

a(n) = n*(3*n-2).
a(n) = (3n-2)*(3n-1)*(3n)/((3n-1) + (3n-2) + (3n)), i.e., (the product of three consecutive numbers)/(their sum). a(1) = 1*2*3/(1+2+3), a(2) = 4*5*6/(4+5+6), etc. - Amarnath Murthy, Aug 29 2002
E.g.f.: exp(x)*(x+3*x^2). - Paul Barry, Jul 23 2003
G.f.: x*(1+5*x)/(1-x)^3. Simon Plouffe in his 1992 dissertation
a(n) = Sum_{k=1..n} (5*n - 4*k). - Paul Barry, Sep 06 2005
a(n) = n + 6*A000217(n-1). - Floor van Lamoen, Oct 14 2005
a(n) = C(n+1,2) + 5*C(n,2).
Starting (1, 8, 21, 40, 65, ...) = binomial transform of [1, 7, 6, 0, 0, 0, ...]. - Gary W. Adamson, Apr 30 2008
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=0, a(1)=1, a(2)=8. - Jaume Oliver Lafont, Dec 02 2008
a(n) = A000578(n) - A007531(n). - Reinhard Zumkeller, Sep 18 2009
a(n) = a(n-1) + 6*n - 5 (with a(0)=0). - Vincenzo Librandi, Nov 20 2010
a(n) = 2*a(n-1) - a(n-2) + 6. - Ant King, Sep 01 2011
a(n) = A000217(n) + 5*A000217(n-1). - Vincenzo Librandi, Nov 20 2010
a(n) = (A185212(n) - 1) / 4. - Reinhard Zumkeller, Dec 20 2012
a(n) = A174709(6n). - Philippe Deléham, Mar 26 2013
a(n) = (2*n-1)^2 - (n-1)^2. - Ivan N. Ianakiev, Apr 10 2013
a(6*a(n) + 16*n + 1) = a(6*a(n) + 16*n) + a(6*n + 1). - Vladimir Shevelev, Jan 24 2014
a(0) = 0, a(n) = Sum_{k=0..n-1} A005408(A051162(n-1,k)), n >= 1. - L. Edson Jeffery, Jul 28 2014
Sum_{n>=1} 1/a(n) = (sqrt(3)*Pi + 9*log(3))/12 = 1.2774090575596367311949534921... . - Vaclav Kotesovec, Apr 27 2016
From Ilya Gutkovskiy, Jul 29 2016: (Start)
Inverse binomial transform of A084857.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(2*sqrt(3)) = A093766. (End)
a(n) = n * A016777(n-1) = A053755(n) - A000290(n+1). - Bruce J. Nicholson, Aug 10 2017
Product_{n>=2} (1 - 1/a(n)) = 3/4. - Amiram Eldar, Jan 21 2021
P(4k+4,n) = ((k+1)*n - k)^2 - (k*n - k)^2 where P(m,n) is the n-th m-gonal number (a generalization of the Apr 10 2013 formula, a(n) = (2*n-1)^2 - (n-1)^2). - Charlie Marion, Oct 07 2021
From Leo Tavares, Oct 31 2021: (Start)
a(n) = A000290(n) + 4*A000217(n-1). See Square Rays illustration.
a(n) = A000290(n) + A046092(n-1)
a(n) = A000384(n) + 2*A000217(n-1). See Twin Rectangular Rays illustration.
a(n) = A000384(n) + A002378(n-1)
a(n) = A003154(n) - A045944(n-1). See Star Rows illustration. (End)

Extensions

Incorrect example removed by Joerg Arndt, Mar 11 2010

A000566 Heptagonal numbers (or 7-gonal numbers): n*(5*n-3)/2.

Original entry on oeis.org

0, 1, 7, 18, 34, 55, 81, 112, 148, 189, 235, 286, 342, 403, 469, 540, 616, 697, 783, 874, 970, 1071, 1177, 1288, 1404, 1525, 1651, 1782, 1918, 2059, 2205, 2356, 2512, 2673, 2839, 3010, 3186, 3367, 3553, 3744, 3940, 4141, 4347, 4558, 4774, 4995, 5221, 5452, 5688
Offset: 0

Views

Author

Keywords

Comments

Binomial transform of (0, 1, 5, 0, 0, 0, ...). Binomial transform is A084899. - Paul Barry, Jun 10 2003
Row sums of triangle A131413. - Gary W. Adamson, Jul 08 2007
Sequence starting (1, 7, 18, 34, ...) = binomial transform of (1, 6, 5, 0, 0, 0, ...). Also row sums of triangle A131896. - Gary W. Adamson, Jul 24 2007
Also the partial sums of A016861, a zero added in front; therefore a(n) = n (mod 5). - R. J. Mathar, Mar 19 2008
Also sequence found by reading the line from 0, in the direction 0, 7, ..., and the line from 1, in the direction 1, 18, ..., in the square spiral whose edges have length A195013 and whose vertices are the numbers A195014. These parallel lines are the semi-axes perpendicular to the main axis A195015 in the same spiral. - Omar E. Pol, Oct 14 2011
Also sequence found by reading the line from 0, in the direction 0, 7, ... and the parallel line from 1, in the direction 1, 18, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Jul 18 2012
Partial sums give A002413. - Omar E. Pol, Jan 12 2013
The n-th heptagonal number equals the sum of the n consecutive integers starting at 2*n-1; for example, 1, 3+4, 5+6+7, 7+8+9+10, etc. In general, the n-th (2k+1)-gonal number is the sum of the n consecutive integers starting at (k-1)*n - (k-2). When k = 1 and 2, this result generates the triangular numbers, A000217, and the pentagonal numbers, A000326, respectively. - Charlie Marion, Mar 02 2022

Examples

			G.f. = x + 7*x^2 + 18*x^3 + 34*x^4 + 55*x^5 + 81*x^6 + 112*x^7 + ... - _Michael Somos_, Jan 25 2019
		

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 38.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
  • Leonard E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
  • 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. See p. 123.

Crossrefs

a(n)= A093562(n+1, 2), (5, 1)-Pascal column.
Cf. sequences listed in A254963.

Programs

  • Haskell
    a000566 n = n * (5 * (n - 1) + 2) `div` 2
    a000566_list = scanl (+) 0 a016861_list  -- Reinhard Zumkeller, Jun 16 2013
    
  • Magma
    a000566:=func< n | n*(5*n-3) div 2 >; [ a000566(n): n in [0..50] ];
    
  • Maple
    A000566 := proc(n)
            n*(5*n-3)/2 ;
    end proc:
    seq(A000566(n),n=0..30); # R. J. Mathar, Oct 02 2020
  • Mathematica
    Table[n (5n - 3)/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 1, 7}, 50] (* Harvey P. Dale, Oct 13 2011 *)
    Join[{0},Accumulate[Range[1,315,5]]] (* Harvey P. Dale, Mar 26 2016 *)
    (* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[7], n], {n, 0, 48}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
    PolygonalNumber[7,Range[0,50]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 23 2021 *)
  • Maxima
    makelist(n*(5*n-3)/2, n, 0, 20); /* Martin Ettl, Dec 11 2012 */
    
  • PARI
    a(n) = n * (5*n - 3) / 2
    
  • Python
    # Intended to compute the initial segment of the sequence, not isolated terms.
    def aList():
         x, y = 1, 1
         yield 0
         while True:
             yield x
             x, y = x + y + 5, y + 5
    A000566 = aList()
    print([next(A000566) for i in range(49)]) # Peter Luschny, Aug 04 2019
    
  • Python
    [n*(5*n-3)//2 for n in range(50)] # Gennady Eremin, Mar 24 2022

Formula

G.f.: x*(1 + 4*x)/(1 - x)^3. Simon Plouffe in his 1992 dissertation.
a(n) = C(n, 1) + 5*C(n, 2). - Paul Barry, Jun 10 2003
a(n) = Sum_{k = 1..n} (4*n - 3*k). - Paul Barry, Sep 06 2005
a(n) = n + 5*A000217(n-1) - Floor van Lamoen, Oct 14 2005
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for a(0) = 0, a(1) = 1, a(2) = 7. - Jaume Oliver Lafont, Dec 02 2008
a(n+1) = A153126(n) + n mod 2; a(2*n+1) = A033571(n); a(2*(n+1)) = A153127(n) + 1. - Reinhard Zumkeller, Dec 20 2008
40*a(n)+ 9 = A017354(n-1). - Ken Rosenbaum, Dec 02 2009.
a(n) = 2*a(n-1) - a(n-2) + 5, with a(0) = 0 and a(1) = 1. - Mohamed Bouhamida, May 05 2010
a(n) = A000217(n) + 4*A000217(n-1). - Vincenzo Librandi, Nov 20 2010
a(n) = a(n-1) + 5*n - 4, with a(0) = 0. - Vincenzo Librandi, Nov 20 2010
a(n) = A130520(5*n). - Philippe Deléham, Mar 26 2013
a(5*a(n) + 11*n + 1) = a(5*a(n) + 11*n) + a(5*n + 1). - Vladimir Shevelev, Jan 24 2014
Sum_{n>=1} 1/a(n) = sqrt(1 - 2/sqrt(5))*Pi/3 + 5*log(5)/6 - sqrt(5)*log((1 + sqrt(5))/2)/3 = 1.32277925312238885674944226131... . See A244639. - Vaclav Kotesovec, Apr 27 2016
E.g.f.: x*(2 + 5*x)*exp(x)/2. - Ilya Gutkovskiy, Aug 27 2016
From Charlie Marion, May 02 2017: (Start)
a(n+m) = a(n) + 5*n*m + a(m);
a(n-m) = a(n) - 5*n*m + a(m) + 3*m;
a(n) - a(m) = (5*(n + m) - 3)*(n - m)/2.
In general, let P(k,n) be the n-th k-gonal number. Then
P(k,n+m) = P(k,n) + (k - 2)*n*m + P(k,m);
P(k,n-m) = P(k,n) - (k - 2)*n*m + P(k,m) + (k - 4)*m;
P(k,n) - P(k,m) = ((k-2)*(n + m) + 4 - k)*(n - m)/2.
(End)
a(n) = A147875(-n) for all n in Z. - Michael Somos, Jan 25 2019
a(n) = A000217(n-1) + A000217(2*n-1). - Charlie Marion, Dec 19 2019
Product_{n>=2} (1 - 1/a(n)) = 5/7. - Amiram Eldar, Jan 21 2021
a(n) + a(n+1) = (2*n+1)^2 + n^2 - 2*n. In general, if we let P(k,n) = the n-th k-gonal number, then P(k^2-k+1,n)+ P(k^2-k+1,n+1) = ((k-1)*n+1)^2 + (k-2)*(n^2-2*n) = ((k-1)*n+1)^2 + (k-2)*A005563(n-2). When k = 2, this formula reduces to the well-known triangular number formula: T(n) + T(n+1) = (n+1)^2. - Charlie Marion, Jul 01 2021

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A000931 Padovan sequence (or Padovan numbers): a(n) = a(n-2) + a(n-3) with a(0) = 1, a(1) = a(2) = 0.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897, 2513, 3329, 4410, 5842, 7739, 10252, 13581, 17991, 23833, 31572, 41824, 55405, 73396, 97229, 128801, 170625
Offset: 0

Views

Author

Keywords

Comments

Number of compositions of n into parts congruent to 2 mod 3 (offset -1). - Vladeta Jovovic, Feb 09 2005
a(n) is the number of compositions of n into parts that are odd and >= 3. Example: a(10)=3 counts 3+7, 5+5, 7+3. - David Callan, Jul 14 2006
Referred to as N0102 in R. K. Guy's "Anyone for Twopins?" - Rainer Rosenthal, Dec 05 2006
Zagier conjectures that a(n+3) is the maximum number of multiple zeta values of weight n > 1 which are linearly independent over the rationals. - Jonathan Sondow and Sergey Zlobin (sirg_zlobin(AT)mail.ru), Dec 20 2006
Starting with offset 6: (1, 1, 2, 2, 3, 4, 5, ...) = INVERT transform of A106510: (1, 1, -1, 0, 1, -1, 0, 1, -1, ...). - Gary W. Adamson, Oct 10 2008
Starting with offset 7, the sequence 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, ... is called the Fibonacci quilt sequence by Catral et al., in Fib. Q. 2017. - N. J. A. Sloane, Dec 24 2021
Triangle A145462: right border = A000931 starting with offset 6. Row sums = Padovan sequence starting with offset 7. - Gary W. Adamson, Oct 10 2008
Starting with offset 3 = row sums of triangle A146973 and INVERT transform of [1, -1, 2, -2, 3, -3, ...]. - Gary W. Adamson, Nov 03 2008
a(n+5) corresponds to the diagonal sums of "triangle": 1; 1; 1,1; 1,1; 1,2,1; 1,2,1; 1,3,3,1; 1,3,3,1; 1,4,6,4,1; ..., rows of Pascal's triangle (A007318) repeated. - Philippe Deléham, Dec 12 2008
With offset 3: (1, 0, 1, 1, 1, 2, 2, ...) convolved with the tribonacci numbers prefaced with a "1": (1, 1, 1, 2, 4, 7, 13, ...) = the tribonacci numbers, A000073. (Cf. triangle A153462.) - Gary W. Adamson, Dec 27 2008
a(n) is also the number of strings of length (n-8) from an alphabet {A, B} with no more than one A or 2 B's consecutively. (E.g., n = 4: {ABAB,ABBA,BABA,BABB,BBAB} and a(4+8) = 5.) - Toby Gottfried, Mar 02 2010
p(n):=A000931(n+3), n >= 1, is the number of partitions of the numbers {1,2,3,...,n} into lists of length two or three containing neighboring numbers. The 'or' is inclusive. For n=0 one takes p(0)=1. For details see the W. Lang link. There the explicit formula for p(n) (analog of the Binet-de Moivre formula for Fibonacci numbers) is also given. Padovan sequences with different inputs are also considered there. - Wolfdieter Lang, Jun 15 2010
Equals the INVERTi transform of Fibonacci numbers prefaced with three 1's, i.e., (1 + x + x^2 + x^3 + x^4 + 2x^5 + 3x^6 + 5x^7 + 8x^8 + 13x^9 + ...). - Gary W. Adamson, Apr 01 2011
When run backwards gives (-1)^n*A050935(n).
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [0, 0, 1; 1, 0, 1; 0, 1, 0] or of the 3 X 3 matrix [0, 1, 0; 0, 0, 1; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
Figure 4 of Brauchart et al., 2014, shows a way to "visualize the Padovan sequence as cuboid spirals, where the dimensions of each cuboid made up by the previous ones are given by three consecutive numbers in the sequence". - N. J. A. Sloane, Mar 26 2014
a(n) is the number of closed walks from a vertex of a unidirectional triangle containing an opposing directed edge (arc) between the second and third vertices. Equivalently the (1,1) entry of A^n where the adjacency matrix of digraph is A=(0,1,0;0,0,1;1,1,0). - David Neil McGrath, Dec 19 2014
Number of compositions of n-3 (n >= 4) into 2's and 3's. Example: a(12)=5 because we have 333, 3222, 2322, 2232, and 2223. - Emeric Deutsch, Dec 28 2014
The Hoffman (2015) paper "offers significant evidence that the number of quantities needed to generate the weight-n multiple harmonic sums mod p is" a(n). - N. J. A. Sloane, Jun 24 2016
a(n) gives the number of compositions of n-5 into odd parts where the order of the 1's does not matter. For example, a(11)=4 counts the following compositions of 6: (5,1)=(1,5), (3,3), (3,1,1,1)=(1,3,1,1)=(1,1,3,1)=(1,1,1,3), (1,1,1,1,1,1). - Gregory L. Simay, Aug 04 2016
For n > 6, a(n) is the number of maximal matchings in the (n-5)-path graph, maximal independent vertex sets and minimal vertex covers in the (n-6)-path graph, and minimal edge covers in the (n-5)-pan graph and (n-3)-path graphs. - Eric W. Weisstein, Mar 30, Aug 03, and Aug 07 2017
From James Mitchell and Wilf A. Wilson, Jul 21 2017: (Start)
a(2n + 5) + 2n - 4, n > 2, is the number of maximal subsemigroups of the monoid of order-preserving mappings on a set with n elements.
a(n + 6) + n - 3, n > 3, is the number of maximal subsemigroups of the monoid of order-preserving or reversing mappings on a set with n elements.
(End)
Has the property that the largest of any four consecutive terms equals the sum of the two smallest. - N. J. A. Sloane, Aug 29 2017 [David Nacin points out that there are many sequences with this property, such as 1,1,1,2,1,1,1,2,1,1,1,2,... or 2,3,4,5,2,3,4,5,2,3,4,5,... or 2,2,1,3,3, 4,1,4, 5,5,1,6,6, 7,1,7, 8,8,1,9,9, 10,1,10, ... (spaces added for clarity), and a conjecture I made here in 2017 was simply wrong. I have deleted it. - N. J. A. Sloane, Oct 23 2018]
a(n) is also the number of maximal cliques in the (n+6)-path complement graph. - Eric W. Weisstein, Apr 12 2018
a(n+8) is the number of solus bitstrings of length n with no runs of 3 zeros. - Steven Finch, Mar 25 2020
Named after the architect Richard Padovan (b. 1935). - Amiram Eldar, Jun 08 2021
Shannon et al. (2006) credit a French architecture student Gérard Cordonnier with the discovery of these numbers.
For n >= 3, a(n) is the number of sequences of 0s and 1s of length (n-2) that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s. - Yifan Xie, Oct 20 2022
For n >= 2, a(n+5) is the number of ways to tile the 1xn board with dominoes and squares (ie. size 1x1) such that are either none or one squares between dominoes, none or one squares at both ends of the board, and there is at least one domino. For example, for n=6, a(11)=4 since the tilings are |2|2, |22|, 2|2| and 222 (where 2 represents a domino and | a square). - Enrique Navarrete, Aug 31 2024

Examples

			G.f. = 1 + x^3 + x^5 + x^6 + x^7 + 2*x^8 + 2*x^9 + 3*x^10 + 4*x^11 + ...
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4.
  • Minerva Catral, Pari L. Ford, Pamela E. Harris, Steven J. Miller, Dawn Nelson, Zhao Pan, and Huanzhong Xu, Legal Decompositions Arising from Non-positive Linear Recurrences, Fib. Quart., 55:3 (2017), 252-275. [Note that there is an earlier version of this paper, with only five authors, on the arXiv in 2016. Note to editors: do not merge these two citations. - N. J. A. Sloane, Dec 24 2021]
  • Richard K. Guy, "Anyone for Twopins?" in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 10-11.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • A. G. Shannon, P. G. Anderson and A. F. Horadam, Properties of Cordonnier, Perrin and Van der Laan numbers, International Journal of Mathematical Education in Science and Technology, Volume 37:7 (2006), 825-831. See P_n.
  • 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).
  • Ian Stewart, L'univers des nombres, "La sculpture et les nombres", pp. 19-20, Belin-Pour La Science, Paris, 2000.
  • Hans van der Laan, Het plastische getal. XV lessen over de grondslagen van de architectonische ordonnantie. Leiden, E.J. Brill, 1967.
  • Don Zagier, Values of zeta functions and their applications, in First European Congress of Mathematics (Paris, 1992), Vol. II, A. Joseph et al. (eds.), Birkhäuser, Basel, 1994, pp. 497-512.

Crossrefs

The following are basically all variants of the same sequence: A000931, A078027, A096231, A124745, A133034, A134816, A164001, A182097, A228361 and probably A020720. However, each one has its own special features and deserves its own entry.
Closely related to A001608.
Doubling every term gives A291289.

Programs

  • GAP
    a:=[1,0,0];; for n in [4..50] do a[n]:=a[n-2]+a[n-3]; od; a; # G. C. Greubel, Dec 30 2019
    
  • Haskell
    a000931 n = a000931_list !! n
    a000931_list = 1 : 0 : 0 : zipWith (+) a000931_list (tail a000931_list)
    -- Reinhard Zumkeller, Feb 10 2011
    
  • Magma
    I:=[1,0,0]; [n le 3 select I[n] else Self(n-2) + Self(n-3): n in [1..60]]; // Vincenzo Librandi, Jul 21 2015
    
  • Maple
    A000931 := proc(n) option remember; if n = 0 then 1 elif n <= 2 then 0 else procname(n-2)+procname(n-3); fi; end;
    A000931:=-(1+z)/(-1+z^2+z^3); # Simon Plouffe in his 1992 dissertation; gives sequence without five leading terms
    a[0]:=1; a[1]:=0; a[2]:=0; for n from 3 to 50 do a[n]:=a[n-2]+a[n-3]; end do; # Francesco Daddi, Aug 04 2011
  • Mathematica
    CoefficientList[Series[(1-x^2)/(1-x^2-x^3), {x, 0, 50}], x]
    a[0]=1; a[1]=a[2]=0; a[n_]:= a[n]= a[n-2] + a[n-3]; Table[a[n], {n, 0, 50}] (* Robert G. Wilson v, May 04 2006 *)
    LinearRecurrence[{0,1,1}, {1,0,0}, 50] (* Harvey P. Dale, Jan 10 2012 *)
    Table[RootSum[-1 -# +#^3 &, 5#^n -6#^(n+1) +4#^(n+2) &]/23, {n,0,50}] (* Eric W. Weisstein, Nov 09 2017 *)
  • PARI
    Vec((1-x^2)/(1-x^2-x^3) + O(x^50)) \\ Charles R Greathouse IV, Feb 11 2011
    
  • PARI
    {a(n) = if( n<0, polcoeff(1/(1+x-x^3) + x * O(x^-n), -n), polcoeff( (1 - x^2)/(1-x^2-x^3) + x * O(x^n), n))}; /* Michael Somos, Sep 18 2012 */
    
  • Python
    def aupton(nn):
        alst = [1, 0, 0]
        for n in range(3, nn+1): alst.append(alst[n-2]+alst[n-3])
        return alst
    print(aupton(49)) # Michael S. Branicky, Mar 28 2022
  • Sage
    def A000931_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( (1-x^2)/(1-x^2-x^3) ).list()
    A000931_list(50) # G. C. Greubel, Dec 30 2019
    

Formula

G.f.: (1-x^2)/(1-x^2-x^3).
a(n) is asymptotic to r^n / (2*r+3) where r = 1.3247179572447... = A060006, the real root of x^3 = x + 1. - Philippe Deléham, Jan 13 2004
a(n)^2 + a(n+2)^2 + a(n+6)^2 = a(n+1)^2 + a(n+3)^2 + a(n+4)^2 + a(n+5)^2 (Barniville, Question 16884, Ed. Times 1911).
a(n+5) = a(0) + a(1) + ... + a(n).
a(n) = central and lower right terms in the (n-3)-th power of the 3 X 3 matrix M = [0 1 0 / 0 0 1 / 1 1 0]. E.g., a(13) = 7. M^10 = [3 5 4 / 4 7 5 / 5 9 7]. - Gary W. Adamson, Feb 01 2004
G.f.: 1/(1 - x^3 - x^5 - x^7 - x^9 - ...). - Jon Perry, Jul 04 2004
a(n+4) = Sum_{k=0..floor((n-1)/2)} binomial(floor((n+k-2)/3), k). - Paul Barry, Jul 06 2004
a(n+3) = Sum_{k=0..floor(n/2)} binomial(k, n-2k). - Paul Barry, Sep 17 2004, corrected by Greg Dresden and Zi Ye, Jul 06 2021
a(n+3) is diagonal sum of A026729 (as a number triangle), with formula a(n+3) = Sum_{k=0..floor(n/2)} Sum_{i=0..n-k} (-1)^(n-k+i)*binomial(n-k, i)*binomial(i+k, i-k). - Paul Barry, Sep 23 2004
a(n) = a(n-1) + a(n-5) = A003520(n-4) + A003520(n-13) = A003520(n-3) - A003520(n-9). - Henry Bottomley, Jan 30 2005
a(n+3) = Sum_{k=0..floor(n/2)} binomial((n-k)/2, k)(1+(-1)^(n-k))/2. - Paul Barry, Sep 09 2005
The sequence 1/(1-x^2-x^3) (a(n+3)) is given by the diagonal sums of the Riordan array (1/(1-x^3), x/(1-x^3)). The row sums are A000930. - Paul Barry, Feb 25 2005
a(n) = A023434(n-7) + 1 for n >= 7. - David Callan, Jul 14 2006
a(n+5) corresponds to the diagonal sums of A030528. The binomial transform of a(n+5) is A052921. a(n+5) = Sum_{k=0..floor(n/2)} Sum_{k=0..n} (-1)^(n-k+i)*binomial(n-k, i)binomial(i+k+1, 2k+1). - Paul Barry, Jun 21 2004
r^(n-1) = (1/r)*a(n) + r*a(n+1) + a(n+2), where r = 1.32471... is the real root of x^3 - x - 1 = 0. Example: r^8 = (1/r)*a(9) + r*a(10) + a(11) = (1/r)*2 + r*3 + 4 = 9.483909... - Gary W. Adamson, Oct 22 2006
a(n) = (r^n)/(2r+3) + (s^n)/(2s+3) + (t^n)/(2t+3) where r, s, t are the three roots of x^3-x-1. - Keith Schneider (schneidk(AT)email.unc.edu), Sep 07 2007
a(n) = -k*a(n-1) + a(n-2) + (k+1)a(n-2) + k*a(n-4), n > 3, for any value of k. - Gary Detlefs, Sep 13 2010
From Francesco Daddi, Aug 04 2011: (Start)
a(0) + a(2) + a(4) + a(6) + ... + a(2*n) = a(2*n+3).
a(0) + a(3) + a(6) + a(9) + ... + a(3*n) = a(3*n+2)+1.
a(0) + a(5) + a(10) + a(15) + ... + a(5*n) = a(5*n+1)+1.
a(0) + a(7) + a(14) + a(21) + ... + a(7*n) = (a(7*n) + a(7*n+1) + 1)/2. (End)
a(n+3) = Sum_{k=0..floor((n+1)/2)} binomial((n+k)/3,k), where binomial((n+k)/3,k)=0 for noninteger (n+k)/3. - Nikita Gogin, Dec 07 2012
a(n) = A182097(n-3) for n > 2. - Jonathan Sondow, Mar 14 2014
a(n) = the k-th difference of a(n+5k) - a(n+5k-1), k>=1. For example, a(10)=3 => a(15)-a(14) => 2nd difference of a(20)-a(19) => 3rd difference of a(25)-a(24)... - Bob Selcoe, Mar 18 2014
Construct the power matrix T(n,j) = [A^*j]*[S^*(j-1)] where A=(0,0,1,0,1,0,1,...) and S=(0,1,0,0,...) or A063524. [* is convolution operation] Define S^*0=I with I=(1,0,0,...). Then a(n) = Sum_{j=1...n} T(n,j). - David Neil McGrath, Dec 19 2014
If x=a(n), y=a(n+1), z=a(n+2), then x^3 + 2*y*x^2 - z^2*x - 3*y*z*x + y^2*x + y^3 - y^2*z + z^3 = 1. - Alexander Samokrutov, Jul 20 2015
For the sequence shifted by 6 terms, a(n) = Sum_{k=ceiling(n/3)..ceiling(n/2)} binomial(k+1,3*k-n) [Doslic-Zubac]. - N. J. A. Sloane, Apr 23 2017
From Joseph M. Shunia, Jan 21 2020: (Start)
a(2n) = 2*a(n-1)*a(n) + a(n)^2 + a(n+1)^2, for n > 8.
a(2n-1) = 2*a(n)*a(n+1) + a(n-1)^2, for n > 8.
a(2n+1) = 2*a(n+1)*a(n+2) + a(n)^2, for n > 7. (End)
0*a(0) + 1*a(1) + 2*a(2) + ... + n*a(n) = n*a(n+5) - a(n+9) + 2. - Greg Dresden and Zi Ye, Jul 02 2021
From Greg Dresden and Zi Ye, Jul 06 2021: (Start)
2*a(n) = a(n+2) + a(n-5) for n >= 5.
3*a(n) = a(n+4) - a(n-9) for n >= 9.
4*a(n) = a(n+5) - a(n-9) for n >= 9. (End)

Extensions

Edited by Charles R Greathouse IV, Mar 17 2010
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A006995 Binary palindromes: numbers whose binary expansion is palindromic.

Original entry on oeis.org

0, 1, 3, 5, 7, 9, 15, 17, 21, 27, 31, 33, 45, 51, 63, 65, 73, 85, 93, 99, 107, 119, 127, 129, 153, 165, 189, 195, 219, 231, 255, 257, 273, 297, 313, 325, 341, 365, 381, 387, 403, 427, 443, 455, 471, 495, 511, 513, 561, 585, 633, 645, 693, 717, 765, 771, 819, 843
Offset: 1

Views

Author

Keywords

Comments

If b > 1 is a binary palindrome then both (2^(m+1) + 1)*b and 2^(m+1) + 2^m - b are also, where m = floor(log_2(b)). - Hieronymus Fischer, Feb 18 2012
Floor and ceiling: If d > 0 is any natural number, then A206913(d) is the greatest binary palindrome <= d and A206914(d) is the least binary palindrome >= d. - Hieronymus Fischer, Feb 18 2012
The greatest binary palindrome <= the n-th non-binary-palindrome is that binary palindrome with number A154809(n)-n+1. The corresponding formula identity is: A206913(A154809(n)) = A006995(A154809(n)-n+1). - Hieronymus Fischer, Mar 18 2012
From Hieronymus Fischer, Jan 23 2013: (Start)
The number of binary digits of a(n) is A070939(a(n)) = 1 + floor(log_2(n)) + floor(log_2(n/3)), for n > 1.
Also: A070939(a(n)) = A070939(n) + A070939(floor(n/3)) - 1, for n <> 2. (End)
Rajasekaran, Shallit, & Smith show that this is an additive basis of order 4. - Charles R Greathouse IV, Nov 06 2018

Examples

			a(3) = 3, since 3 = 11_2 is the 3rd symmetric binary number;
a(6) = 9, since 9 = 1001_2 is the 6th symmetric binary number.
		

References

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

Crossrefs

See A057148 for the binary representations.
Cf. A178225, A005408, A164126, A154809 (complement).
Even numbers that are not the sum of two terms: A241491, A261678, A262556.
Cf. A145799.
Primes: A016041 and A117697.
Cf. A000051 (a subsequence).

Programs

  • Haskell
    a006995 n = a006995_list !! (n-1)
    a006995_list = 0 : filter ((== 1) . a178225) a005408_list
    -- Reinhard Zumkeller, Oct 21 2011
    
  • Magma
    [n: n in [0..850] | Intseq(n,2) eq Reverse(Intseq(n,2))];  // Bruno Berselli, Aug 29 2011
    
  • Maple
    dmax:= 15; # to get all terms with at most dmax binary digits
    revdigs:= proc(n)
      local L, Ln, i;
      L:= convert(n,base,2);
      Ln:= nops(L);
      add(L[i]*2^(Ln-i),i=1..Ln);
    end proc;
    A:= {0,1}:
    for d from 2 to dmax do
      if d::even then
        A:= A union {seq(2^(d/2)*x + revdigs(x),x=2^(d/2-1)..2^(d/2)-1)}
      else
        m:= (d-1)/2;
        B:={seq(2^(m+1)*x + revdigs(x),x=2^(m-1)..2^m-1)};
        A:= A union B union map(`+`,B,2^m)
      fi
    od:
    A;  # Robert Israel, Aug 17 2014
  • Mathematica
    palQ[n_Integer, base_Integer] := Module[{idn=IntegerDigits[n, base]}, idn==Reverse[idn]]; Select[Range[1000], palQ[ #, 2]&]
    Select[ Range[0, 1000], # == IntegerReverse[#, 2] &] (* Robert G. Wilson v, Feb 24 2018 *)
    Select[Range[0, 1000], PalindromeQ[IntegerDigits[#, 2]]&] (* Jean-François Alcover, Mar 01 2018 *)
  • PARI
    for(n=0,999,n-subst(Polrev(binary(n)),x,2)||print1(n,",")) \\ Thomas Buchholz, Aug 16 2014
    
  • PARI
    for(n=0,10^3, my(d=digits(n,2)); if(d==Vecrev(d), print1(n,", "))); \\ Joerg Arndt, Aug 17 2014
    
  • PARI
    is_A006995(n)=Vecrev(n=binary(n))==n \\ M. F. Hasler, Feb 23 2018
    
  • PARI
    A006995(n,m=logint(n,2),c=1<<(m-1),a,d)={if(n>=3*c,a=n-3*c;d=2*c^2,a=n-2*c;n%2*c+d=c^2)+sum(k=1,m-2^(n<3*c),if(bittest(a,m-1-k),1<>k))+(n>2)} \\ Based on Fischer's smalltalk program. - M. F. Hasler, Feb 23 2018
    
  • Python
    from itertools import count, islice, product
    def bin_pals(): # generator of binary palindromes in base 10
        yield from [0, 1]
        digits, midrange = 2, [[""], ["0", "1"]]
        for digits in count(2):
            for p in product("01", repeat=digits//2-1):
                left = "1"+"".join(p)
                for middle in midrange[digits%2]:
                    yield int(left + middle + left[::-1], 2)
    print(list(islice(bin_pals(), 58))) # Michael S. Branicky, Jan 09 2023
    
  • Python
    def A006995(n):
        if n == 1: return 0
        a = 1<<(l:=n.bit_length()-2)
        m = a|(n&a-1)
        return (m<Chai Wah Wu, Jun 10 2024
  • Sage
    def palgenbase2(): # generator of palindromes in base 2
        yield 0
        x, n, n2 = 1, 1, 2
        while True:
            for y in range(n,n2):
                s = format(y,'b')
                yield int(s+s[-2::-1],2)
            for y in range(n,n2):
                s = format(y,'b')
                yield int(s+s[::-1],2)
            x += 1
            n *= 2
            n2 *= 2 # Chai Wah Wu, Jan 07 2015
    
  • Sage
    [n for n in (0..843) if Word(n.digits(2)).is_palindrome()] # Peter Luschny, Sep 13 2018
    
  • Smalltalk
    A006995
    "Answer the n-th binary palindrome
    (nonrecursive implementation)"
    | m n a b c d k2 |
    n := self.
    n = 1 ifTrue: [^0].
    n = 2 ifTrue: [^1].
    m := n integerFloorLog: 2.
    c := 2 raisedToInteger: m - 1.
    n >= (3 * c)
      ifTrue:
       [a := n - (3 * c).
       d := 2 * c * c.
       b := d + 1.
       k2 := 1.
       1 to: m - 1
        do:
         [:k |
         k2 := 2 * k2.
         b := b + (a * k2 // c \\ 2 * (k2 + (d // k2)))]]
      ifFalse:
       [a := n - (2 * c).
       d := c * c.
       b := d + 1 + (n \\ 2 * c).
       k2 := 1.
       1 to: m - 2
        do:
         [:k |
         k2 := 2 * k2.
         b := b + (a * k2 // c \\ 2 * (k2 + (d // k2)))]].
    ^b // by Hieronymus Fischer, Feb 15 2013
    

Formula

A178225(a(n)) = 1; union of A048700 and A048701. - Reinhard Zumkeller, Oct 21 2011
From Hieronymus Fischer, Dec 31 2008, Jan 10 2012, Feb 18 2012: (Start)
Written as a decimal, a(10^n) has 2*n digits. For n > 1, the decimal expansion of a(10^n) starts with 22..., 23... or 24...:
a(1000) = 249903,
a(10^4) = 24183069,
a(10^5) = 2258634081,
a(10^6) = 249410097687,
a(10^7) = 24350854001805,
a(10^8) = 2229543293296319,
a(10^9) = 248640535848971067,
a(10^10)= 24502928886295666773.
Inequality: (2/9)*n^2 < a(n) < (1/4)*(n+1)^2, if n > 1.
lim sup_{n -> oo} a(n)/n^2 = 1/4, lim inf_{n -> oo} a(n)/n^2 = 2/9.
For n >= 2, a(2^n-1) = 2^(2n-2) - 1; a(2^n) = 2^(2n-2) + 1;
a(2^n+1) = 2^(2n-2) + 2^(n-1) + 1; a(2^n + 2^(n-1)) = 2^(2n-1) + 1.
Recursion for n > 2: a(n) = 2^(2k-q) + 1 + 2^p*a(m), where k = floor(log_2(n-1)), and p, q and m are determined as follows:
Case 1: If n = 2^(k+1), then p = 0, q = 0, m = 1;
Case 2: If 2^k < n < 2^k+2^(k-1), then p = k-floor(log_2(i))-1 with i = n-2^k, q = 2, m = 2^floor(log_2(i)) + i;
Case 3: If n = 2^k + 2^(k-1), then p = 0, q = 1, m = 1;
Case 4: If 2^k + 2^(k-1) < n < 2^(k+1), then p = k-floor(log_2(j))-1 with j = n-2^k-2^(k-1), q = 1, m = 2*2^floor(log_2(j))+j.
Non-recursive formula:
Let n >= 3, m = floor(log_2(n)), p = floor((3*2^(m-1)-1)/n), then
a(n) = 2^(2*m-1-p) + 1 + p*(1-(-1)^n)*2^(m-1-p) + sum_{k=1 .. m-1-p} (floor((n-(3-p)*2^(m-1))/2^(m-1-k)) mod 2)*(2^k+2^(2*m-1-p-k)). [Typo at the last exponent of the third sum term eliminated by the author, Sep 05 2018]
a(n) = 2^(2*m-2) + 1 + 2*floor((n-2^m)/2^(m-1)) + 2^(m-1)*floor((1/2)*min(n+1-2^m,2^(m-1)+1)) + 3*2^(m-1)*floor((1/2)*max(n+1-3*2^(m-1),0)) + 3*sum_{j=2 .. m-1} floor((n+2^(j-1)-2^m)/2^j)*2^(m-j). [Seems correct for n > 3. - The Editors]
Inversion formula: The index of any binary palindrome b = a(n) > 0 is n = palindromicIndex(b) = ((5-(-1)^m)/2 + Sum_{k=1..[m/2]} ([b/2^k] mod 2)/2^k)*2^[m/2], where [.] = floor(.) and m = [log_2(b)].
(End)
G.f.: g(x) = x^2 + 3x^3 + sum_{j=1..oo}( 3*2^j*(1-x^floor((j+1)/2))/(1-x)*x^((1/2)-floor((j+1)/2)) + f_j(x) - f_j(1/x))*x^(2*2^floor(j/2)+3*2^floor((j-1)/2)-(1/2)), where the f_j(x) are defined as follows:
f_1(x) = x^(1/2), and for j > 1,
f_j(x) = x^(1/2)*sum_{i=0..2^floor((j-1)/2)-1}((3+(1/2)*sum_{k=1..floor((j-1)/2)}(1-(-1)^floor(2i/2^k))*b(j,k))*x^i), where b(j,k) = 2^(floor((j-1)/2)-k)*((3+(-1)^j)*2^(2*k+1)+4) for k > 1, and b(j,1) = (2+(-1)^j)*2^(floor((j-1)/2)+1). - Hieronymus Fischer, Apr 04 2012
A044051(n) = (a(n)+1)/2 for n > 0. - Reinhard Zumkeller, Apr 20 2015
A145799(a(n)) = a(n). - Reinhard Zumkeller, Sep 24 2015
Sum_{n>=2} 1/a(n) = A244162. - Amiram Eldar, Oct 17 2020

Extensions

Edited and extended by Hieronymus Fischer, Feb 21 2012
Edited by M. F. Hasler, Feb 23 2018

A006370 The Collatz or 3x+1 map: a(n) = n/2 if n is even, 3n + 1 if n is odd.

Original entry on oeis.org

0, 4, 1, 10, 2, 16, 3, 22, 4, 28, 5, 34, 6, 40, 7, 46, 8, 52, 9, 58, 10, 64, 11, 70, 12, 76, 13, 82, 14, 88, 15, 94, 16, 100, 17, 106, 18, 112, 19, 118, 20, 124, 21, 130, 22, 136, 23, 142, 24, 148, 25, 154, 26, 160, 27, 166, 28, 172, 29, 178, 30, 184, 31, 190, 32, 196, 33
Offset: 0

Views

Author

Keywords

Comments

The 3x+1 or Collatz problem is as follows: start with any number n. If n is even, divide it by 2, otherwise multiply it by 3 and add 1. Do we always reach 1? This is an unsolved problem. It is conjectured that the answer is yes.
The Krasikov-Lagarias paper shows that at least N^0.84 of the positive numbers < N fall into the 4-2-1 cycle of the 3x+1 problem. This is far short of what we think is true, that all positive numbers fall into this cycle, but it is a step. - Richard C. Schroeppel, May 01 2002
Also A001477 and A016957 interleaved. - Omar E. Pol, Jan 16 2014, updated Nov 07 2017
a(n) is the image of a(2*n) under the 3*x+1 map. - L. Edson Jeffery, Aug 17 2014
The positions of powers of 2 in this sequence are given in A160967. - Federico Provvedi, Oct 06 2021
If displayed as a rectangular array with six columns, the columns are A008585, A350521, A016777, A082286, A016789, A350522 (see example). - Omar E. Pol, Jan 03 2022

Examples

			G.f. = 4*x + x^2 + 10*x^3 + 2*x^4 + 16*x^5 + 3*x^6 + 22*x^7 + 4*x^8 + 28*x^9 + ...
From _Omar E. Pol_, Jan 03 2022: (Start)
Written as a rectangular array with six columns read by rows the sequence begins:
   0,   4,  1,  10,  2,  16;
   3,  22,  4,  28,  5,  34;
   6,  40,  7,  46,  8,  52;
   9,  58, 10,  64, 11,  70;
  12,  76, 13,  82, 14,  88;
  15,  94, 16, 100, 17, 106;
  18, 112, 19, 118, 20, 124;
  21, 130, 22, 136, 23, 142;
  24, 148, 25, 154, 26, 160;
  27, 166, 28, 172, 29, 178;
  30, 184, 31, 190, 32, 196;
...
(End)
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, E16.
  • J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

A006577 gives number of steps to reach 1.
Column k=1 of A347270, n >= 1.

Programs

  • Haskell
    a006370 n | m /= 0    = 3 * n + 1
              | otherwise = n' where (n',m) = divMod n 2
    -- Reinhard Zumkeller, Oct 07 2011
    
  • Magma
    [(1/4)*(7*n+2-(-1)^n*(5*n+2)): n in [1..70]]; // Vincenzo Librandi, Dec 20 2016
  • Maple
    f := n-> if n mod 2 = 0 then n/2 else 3*n+1; fi;
    A006370:=(4+z+2*z**2)/(z-1)**2/(1+z)**2; # Simon Plouffe in his 1992 dissertation; uses offset 0
  • Mathematica
    f[n_]:=If[EvenQ[n],n/2,3n+1];Table[f[n],{n,50}] (* Geoffrey Critzer, Jun 29 2013 *)
    LinearRecurrence[{0,2,0,-1},{4,1,10,2},70] (* Harvey P. Dale, Jul 19 2016 *)
  • PARI
    for(n=1,100,print1((1/4)*(7*n+2-(-1)^n*(5*n+2)),","))
    
  • PARI
    A006370(n)=if(n%2,3*n+1,n/2) \\ Michael B. Porter, May 29 2010
    
  • Python
    def A006370(n):
        q, r = divmod(n, 2)
        return 3*n+1 if r else q # Chai Wah Wu, Jan 04 2015
    

Formula

G.f.: (4*x+x^2+2*x^3) / (1-x^2)^2.
a(n) = (1/4)*(7*n+2-(-1)^n*(5*n+2)). - Benoit Cloitre, May 12 2002
a(n) = ((n mod 2)*2 + 1)*n/(2 - (n mod 2)) + (n mod 2). - Reinhard Zumkeller, Sep 12 2002
a(n) = A014682(n+1) * A000034(n). - R. J. Mathar, Mar 09 2009
a(n) = a(a(2*n)) = -A001281(-n) for all n in Z. - Michael Somos, Nov 10 2016
E.g.f.: (2 + x)*sinh(x)/2 + 3*x*cosh(x). - Ilya Gutkovskiy, Dec 20 2016
From Federico Provvedi, Aug 17 2021: (Start)
Dirichlet g.f.: (1-2^(-s))*zeta(s) + (3-5*2^(-s))*zeta(s-1).
a(n) = ( a(n+2k) + a(n-2k) ) / 2, for every integer k. (End)
a(n) + a(n+1) = A047374(n+1). - Leo Ortega, Aug 22 2025

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Apr 27 2001
Zero prepended and new Name from N. J. A. Sloane at the suggestion of M. F. Hasler, Nov 06 2017
Previous Showing 41-50 of 852 results. Next