cp's OEIS Frontend

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

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A002623 Expansion of 1/((1-x)^4*(1+x)).

Original entry on oeis.org

1, 3, 7, 13, 22, 34, 50, 70, 95, 125, 161, 203, 252, 308, 372, 444, 525, 615, 715, 825, 946, 1078, 1222, 1378, 1547, 1729, 1925, 2135, 2360, 2600, 2856, 3128, 3417, 3723, 4047, 4389, 4750, 5130, 5530, 5950, 6391, 6853, 7337, 7843, 8372, 8924, 9500
Offset: 0

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Author

Keywords

Comments

Also a(n) is the number of nondegenerate triangles that can be made from rods of lengths 1 to n+1. - Alfred Bruckstein; corrected by Hans Rudolf Widmer, Nov 02 2023
Also number of circumscribable (or escrible) quadrilaterals that can be made from rods of length 1,2,3,4,...,n. - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr)
Also number of 2 X n binary matrices up to row and column permutation (see the link: Binary matrices up to row and column permutations). - Vladeta Jovovic
Also partial sum of alternate triangular numbers (1, 3, 1+6, 3+10, 1+6+15, 3+10+21, etc.); and also number of triangles pointing in opposite direction to largest triangle in triangular matchstick arrangement of side n+2 (cf. A002717, also the Larsen article). - Henry Bottomley, Aug 08 2000
Ordered union of A002412(n+1) and A016061(n+1). - Lekraj Beedassy, Oct 13 2003
Also Molien series for certain 4-D representation of cyclic group of order 2. - N. J. A. Sloane, Jun 12 2004
From Radu Grigore (radugrigore(AT)gmail.com), Jun 19 2004: (Start)
a(n) = floor( (n+2)*(n+4)*(2n+3) / 24 ). E.g., a(2) = floor(4*6*7/24) = 7 because there are 7 upside down triangles (6 of size one and 1 of size two) in the matchstick figure:
/\
/\/\
/\/\/\
/\/\/\/\
(End)
Number of non-congruent non-parallelogram trapezoids with positive integer sides (trapezints) and perimeter 2n+5. Also with perimeter 2n+8. - Michael Somos, May 12 2005
a(n) = A108561(n+4,n) for n > 0. - Reinhard Zumkeller, Jun 10 2005
Also number of nonisomorphic planes with n points and 2 lines. E.g., a(0)=1 because with no points, we just have two empty lines. a(1)=3 because the one point may belong to 0, 1 or 2 lines. a(2)=7 because there are 7 ways to determine which of 2 points belong to which of 2 lines, up to isomorphism, i.e., up to a bijection f on the sets of points and a bijection g on the sets of lines, such that A belongs to a iff f(A) belongs to g(a). - Bjorn Kjos-Hanssen (bjorn(AT)math.uconn.edu), Nov 10 2005
a(n-2) is the number of ways to pick two non-overlapping subwords of equal nonzero length from a word of length n. E.g., a(5-2)=a(3)=13 since the word 12345 of length 5 has the following subword pairs: 1,2; 1,3; 1,4; 1,5; 2,3; 2,4; 2,5; 3,4; 3,5; 4,5; 12,34; 12,45; 23,45. - Michael Somos, Oct 22 2006
Partial sums of A002620. - G.H.J. van Rees (vanrees(AT)cs.umanitoba.ca), Feb 16 2007
From Philippe LALLOUET (philip.lallouet(AT)orange.fr), Oct 19 2007: (Start)
Also number of squares of any size in a staircase of n steps built with unit squares:
||__
||__|
||__||
For a staircase of 3 steps 6 squares of size 1 and 1 square of size 2, hence c(3)=7.
Columns sums of:
1 3 6 10 15 21 28 ...
1 3 6 10 15 ...
1 3 6 ...
1 ...
---------------------
1 3 7 13 22 34 50 ...
(End)
a(n) = sum of row n+1 of triangle A134446. Also, binomial transform of [1, 2, 2, 0, 1, -2, 4, -8, 16, -32, ...]. - Gary W. Adamson, Oct 25 2007
Let b(n) be the number of 4-tuples (w,x,y,z) having all terms in {1,...,n} and 2w=x+y+z+n; then b(n+3) = a(n) for n >= 0. - Clark Kimberling, May 08 2012
a(n) is the number of 3-tuples (w,x,y) having all terms in {0,...,n} and w >= x+y and x <= y. - Clark Kimberling, Jun 04 2012
Also, number of unlabeled bipartite graphs with two left vertices and n right vertices. - Yavuz Oruc, Jan 14 2018
Also number of triples (x,y,z) with 0 < x <= y <= z <= n + 1, x + y > z. - Ralf Steiner, Feb 06 2020
Bisections A002412 and A016061: a(2*k) = k*(k+1)*(4*k-1)/3! and a(2*k+1) = (k+1)*(k+2)*(4*k+9)/3!, for k >= 0. See the Woolhouse link, II. Solution by Stephen Watson, p. 65, with index shifts. - Mo Li, Apr 02 2020
Also, Wiener index of the square of the path graph P_(n+2). - Allan Bickle, Aug 01 2020
Maximum Wiener index of all maximal 2-degenerate graphs with n+2 vertices. (A maximal 2-degenerate graph can be constructed from a 2-clique by iteratively adding a new 2-leaf (vertex of degree 2) adjacent to two existing vertices.) The extremal graphs are squares of paths, so the bound also applies to 2-trees and maximal outerplanar graphs. - Allan Bickle, Sep 15 2022

Examples

			G.f. = 1 + 3*x + 7*x^2 + 13*x^3 + 22*x^4 + 34*x^5 + 50*x^6 + 70*x^7 + 95*x^8 + ...
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 74, Problem 7.
  • P. Diaconis, R. L. Graham and B. Sturmfels, Primitive partition identities, in Combinatorics: Paul Erdős is Eighty, Vol. 2, Bolyai Soc. Math. Stud., 2, 1996, pp. 173-192.
  • H. Gupta, Partitions of j-partite numbers into twelve or a smaller number of parts. Collection of articles dedicated to Professor P. L. Bhatnagar on his sixtieth birthday. Math. Student 40 (1972), 401-441 (1974).
  • I. Siap, Linear codes over F_2 + u*F_2 and their complete weight enumerators, in Codes and Designs (Ohio State, May 18, 2000), pp. 259-271. De Gruyter, 2002.
  • 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. A002620 (first differences), A000292, A001752 (partial sums), A062109 (binomial transf.).
Bisections A002412, A016061.
Cf. also A002717 (a companion sequence), A002727, A006148, A057524, A134446, A014125, A122046, A122047.
The maximum Wiener index of all maximal k-degenerate graphs for k=1..6 are given in A000292, A002623 (this sequence), A014125, A122046, A122047, A175724, respectively.

Programs

  • Maple
    A002623 := n->(1/16)*(1+(-1)^n)+(n+1)/8+binomial(n+2,2)/4+binomial(n+3,3)/2;
    seq( ((2*n+3)*(n+2)*(n+1)/6-floor((n+2)/2))/4,n=1..47); # Lewis
    a := n -> ((-1)^n*3 + 45 + 68*n + 30*n^2 + 4*n^3) / 48:
    seq(a(n), n=0..46); # Peter Luschny, Jan 22 2018
  • Mathematica
    CoefficientList[Series[1/((1-x)^3(1-x^2)),{x,0,50}],x] (* or *) LinearRecurrence[{3,-2,-2,3,-1},{1,3,7,13,22},50] (* Harvey P. Dale, Jul 19 2011 *)
    Table[((2 n^3 + 15 n^2 + 34 n + 45 / 2 + (3/2) (-1)^n) / 24), {n, 0, 100}] (* Vincenzo Librandi, Jan 15 2018 *)
    a[ n_] := Floor[(n + 2)*(n + 4)*(2*n + 3)/24]; (* Michael Somos, Feb 19 2024 *)
  • PARI
    {a(n) = (8 + 34/3*n + 5*n^2 + 2/3*n^3) \ 8}; /* Michael Somos, Sep 04 1999 */
    
  • PARI
    x='x+O('x^50); Vec(1/((1 - x)^3 * (1 - x^2))) \\ Indranil Ghosh, Apr 04 2017
    
  • Python
    def A002623(n): return ((n+2)*(n+4)*((n<<1)+3)>>3)//3 # Chai Wah Wu, Mar 25 2024

Formula

a(n+1) = a(n) + {(k-1)*k if n=2*k} or {k*k if n=2*k+1}.
a(n)+a(n+1) = A000292(n+1).
a(n) = a(n-2) + A000217(n+1) = A002717(n+2) - A000292(n+1).
Also: a(n) = C(n+3, 3) - a(n-1) with a(0)=1. - Labos Elemer, Apr 26 2003
From Paul Barry, Jul 01 2003: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k)*C(k+3,3).
The signed version 1, -3, 7, ... has the formula:
a(n) = (4*n^3 + 30*n^2 + 68*n + 45)*(-1)^n/48 + 1/16.
This is the partial sums of the signed version of A000292. (End)
From Paul Barry, Jul 21 2003: (Start)
a(n) = Sum_{k=0..n} floor((k+2)^2/4).
a(n) = Sum_{k=0..n} Sum_{j=0..k} Sum_{i=0..j} (1+(-1)^i)/2. (End)
a(n) = a(n - 2) + (n*(n - 1))/2, with n>2, a(1)=0, a(2)=1; a(n) = (4*n^3+6*n^2-4*n+3*(-1)^n-3)/48, with offset 2. - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004 (formula simplified by Bruno Berselli, Aug 29 2013)
a(n) = ((2*n+3)*(n+2)*(n+1)/6-floor((n+2)/2))/4, with offset 1. - Jerry W. Lewis (JLewis(AT)wyeth.com), Mar 23 2005
a(n) = 2*a(n-1) - a(n-2) + 1 + floor(n/2). - Bjorn Kjos-Hanssen (bjorn(AT)math.uconn.edu), Nov 10 2005
A002620(n+3) = a(n+1) - a(n). - Michael Somos, Sep 04 1999
Euler transform of length 2 sequence [ 3, 1]. - Michael Somos, Sep 04 2006
a(n) = -a(-5-n) for all n in Z. - Michael Somos, Sep 04 2006
Let P(i,k) be the number of integer partitions of n into k parts, then with k=2 we have a(n) = sum_{m=1}^{n} sum_{i=k}^{m} P(i,k). For k=1 we get A000217 = triangular numbers. - Thomas Wieder, Feb 18 2007
a(n) = (n+(3+(-1)^n)/2)*(n+(7+(-1)^n)/2)*(2*n+5-2*(-1)^n)/24. - Philippe LALLOUET (philip.lallouet(AT)orange.fr), Oct 19 2007 (corrected by Bruno Berselli, Aug 30 2013)
From Johannes W. Meijer, May 20 2011: (Start)
a(n) = A006918(n+1) + A006918(n).
a(n) = A058187(n-2) + 2*A058187(n-1) + A058187(n). (End)
a(0)=1, a(1)=3, a(2)=7, a(3)=13, a(4)=22; for n > 4, a(n) = 3*a(n-1) - 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - a(n-5). - Harvey P. Dale, Jul 19 2011
a(n) = Sum_{i=0..n+2} floor(i/2)*ceiling(i/2). - Bruno Berselli, Aug 30 2013
a(n) = 15/16 + (1/16)*(-1)^n + (17/12)*n + (5/8)*n^2 + (1/12)*n^3. - Robert Israel, Jul 07 2014
a(n) = Sum_{i=0..n+2} (n+1-i)*floor(i/2+1). - Bruno Berselli, Apr 04 2017
a(n) = 1 + floor((2*n^3 + 15*n^2 + 34*n) / 24). - Allan Bickle, Aug 01 2020
E.g.f.: ((24 + 51*x + 21*x^2 + 2*x^3)*cosh(x) + (21 + 51*x + 21*x^2 + 2*x^3)*sinh(x))/24. - Stefano Spezia, Jun 02 2021

A002817 Doubly triangular numbers: a(n) = n*(n+1)*(n^2+n+2)/8.

Original entry on oeis.org

0, 1, 6, 21, 55, 120, 231, 406, 666, 1035, 1540, 2211, 3081, 4186, 5565, 7260, 9316, 11781, 14706, 18145, 22155, 26796, 32131, 38226, 45150, 52975, 61776, 71631, 82621, 94830, 108345, 123256, 139656, 157641, 177310, 198765, 222111, 247456, 274911, 304590
Offset: 0

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Comments

Number of inequivalent ways to color vertices of a square using <= n colors, allowing rotations and reflections. Group is dihedral group D_8 of order 8 with cycle index (1/8)*(x1^4 + 2*x4 + 3*x2^2 + 2*x1^2*x2); setting all x_i = n gives the formula a(n) = (1/8)*(n^4 + 2*n + 3*n^2 + 2*n^3).
Number of semi-magic 3 X 3 squares with a line sum of n-1. That is, 3 X 3 matrices of nonnegative integers such that row sums and column sums are all equal to n-1. - [Gupta, 1968, page 653; Bell, 1970, page 279]. - Peter Bertok (peter(AT)bertok.com), Jan 12 2002. See A005045 for another version.
Also the coefficient h_2 of x^{n-3} in the shelling polynomial h(x)=h_0*x^n-1 + h_1*x^n-2 + h_2*x^n-3 + ... + h_n-1 for the independence complex of the cycle matroid of the complete graph K_n on n vertices (n>=2) - Woong Kook (andrewk(AT)math.uri.edu), Nov 01 2006
If X is an n-set and Y a fixed 3-subset of X then a(n-4) is equal to the number of 5-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
Starting with offset 1 = binomial transform of [1, 5, 10, 9, 3, 0, 0, 0, ...]. - Gary W. Adamson, Aug 05 2009
Starting with "1" = row sums of triangle A178238. - Gary W. Adamson, May 23 2010
The equation n*(n+1)*(n^2 + n + 2)/8 may be arrived at by solving for x in the following equality: (n^2+n)/2 = (sqrt(8x+1)-1)/2. - William A. Tedeschi, Aug 18 2010
Partial sums of A006003. - Jeremy Gardiner, Jun 23 2013
Doubly triangular numbers are revealed in the sums of row sums of Floyd's triangle.
1, 1+5, 1+5+15, ...
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
- Tony Foster III, Nov 14 2015
From Jaroslav Krizek, Mar 04 2017: (Start)
For n>=1; a(n) = sum of the different sums of elements of all the nonempty subsets of the sets of numbers from 1 to n.
Example: for n = 6; nonempty subsets of the set of numbers from 1 to 3: {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}; sums of elements of these subsets: 1, 2, 3, 3, 4, 5, 6; different sums of elements of these subsets: 1, 2, 3, 4, 5, 6; a(3) = (1+2+3+4+5+6) = 21, ... (End)
a(n) is also the number of 4-cycles in the (n+4)-path complement graph. - Eric W. Weisstein, Apr 11 2018

Examples

			G.f. = x + 6*x^2 + 21*x^3 + 55*x^4 + 120*x^5 + 231*x^6 + 406*x^7 + 666*x^8 + ...
		

References

  • A. Björner, The homology and shellability of matroids and geometric lattices, in Matroid Applications (ed. N. White), Encyclopedia of Mathematics and Its Applications, 40, Cambridge Univ. Press 1992.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 124, #25, Q(3,r).
  • 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).
  • R. P. Stanley, Enumerative Combinatorics I, p. 292.

Crossrefs

Cf. A006003 (first differences), A165211 (mod 2).
Multiple triangular: A000217, A064322, A066370.
Cf. A006528 (square colorings).
Cf. A236770 (see crossrefs).
Row n=3 of A257493 and row n=2 of A331436 and A343097.
Cf. A000332.
Cf. A000292 (3-cycle count of \bar P_{n+4}), A060446 (5-cycle count of \bar P_{n+3}), A302695 (6-cycle count of \bar P_{n+5}).

Programs

  • Maple
    A002817 := n->n*(n+1)*(n^2+n+2)/8;
  • Mathematica
    a[ n_] := n (n + 1) (n^2 + n + 2) / 8; (* Michael Somos, Jul 24 2002 *)
    LinearRecurrence[{5,-10,10,-5,1}, {0,1,6,21,55},40] (* Harvey P. Dale, Jul 18 2011 *)
    nn=50;Join[{0},With[{c=(n(n+1))/2},Flatten[Table[Take[Accumulate[Range[ (nn(nn+1))/2]], {c,c}],{n,nn}]]]] (* Harvey P. Dale, Mar 19 2013 *)
  • PARI
    {a(n) = n * (n+1) * (n^2 + n + 2) / 8}; /* Michael Somos, Jul 24 2002 */
    
  • PARI
    concat(0, Vec(x*(1+x+x^2)/(1-x)^5 + O(x^50))) \\ Altug Alkan, Nov 15 2015
    
  • Python
    def A002817(n): return (m:=n*(n+1))*(m+2)>>3 # Chai Wah Wu, Aug 30 2024

Formula

a(n) = 3*binomial(n+2, 4) + binomial(n+1, 2).
G.f.: x*(1 + x + x^2)/(1-x)^5. - Simon Plouffe (in his 1992 dissertation); edited by N. J. A. Sloane, May 13 2008
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 3. - Warut Roonguthai, Dec 13 1999
a(n) = 5a(n-1) - 10a(n-2) + 10a(n-3) - 5a(n-4) + a(n-5) = A000217(A000217(n)). - Ant King, Nov 18 2010
a(n) = Sum(Sum(1 + Sum(3*n))). - Xavier Acloque, Jan 21 2003
a(n) = A000332(n+1) + A000332(n+2) + A000332(n+3), with A000332(n) = binomial(n, 4). - Mitch Harris, Oct 17 2006 and Bruce J. Nicholson, Oct 22 2017
a(n) = Sum_{i=1..C(n,2)} i = C(C(n,2) + 1, 2) = A000217(A000217(n+1)). - Enrique Pérez Herrero, Jun 11 2012
Euler transform of length 3 sequence [6, 0, -1]. - Michael Somos, Nov 19 2015
E.g.f.: x*(8 + 16*x + 8*x^2 + x^3)*exp(x)/8. - Ilya Gutkovskiy, Apr 26 2016
Sum_{n>=1} 1/a(n) = 6 - 4*Pi*tanh(sqrt(7)*Pi/2)/sqrt(7) = 1.25269064911978447... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(n)*A000124(n)/2.
a(n) = ((n-1)^4 + 3*(n-1)^3 + 2*(n-1)^2 + 2*n))/8. - Bruce J. Nicholson, Apr 05 2017
a(n) = (A016754(n)+ A007204(n)- 2) / 32. - Bruce J. Nicholson, Apr 14 2017
a(n) = a(-1-n) for all n in Z. - Michael Somos, Apr 17 2017
a(n) = T(T(n)) where T are the triangular numbers A000217. - Albert Renshaw, Jan 05 2020
a(n) = 2*n^2 - n + 6*binomial(n, 3) + 3*binomial(n, 4). - Ryan Jean, Mar 20 2021
a(n) = (A008514(n) - 1)/16. - Charlie Marion, Dec 20 2024

Extensions

More terms from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999

A005232 Expansion of (1-x+x^2)/((1-x)^2*(1-x^2)*(1-x^4)).

Original entry on oeis.org

1, 1, 3, 4, 8, 10, 16, 20, 29, 35, 47, 56, 72, 84, 104, 120, 145, 165, 195, 220, 256, 286, 328, 364, 413, 455, 511, 560, 624, 680, 752, 816, 897, 969, 1059, 1140, 1240, 1330, 1440, 1540, 1661, 1771, 1903, 2024, 2168, 2300, 2456, 2600, 2769, 2925, 3107, 3276
Offset: 0

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Author

Keywords

Comments

Also number of n X 2 binary matrices under row and column permutations and column complementations (if offset is 0).
Also Molien series for certain 4-D representation of dihedral group of order 8.
With offset 4, number of bracelets (turnover necklaces) of n-bead of 2 colors with 4 red beads. - Washington Bomfim, Aug 27 2008
From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of 4 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in case k=4 (see our comment to A032279). (End)
Number of 2 X 2 matrices with nonnegative integer values totaling n under row and column permutations. - Gabriel Burns, Nov 08 2016
From Petros Hadjicostas, Jan 12 2019: (Start)
By "necklace", Vladimir Shevelev (above) means "turnover necklace", i.e., a bracelet. Zagaglia Salvi (1999) also uses this terminology: she calls a bracelet "necklace" and a necklace "cycle".
According to Cyvin et al. (1997), the sequence (a(n): n >= 0) consists of "the total numbers of isomers of polycyclic conjugated hydrocarbons with q + 1 rings and q internal carbons in one ring (class Q_q)", where q = 4 and n is the hydrogen content (i.e., we count certain isomers of C_{n+2*q} H_n with q = 4 and n >= 0). (End)

Examples

			G.f. = 1 + x + 3*x^2 + 4*x^3 + 8*x^4 + 10*x^5 + 16*x^6 + 20*x^7 + 29*x^8 + ...
There are 8 4 X 2 matrices up to row and column permutations and column complementations:
  [1 1] [1 0] [1 0] [0 1] [0 1] [0 1] [0 1] [0 0]
  [1 1] [1 1] [1 0] [1 0] [1 0] [1 0] [0 1] [0 1]
  [1 1] [1 1] [1 1] [1 1] [1 0] [1 0] [1 0] [1 0]
  [1 1] [1 1] [1 1] [1 1] [1 1] [1 0] [1 0] [1 1].
There are 8 2 X 2 matrices of nonnegative integers totaling 4 up to row and column permutations:
  [4 0] [3 1] [2 2] [2 1] [2 1] [3 0] [2 0] [1 1]
  [0 0] [0 0] [0 0] [0 1] [1 0] [1 0] [2 0] [1 1].
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Row n=2 of A343875.
Column k=4 of A052307.

Programs

  • Maple
    A005232:=-(-1-z-2*z**3+2*z**2+z**7-2*z**6+2*z**4)/(z**2+1)/(1+z)**2/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence apart from an initial 1
  • Mathematica
    k = 4; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    CoefficientList[ Series[(1 - x + x^2)/((1 - x)^2(1 - x^2)(1 - x^4)), {x, 0, 51}], x] (* Robert G. Wilson v, Mar 29 2006 *)
    LinearRecurrence[{2,0,-2,2,-2,0,2,-1},{1,1,3,4,8,10,16,20},60] (* Harvey P. Dale, Oct 24 2012 *)
    k=4 (* Number of red beads in bracelet problem *); CoefficientList[Series[(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)
  • PARI
    {a(n) = (n^3 + 9*n^2 + (32-9*(n%2))*n + [48, 15, 36, 15][n%4+1]) / 48}; \\ Michael Somos, Feb 01 2007
    
  • PARI
    {a(n) = my(s=1); if( n<-5, n = -6-n; s=-1); if( n<0, 0, s * polcoeff( (1 - x + x^2) / ((1 - x)^2 * (1 - x^2) * (1 - x^4)) + x * O(x^n), n))}; \\ Michael Somos, Feb 01 2007
    
  • PARI
    a(n) = round((n^3 +9*n^2 +(32-9*(n%2))*n)/48 +0.6) \\ Washington Bomfim, Jul 17 2008
    
  • PARI
    a(n) = ceil((n+1)*(2*n^2+16*n+39+9*(-1)^n)/96) \\ Tani Akinari, Aug 23 2013
    
  • Python
    a=lambda n: sum((k//2+1)*((n-k)//2+1) for k in range((n-1)//2+1))+(n+1)%2*(((n//4+1)*(n//4+2))//2)  # Gabriel Burns, Nov 08 2016

Formula

G.f.: (1+x^3)/((1-x)*(1-x^2)^2*(1-x^4)).
G.f.: (1/8)*(1/(1-x)^4+3/(1-x^2)^2+2/(1-x)^2/(1-x^2)+2/(1-x^4)). - Vladeta Jovovic, Aug 05 2000
Euler transform of length 6 sequence [ 1, 2, 1, 1, 0, -1 ]. - Michael Somos, Feb 01 2007
a(2n+1) = A006918(2n+2)/2;
a(2n) = (A006918(2n+1) + A008619(n))/2.
a(n) = -a(-6 - n) for all n in Z. - Michael Somos, Feb 05 2011
From Vladimir Shevelev, Apr 22 2011: (Start)
if n == 0 (mod 4), then a(n) = n*(n^2-3*n+8)/48;
if n == 1, 3 (mod 4), then a(n) = (n^2-1)*(n-3)/48;
if n == 2 (mod 4), then a(n) = (n-2)*(n^2-n+6)/48. (End)
a(n) = 2*a(n-1) - 2*a(n-3) + 2*a(n-4) - 2*a(n-5) + 2*a(n-7) - a(n-8) with a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 4, a(4) = 8, a(5) = 10, a(6) = 16, a(7) = 20. - Harvey P. Dale, Oct 24 2012
a(n) = ((n+3)*(2*n^2+12*n+19+9*(-1)^n) + 6*(-1)^((2*n-1+(-1)^n)/4)*(1+(-1)^n))/96. - Luce ETIENNE, Mar 16 2015
a(n) = |A128498(n)| + |A128498(n-3)|. - R. J. Mathar, Jun 11 2019

Extensions

Sequence extended by Christian G. Bower
Showing 1-3 of 3 results.