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.

Showing 1-10 of 14 results. Next

A005380 Expansion of 1 / Product_{k>=1} (1-x^k)^(k+1).

Original entry on oeis.org

1, 2, 6, 14, 33, 70, 149, 298, 591, 1132, 2139, 3948, 7199, 12894, 22836, 39894, 68982, 117948, 199852, 335426, 558429, 922112, 1511610, 2460208, 3977963, 6390942, 10206862, 16207444, 25596941, 40214896, 62868772, 97814358
Offset: 0

Views

Author

Keywords

Comments

Also, a(n) = number of partitions of the integer n where there are k+1 different kinds of part k for k = 1, 2, 3, ....
Also, a(n) = number of partitions of n objects of 2 colors. These are set partitions, the n objects are not labeled but colored, using two colors. For each subset of size k there are k+1 different possibilities, i=0..k white and k-i black objects.
Also, a(n) = number of simple unlabeled graphs with n nodes of 2 colors whose components are complete graphs. - Geoffrey Critzer, Sep 27 2012

Examples

			We represent each summand, k, in a partition of n as k identical objects. Then we color each object. We have no regard for the order of the colored objects.
a(3) = 14 because we have:  www; wwb; wbb; bbb; ww + w; ww + b;  wb + w; wb + b; bb + w; bb + b; w + w + w; w + w + b; w + b + b; b + b + b, where the 2 colors are black b and white w. - _Geoffrey Critzer_, Sep 27 2012
a(3) = 14 because we have:  3; 3'; 3''; 3'''; 2 + 1; 2 + 1';  2' + 1; 2' + 1'; 2'' + 1; 2'' + 1'; 1 + 1 + 1; 1 + 1 + 1'; 1 + 1' + 1'; 1' + 1' + 1', where a part k of different sorts is given as k, k', k'', etc. - _Joerg Arndt_, Mar 09 2015
From _Alois P. Heinz_, Mar 09 2015: (Start)
The a(4) = 33 = 5 + 9 + 6 + 8 + 5 partitions of 4 objects of 2 colors are:
5 partitions for the integer partition of 4 = 1 + 1 + 1 + 1:
  01: {{b}, {b}, {b}, {b}}
  02: {{b}, {b}, {b}, {w}}
  03: {{b}, {b}, {w}, {w}}
  04: {{b}, {w}, {w}, {w}}
  05: {{w}, {w}, {w}, {w}}
9 partitions for the integer partition of 4 = 1 + 1 + 2:
  06: {{b}, {b}, {b,b}}
  07: {{b}, {w}, {b,b}}
  08: {{w}, {w}, {b,b}}
  09: {{b}, {b}, {w,b}}
  10: {{b}, {w}, {w,b}}
  11: {{w}, {w}, {w,b}}
  12: {{b}, {b}, {w,w}}
  13: {{b}, {w}, {w,w}}
  14: {{w}, {w}, {w,w}}
6 partitions for the integer partition of 4 = 2 + 2:
  15: {{b,b}, {b,b}}
  16: {{b,b}, {w,b}}
  17: {{b,b}, {w,w}}
  18: {{w,b}, {w,b}}
  19: {{w,b}, {w,w}}
  20: {{w,w}, {w,w}}
8 partitions for the integer partition of 4 = 1 + 3:
  21: {{b}, {b,b,b}}
  22: {{w}, {b,b,b}}
  23: {{b}, {w,b,b}}
  24: {{w}, {w,b,b}}
  25: {{b}, {w,w,b}}
  26: {{w}, {w,w,b}}
  27: {{b}, {w,w,w}}
  28: {{w}, {w,w,w}}
5 partitions for the integer partition of 4 = 4:
  29: {{b,b,b,b}}
  30: {{w,b,b,b}}
  31: {{w,w,b,b}}
  32: {{w,w,w,b}}
  33: {{w,w,w,w}}
Some see number partitions, others see set partitions, ...
(End)
It is obvious from the example of _Alois P. Heinz_ that a(n) enumerates multi-set partitions of a multi-set of n elements of two kinds. In the case that there is only one kind, this reduces to the usual case of numerical partitions. In the case that all the n elements are distinct, then this reduces to the case of set partitions. - _Michael Somos_, Mar 09 2015
There are a(3) = 14 plane partitions of 6 with trace 3; of 7 with trace 4; of 8 with trace 5; etc. See a formula above with the Stanley Exercise 7.99. - _Wolfdieter Lang_, Mar 09 2015
From _Daniel Forgues_, Mar 09 2015: (Start)
The a(3) = 14 = 4 + 6 + 4 partitions of 3 objects of 2 colors are:
4 partitions for the integer partition of 3 = 1 + 1 + 1:
  01: {{b}, {b}, {b}}
  02: {{b}, {b}, {w}}
  03: {{b}, {w}, {w}}
  04: {{w}, {w}, {w}}
6 partitions for the integer partition of 3 = 1 + 2:
  05: {{b}, {b,b}}
  06: {{w}, {b,b}}
  07: {{b}, {w,b}}
  08: {{w}, {w,b}}
  09: {{b}, {w,w}}
  10: {{w}, {w,w}}
4 partitions for the integer partition of 3 = 3:
  11: {{b,b,b}}
  12: {{w,b,b}}
  13: {{w,w,b}}
  14: {{w,w,w}}
The a(2) = 6 = 3 + 3 partitions of 2 objects of 2 colors are:
3 partitions for the integer partition of 2 = 1 + 1:
  01: {{b}, {b}}
  02: {{b}, {w}}
  03: {{w}, {w}}
3 partitions for the integer partition of 2 = 2:
  04: {{b,b}}
  05: {{w,b}}
  06: {{w,w}}
The a(1) = 2 partitions of 1 object of 2 colors are:
2 partitions for the integer partition of 1 = 1:
  01: {{b}}
  02: {{w}}
a(0) = 1: the empty partition, since empty sum is 0.
Triangle(sort of, since n_th row has p(n) = A000041 terms):
  1:  2
  2:  3, 3
  3:  4, 6, 4
  4:  5, 9, 6, 8, 5
  5:  6, ?, ?, ?, ?, ?, 6
  6:  7, ?, ?, ?, ?, ?, ?, ?, ?, ?, 7
Can we find a recurrence relation? (End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Exercise 7.99, p. 484 and pp. 548-549.

Crossrefs

Row sums of A054225.
Column k=2 of A075196.

Programs

  • Maple
    mul( (1-x^i)^(-i-1),i=1..80); series(%,x,80); seriestolist(%);
    # second Maple program:
    with(numtheory): etr:= proc(p) local b; b:=proc(n) option remember; local d,j; if n=0 then 1 else add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n)/n fi end end: a:=etr(n-> n+1): seq(a(n), n=0..40); # Alois P. Heinz, Sep 08 2008
  • Mathematica
    max = 31; f[x_] = Product[ 1/(1-x^k)^(k+1), {k, 1, max}]; CoefficientList[ Series[ f[x], {x, 0, max}], x] (* Jean-François Alcover, Nov 08 2011, after g.f. *)
    etr[p_] := Module[{b}, b[n_] := b[n] = Module[{d, j}, If[n==0, 1, Sum[ Sum[ d*p[d], {d, Divisors[j]}]*b[n-j], {j, 1, n}]/n]]; b]; a = etr[#+1&]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 23 2015, after Alois P. Heinz *)
  • PARI
    a(n)=polcoeff(prod(i=1,n,(1-x^i+x*O(x^n))^-(i+1)),n)

Formula

EULER transform of b(n) = n+1.
a(n) ~ Zeta(3)^(13/36) * exp(1/12 - Pi^4/(432*Zeta(3)) + Pi^2 * n^(1/3) / (3*2^(4/3)*Zeta(3)^(1/3)) + 3*Zeta(3)^(1/3) * n^(2/3) / 2^(2/3)) / (A * 2^(23/36) * 3^(1/2) * Pi * n^(31/36)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant and Zeta(3) = A002117 = 1.202056903... . - Vaclav Kotesovec, Mar 07 2015
a(n) = A089353(n+m, m), n >= 1, for each m >= n. a(0) =1. See the Stanley reference, Exercise 7.99. - Wolfdieter Lang, Mar 09 2015
G.f.: exp(Sum_{k>=1} (sigma_1(k) + sigma_2(k))*x^k/k). - Ilya Gutkovskiy, Aug 11 2018

Extensions

Edited by Christian G. Bower, Sep 07 2002
New name from Joerg Arndt, Mar 09 2015
Restored 1995 name. - N. J. A. Sloane, Mar 09 2015

A255835 G.f.: Product_{k>=1} (1+x^k)^(2*k-1).

Original entry on oeis.org

1, 1, 3, 8, 15, 34, 67, 133, 255, 486, 901, 1649, 2984, 5312, 9373, 16342, 28221, 48283, 81928, 137858, 230278, 381919, 629156, 1029933, 1675856, 2711288, 4362575, 6983196, 11122327, 17630798, 27820283, 43706461, 68375137, 106534093, 165340844, 255643289
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 07 2015

Keywords

Crossrefs

Programs

  • Mathematica
    nmax=50; CoefficientList[Series[Product[(1+x^k)^(2*k-1),{k,1,nmax}],{x,0,nmax}],x]

Formula

a(n) ~ Zeta(3)^(1/6) * exp(-Pi^4 / (2592*Zeta(3)) - Pi^2 * n^(1/3) / (12*(3*Zeta(3))^(1/3)) + 3^(4/3)/2 * Zeta(3)^(1/3) * n^(2/3)) / (2^(1/6) * 3^(1/3) * sqrt(Pi) * n^(2/3)), where Zeta(3) = A002117.
G.f.: exp(Sum_{k>=1} (-1)^(k+1)*x^k*(1 + x^k)/(k*(1 - x^k)^2)). - Ilya Gutkovskiy, Jun 07 2018

A054242 Triangle read by rows: row n (n>=0) gives the number of partitions of (n,0), (n-1,1), (n-2,2), ..., (0,n) respectively into sums of distinct pairs.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 3, 2, 2, 5, 5, 5, 2, 3, 7, 9, 9, 7, 3, 4, 10, 14, 17, 14, 10, 4, 5, 14, 21, 27, 27, 21, 14, 5, 6, 19, 31, 42, 46, 42, 31, 19, 6, 8, 25, 44, 64, 74, 74, 64, 44, 25, 8, 10, 33, 61, 93, 116, 123, 116, 93, 61, 33, 10
Offset: 0

Views

Author

Marc LeBrun, Feb 08 2000 and Jul 01 2003

Keywords

Comments

By analogy with ordinary partitions into distinct parts (A000009). The empty partition gives T(0,0)=1 by definition. A054225 and A201376 give pair partitions with repeats allowed.
Also number of partitions into pairs which are not both even.
In the paper by S. M. Luthra: "Partitions of bipartite numbers when the summands are unequal", the square table on page 370 contains an errors. In the formula (6, p. 372) for fixed m there should be factor 1/m!. The correct asymptotic formula is q(m, n) ~ (sqrt(12*n)/Pi)^m * exp(Pi*sqrt(n/3)) / (4*3^(1/4)*m!*n^(3/4)). The same error is also in article by F. C. Auluck (see A054225). - Vaclav Kotesovec, Feb 02 2016

Examples

			The second row (n=1) is 1,1 since (1,0) and (0,1) each have a single partition.
The third row (n=2) is 1, 2, 1 from (2,0), (1,1) or (1,0)+(0,1), (0,2).
In the fourth row, T(1,3)=5 from (1,3), (0,3)+(1,0), (0,2)+(1,1), (0,2)+(0,1)+(1,0), (0,1)+(1,2).
The triangle begins:
  1;
  1,  1;
  1,  2,  1;
  2,  3,  3,  2;
  2,  5,  5,  5,  2;
  3,  7,  9,  9,  7,  3;
  4, 10, 14, 17, 14, 10,  4;
  5, 14, 21, 27, 27, 21, 14,  5;
  6, 19, 31, 42, 46, 42, 31, 19,  6;
  8, 25, 44, 64, 74, 74, 64, 44, 25, 8;
  ...
		

Crossrefs

See A201377 for the same triangle formatted in a different way.
The outer diagonals are T(n,0) = T(n,n) = A000009(n).
Cf. A054225.
T(2*n,n) = A219554(n). Row sums give A219555. - Alois P. Heinz, Nov 22 2012

Programs

  • Haskell
    see Zumkeller link.
  • Mathematica
    max = 10; f[x_, y_] := Product[1 + x^n*y^k, {n, 0, max}, {k, 0, max}]/2; se = Series[f[x, y], {x, 0, max}, {y, 0, max}] ; coes = CoefficientList[ se, {x, y}]; t[n_, k_] := coes[[n-k+1, k+1]]; Flatten[ Table[ t[n, k], {n, 0, max}, {k, 0, n}]] (* Jean-François Alcover, Dec 06 2011 *)

Formula

G.f.: (1/2)*Product(1+x^i*y^j), i, j>=0.

Extensions

Entry revised by N. J. A. Sloane, Nov 30 2011, to incorporate corrections provided by Reinhard Zumkeller, who also contributed the alternative version A201377.

A052812 A simple grammar: power set of pairs of sequences.

Original entry on oeis.org

1, 0, 1, 2, 3, 6, 9, 16, 24, 42, 63, 102, 157, 244, 373, 570, 858, 1290, 1930, 2858, 4228, 6208, 9084, 13216, 19175, 27666, 39804, 57020, 81412, 115820, 164264, 232178, 327220, 459796, 644232, 900214, 1254554, 1743896, 2418071, 3344896, 4616026
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Number of partitions of n objects of two colors into distinct parts, where each part must contain at least one of each color. - Franklin T. Adams-Watters, Dec 28 2006

Crossrefs

Programs

  • Maple
    spec := [S,{B=Sequence(Z,1 <= card),C=Prod(B,B),S= PowerSet(C)},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    nmax=50; CoefficientList[Series[Product[(1+x^k)^(k-1),{k,1,nmax}],{x,0,nmax}],x] (* Vaclav Kotesovec, Mar 07 2015 *)

Formula

G.f.: exp(Sum((-1)^(j[1]+1)*(x^j[1])^2/(x^j[1]-1)^2/j[1], j[1]=1 .. infinity))
G.f.: Product_{k>=1} (1+x^k)^(k-1). - Vladeta Jovovic, Sep 17 2002
Weigh transform of b(n) = n-1. - Franklin T. Adams-Watters, Dec 28 2006
a(n) ~ Zeta(3)^(1/6) * exp(-Pi^4/(1296*Zeta(3)) - Pi^2 * n^(1/3) / (3^(4/3) * 2^(5/3) * Zeta(3)^(1/3)) + (3/2)^(4/3) * Zeta(3)^(1/3) * n^(2/3)) / (2^(1/4) * 3^(1/3) * n^(2/3) * sqrt(Pi)), where Zeta(3) = A002117. - Vaclav Kotesovec, Mar 07 2015

Extensions

More terms from Vladeta Jovovic, Sep 17 2002

A255834 G.f.: Product_{k>=1} (1+x^k)^(2*k+1).

Original entry on oeis.org

1, 3, 8, 23, 55, 129, 291, 627, 1317, 2697, 5398, 10589, 20421, 38743, 72452, 133724, 243792, 439496, 784070, 1385195, 2424971, 4209094, 7247141, 12383496, 21008559, 35398548, 59259781, 98595110, 163077878, 268221706, 438791204, 714142139, 1156552537
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 07 2015

Keywords

Crossrefs

Programs

  • Mathematica
    nmax=50; CoefficientList[Series[Product[(1+x^k)^(2*k+1),{k,1,nmax}],{x,0,nmax}],x]

Formula

a(n) ~ Zeta(3)^(1/6) * exp(-Pi^4 / (2592*Zeta(3)) + Pi^2 * n^(1/3) / (12*(3*Zeta(3))^(1/3)) + 3^(4/3)/2*Zeta(3)^(1/3) * n^(2/3)) / (2^(7/6)* 3^(1/3) * sqrt(Pi) * n^(2/3)), where Zeta(3) = A002117.

A255836 G.f.: Product_{k>=1} (1+x^k)^(3*k+1).

Original entry on oeis.org

1, 4, 13, 42, 117, 310, 785, 1896, 4433, 10062, 22248, 48080, 101821, 211682, 432795, 871520, 1730491, 3391894, 6568996, 12580316, 23841774, 44742634, 83193865, 153347110, 280336704, 508499474, 915540681, 1636805438, 2906642396, 5128530946, 8993376689
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 07 2015

Keywords

Crossrefs

Programs

  • Mathematica
    nmax=50; CoefficientList[Series[Product[(1+x^k)^(3*k+1),{k,1,nmax}],{x,0,nmax}],x]

Formula

a(n) ~ Zeta(3)^(1/6) * exp(-Pi^4 / (3888*Zeta(3)) + Pi^2 * n^(1/3) / (6^(5/3) * Zeta(3)^(1/3)) + 3^(5/3)/2^(4/3) * Zeta(3)^(1/3) * n^(2/3)) / (2^(17/12) * 3^(1/6) * sqrt(Pi) * n^(2/3)), where Zeta(3) = A002117.

A261451 Expansion of Product_{k>=1} ((1+x^k)/(1-x^k))^(k+1).

Original entry on oeis.org

1, 4, 14, 44, 124, 328, 824, 1980, 4590, 10320, 22584, 48268, 101016, 207432, 418704, 832032, 1629764, 3150280, 6014998, 11354084, 21204488, 39206168, 71811256, 130369900, 234704360, 419195412, 743085912, 1307823672, 2286094704, 3970174648, 6852048368
Offset: 0

Views

Author

Vaclav Kotesovec, Aug 19 2015

Keywords

Comments

Convolution of A005380 and A219555.

Crossrefs

Programs

  • Mathematica
    nmax = 40; CoefficientList[Series[Product[((1+x^k)/(1-x^k))^(k+1), {k, 1, nmax}], {x, 0, nmax}], x]

Formula

a(n) ~ (7*Zeta(3))^(13/36) * exp(1/12 - Pi^4/(336*Zeta(3)) + Pi^2 * n^(1/3) / (2^(5/3) * (7*Zeta(3))^(1/3)) + 3/2 * ((7*Zeta(3))/2)^(1/3) * n^(2/3)) / (A * 2^(35/18) * 3^(1/2) * Pi * n^(31/36)), where Zeta(3) = A002117 and A = A074962 is the Glaisher-Kinkelin constant.

A338645 Expansion of Product_{k>=1} (1 + x^k)^binomial(k+2,2).

Original entry on oeis.org

1, 3, 9, 29, 78, 207, 526, 1284, 3054, 7084, 16071, 35748, 78167, 168195, 356754, 746772, 1544145, 3157056, 6387114, 12795366, 25397760, 49977262, 97542936, 188912466, 363196750, 693424803, 1315161528, 2478648920, 4643337213, 8648452782, 16019345259, 29515269060, 54104712129
Offset: 0

Views

Author

Ilya Gutkovskiy, May 09 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 32; CoefficientList[Series[Product[(1 + x^k)^Binomial[k + 2, 2], {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d Binomial[d + 2, 2], {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 32}]

Formula

a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} ( Sum_{d|k} (-1)^(k/d+1) * A027480(d) ) * a(n-k).
a(n) ~ (7/15)^(1/8) * 2^(-21/8) * n^(-5/8) * exp((2/3)*(7/15)^(1/4)*Pi * n^(3/4) + 9*sqrt(15/7)*zeta(3) * sqrt(n) / (2*Pi^2) + ((5/7)^(1/4)*Pi / (2*3^(3/4)) - 1215*(15/7)^(1/4)*zeta(3)^2 / (28*Pi^5)) * n^(1/4) + 54675*zeta(3)^3 / (98*Pi^8) - 45*zeta(3) / (28*Pi^2)). - Vaclav Kotesovec, May 12 2021

A343200 Expansion of Product_{k>=1} (1 + x^k)^binomial(k+3,3).

Original entry on oeis.org

1, 4, 16, 64, 221, 736, 2338, 7132, 21093, 60652, 170172, 467140, 1257571, 3325824, 8654576, 22189340, 56116043, 140122760, 345769094, 843827436, 2038017983, 4874329024, 11550814704, 27134195608, 63215468883, 146120097736, 335227455982, 763592477104, 1727482413548
Offset: 0

Views

Author

Ilya Gutkovskiy, May 09 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 28; CoefficientList[Series[Product[(1 + x^k)^Binomial[k + 3, 3], {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d Binomial[d + 3, 3], {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 28}]

Formula

a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} ( Sum_{d|k} (-1)^(k/d+1) * A033488(d) ) * a(n-k).
a(n) ~ (3*zeta(5))^(1/10) / (2^(7/10) * 5^(2/5) * sqrt(Pi) * n^(3/5)) * exp(-469*log(2)/720 - 2401*Pi^16 / (656100000000*zeta(5)^3) + 539*Pi^8*zeta(3) / (8100000*zeta(5)^2) - 7*Pi^6 / (27000*zeta(5)) - 121*zeta(3)^2 / (600*zeta(5)) + (343*Pi^12 / (303750000 * 2^(3/5) * 15^(1/5) * zeta(5)^(11/5)) - 77*Pi^4*zeta(3) / (4500 * 2^(3/5) * 15^(1/5) * zeta(5)^(6/5)) + Pi^2 / (6*2^(3/5) * (15*zeta(5))^(1/5))) * n^(1/5) + (-49*Pi^8 / (270000 * 2^(1/5) * 15^(2/5) * zeta(5)^(7/5)) + 11*zeta(3) / (4*2^(1/5) * (15*zeta(5))^(2/5))) * n^(2/5) + (7*Pi^4 / (90*2^(4/5) * (15*zeta(5))^(3/5))) * n^(3/5) + (5*(15*zeta(5))^(1/5) / (4*2^(2/5))) * n^(4/5)). - Vaclav Kotesovec, May 12 2021

A255837 G.f.: Product_{k>=1} (1+x^k)^(3*k+2).

Original entry on oeis.org

1, 5, 18, 61, 182, 506, 1338, 3369, 8172, 19197, 43833, 97636, 212748, 454461, 953505, 1968095, 4001627, 8024295, 15885484, 31074351, 60111277, 115071431, 218126868, 409662895, 762679151, 1408172844, 2579599582, 4690277001, 8467363674, 15182486586
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 07 2015

Keywords

Comments

In general, if g.f. = Product_{k>=1} (1+x^k)^(m*k+c), m > 0, then a(n) ~ (m*Zeta(3))^(1/6) * exp(-c^2 * Pi^4 / (1296*m*Zeta(3)) + (c * Pi^2 * n^(1/3)) / (2^(5/3) * 3^(4/3) * (m*Zeta(3))^(1/3)) + 3^(4/3) * (m*Zeta(3))^(1/3) * n^(2/3) / 2^(4/3)) / (2^(m/12 + c/2 + 2/3) * 3^(1/3) * sqrt(Pi) * n^(2/3)). - Vaclav Kotesovec, Mar 08 2015

Crossrefs

Cf. A026007 (k), A219555 (k+1), A052812 (k-1), A255834 (2*k+1), A255835 (2*k-1), A255836 (3*k+1).
Cf. A255803.

Programs

  • Mathematica
    nmax=50; CoefficientList[Series[Product[(1+x^k)^(3*k+2),{k,1,nmax}],{x,0,nmax}],x]

Formula

a(n) ~ Zeta(3)^(1/6) * exp(-Pi^4/(972*Zeta(3)) + Pi^2 * n^(1/3) / (2^(2/3) * 3^(5/3) * Zeta(3)^(1/3)) + 3^(5/3)/2^(4/3) * Zeta(3)^(1/3) * n^(2/3)) / (2^(23/12) * 3^(1/6) * sqrt(Pi) * n^(2/3)), where Zeta(3) = A002117.
Showing 1-10 of 14 results. Next