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 13 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

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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

A075196 Table T(n,k) by antidiagonals: T(n,k) = number of partitions of n balls of k colors.

Original entry on oeis.org

1, 2, 2, 3, 6, 3, 4, 12, 14, 5, 5, 20, 38, 33, 7, 6, 30, 80, 117, 70, 11, 7, 42, 145, 305, 330, 149, 15, 8, 56, 238, 660, 1072, 906, 298, 22, 9, 72, 364, 1260, 2777, 3622, 2367, 591, 30, 10, 90, 528, 2198, 6174, 11160, 11676, 6027, 1132, 42, 11, 110, 735, 3582, 12292, 28784, 42805, 36450, 14873, 2139, 56
Offset: 1

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Author

Christian G. Bower, Sep 07 2002

Keywords

Comments

For k>=1, n->infinity is log(T(n,k)) ~ (1+1/k) * k^(1/(k+1)) * Zeta(k+1)^(1/(k+1)) * n^(k/(k+1)). - Vaclav Kotesovec, Mar 08 2015

Examples

			Square array T(n,k) begins:
  1,  2,   3,    4,    5, ...
  2,  6,  12,   20,   30, ...
  3, 14,  38,   80,  145, ...
  5, 33, 117,  305,  660, ...
  7, 70, 330, 1072, 2777, ...
		

Crossrefs

Rows 1-3: A000027, A002378, A162147.
Main diagonal: A075197.
Cf. A255903.

Programs

  • Maple
    with(numtheory):
    A:= proc(n, k) option remember; local d, j;
          `if`(n=0, 1, add(add(d*binomial(d+k-1, k-1),
           d=divisors(j)) *A(n-j, k), j=1..n)/n)
        end:
    seq(seq(A(n, 1+d-n), n=1..d), d=1..12);  # Alois P. Heinz, Sep 26 2012
  • Mathematica
    Transpose[Table[nn=6;p=Product[1/(1- x^i)^Binomial[i+n,n],{i,1,nn}];Drop[CoefficientList[Series[p,{x,0,nn}],x],1],{n,0,nn}]]//Grid  (* Geoffrey Critzer, Sep 27 2012 *)

Formula

T(n,k) = Sum_{i=0..k} C(k,i) * A255903(n,i). - Alois P. Heinz, Mar 10 2015

A253289 G.f.: Product_{k>=1} 1/(1-x^k)^(2*k-1).

Original entry on oeis.org

1, 1, 4, 9, 22, 46, 103, 208, 431, 849, 1671, 3195, 6079, 11321, 20937, 38146, 68931, 123121, 218212, 383019, 667425, 1153544, 1980268, 3375394, 5717773, 9624541, 16108496, 26807662, 44379189, 73089219, 119789926, 195401275, 317309532, 513025167, 826000651
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 07 2015

Keywords

Comments

a(n) is the number of partitions of n where there are 2*k-1 sorts of parts k. - Joerg Arndt, Aug 15 2020

Crossrefs

Programs

  • Maple
    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-> 2*n-1): seq(a(n), n=0..50); # after Alois P. Heinz
    with(numtheory):
    series(exp(add((2*sigma[2](k) - sigma[1](k))*x^k/k, k = 1..30)), x, 31):
    seq(coeftayl(%, x = 0, n), n = 0..30); # Peter Bala, Jan 16 2025
  • Mathematica
    nmax=50; CoefficientList[Series[Product[1/(1-x^k)^(2*k-1),{k,1,nmax}],{x,0,nmax}],x]
    (* Using EulerTransforms from 'Transforms'. *)
    Prepend[EulerTransform[Table[2k + 1, {k, 0, 20}]], 1] (* Peter Luschny, Aug 15 2020 *)

Formula

a(n) ~ 2^(1/9) * Zeta(3)^(1/18) * exp(1/6 - Pi^4/(864*Zeta(3)) - Pi^2 * n^(1/3) / (3 * 2^(5/3) * Zeta(3)^(1/3)) + 3 * (Zeta(3)/2)^(1/3) * n^(2/3)) / (A^2 * 3^(1/2) * n^(5/9)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant and Zeta(3) = A002117 = 1.202056903... .
G.f.: exp(Sum_{k>=1} x^k*(1 + x^k)/(k*(1 - x^k)^2)). - Ilya Gutkovskiy, Jun 07 2018
Euler transform of A005408 (the odd numbers). - Georg Fischer, Aug 15 2020
G.f.: exp(Sum_{k >= 1} (2*sigma_2(k) - sigma_1(k))*x^k/k) = 1 + x + 4*x^2 + 9*x^3 + 22*x^4 + .... - Peter Bala, Jan 16 2025

A255050 G.f.: Product_{j>=1} 1/(1-x^j)^binomial(j+3,3).

Original entry on oeis.org

1, 4, 20, 80, 305, 1072, 3622, 11676, 36450, 110240, 324936, 935076, 2635338, 7285560, 19795370, 52930360, 139462956, 362471020, 930186694, 2358867240, 5915606398, 14680528648, 36073675792, 87816701332, 211891552280, 506981067168, 1203337174120, 2834401172172
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 08 2015

Keywords

Comments

Number of partitions of n unlabeled objects of 4 colors. - Peter Dolland, Feb 20 2025

Crossrefs

Column k=4 of A075196.

Programs

  • Maple
    with(numtheory):
    a:= proc(n) option remember; local d, j; `if`(n=0, 1,
          add(add(d*binomial(d+3, 3), d=divisors(j))*a(n-j), j=1..n)/n)
        end:
    seq(a(n), n=0..50); # after Alois P. Heinz
  • Mathematica
    nmax=50; CoefficientList[Series[Product[1/(1-x^j)^Binomial[j+3,3],{j,1,nmax}],{x,0,nmax}],x]

Formula

G.f.: Product_{j>=1} 1/(1-x^j)^C(j+3,3).
a(n) ~ Zeta(5)^(829/3600) * exp(11/72 - Zeta(3)/(4*Pi^2) + Zeta'(-3)/6 - 121*Zeta(3)^2 / (360*Zeta(5)) - Pi^6/(1800*Zeta(5)) + 11*Pi^8*Zeta(3) / (108000*Zeta(5)^2) - Pi^16/(194400000*Zeta(5)^3) + Pi^2 * n^(1/5)/ (6*2^(2/5) * Zeta(5)^(1/5)) - 11*Pi^4 * Zeta(3) * n^(1/5) / (900*2^(2/5)*Zeta(5)^(6/5)) + Pi^12 * n^(1/5) / (1350000 * 2^(2/5) * Zeta(5)^(11/5)) + 11*Zeta(3) * n^(2/5) / (6*2^(4/5) * Zeta(5)^(2/5)) - Pi^8 * n^(2/5) / (9000 * 2^(4/5) * Zeta(5)^(7/5)) + Pi^4 * n^(3/5) / (90 * 2^(1/5) * Zeta(5)^(3/5)) + 5 * Zeta(5)^(1/5) * n^(4/5) / 2^(8/5)) / (A^(11/6) * 2^(971/1800) * 5^(1/2) * Pi * n^(2629/3600)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant, Zeta(3) = A002117 = 1.202056903..., Zeta(5) = A013663 = 1.036927755... and Zeta'(-3) = ((gamma + log(2*Pi) - 11/6)/30 - 3*Zeta'(4)/Pi^4)/4 = 0.0053785763577743... .
EULER transform of 1, 4, 10, 20, 35, 56, 84, ... (= A000292(n+1)). - Peter Dolland, Feb 20 2025

A120844 Number of multi-trace BPS operators for the quiver gauge theory of the orbifold C^2/Z_2.

Original entry on oeis.org

1, 3, 11, 32, 90, 231, 576, 1363, 3141, 7003, 15261, 32468, 67788, 138892, 280103, 556302, 1089991, 2108332, 4030649, 7620671, 14261450, 26431346, 48544170, 88393064, 159654022, 286149924, 509137464, 899603036, 1579014769
Offset: 0

Views

Author

Amihay Hanany (hanany(AT)mit.edu), Aug 25 2006

Keywords

Crossrefs

Programs

  • Maple
    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-> 2*n+1): seq(a(n), n=0..50); # Vaclav Kotesovec, Mar 06 2015 after Alois P. Heinz
    # alternative program
    with(numtheory):
    series(exp(add((2*sigma[2](k) + sigma[1](k))*x^k/k, k = 1..30)), x, 31):
    seq(coeftayl(%, x = 0, n), n = 0..30); # Peter Bala, Jan 16 2025
  • Mathematica
    nmax=50; CoefficientList[Series[Product[1/(1-x^k)^(2*k+1),{k,1,nmax}],{x,0,nmax}],x] (* Vaclav Kotesovec, Feb 27 2015 *)

Formula

G.f.: exp( Sum_{n>0} (3*x^n - x^(2*n)) / (n*(1-x^n)^2) ).
a(n) ~ Zeta(3)^(7/18) * exp(1/6 - Pi^4/(864*Zeta(3)) + Pi^2 * n^(1/3)/(3 * 2^(5/3) * Zeta(3)^(1/3)) + 3 * (Zeta(3)/2)^(1/3) * n^(2/3)) / (A^2 * 2^(2/9) * 3^(1/2) * Pi * n^(8/9)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant and Zeta(3) = A002117 = 1.202056903... . - Vaclav Kotesovec, Mar 07 2015
From Peter Bala, Jan 16 2025: (Start)
G.f.: 1/Product_{k >= 1} (1 - x^k)^(2*k+1).
G.f.: exp(Sum_{k >= 1} (2*sigma_2(k) + sigma_1(k))*x^k/k) = 1 + 3*x + 11*x^2 + 32*x^3 + 90*x^4 + 231*x^5 + .... (End)

A363601 Number of partitions of n where there are k^2 - 1 kinds of parts k.

Original entry on oeis.org

1, 0, 3, 8, 21, 48, 126, 288, 693, 1568, 3570, 7896, 17417, 37632, 80823, 171192, 359733, 747936, 1543192, 3155760, 6407037, 12909024, 25835649, 51359136, 101470854, 199264128, 389096028, 755591256, 1459643343, 2805471984, 5366161740, 10216161336, 19362398580
Offset: 0

Views

Author

Seiichi Manyama, Jun 10 2023

Keywords

Crossrefs

Programs

  • Maple
    with(numtheory):
    series(exp(add((sigma[3](k) - sigma[1](k))*x^k/k, k = 1..50)), x, 51):
    seq(coeftayl(%, x = 0, n), n = 0..50); # Peter Bala, Jan 16 2025
  • PARI
    my(N=40, x='x+O('x^N)); Vec(1/prod(k=1, N, (1-x^k)^(k^2-1)))

Formula

G.f.: 1/Product_{k>=1} (1-x^k)^(k^2-1).
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} A092348(k) * a(n-k).
G.f.: exp(Sum_{k >= 1} (sigma_3(k) - sigma_1(k))*x^k/k) = 1 + 3*x^2 + 8*x^3 + 21*x^4 + 48*x^5 + .... - Peter Bala, Jan 16 2025

Extensions

Name suggested by Joerg Arndt, Jun 11 2023

A255802 G.f.: Product_{k>=1} 1/(1-x^k)^(2*k+3).

Original entry on oeis.org

1, 5, 22, 79, 259, 777, 2201, 5911, 15239, 37865, 91224, 213741, 488759, 1093173, 2396934, 5160756, 10928181, 22787949, 46848176, 95046026, 190466354, 377295743, 739319876, 1433974869, 2754597217, 5243308562, 9894376295, 18517966608, 34386781020, 63378252332
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 07 2015

Keywords

Crossrefs

Programs

  • Maple
    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-> 2*n+3): seq(a(n), n=0..50); # after Alois P. Heinz
    with(numtheory):
    series(exp(add((2*sigma[2](k) + 3*sigma[1](k))*x^k/k, k = 1..30)), x, 31):
    seq(coeftayl(%, x = 0, n), n = 0..30); # Peter Bala, Jan 16 2025
  • Mathematica
    nmax=50; CoefficientList[Series[Product[1/(1-x^k)^(2*k+3),{k,1,nmax}],{x,0,nmax}],x]

Formula

a(n) ~ Zeta(3)^(13/18) * exp(1/6 - Pi^4/(96*Zeta(3)) + Pi^2 * n^(1/3) / (2^(5/3) * Zeta(3)^(1/3)) + 3 * (Zeta(3)/2)^(1/3) * n^(2/3)) / (A^2 * 2^(5/9) * 3^(1/2) * Pi^2 * n^(11/9)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant and Zeta(3) = A002117 = 1.202056903... .
G.f.: exp(Sum_{k >= 1} (2*sigma_2(k) + 3*sigma_1(k))*x^k/k) = 1 + 5*x + 22*x^2 + 29*x^3 + 777*x^4 + .... - Peter Bala, Jan 16 2025

A264923 G.f.: 1 / Product_{n>=0} (1 - x^(n+3))^((n+1)*(n+2)/2!).

Original entry on oeis.org

1, 0, 0, 1, 3, 6, 11, 18, 33, 57, 105, 183, 330, 567, 990, 1693, 2904, 4917, 8343, 14010, 23511, 39171, 65100, 107592, 177352, 290931, 475905, 775381, 1259637, 2039094, 3291613, 5296467, 8499339, 13599292, 21702795, 34541724, 54839894, 86847255, 137212197, 216274466, 340129773, 533726442, 835732774, 1305877914, 2036369010
Offset: 0

Views

Author

Paul D. Hanna, Nov 28 2015

Keywords

Comments

Number of partitions of n objects of 3 colors, where each part must contain at least one of each color. [Conjecture - see comment by Franklin T. Adams-Watters in A052847].

Examples

			G.f.: A(x) = 1 + x^3 + 3*x^4 + 6*x^5 + 11*x^6 + 18*x^7 + 33*x^8 + 57*x^9 + 105*x^10 +...
where
1/A(x) = (1-x^3) * (1-x^4)^3 * (1-x^5)^6 * (1-x^6)^10 * (1-x^7)^15 * (1-x^8)^21 * (1-x^9)^28 * (1-x^10)^36 * (1-x^11)^45 *...
Also,
log(A(x)) = (x/(1-x))^3 + (x^2/(1-x^2))^3/2 + (x^3/(1-x^3))^3/3 + (x^4/(1-x^4))^3/4 + (x^5/(1-x^5))^3/5 + (x^6/(1-x^6))^3/6 +...
		

Crossrefs

Programs

  • Mathematica
    nmax = 50; CoefficientList[Series[Product[1/(1-x^k)^((k-2)*(k-1)/2), {k,1,nmax}], {x,0,nmax}], x] (* Vaclav Kotesovec, Dec 09 2015 *)
  • PARI
    {a(n) = my(A=1); A = prod(k=0,n, 1/(1 - x^(k+3) +x*O(x^n) )^((k+1)*(k+2)/2) ); polcoeff(A,n)}
    for(n=0,50,print1(a(n),", "))
    
  • PARI
    {a(n) = my(A=1); A = exp( sum(k=1,n+1, (x^k/(1 - x^k))^3 /k +x*O(x^n) ) ); polcoeff(A,n)}
    for(n=0,50,print1(a(n),", "))
    
  • PARI
    {L(n) = sumdiv(n,d, d*(d-1)*(d-2)/2! )}
    {a(n) = my(A=1); A = exp( sum(k=1,n+1, L(k) * x^k/k +x*O(x^n) ) ); polcoeff(A,n)}
    for(n=0,50,print1(a(n),", "))

Formula

G.f.: exp( Sum_{n>=1} ( x^n/(1-x^n) )^3 /n ).
G.f.: exp( Sum_{n>=1} L(n) * x^n/n ), where L(n) = Sum_{d|n} d*(d-1)*(d-2)/2!.
a(n) ~ Pi^(3/8) / (2^(55/32) * 15^(7/32) * n^(23/32)) * exp(29*Zeta(3)/(8*Pi^2) - log(2*Pi)/2 - 3*Zeta'(-1)/2 - 2025*Zeta(3)^3/(2*Pi^8) + (5^(1/4)*Pi/6^(3/4) - 135*15^(1/4)*Zeta(3)^2/(2^(7/4)*Pi^5)) * n^(1/4) - 3*sqrt(15*n/2)*Zeta(3)/Pi^2 + 2^(7/4)*Pi/(3*15^(1/4)) * n^(3/4)). - Vaclav Kotesovec, Dec 09 2015

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

A382045 Triangle read by rows: T(n,k) is the number of partitions of a 3-colored set of n objects into at most k parts with 0 <= k <= n.

Original entry on oeis.org

1, 0, 3, 0, 6, 12, 0, 10, 28, 38, 0, 15, 66, 102, 117, 0, 21, 126, 249, 309, 330, 0, 28, 236, 562, 788, 878, 906, 0, 36, 396, 1167, 1845, 2205, 2331, 2367, 0, 45, 651, 2292, 4128, 5289, 5814, 5982, 6027, 0, 55, 1001, 4272, 8703, 12106, 13881, 14602, 14818, 14873, 0, 66, 1512, 7608, 17634, 26616, 32088, 34608, 35556, 35826, 35892
Offset: 0

Views

Author

Peter Dolland, Mar 13 2025

Keywords

Comments

The 1-color case is Euler's table A026820.
The 2-color case is A381891.

Examples

			Triangle starts:
 0 : [1]
 1 : [0,  3]
 2 : [0,  6,   12]
 3 : [0, 10,   28,   38]
 4 : [0, 15,   66,  102,   117]
 5 : [0, 21,  126,  249,   309,   330]
 6 : [0, 28,  236,  562,   788,   878,   906]
 7 : [0, 36,  396, 1167,  1845,  2205,  2331,  2367]
 8 : [0, 45,  651, 2292,  4128,  5289,  5814,  5982,  6027]
 9 : [0, 55, 1001, 4272,  8703, 12106, 13881, 14602, 14818, 14873]
10 : [0, 66, 1512, 7608, 17634, 26616, 32088, 34608, 35556, 35826, 35892]
...
		

Crossrefs

Main diagonal gives A217093.

Programs

  • Maple
    b:= proc(n, i) option remember; expand(`if`(n=0, 1, `if`(i<1, 0, add(
          b(n-i*j, min(n-i*j, i-1))*binomial(i*(i+3)/2+j, j)*x^j, j=0..n/i))))
        end:
    T:= proc(n, k) option remember;
         `if`(k<0, 0, T(n, k-1)+coeff(b(n$2), x, k))
        end:
    seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Mar 13 2025
  • Mathematica
    b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, Sum[b[n - i*j, Min[n - i*j, i - 1]]*Binomial[i*(i + 3)/2 + j, j]*x^j, {j, 0, n/i}]]]];
    T[n_, k_] := T[n, k] = If[k < 0, 0, T[n, k-1] + Coefficient[b[n, n], x, k]];
    Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Mar 21 2025, after Alois P. Heinz *)
  • Python
    from sympy import binomial
    from sympy.utilities.iterables import partitions
    from sympy.combinatorics.partitions import IntegerPartition
    colors = 3 - 1   # the number of colors - 1
    def t_row( n):
        if n == 0 : return [1]
        t = list( [0] * n)
        for p in partitions( n):
            p = IntegerPartition( p).as_dict()
            fact = 1
            s = 0
            for k in p :
                s += p[k]
                fact *= binomial( binomial( k + colors, colors) + p[k] - 1, p[k])
            if s > 0 :
                t[s - 1] += fact
        for i in range( n - 1):
            t[i+1] += t[i]
        return [0] + t

Formula

T(n,1) = binomial(n + 2, 2) = A000217(n + 1) for n >= 1.
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