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 1-7 of 7 results.

A294591 Expansion of Product_{k>=1} 1/((1 - x^(2*k-1))^(k*(3*k-1)/2)*(1 - x^(2*k))^(k*(3*k+1)/2)).

Original entry on oeis.org

1, 1, 3, 8, 18, 40, 88, 184, 384, 783, 1573, 3110, 6087, 11745, 22450, 42466, 79597, 147890, 272632, 498696, 905846, 1634270, 2929804, 5220581, 9249440, 16297659, 28567571, 49825296, 86487331, 149438681, 257077485, 440378787, 751313413, 1276765557, 2161511352
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 05 2017

Keywords

Comments

Euler transform of the generalized pentagonal numbers (A001318).

Crossrefs

Programs

  • Mathematica
    nmax = 34; CoefficientList[Series[Product[1/((1 - x^(2 k - 1))^(k (3 k - 1)/2) (1 - x^(2 k))^(k (3 k + 1)/2)), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d Ceiling[d/2] Ceiling[(3 d + 1)/2]/2, {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 34}]

Formula

G.f.: Product_{k>=1} 1/(1 - x^k)^A001318(k).
a(n) ~ exp(Pi * 2^(5/4) / (3*5^(1/4)) * n^(3/4) + 3*Zeta(3) * sqrt(5*n) / (2^(3/2) * Pi^2) + (Pi/48 - 45*Zeta(3)^2 / (8*Pi^5)) * (5*n/2)^(1/4) + 225*Zeta(3)^3 / (8*Pi^8) - 11*Zeta(3) / (64*Pi^2))/ (2^(95/48) * 5^(1/8) * n^(5/8)). - Vaclav Kotesovec, Nov 07 2017

A294667 Expansion of Product_{k>=1} 1/(1 - x^k)^(k*(3*k+1)/2).

Original entry on oeis.org

1, 2, 10, 33, 110, 332, 997, 2829, 7889, 21299, 56400, 146028, 371681, 929498, 2290296, 5562369, 13336036, 31583177, 73957845, 171342592, 393018517, 893000610, 2011039286, 4490680381, 9947577333, 21867539862, 47721817473, 103420870299, 222641160569
Offset: 0

Views

Author

Vaclav Kotesovec, Nov 06 2017

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 30; CoefficientList[Series[Product[1/(1-x^k)^(k*(3*k+1)/2), {k, 1, nmax}], {x, 0, nmax}], x]

Formula

a(n) ~ exp(Pi * 2^(7/4) * n^(3/4) / (3*5^(1/4)) + Zeta(3) * sqrt(5*n) / (sqrt(2) * Pi^2) - 5^(5/4) * Zeta(3)^2 * n^(1/4) / (2^(7/4) * Pi^5) + (25 * Zeta(3)^3) / (6*Pi^8) - 3*Zeta(3) / (8*Pi^2) + 1/24) * Pi^(1/24) / (sqrt(A) * 2^(157/96) * 5^(13/96) * n^(61/96)), where A is the Glaisher-Kinkelin constant A074962.

A274998 Expansion of Product_{k>=1} 1/(1 - x^k)^(k*(3*k-2)).

Original entry on oeis.org

1, 1, 9, 30, 106, 339, 1106, 3355, 10102, 29358, 83908, 234394, 644286, 1739933, 4631675, 12153197, 31485413, 80576160, 203902261, 510490213, 1265353568, 3106771717, 7559844833, 18239351931, 43650061720, 103657177941, 244346681972, 571930478187, 1329655624297, 3071230379625, 7049750442386, 16085170634548, 36489192684910
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 30 2016

Keywords

Comments

Euler transform of the octagonal numbers (A000567).

Crossrefs

Programs

  • Maple
    with(numtheory):
    a:= proc(n) option remember; `if`(n=0, 1, add(add(
          d^2*(3*d-2), d=divisors(j))*a(n-j), j=1..n)/n)
        end:
    seq(a(n), n=0..35);  # Alois P. Heinz, Dec 02 2016
  • Mathematica
    nmax=32; CoefficientList[Series[Product[1/(1 - x^k)^(k (3 k - 2)), {k, 1, nmax}], {x, 0, nmax}], x]
  • Python
    from sympy import divisors
    from sympy.core.cache import cacheit
    @cacheit
    def a(n): return 1 if n==0 else sum(sum(d**2*(3*d - 2) for d in divisors(j))*a(n - j) for j in range(1, n + 1))//n
    print([a(n) for n in range(51)]) # Indranil Ghosh, Aug 06 2017, after Maple code

Formula

G.f.: Product_{k>=1} 1/(1 - x^k)^(k*(3*k-2)).
a(n) ~ exp(4*Pi*n^(3/4) / (3*5^(1/4)) - 2*Zeta(3) * sqrt(5*n) / Pi^2 - 10*Zeta(3)^2 * (5*n)^(1/4) / Pi^5 - 200*Zeta(3)^3 / (3*Pi^8) - 3*Zeta(3) / (4*Pi^2) - 1/6) * A^2 / (2^(3/2) * 5^(1/12) * Pi^(1/6) * n^(7/12)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Nov 08 2017

A318118 a(n) = [x^n] exp(Sum_{k>=1} x^k*(1 + (n - 3)*x^k)/(k*(1 - x^k)^3)).

Original entry on oeis.org

1, 1, 3, 10, 40, 150, 616, 2456, 10102, 41400, 171526, 712111, 2972115, 12434993, 52195414, 219567909, 925704792, 3909841659, 16541598215, 70085877919, 297347922785, 1263046810334, 5370930049915, 22861883482838, 97402827429118, 415332438952517, 1772380322197432
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 18 2018

Keywords

Comments

For n > 2, a(n) is the n-th term of the Euler transform of n-gonal numbers.

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[Exp[Sum[x^k (1 + (n - 3) x^k)/(k (1 - x^k)^3), {k, 1, n}]], {x, 0, n}], {n, 0, 26}]

Formula

a(n) ~ c * d^n / sqrt(n), where d = 4.3505530790182509701639869563721679988879373943131559534408716195123... and c = 0.2276354216252041005336767937139336687746108521151301186102034... - Vaclav Kotesovec, Aug 18 2018

A294669 Expansion of Product_{k>=1} 1/(1 - x^(2*k-1))^(k*(3*k-1)/2).

Original entry on oeis.org

1, 1, 1, 6, 6, 18, 33, 55, 115, 185, 373, 604, 1113, 1903, 3251, 5678, 9350, 16153, 26420, 44561, 72912, 120150, 196329, 317988, 516881, 827778, 1333570, 2120492, 3381947, 5347513, 8447482, 13285450, 20813814, 32547272, 50638328, 78707858, 121738479
Offset: 0

Views

Author

Vaclav Kotesovec, Nov 06 2017

Keywords

Crossrefs

Programs

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

Formula

a(n) ~ exp(2*Pi * n^(3/4) / (3*5^(1/4)) + Zeta(3) * sqrt(5*n) / Pi^2 + 5^(1/4) * (Pi/48 - 5*Zeta(3)^2 / Pi^5) * n^(1/4) + 100*Zeta(3)^3 / (3*Pi^8) + 17*Zeta(3) / (96*Pi^2) - 1/24) * sqrt(A) / (2^(101/48) * 5^(11/96) * Pi^(1/24) * n^(59/96)), where A is the Glaisher-Kinkelin constant A074962.

A294692 Expansion of Product_{k>=1} 1 / (1 - x^k)^(k*(3*k + 2)).

Original entry on oeis.org

1, 5, 31, 148, 667, 2754, 10823, 40393, 145085, 502780, 1690603, 5530649, 17658430, 55141520, 168751779, 506933980, 1496999360, 4350994324, 12460305177, 35192973824, 98116587875, 270220568883, 735668636567, 1981082952258, 5279879097853, 13933764841202
Offset: 0

Views

Author

Vaclav Kotesovec, Nov 07 2017

Keywords

Crossrefs

Programs

  • Maple
    N:= 50:
    S:= series(mul(1/(1-x^k)^(k*(3*k+2)), k=1..N),x,N+1):
    seq(coeff(S,x,n),n=0..N); # Robert Israel, Nov 07 2017
  • Mathematica
    nmax = 30; CoefficientList[Series[Product[1/(1-x^k)^(k*(3*k+2)), {k, 1, nmax}], {x, 0, nmax}], x]

Formula

a(n) ~ exp(4*Pi*n^(3/4) / (3*5^(1/4)) + 2*Zeta(3) * sqrt(5*n) / Pi^2 - 2*5^(5/4) * Zeta(3)^2 * n^(1/4) / Pi^5 + 200*Zeta(3)^3 / (3*Pi^8) - 3*Zeta(3) / (4*Pi^2) + 1/6) * Pi^(1/6) / (A^2 * 2^(3/2) * 5^(1/6) * n^(2/3)), where A is the Glaisher-Kinkelin constant A074962.

A317017 Expansion of Product_{k>=1} 1/(1 - x^k)^((3*k+1)*binomial(k+2,3)/4).

Original entry on oeis.org

1, 1, 8, 33, 126, 441, 1571, 5338, 17900, 58359, 187134, 588966, 1826537, 5580784, 16831549, 50135506, 147650112, 430187724, 1240908651, 3545808444, 10042128414, 28201458999, 78567720054, 217225969695, 596254164090, 1625343030654, 4401332943214, 11843216471115, 31674767502610
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 19 2018

Keywords

Comments

Euler transform of A001296.

Crossrefs

Programs

  • Maple
    a:=series(mul(1/(1-x^k)^((3*k+1)*binomial(k+2,3)/4),k=1..100),x=0,29): seq(coeff(a,x,n),n=0..28); # Paolo P. Lava, Apr 02 2019
  • Mathematica
    nmax = 28; CoefficientList[Series[Product[1/(1 - x^k)^((3 k + 1) Binomial[k + 2, 3]/4), {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 28; CoefficientList[Series[Exp[Sum[x^k (1 + 2 x^k)/(k (1 - x^k)^5), {k, 1, nmax}]], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d^2 (d + 1) (d + 2) (3 d + 1)/24, {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 28}]

Formula

G.f.: Product_{k>=1} 1/(1 - x^k)^A001296(k).
G.f.: exp(Sum_{k>=1} x^k*(1 + 2*x^k)/(k*(1 - x^k)^5)).
a(n) ~ Pi^(1/288)/(2 * 3^(577/864) * 7^(145/1728) * n^(1009/1728)) * exp(1/144 - (1/12-Zeta'(-1))/12 - (11 * Zeta(3))/(80 * Pi^2) + (1383 * Zeta(5))/(640 * Pi^4) + (11025 * Zeta(3) * Zeta(5)^2)/(2 * Pi^12) - (694575 * Zeta(5)^3)/(2 * Pi^14) + (13127467500 * Zeta(5)^5)/Pi^24 + (5 * Zeta'(-3))/12 + ((-21 * 3^(1/3) * 7^(1/6) * Pi)/6400 - (35 * 3^(1/3) * 7^(1/6) * Zeta(3) * Zeta(5))/(2 * Pi^7) + (15435 * 3^(1/3) * 7^(1/6) * Zeta(5)^2)/(16 * Pi^9) - (175573125 * 3^(1/3) * 7^(1/6) * Zeta(5)^4)/(4 * Pi^19)) * n^(1/6) + (((7/3)^(1/3) * Zeta(3))/(4 * Pi^2) - (21 * 3^(2/3) * 7^(1/3) * Zeta(5))/(8 * Pi^4) + (147000 * 3^(2/3) * 7^(1/3) * Zeta(5)^3)/Pi^14) * n^(1/3) + ((sqrt(7) * Pi)/40 - (1575 * sqrt(7) * Zeta(5)^2)/Pi^9) * sqrt(n) + ((15 * 3^(1/3) * 7^(2/3) * Zeta(5))/(2 * Pi^4)) * n^(2/3) + ((2 * 3^(2/3) * Pi)/(5 * 7^(1/6))) * n^(5/6)). - Vaclav Kotesovec, Jul 28 2018
Showing 1-7 of 7 results.