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.

A343363 Expansion of Product_{k>=1} (1 + x^k)^(6^(k-1)).

Original entry on oeis.org

1, 1, 6, 42, 267, 1743, 11234, 72470, 466251, 2996883, 19234836, 123315828, 789682546, 5051601010, 32282443044, 206104519572, 1314652656453, 8378283675645, 53350205335626, 339445117302366, 2158091256282273, 13710402587540469, 87040883294333382, 552205562345916570
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 12 2021

Keywords

Crossrefs

Programs

  • Maple
    N:= 100: # for a(0)..a(N)
    G:= mul((1+x^k)^(6^(k-1)),k=1..N):
    S:= series(G,x,N+1):
    seq(coeff(S,x,k),k=0..N); # Robert Israel, Apr 12 2021
  • Mathematica
    nmax = 23; CoefficientList[Series[Product[(1 + x^k)^(6^(k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d 6^(d - 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 23}]
  • PARI
    seq(n)={Vec(prod(k=1, n, (1 + x^k + O(x*x^n))^(6^(k-1))))} \\ Andrew Howroyd, Apr 12 2021

Formula

a(n) ~ exp(sqrt(2*n/3) - 1/12 - c/6) * 6^(n - 1/4) / (2*sqrt(Pi)*n^(3/4)), where c = Sum_{j>=2} (-1)^j / (j * (6^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021

A343361 Expansion of Product_{k>=1} (1 + x^k)^(4^(k-1)).

Original entry on oeis.org

1, 1, 4, 20, 86, 390, 1724, 7644, 33697, 148401, 651584, 2855840, 12491276, 54540636, 237733768, 1034610232, 4495832776, 19508749928, 84540638312, 365888222552, 1581630245756, 6829047398156, 29453496620000, 126898489491904, 546183557447366, 2348560270762006, 10089340886428928
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 12 2021

Keywords

Crossrefs

Programs

  • Maple
    h:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(h(n-i*j, i-1)*binomial(4^(i-1), j), j=0..n/i)))
        end:
    a:= n-> h(n$2):
    seq(a(n), n=0..26);  # Alois P. Heinz, Apr 12 2021
  • Mathematica
    nmax = 26; CoefficientList[Series[Product[(1 + x^k)^(4^(k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d 4^(d - 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 26}]
  • PARI
    seq(n)={Vec(prod(k=1, n, (1 + x^k + O(x*x^n))^(4^(k-1))))} \\ Andrew Howroyd, Apr 12 2021

Formula

a(n) ~ exp(sqrt(n) - 1/8 - c/4) * 2^(2*n - 3/2) / (sqrt(Pi)*n^(3/4)), where c = Sum_{j>=2} (-1)^j / (j * (4^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021

A343362 Expansion of Product_{k>=1} (1 + x^k)^(5^(k-1)).

Original entry on oeis.org

1, 1, 5, 30, 160, 885, 4810, 26185, 142005, 769305, 4159301, 22455876, 121057525, 651737675, 3504241650, 18818709130, 100945053055, 540885242825, 2895159035375, 15481318817450, 82704855762375, 441427664993275, 2354020475714775, 12542918682786300, 66778882780674975
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 12 2021

Keywords

Crossrefs

Programs

  • Maple
    h:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(h(n-i*j, i-1)*binomial(5^(i-1), j), j=0..n/i)))
        end:
    a:= n-> h(n$2):
    seq(a(n), n=0..24);  # Alois P. Heinz, Apr 12 2021
  • Mathematica
    nmax = 24; CoefficientList[Series[Product[(1 + x^k)^(5^(k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d 5^(d - 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 24}]
  • PARI
    seq(n)={Vec(prod(k=1, n, (1 + x^k + O(x*x^n))^(5^(k-1))))} \\ Andrew Howroyd, Apr 12 2021

Formula

a(n) ~ exp(2*sqrt(n/5) - 1/10 - c/5) * 5^(n - 1/4) / (2*sqrt(Pi)*n^(3/4)), where c = Sum_{j>=2} (-1)^j / (j * (5^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021

A343364 Expansion of Product_{k>=1} (1 + x^k)^(7^(k-1)).

Original entry on oeis.org

1, 1, 7, 56, 413, 3108, 23163, 172711, 1285256, 9556603, 70980000, 526711507, 3904946864, 28926003505, 214095348671, 1583389916081, 11701578676851, 86415267247743, 637732279701496, 4703270177738076, 34664585073280204, 255332979654402524, 1879629724498860397, 13829015594546304600
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 12 2021

Keywords

Crossrefs

Programs

  • Maple
    h:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(h(n-i*j, i-1)*binomial(7^(i-1), j), j=0..n/i)))
        end:
    a:= n-> h(n$2):
    seq(a(n), n=0..23);  # Alois P. Heinz, Apr 12 2021
  • Mathematica
    nmax = 23; CoefficientList[Series[Product[(1 + x^k)^(7^(k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d 7^(d - 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 23}]
  • PARI
    seq(n)={Vec(prod(k=1, n, (1 + x^k + O(x*x^n))^(7^(k-1))))} \\ Andrew Howroyd, Apr 12 2021

Formula

a(n) ~ exp(2*sqrt(n/7) - 1/14 - c/7) * 7^(n - 1/4) / (2*sqrt(Pi)*n^(3/4)), where c = Sum_{j>=2} (-1)^j / (j * (7^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021

A343365 Expansion of Product_{k>=1} (1 + x^k)^(8^(k-1)).

Original entry on oeis.org

1, 1, 8, 72, 604, 5148, 43544, 368408, 3112262, 26273542, 221605240, 1867736120, 15730022540, 132385106956, 1113413229000, 9358220560136, 78606905495809, 659886123312449, 5536404584185376, 46424396382193376, 389074608184431328, 3259085506224931424, 27286163457927575200
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 12 2021

Keywords

Crossrefs

Programs

  • Maple
    h:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(h(n-i*j, i-1)*binomial(8^(i-1), j), j=0..n/i)))
        end:
    a:= n-> h(n$2):
    seq(a(n), n=0..22);  # Alois P. Heinz, Apr 12 2021
  • Mathematica
    nmax = 22; CoefficientList[Series[Product[(1 + x^k)^(8^(k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d 8^(d - 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 22}]
  • PARI
    seq(n)={Vec(prod(k=1, n, (1 + x^k + O(x*x^n))^(8^(k-1))))} \\ Andrew Howroyd, Apr 12 2021

Formula

a(n) ~ exp(sqrt(n/2) - 1/16 - c/8) * 2^(3*n - 7/4) / (sqrt(Pi)*n^(3/4)), where c = Sum_{j>=2} (-1)^j / (j * (8^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021

A343366 Expansion of Product_{k>=1} (1 + x^k)^(9^(k-1)).

Original entry on oeis.org

1, 1, 9, 90, 846, 8055, 76224, 721389, 6819192, 64422126, 608173020, 5737815756, 54100140735, 509794737636, 4801164836634, 45192001954005, 425156458320783, 3997756503852489, 37572655020653089, 352957677187938076, 3314174696310855888, 31105460092251410001, 291818245344169918725
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 12 2021

Keywords

Comments

In general, if m > 1 and g.f. = Product_{k>=1} (1 + x^k)^(m^(k-1)), then a(n, m) ~ exp(2*sqrt(n/m) - 1/(2*m) - c(m)/m) * m^(n - 1/4) / (2*sqrt(Pi)*n^(3/4)), where c(m) = Sum_{j>=2} (-1)^j / (j * (m^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021

Crossrefs

Programs

  • Maple
    h:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(h(n-i*j, i-1)*binomial(9^(i-1), j), j=0..n/i)))
        end:
    a:= n-> h(n$2):
    seq(a(n), n=0..22);  # Alois P. Heinz, Apr 12 2021
  • Mathematica
    nmax = 22; CoefficientList[Series[Product[(1 + x^k)^(9^(k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d 9^(d - 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 22}]
  • PARI
    seq(n)={Vec(prod(k=1, n, (1 + x^k + O(x*x^n))^(9^(k-1))))} \\ Andrew Howroyd, Apr 12 2021

Formula

a(n) ~ exp(2*sqrt(n/9) - 1/18 - c/9) * 9^(n - 1/4) / (2*sqrt(Pi)*n^(3/4)), where c = Sum_{j>=2} (-1)^j / (j * (9^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021

A343331 Expansion of Product_{k>=1} (1 + x^k)^(10^(k-1)).

Original entry on oeis.org

1, 1, 10, 110, 1145, 12045, 126070, 1319570, 13798710, 144217910, 1506406702, 15726571002, 164096557935, 1711386871635, 17839701265570, 185876723016390, 1935830424374840, 20152131324766520, 209696974024339610, 2181155691766631710, 22678274833738085501, 235704268837407670401
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 12 2021

Keywords

Comments

In general, if m > 1 and g.f. = Product_{k>=1} (1 + x^k)^(m^(k-1)), then a(n, m) ~ exp(2*sqrt(n/m) - 1/(2*m) - c(m)/m) * m^(n - 1/4) / (2*sqrt(Pi)*n^(3/4)), where c(m) = Sum_{j>=2} (-1)^j / (j * (m^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021

Crossrefs

Programs

  • Maple
    h:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(h(n-i*j, i-1)*binomial(10^(i-1), j), j=0..n/i)))
        end:
    a:= n-> h(n$2):
    seq(a(n), n=0..21);  # Alois P. Heinz, Apr 12 2021
  • Mathematica
    nmax = 21; CoefficientList[Series[Product[(1 + x^k)^(10^(k - 1)), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d 10^(d - 1), {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 21}]

Formula

a(n) ~ exp(sqrt(2*n/5) - 1/20 - c/10) * 10^(n - 1/4) / (2*sqrt(Pi)*n^(3/4)), where c = Sum_{j>=2} (-1)^j / (j * (10^(j-1) - 1)). - Vaclav Kotesovec, Apr 13 2021
Showing 1-7 of 7 results.