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 1-4 of 4 results.

A378484 Expansion of (Sum_{k>=0} binomial(4*k,k) * x^k)^4.

Original entry on oeis.org

1, 16, 208, 2480, 28176, 310336, 3344688, 35472672, 371570320, 3853862080, 39650662720, 405221752112, 4117879215472, 41643345090240, 419362920305952, 4207604570770752, 42079232716865424, 419609034657373120, 4173470598366784960, 41413032430984848832, 410071444666659404352
Offset: 0

Views

Author

Seiichi Manyama, Nov 28 2024

Keywords

Links

Crossrefs

Cf. A000302, A378483.
Cf. A005810, A078995.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Sum[Binomial[4*k,k] * x^k, {k, 0, nmax}]^4, {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 19 2025 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(sum(k=0, N, binomial(4*k, k)*x^k)^4)

Formula

a(n) = Sum_{i+j+k+l=n, i,j,k,l >= 0} binomial(4*i,i) * binomial(4*j,j) * binomial(4*k,k) * binomial(4*l,l).
G.f.: B(x)^4 where B(x) is the g.f. of A005810.
27*a(n) - 256*a(n-1) = 18*A078995(n) + 8*A005810(n) for n > 0.
a(n) ~ n * 2^(8*n + 2) / 3^(3*n + 2) * (1 + 2^(7/2)/(3^(3/2)*sqrt(Pi*n))). - Vaclav Kotesovec, Jul 19 2025

A386365 Expansion of e.g.f. (Sum_{k>=0} binomial(3*k,k) * x^k)^(1/3).

Original entry on oeis.org

1, 1, 8, 118, 2536, 71600, 2504320, 104482000, 5063797760, 279579704320, 17322126976000, 1190107376057600, 89795437443712000, 7381088691027251200, 656522501105485414400, 62825541269686074880000, 6436003096247592964096000, 702751431134395346063360000
Offset: 0

Views

Author

Seiichi Manyama, Jul 19 2025

Keywords

Crossrefs

Cf. A000142, A007696, A005809, A386366.
Cf. A378483.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Sum[Binomial[3*k,k] * x^k, {k, 0, nmax}]^(1/3), {x, 0, nmax}], x] * Range[0,nmax]! (* Vaclav Kotesovec, Jul 19 2025 *)
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N, binomial(3*k, k)*x^k)^(1/3)))

Formula

a(n) ~ Pi * 3^(3*n - 1/3) * n^(n - 1/3) / (Gamma(1/3)^2 * exp(n) * 2^(2*n - 1/2)). - Vaclav Kotesovec, Jul 19 2025
E.g.f.: B(x)^(1/3) where B(x) is the g.f. of A005809. - Georg Fischer, Jul 21 2025

A378503 Expansion of (Sum_{k>=0} binomial(4*k,k) * x^k)^3.

Original entry on oeis.org

1, 12, 132, 1396, 14436, 147120, 1483996, 14854968, 147821604, 1464031120, 14443875984, 142042418004, 1393053544508, 13630170286224, 133092301736232, 1297274743175856, 12624909478998948, 122692158505386960, 1190859983017752880, 11545524234978791952, 111820579340839270416
Offset: 0

Views

Author

Seiichi Manyama, Nov 28 2024

Keywords

Crossrefs

Cf. A005810, A078995, A337291, A378484.
Cf. A002457, A378483.

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(sum(k=0, N, binomial(4*k, k)*x^k)^3)

Formula

a(n) = Sum_{i+j+k=n, i,j,k >= 0} binomial(4*i,i) * binomial(4*j,j) * binomial(4*k,k).
G.f.: B(x)^3 where B(x) is the g.f. of A005810.
27*a(n) - 256*a(n-1) = 18*A005810(n) - A337291(n) for n > 0.

A378504 Expansion of (Sum_{k>=0} binomial(3*k,k) * x^k)^4.

Original entry on oeis.org

1, 12, 114, 984, 8055, 63744, 492702, 3742704, 28053423, 208057260, 1529802648, 11168142048, 81041199876, 585045970992, 4204705925670, 30101448952032, 214756404746031, 1527491122906212, 10834911076417458, 76666402505673720, 541277205506059743
Offset: 0

Views

Author

Seiichi Manyama, Nov 28 2024

Keywords

Links

Crossrefs

Cf. A005809, A006256, A378483.
Cf. A002697, A378484.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Sum[Binomial[3*k, k] * x^k, {k, 0, nmax}]^4, {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 23 2025 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(sum(k=0, N, binomial(3*k, k)*x^k)^4)

Formula

a(n) = Sum_{i+j+k+l=n, i,j,k,l >= 0} binomial(3*i,i) * binomial(3*j,j) * binomial(3*k,k) * binomial(3*l,l).
G.f.: B(x)^4 where B(x) is the g.f. of A005809.
4*a(n) - 27*a(n-1) = 3*A006256(n) + A005809(n) for n > 0.
Sum_{n >= 0} a(n) * z^n / (1+z)^(3*n) = (1+z)^4 / (1-2*z)^4. - Marko Riedel, Jul 22 2025
From Vaclav Kotesovec, Jul 23 2025: (Start)
Recurrence: 8*(n-1)*n*(2*n - 1)*a(n) = 6*(n-1)*(36*n^2 - 9*n - 5)*a(n-1) - 81*n*(3*n - 2)*(3*n - 1)*a(n-2).
a(n) ~ n * 3^(3*n+2) / 2^(2*n+4). (End)
Showing 1-4 of 4 results.

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