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 1-3 of 3 results.

A349534 G.f. A(x) satisfies A(x) = 1 / ((1 - 3 * x) * (1 - x * A(x)^2)).

Original entry on oeis.org

1, 4, 21, 138, 1063, 9075, 82770, 789204, 7766721, 78267306, 803447526, 8371413999, 88300495746, 941004684748, 10116276976218, 109578418285452, 1194764348642313, 13102287157827918, 144422108994233625, 1599198859915070640, 17780781456147340764
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 21 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = 1/((1 - 3 x) (1 - x A[x]^2)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    a[n_] := a[n] = 3^n + Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 20}]
    Table[Sum[Binomial[n + k, 2 k] Binomial[3 k, k] 3^(n - k)/(2 k + 1), {k, 0, n}], {n, 0, 20}]

Formula

a(n) = 3^n + Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
a(n) = Sum_{k=0..n} binomial(n+k,2*k) * binomial(3*k,k) * 3^(n-k) / (2*k+1).
a(n) = 3^n*F([1/3, 2/3, -n, 1+n], [1/2, 1, 3/2], -3^2/2^4), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 21 2021
a(n) ~ sqrt(5/Pi) * 3^(n-1) * 4^n / n^(3/2). - Vaclav Kotesovec, Nov 22 2021

A349535 G.f. A(x) satisfies A(x) = 1 / ((1 - 4 * x) * (1 - x * A(x)^2)).

Original entry on oeis.org

1, 5, 31, 232, 2007, 19213, 196912, 2114360, 23467039, 266945359, 3094963919, 36433302124, 434269638196, 5230582604476, 63560612650120, 778281652963504, 9593248679529551, 118938874458415987, 1482248642350235245, 18557299563105608336
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 21 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 19; A[] = 0; Do[A[x] = 1/((1 - 4 x) (1 - x A[x]^2)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    a[n_] := a[n] = 4^n + Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 19}]
    Table[Sum[Binomial[n + k, 2 k] Binomial[3 k, k] 4^(n - k)/(2 k + 1), {k, 0, n}], {n, 0, 19}]

Formula

a(n) = 4^n + Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
a(n) = Sum_{k=0..n} binomial(n+k,2*k) * binomial(3*k,k) * 4^(n-k) / (2*k+1).
a(n) = 4^n*F([1/3, 2/3, -n, 1+n], [1/2, 1, 3/2], -3^3/2^6), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 21 2021
a(n) ~ 273^(1/4) * (59 + 3*sqrt(273))^(n + 1/2) / (9 * sqrt(Pi) * n^(3/2) * 2^(3*n + 5/2)). - Vaclav Kotesovec, Nov 22 2021

A364641 G.f. satisfies A(x) = 1/(1 - 2*x) - x*A(x)^3.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 10, 16, 31, 59, 101, 206, 376, 692, 1408, 2528, 4943, 9767, 17755, 35950, 68659, 129029, 262758, 490832, 958948, 1920580, 3581020, 7203080, 14054600, 26665160, 54195040, 103450560, 201749935, 406617695, 769870535, 1539785150, 3042812185
Offset: 0

Views

Author

Seiichi Manyama, Jul 31 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, (-1)^k*2^(n-k)*binomial(n+k, 2*k)*binomial(3*k, k)/(2*k+1));

Formula

a(n) = Sum_{k=0..n} (-1)^k * 2^(n-k) * binomial(n+k,2*k) * binomial(3*k,k) / (2*k+1).
Showing 1-3 of 3 results.