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.

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A348957 G.f. A(x) satisfies A(x) = (1 + x * A(-x)) / (1 - x * A(x)).

Original entry on oeis.org

1, 2, 2, 10, 18, 98, 210, 1194, 2786, 16258, 39906, 236938, 601458, 3615330, 9399858, 57024426, 150947010, 922283522, 2475603138, 15212318730, 41290579410, 254909413218, 698230131858, 4327273358250, 11943274468770, 74260741616514, 206279837823650, 1286199407132554
Offset: 0

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Author

Ilya Gutkovskiy, Nov 05 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 27; A[] = 0; Do[A[x] = (1 + x A[-x])/(1 - x A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    a[0] = 1; a[n_] := a[n] = -(-1)^n a[n - 1] + Sum[a[k] a[n - k - 1], {k, 0, n - 1}]; Table[a[n], {n, 0, 27}]
    CoefficientList[y/.AsymptoticSolve[y-y^2+x(1+y^3)==0,y->1,{x,0,27}][[1]],x] (* Alexander Burstein, Nov 26 2021 *)

Formula

a(0) = 1; a(n) = -(-1)^n * a(n-1) + Sum_{k=0..n-1} a(k) * a(n-k-1).
a(n) ~ c * 3^(3*n/4) * (1 + sqrt(3))^n / (sqrt(2*Pi) * n^(3/2) * 2^(n/2)), where c = 3^(1/4) if n is even and c = (1 + sqrt(3))/sqrt(2) if n is odd. - Vaclav Kotesovec, Nov 14 2021
From Alexander Burstein, Nov 26 2021: (Start)
G.f.: A(-x) = 1/A(x).
G.f.: A(x) = 1 + x*(1+A(x)^3)/A(x). (End)
a(n) = (-1)^(n-1) * (1/n) * Sum_{k=0..n} binomial(n,k) * binomial(2*n-3*k-2,n-1) for n > 0. - Seiichi Manyama, Apr 11 2024

A371562 G.f. A(x) satisfies A(x) = 1 + x/A(x)^3 * (1 + A(x)^5).

Original entry on oeis.org

1, 2, -2, 30, -166, 1514, -12474, 114006, -1050830, 10005138, -96772786, 951500686, -9469982966, 95267209850, -966979784554, 9891522355270, -101866781649310, 1055294818173474, -10989809960251490, 114983445265899774, -1208092406024272710
Offset: 0

Views

Author

Seiichi Manyama, Apr 13 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = if(n==0, 1, (-1)^(n-1)*sum(k=0, n, binomial(n, k)*binomial(4*n-5*k-2, n-1))/n);

Formula

a(n) = (-1)^(n-1) * (1/n) * Sum_{k=0..n} binomial(n,k) * binomial(4*n-5*k-2,n-1) for n > 0.
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