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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A360025 a(n) = Sum_{k=0..floor(n/3)} (-1)^k * binomial(n-2*k,k) * Catalan(k).

Original entry on oeis.org

1, 1, 1, 0, -1, -2, -1, 2, 7, 9, 3, -16, -39, -43, 9, 126, 247, 199, -213, -984, -1555, -756, 2525, 7518, 9593, 559, -24899, -56216, -55241, 33150, 225879, 407194, 273199, -529745, -1938549, -2822128, -833219, 6083986, 15904733, 18288966, -4172187, -61154333
Offset: 0

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Author

Seiichi Manyama, Jan 22 2023

Keywords

Links

Crossrefs

Cf. A360024, A360026, A360027.
Cf. A000108, A216604, A349047.

Programs

  • PARI
    a(n) = sum(k=0, n\3, (-1)^k*binomial(n-2*k, k)*binomial(2*k, k)/(k+1));
  • PARI
    my(N=50, x='x+O('x^N)); Vec(2/(1-x+sqrt((1-x)^2+4*x^3*(1-x))))

Formula

a(n) = 1 - Sum_{k=0..n-3} a(k) * a(n-k-3).
G.f. A(x) satisfies: A(x) = 1/(1-x) - x^3 * A(x)^2.
G.f.: 2 / ( 1-x + sqrt((1-x)^2 + 4*x^3*(1-x)) ).
D-finite with recurrence +(n+3)*a(n) +2*(-n-2)*a(n-1) +(n+1)*a(n-2) +2*(2*n-3)*a(n-3) +4*(-n+2)*a(n-4)=0. - R. J. Mathar, Jan 25 2023

A349048 G.f. A(x) satisfies: A(x) = 1 / (1 - x + x^4 * A(x)).

Original entry on oeis.org

1, 1, 1, 1, 0, -2, -5, -9, -12, -10, 3, 35, 91, 163, 215, 163, -136, -858, -2107, -3675, -4639, -2879, 5161, 23741, 54910, 91988, 108843, 47483, -186582, -700420, -1527461, -2440985, -2656442, -507076, 6617735, 21456279, 44213835, 67037683, 65541879, -9699085, -232548686
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 06 2021

Keywords

Links

Crossrefs

Cf. A000108, A007440, A023426, A100223, A214649, A343773, A349047.

Programs

  • Mathematica
    nmax = 40; A[] = 0; Do[A[x] = 1/(1 - x + x^4 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = 1; a[n_] := a[n] = a[n - 1] - Sum[a[k] a[n - k - 4], {k, 0, n - 4}]; Table[a[n], {n, 0, 40}]
Table[Sum[(-1)^k Binomial[n - 2 k, 2 k] CatalanNumber[k], {k, 0, Floor[n/4]}], {n, 0, 40}]

Formula

G.f.: (-1 + x + sqrt((1 - x)^2 + 4*x^4)) / (2*x^4).
a(0) = 1; a(n) = a(n-1) - Sum_{k=0..n-4} a(k) * a(n-k-4).
a(n) = Sum_{k=0..floor(n/4)} (-1)^k * binomial(n-2*k,2*k) * Catalan(k).
a(n) = F([(1-n)/4, (2-n)/4, (3-n)/4, -n/4], [2, (1-n)/2, -n/2], -64), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 07 2021
Showing 1-2 of 2 results.

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