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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A364591 G.f. satisfies A(x) = 1/(1-x) + x^4*A(x)^4.

Original entry on oeis.org

1, 1, 1, 1, 2, 5, 11, 21, 40, 85, 197, 457, 1028, 2289, 5193, 12069, 28338, 66445, 155563, 365701, 865815, 2061133, 4919431, 11758741, 28165412, 67657225, 162977081, 393445865, 951438682, 2304494349, 5591221729, 13588455861, 33075115578, 80616857525, 196742749155
Offset: 0

Views

Author

Seiichi Manyama, Jul 29 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\4, binomial(n-k, 3*k)*binomial(4*k, k)/(3*k+1));

Formula

a(n) = Sum_{k=0..floor(n/4)} binomial(n-k,3*k) * binomial(4*k,k) / (3*k+1).

A385767 G.f. A(x) satisfies A(x) = 1/((1 - x) * (1 - x*A(x)^3 - x^2*A(x)^2*A'(x))).

Original entry on oeis.org

1, 2, 11, 103, 1240, 17405, 272647, 4652676, 85204285, 1657791964, 34030090459, 733238701637, 16520229963511, 388058679087053, 9481616930642904, 240524381652918706, 6324953229391777117, 172191111285984106951, 4847629590517906310392, 140987258808372483601766
Offset: 0

Views

Author

Seiichi Manyama, Jul 09 2025

Keywords

Crossrefs

Programs

  • Mathematica
    terms = 20; A[] = 0; Do[A[x] = 1/((1-x)*(1-x*A[x]^3-x^2*A[x]^2*A'[x])) + O[x]^terms // Normal, terms]; CoefficientList[A[x], x] (* Stefano Spezia, Jul 09 2025 *)

Formula

a(n) = 1 + Sum_{i, j, k, l>=0 and i+j+k+l=n-1} (i+1) a(i) * a(j) * a(k) * a(l).

A364588 G.f. satisfies A(x) = 1/(1-x) + x^2*A(x)^4.

Original entry on oeis.org

1, 1, 2, 5, 15, 49, 170, 613, 2275, 8629, 33301, 130333, 516077, 2063685, 8321892, 33803161, 138181521, 568031297, 2346668400, 9737766513, 40569611691, 169632827345, 711611670532, 2994165070045, 12632782541053, 53433933353885, 226540298098019
Offset: 0

Views

Author

Seiichi Manyama, Jul 29 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\2, binomial(n+k, 3*k)*binomial(4*k,k)/(3*k+1));

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n+k,3*k) * binomial(4*k,k) / (3*k+1).
Previous Showing 11-13 of 13 results.