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.

A161797 G.f. satisfies: A(x) = 1/(1 - x/(1 - x*A(x))^3).

Original entry on oeis.org

1, 1, 4, 16, 71, 336, 1660, 8464, 44207, 235306, 1271807, 6961307, 38508659, 214950425, 1209170536, 6848080767, 39014400171, 223439516338, 1285660965508, 7428738358924, 43087099589998, 250766507928988, 1464026402082801
Offset: 0

Views

Author

Paul D. Hanna, Jun 19 2009

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]/(n-k+1) * Binomial[n+2*k-1,n-k], {k,0,n}], {n,0,30}] (* Vaclav Kotesovec, Nov 18 2017 *)
  • PARI
    {a(n,m=1)=sum(k=0,n,binomial(n+m-1,k)*m/(n-k+m)*binomial(n+2*k-1,n-k))}

Formula

a(n) = Sum_{k=0..n} C(n,k)/(n-k+1) * C(n+2*k-1,n-k).
Let A(x)^m = Sum_{n>=0} a(n,m)*x^n then
a(n,m) = Sum_{k=0..n} C(n+m-1,k)*m/(n-k+m) * C(n+2*k-1,n-k).
G.f.: A(x) = (1/x)*serreverse[x/(1 + x/(1 - x)^3)].
Recurrence: 3*(n+1)*(3*n - 2)*(3*n + 2)*(2145*n^4 - 14355*n^3 + 33844*n^2 - 32668*n + 10380)*a(n) = 3*(115830*n^7 - 833085*n^6 + 2195691*n^5 - 2521863*n^4 + 998671*n^3 + 259048*n^2 - 263292*n + 41520)*a(n-1) + 3*(n-2)*(19305*n^6 - 129195*n^5 + 315651*n^4 - 367201*n^3 + 219176*n^2 - 66584*n + 7608)*a(n-2) + 3*(n-3)*(n-2)*(64350*n^5 - 334125*n^4 + 546005*n^3 - 255608*n^2 - 71320*n + 54328)*a(n-3) - 23*(n-4)*(n-3)*(n-2)*(2145*n^4 - 5775*n^3 + 3649*n^2 + 535*n - 654)*a(n-4). - Vaclav Kotesovec, Nov 18 2017
a(n) ~ sqrt(sqrt((159 + 100*sqrt(3))/13) - 2 - 5/sqrt(3)) * (3 + 2*sqrt(3) + sqrt(153 + 100*sqrt(3))/3)^(n+1) / (sqrt(Pi) * n^(3/2) * 2^(n + 5/2)). - Vaclav Kotesovec, Nov 18 2017

A321798 G.f. satisfies: A(x) = 1/(1 - x/(1 - x*A(x))^4).

Original entry on oeis.org

1, 1, 5, 23, 117, 636, 3607, 21106, 126489, 772468, 4789844, 30075937, 190851839, 1222000222, 7885041530, 51222338580, 334720178969, 2198755865424, 14511029102232, 96169424666028, 639757737711300, 4270520564506069, 28595671605541357, 192025292117465445, 1292866976587651519
Offset: 0

Views

Author

Ludovic Schwob, Nov 19 2018

Keywords

Crossrefs

Programs

  • GAP
    List([0..25],n->Sum([0..n],k->Binomial(n,k)/(n-k+1)*Binomial(n+3*k-1,n-k))); # Muniru A Asiru, Nov 24 2018
  • Mathematica
    a[n_] := Sum[Binomial[n, k] * Binomial[n + 3k - 1, n - k]/(n - k + 1), {k, 0,
    n}]; Array[a, 20, 0] (* Amiram Eldar, Nov 19 2018 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(n+3*k-1, n-k)/(n-k+1)); \\ Michel Marcus, Nov 19 2018
    

Formula

a(n) = Sum_{k=0..n} (C(n,k)/(n-k+1)) * C(n+3*k-1,n-k).
a(n) ~ sqrt((1 - r*s)*(1 + 3*r*s)/(8*Pi*(8*s - 3))) / (n^(3/2) * r^(n+1)), where r = 0.139684805934917057093949761392656080860096066578... and s = 1.76437708701490464570032194388560298744432681226... are real roots of the system of equations s*(1 - r/(1 - r*s)^4) = 1, 4*r^2*s^2 = (1 - r*s)^5. - Vaclav Kotesovec, Nov 21 2018

A365088 G.f. satisfies A(x) = 1 + x*A(x) / (1 + x*A(x))^5.

Original entry on oeis.org

1, 1, -4, 1, 46, -129, -405, 3319, -1617, -59258, 199541, 642170, -6038395, 3886091, 119884973, -440626784, -1367688245, 14055527190, -11043763380, -290488387366, 1137260033731, 3336325340735, -36966844508130, 34098313310315, 776097820004580
Offset: 0

Views

Author

Seiichi Manyama, Aug 21 2023

Keywords

Crossrefs

Programs

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

Formula

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

A365115 G.f. satisfies A(x) = 1 + x / (1 - x*A(x))^5.

Original entry on oeis.org

1, 1, 5, 20, 90, 440, 2236, 11720, 62960, 344690, 1916170, 10787762, 61380770, 352410760, 2039099640, 11878519460, 69608606348, 410056995475, 2426936098575, 14424334077975, 86055337016695, 515170271387970, 3093724519080210, 18631778892165080
Offset: 0

Views

Author

Seiichi Manyama, Aug 22 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n, s=5) = sum(k=0, n, binomial(n-k+1, k)*binomial(n+(s-1)*k-1, n-k)/(n-k+1));

Formula

If g.f. satisfies A(x) = 1 + x/(1 - x*A(x))^s, then a(n) = Sum_{k=0..n} binomial(n-k+1,k) * binomial(n+(s-1)*k-1,n-k)/(n-k+1).
Showing 1-4 of 4 results.