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.

A368524 a(n) = Sum_{k=1..n} k^2 * n^(n-k).

Original entry on oeis.org

0, 1, 6, 30, 180, 1455, 15666, 213500, 3521736, 68101245, 1508916310, 37661140506, 1045012524348, 31900040161899, 1062139933257690, 38299757176168440, 1486670929792295696, 61800664096000744569, 2738952078516469743678, 128909373997071187219990
Offset: 0

Views

Author

Seiichi Manyama, Dec 28 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=1, n, k^2*n^(n-k));

Formula

a(n) = [x^n] x * (1+x)/((1-n*x) * (1-x)^3).
a(n) = n * (n+1) * (n^n - n^2 + n - 1)/(n-1)^3 for n > 1.

A368527 a(n) = Sum_{k=1..n} k^3 * n^k.

Original entry on oeis.org

0, 1, 34, 804, 18244, 434205, 11138766, 310151632, 9370253320, 306232628625, 10783859167810, 407523041660196, 16461877678462668, 708207095198943613, 32338800248010936694, 1562509380160144645440, 79657105206246202521616
Offset: 0

Views

Author

Seiichi Manyama, Dec 28 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=1, n, k^3*n^k);

Formula

a(n) = [x^n] n*x * (1+4*n*x+(n*x)^2)/((1-x) * (1-n*x)^4).
a(n) = n * (n^n * (n^6-3*n^5+8*n^3-4*n^2-7*n-1) + n^2 + 4*n + 1)/(n-1)^4 for n > 1.

A368530 a(n) = Sum_{k=1..n} k^3 * 4^(n-k).

Original entry on oeis.org

0, 1, 12, 75, 364, 1581, 6540, 26503, 106524, 426825, 1708300, 6834531, 27339852, 109361605, 437449164, 1749800031, 6999204220, 27996821793, 111987293004, 447949178875, 1791796723500, 7167186903261, 28668747623692, 114674990506935, 458699962041564
Offset: 0

Views

Author

Seiichi Manyama, Dec 29 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=1, n, k^3*4^(n-k));

Formula

G.f.: x * (1+4*x+x^2)/((1-4*x) * (1-x)^4).
a(n) = 8*a(n-1) - 22*a(n-2) + 28*a(n-3) - 17*a(n-4) + 4*a(n-5).
a(n) = (11*4^(n+1) - (9*n^3 + 36*n^2 + 60*n + 44))/27.
a(0) = 0; a(n) = 4*a(n-1) + n^3.
Showing 1-3 of 3 results.