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.

A377446 E.g.f. satisfies A(x) = 1/(1 + A(x) * log(1 - x))^3.

Original entry on oeis.org

1, 3, 33, 642, 18312, 694242, 32960910, 1883757264, 126015816624, 9664169177136, 836144408644560, 80584887639709296, 8562470160920850144, 994509363776066618256, 125361725299037966371824, 17045520282440541500805504, 2486876242354800277464657792, 387527092638347603108279296512
Offset: 0

Views

Author

Seiichi Manyama, Oct 28 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = 3*sum(k=0, n, (4*k+2)!/(3*k+3)!*abs(stirling(n, k, 1)));

Formula

E.g.f.: B(x)^3, where B(x) is the e.g.f. of A377447.
a(n) = 3 * Sum_{k=0..n} (4*k+2)!/(3*k+3)! * |Stirling1(n,k)|.

A377449 E.g.f. satisfies A(x) = 1/(1 + A(x) * log(1 - x))^4.

Original entry on oeis.org

1, 4, 56, 1388, 50444, 2436176, 147308248, 10720410984, 913099165080, 89150817350880, 9819313409197632, 1204676163038931744, 162935364815509750368, 24088567621306193343360, 3864931159784777490964608, 668886871993798772730203136, 124209455281616641852564586496
Offset: 0

Views

Author

Seiichi Manyama, Oct 28 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = 4*sum(k=0, n, (5*k+3)!/(4*k+4)!*abs(stirling(n, k, 1)));

Formula

E.g.f.: B(x)^4, where B(x) is the e.g.f. of A377448.
a(n) = 4 * Sum_{k=0..n} (5*k+3)!/(4*k+4)! * |Stirling1(n,k)|.

A377692 E.g.f. satisfies A(x) = (1 - log(1 - x) * A(x))^2.

Original entry on oeis.org

1, 2, 12, 118, 1634, 29408, 654040, 17362056, 536410200, 18922946928, 750902659200, 33118793900784, 1607673329621712, 85192554602094912, 4894219487974911552, 303021216528999244416, 20116223556200658052992, 1425479651299747192856832, 107400336067263661850548224
Offset: 0

Views

Author

Seiichi Manyama, Nov 04 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = 2*sum(k=0, n, (2*k+1)!/(k+2)!*abs(stirling(n, k, 1)));

Formula

E.g.f.: 4/(1 + sqrt(1 + 4*log(1-x)))^2.
E.g.f.: B(x)^2, where B(x) is the e.g.f. of A052803.
a(n) = 2 * Sum_{k=0..n} (2*k+1)!/(k+2)! * |Stirling1(n,k)|.
a(n) ~ 2^(7/2) * n^(n-1) / ((exp(1/4) - 1)^(n - 1/2) * exp(3*n/4)). - Vaclav Kotesovec, Aug 27 2025
Showing 1-3 of 3 results.