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.

A366037 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^5 / (1 - 5 * A(x)).

Original entry on oeis.org

0, 1, 10, 160, 3110, 67155, 1548526, 37346040, 930513870, 23765376580, 618871054120, 16370119905880, 438628647940730, 11880264846822610, 324739360804852980, 8946782070689651280, 248184394985913218910, 6926162613387923126700, 194320992885495965332600, 5477763483026946993808960, 155070883903415687652796120
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 26 2023

Keywords

Comments

Reversion of g.f. for 4-dimensional figurate numbers A002419 (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^5/(1 - 5 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 5 x)/(1 + x)^5, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[5 n, n - k - 1] 5^k, {k, 0, n - 1}], {n, 1, 20}]]

Formula

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(5*n,n-k-1) * 5^k for n > 0.
a(n) ~ sqrt((5168 - 869*sqrt(34)) / (17*Pi)) * (22 - sqrt(34))^(5*n) / (2 * n^(3/2) * 3^(3*n + 3/2) * 5^(4*n + 1) * (11*sqrt(34) - 62)^n). - Vaclav Kotesovec, Sep 27 2023

A365668 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^5 / (1 - 2 * A(x)).

Original entry on oeis.org

0, 1, 7, 73, 905, 12354, 179305, 2715192, 42414021, 678476755, 11058588574, 182999237590, 3066447596459, 51926183715280, 887204891847960, 15276037569668880, 264797324173666845, 4617195655522976361, 80930337327794271445, 1425171253004955494215, 25202145191953299213490
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 26 2023

Keywords

Comments

Reversion of g.f. for 4-dimensional figurate numbers A001296 (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^5/(1 - 2 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 2 x)/(1 + x)^5, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[5 n, n - k - 1] 2^k, {k, 0, n - 1}], {n, 1, 20}]]

Formula

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(5*n,n-k-1) * 2^k for n > 0.
a(n) ~ sqrt(32 - 19*sqrt(5/2)) * 3^(4*n - 3/2) * 5^(3*n) / (sqrt(Pi) * n^(3/2) * 2^(2*n + 3/2) * (25 + 34*sqrt(10))^n). - Vaclav Kotesovec, Sep 27 2023

A366035 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^5 / (1 - 3 * A(x)).

Original entry on oeis.org

0, 1, 8, 98, 1440, 23389, 404712, 7314724, 136476912, 2608808180, 50828498336, 1005682252458, 20152470321984, 408149824237302, 8341496306085040, 171812412714350280, 3562961488550366480, 74328284438252301996, 1558783863783469298016, 32844108784368485209320, 694957689921176181019520
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 26 2023

Keywords

Comments

Reversion of g.f. for 4-dimensional figurate numbers A002417 (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^5/(1 - 3 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 3 x)/(1 + x)^5, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[5 n, n - k - 1] 3^k, {k, 0, n - 1}], {n, 1, 20}]]

Formula

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(5*n,n-k-1) * 3^k for n > 0.
a(n) ~ 2^(4*n - 1) * 5^(5*n + 1/2) / (sqrt(Pi) * n^(3/2) * 3^(7*n + 5/2)). - Vaclav Kotesovec, Sep 27 2023
Showing 1-3 of 3 results.