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 1-5 of 5 results.

A366016 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^4 / (1 - 4 * A(x)).

Original entry on oeis.org

0, 1, 8, 102, 1580, 27193, 499828, 9609372, 190869948, 3886281300, 80681111940, 1701418017390, 36345240847188, 784821812522062, 17103169093916120, 375670490644949624, 8308349385885678684, 184856293637482503660, 4134886240989315235840, 92928784113832360511800, 2097399158679611824619120
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 26 2023

Keywords

Comments

Reversion of g.f. for heptagonal pyramidal numbers (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^4/(1 - 4 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 4 x)/(1 + x)^4, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[4 n, n - k - 1] 4^k, {k, 0, n - 1}], {n, 1, 20}]]

Formula

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(4*n,n-k-1) * 4^k for n > 0.
a(n) ~ sqrt(163 - 1521/sqrt(89)) * (4933 + 801*sqrt(89))^n / (sqrt(Pi) * n^(3/2) * 2^(9*n + 9/2)). - Vaclav Kotesovec, Sep 27 2023

A366014 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^4 / (1 - 2 * A(x)).

Original entry on oeis.org

0, 1, 6, 54, 580, 6873, 86688, 1141500, 15512220, 215928900, 3063184410, 44124882750, 643692232404, 9490176205006, 141184118174640, 2116751269990968, 31951313566227228, 485159929343783532, 7405637373574690968, 113572576254948487800, 1749075343256441443320
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 26 2023

Keywords

Comments

Reversion of g.f. for pentagonal pyramidal numbers (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^4/(1 - 2 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 2 x)/(1 + x)^4, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[4 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(4*n,n-k-1) * 2^k for n > 0.

A366017 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^4 / (1 - 5 * A(x)).

Original entry on oeis.org

0, 1, 9, 132, 2365, 47169, 1005564, 22431720, 517122117, 12222124035, 294569159313, 7212098118888, 178877944712844, 4484938858752940, 113488477622130600, 2894560146756466320, 74335973069605120725, 1920587845828953301479, 49886703842977713177723, 1301959618949870922531300, 34123873581608909988904245
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 26 2023

Keywords

Comments

Reversion of g.f. for octagonal pyramidal numbers (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^4/(1 - 5 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 5 x)/(1 + x)^4, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[4 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(4*n,n-k-1) * 5^k for n > 0.

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

A366204 a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(4*n,n-k-1) * (n-3)^k.

Original entry on oeis.org

1, 3, 22, 305, 6873, 223300, 9609372, 517122117, 33450100420, 2528420918595, 218708219876094, 21304932729509468, 2307805461194581390, 275157252809857575960, 35806664475402303854328, 5049845899886455033320237, 767208489677203200554103660, 124917404793477227061928480153
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 04 2023

Keywords

Comments

a(n) is the coefficient of x^n in expansion of series reversion of g.f. for n-gonal pyramidal numbers (with signs).

Crossrefs

Programs

  • Mathematica
    Unprotect[Power]; 0^0 = 1; Table[1/n Sum[Binomial[n + k - 1, k] Binomial[4 n, n - k - 1] (n - 3)^k, {k, 0, n - 1}], {n, 1, 18}]
    Table[Binomial[4 n, n - 1] Hypergeometric2F1[1 - n, n, 3 n + 2, 3 - n]/n, {n, 1, 18}]
    Table[SeriesCoefficient[InverseSeries[Series[x (1 - (n - 3) x)/(1 + x)^4, {x, 0, n}], x], {x, 0, n}], {n, 1, 18}]

Formula

a(n) = [x^n] Series_Reversion( x * (1 - (n - 3) * x) / (1 + x)^4 ).
Showing 1-5 of 5 results.