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.

A365816 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^3 / (1 - 3 * A(x)).

Original entry on oeis.org

0, 1, 6, 57, 658, 8442, 115692, 1658505, 24565530, 372999198, 5774883348, 90821581578, 1446901409268, 23301338376916, 378711707274072, 6203898306232233, 102329366764727658, 1698047225583890550, 28327664136201303300, 474821679792884860590, 7992739387298462213340
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 25 2023

Keywords

Comments

Reversion of g.f. for hexagonal numbers (with signs).

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^3/(1 - 3 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 3 x)/(1 + x)^3, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[3 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(3*n,n-k-1) * 3^k for n > 0.
a(n) ~ 6^(3*n + 1/2) / (sqrt((481 + 133*sqrt(13))*Pi) * n^(3/2) * (13*sqrt(13) - 35)^n). - Vaclav Kotesovec, Sep 26 2023

A365817 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^3 / (1 - 4 * A(x)).

Original entry on oeis.org

0, 1, 7, 80, 1119, 17437, 290532, 5066364, 91311055, 1687341227, 31797227631, 608727899936, 11805599569092, 231454163924700, 4579765707561240, 91340133073920420, 1834295500622405295, 37059418988408887015, 752741444501505866325, 15362331852042084534240, 314860558967057266779495
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 25 2023

Keywords

Comments

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

Crossrefs

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^3/(1 - 4 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
    CoefficientList[InverseSeries[Series[x (1 - 4 x)/(1 + x)^3, {x, 0, 20}], x], x]	
    Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[3 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(3*n,n-k-1) * 4^k for n > 0.
a(n) ~ 5 * (81 + 21*sqrt(21))^n / (sqrt((427 + 93*sqrt(21))*Pi) * n^(3/2) * 2^(3*n + 3/2)). - Vaclav Kotesovec, Sep 26 2023

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.

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

A366203 a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(3*n,n-k-1) * (n-3)^k.

Original entry on oeis.org

1, 2, 12, 156, 3507, 115692, 5066364, 276943568, 18152243967, 1387267590540, 121106707350928, 11889022355301672, 1296359140925188212, 155440199716271334648, 20327081449263918542412, 2879054747404226046119448, 439060192463001381367975215, 71727764882350305085962745740
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 numbers (with signs).

Crossrefs

Programs

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

Formula

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