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 1-4 of 4 results.

A216831 a(n) = Sum_{k=0..n} binomial(n,k)^3 * k!.

Original entry on oeis.org

1, 2, 11, 88, 905, 11246, 162607, 2668436, 48830273, 983353690, 21570885011, 511212091952, 13001401709881, 352856328962918, 10170853073795975, 310093415465876716, 9964607161173899777, 336439048405066012466, 11902368222382731461083, 440122520333417057761160
Offset: 0

Views

Author

Vaclav Kotesovec, Sep 17 2012

Keywords

Links

Crossrefs

Cf. A000522, A002720, A000172, A119400, A206178, A021009.

Programs

  • Mathematica
    Table[Sum[Binomial[n, k]^3*k!, {k, 0, n}], {n, 0, 25}]
Table[HypergeometricPFQ[{-n, -n, -n}, {1}, -1], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 28 2016 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)^3 * k!); \\ Michel Marcus, May 04 2021

Formula

Recurrence: (8*n^2+31*n+21)*a(n+3) - (24*n^3+157*n^2+308*n+162)*a(n+2) + (24*n^4+117*n^3+178*n^2+71*n-18)*a(n+1) - (8*n^2+31*n+30)*(n+1)^3*a(n) = 0.
a(n) ~ n^(n-1/6)/(sqrt(6*Pi)*exp(n+n^(1/3)-3*n^(2/3)-1/3)). - Vaclav Kotesovec, Sep 30 2012
a(n) = hypergeom([-n, -n, -n], [1], -1). - Vladimir Reshetnikov, Sep 28 2016
a(n) = Sum_{k=0..n} binomial(n, k)*|A021009(n, k)|. - Peter Luschny, May 04 2021
Sum_{n>=0} a(n) * x^n / n!^3 = BesselI(0,2*sqrt(x)) * Sum_{n>=0} x^n / n!^3. - Ilya Gutkovskiy, Jun 19 2022

A385750 a(n) = Sum_{k=0..n} Stirling2(n,k) * (n!/k!)^2.

Original entry on oeis.org

1, 1, 5, 64, 1681, 78651, 5891041, 653545390, 101785047169, 21431911982437, 5927319770834701, 2101574777340578156, 935265924020629176625, 512945332353359967175999, 341342159773993944429746793, 272012935493149854994361194426, 256689188247205271953044107166721, 284051735653584424779666013789038985
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 08 2025

Keywords

Crossrefs

Cf. A000110, A064618, A119392, A119400, A385751, A385752.

Programs

  • Mathematica
    Table[Sum[StirlingS2[n, k] (n!/k!)^2, {k, 0, n}], {n, 0, 17}]
nmax = 17; CoefficientList[Series[Sum[(Exp[x] - 1)^k/k!^3, {k, 0, nmax}], {x, 0, nmax}], x] Range[0, nmax]!^3

Formula

Sum_{n>=0} a(n) * x^n / n!^2 = Sum_{k>=0} (x^k / k!^2) * Product_{j=1..k} 1 / (1 - j*x).
Sum_{n>=0} a(n) * x^n / n!^3 = Sum_{k>=0} (exp(x) - 1)^k / k!^3.

A119401 a(n) = Sum_{k=0..n} (-1)^(n-k)*(n!/k!)^2*binomial(n,k).

Original entry on oeis.org

1, 0, -3, 46, -927, 25476, -922715, 42240402, -2337537279, 147901509928, -9689806983699, 464655683171670, 44744831894861857, -27559636076854374804, 9449663596631181414933, -3046842389019074859527174, 1013788651063121586526459905
Offset: 0

Views

Author

Vladeta Jovovic, Jul 25 2006

Keywords

Crossrefs

Cf. A119400.

Programs

  • Mathematica
    Table[Sum[(-1)^(n - k)*(n!/k!)^2*Binomial[n, k], {k, 0, n}], {n, 0, 16}] (* Stefan Steinerberger, Jun 17 2007 *)

Formula

Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0,2*sqrt(x/(1+x)))/(1+x).
a(n) = -3*(n-1)*n*a(n-1) - 3*(n-1)^4*a(n-2) - (n-2)^3*(n-1)^3*a(n-3). - Vaclav Kotesovec, Jun 08 2019
a(n) = Sum_{k=0..n} (-1)^k*(k!)^2*binomial(n,k)^3. - Ridouane Oudra, Jul 11 2025

Extensions

More terms from Stefan Steinerberger, Jun 17 2007

A336249 a(n) = (n!)^n * Sum_{k=0..n} 1 / ((k!)^n * (n-k)!).

Original entry on oeis.org

1, 2, 7, 172, 79745, 1375363126, 1445639634946657, 136511607703654177490168, 1597074319746489837872943936307201, 3049096207067719868011671739966873049880826186, 1209808678412717193052533393657339738066086793611743000000001
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 14 2020

Keywords

Crossrefs

Cf. A000079, A002720, A036740, A119400.

Programs

  • Mathematica
    Table[(n!)^n Sum[1/((k!)^n (n - k)!), {k, 0, n}], {n, 0, 10}]
Table[(n!)^n SeriesCoefficient[Exp[x] Sum[x^k/(k!)^n, {k, 0, n}], {x, 0, n}], {n, 0, 10}]
  • PARI
    a(n) = (n!)^n * sum(k=0, n, 1 / ((k!)^n * (n-k)!)); \\ Michel Marcus, Jul 14 2020

Formula

a(n) = (n!)^n * [x^n] exp(x) * Sum_{k>=0} x^k / (k!)^n.
a(n) ~ (2*Pi)^((n-1)/2) * n^(n^2 - n/2 + 1/2) / exp(n*(n-1) - 1/12). - Vaclav Kotesovec, Jul 14 2020
Showing 1-4 of 4 results.

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