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.

A187021 Coefficient of x^n in (1 + (n+1)*x + n*x^2)^n.

Original entry on oeis.org

1, 2, 13, 136, 1921, 33876, 712909, 17383584, 481003009, 14869654300, 507406003501, 18928740714192, 765897591633409, 33392080668673832, 1559976990077534253, 77717020110946293376, 4111810085670587224065, 230190619432401207833004, 13591965974806603671569101
Offset: 0

Views

Author

Emanuele Munarini, Mar 02 2011

Keywords

Crossrefs

Main diagonal of A307883.

Programs

  • Magma
    P:=PolynomialRing(Integers()); [ Coefficients((1+(n+1)*x+n*x^2)^n)[n+1]: n in [0..22] ]; // Klaus Brockhaus, Mar 03 2011
    
  • Maple
    A187021:= n -> simplify( n^(n/2)*GegenbauerC(n, -n, -(n+1)/(2*sqrt(n))) );
    1, seq(A187021(n), n = 1..30); # G. C. Greubel, May 31 2020
    a := n -> hypergeom([-n, -n], [1], n):
    seq(simplify(a(n)), n=0..18); # Peter Luschny, Dec 22 2020
  • Mathematica
    Flatten[{1,Table[Sum[Binomial[n,k]^2*n^k,{k,0,n}],{n,1,20}]}] (* Vaclav Kotesovec, Apr 17 2014 *)
    Table[If[n==0, 1, Simplify[n^(n/2)*GegenbauerC[n, -n, -(n+1)/(2 Sqrt[n])]]], {n, 0, 30}] (* Emanuele Munarini, Oct 20 2016 *)
  • Maxima
    a(n):=coeff(expand((1+(n+1)*x+n*x^2)^n),x,n);
    makelist(a(n),n,0,20);
    
  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)^2*n^k)} \\ Paul D. Hanna, Mar 29 2011
    
  • Sage
    [1]+[ n^(n/2)*gegenbauer(n, -n, -(n+1)/(2*sqrt(n))) for n in (1..30)] # G. C. Greubel, May 31 2020

Formula

a(n) = [x^n] (1 + (n+1)*x + n*x^2)^n.
a(n) = n^(n/2)*GegenbauerPoly(n,-n,-(n+1)/(2*sqrt(n))). - Emanuele Munarini, Oct 20 2016
a(n) = Sum_{k=0..n} binomial(n,k)^2 * n^k. - Paul D. Hanna, Mar 29 2011
a(n) ~ n^(n-1/4) * exp(2*sqrt(n)-1) / (2*sqrt(Pi)). - Vaclav Kotesovec, Apr 17 2014
a(n) = n! * [x^n] exp((n + 1)*x) * BesselI(0,2*sqrt(n)*x). - Ilya Gutkovskiy, May 31 2020
a(n) = hypergeom([-n, -n], [1], n). - Peter Luschny, Dec 22 2020

A241247 a(n) = Sum_{k=0..n} n^k * binomial(n,k)^3.

Original entry on oeis.org

2, 21, 352, 8065, 231876, 7951069, 314931968, 14095941633, 701590424500, 38358147922501, 2281458125531520, 146469277526152321, 10084388675810865248, 740560093656498673965, 57738578482070455269376, 4760258648137662340202497, 413561386818608994516491316
Offset: 1

Views

Author

Vaclav Kotesovec, Apr 18 2014

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[n^k*Binomial[n,k]^3,{k,0,n}],{n,1,20}]
    Table[HypergeometricPFQ[{-n,-n,-n},{1,1},-n],{n,1,20}]
  • PARI
    a(n) = sum(k=0, n, n^k*binomial(n,k)^3); \\ Michel Marcus, Jul 11 2020

Formula

a(n) ~ exp(1 - 3*n^(1/3)/2 + 3*n^(2/3)) * n^(n-2/3) / (2*Pi*sqrt(3)) * (1 + 5/(4*n^(1/3))).

A336188 a(n) = Sum_{k=0..n} n^k * binomial(n,k)^n.

Original entry on oeis.org

1, 2, 13, 352, 38401, 16971876, 29359436149, 207003074670848, 5679112509686022145, 636468045901197095750500, 277939985126193076692203962501, 494649880078824954885176565423811200, 3447375085398645453825889951638344722092289, 97424105704407389799712313421357308088296084669504
Offset: 0

Views

Author

Seiichi Manyama, Jul 11 2020

Keywords

Crossrefs

Programs

  • Magma
    [(&+[n^j*Binomial(n,j)^n: j in [0..n]]): n in [0..20]]; // G. C. Greubel, Aug 26 2022
    
  • Mathematica
    Unprotect[Power]; 0^0 = 1; a[n_] := Sum[n^k * Binomial[n, k]^n, {k, 0, n} ]; Array[a, 14, 0] (* Amiram Eldar, Jul 11 2020 *)
  • PARI
    {a(n) = sum(k=0, n, n^k*binomial(n, k)^n)}
    
  • SageMath
    [sum(n^j*binomial(n,j)^n for j in (0..n)) for n in (0..20)] # G. C. Greubel, Aug 26 2022

Formula

Let f(n) = 2^((n+1)*(2*n-1)/2) * n^(log(n)/8) / Pi^((n-1)/2). For sufficiently large n 0.7675... < a(n)/f(n) < 0.7900... - Vaclav Kotesovec, Jul 11 2020
The above bounds of Vaclav Kotesovec can be recast as: |a(n)/f(n) - exp(-1/4)| <= (3*Pi)^(-2) for sufficiently large n. - Peter Luschny, Jul 12 2020
a(n) ~ exp(-1/4) * QPochhammer(exp(-4)) * QPochhammer(-n*exp(-2), exp(-4)) * 2^(n^2 + n/2) / Pi^(n/2) if n is even and a(n) ~ exp(-3/4) * QPochhammer(exp(-4)) * QPochhammer(-n*exp(-4), exp(-4)) * 2^(n^2 + n/2) * sqrt(n) / Pi^(n/2) if n is odd. - Vaclav Kotesovec, Jul 13 2020
Showing 1-3 of 3 results.