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.

A224732 G.f.: exp( Sum_{n>=1} binomial(2*n,n)^n * x^n/n ).

Original entry on oeis.org

1, 2, 20, 2704, 6008032, 203263062688, 103724721990326528, 801185400238209125917312, 94088900962948953837864576996352, 168691065596220817138271126002845218561536, 4634314586972355372645450331391809316221983940020224
Offset: 0

Views

Author

Paul D. Hanna, Apr 16 2013

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 20*x^2 + 2704*x^3 + 6008032*x^4 + 203263062688*x^5 +...
where
log(A(x)) = 2*x + 6^2*x^2/2 + 20^3*x^3/3 + 70^4*x^4/4 + 252^5*x^5/5 + 924^6*x^6/6 + 3432^7*x^7/7 + 12870^8*x^8/8 +...+ A000984(n)^n*x^n/n +...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,binomial(2*k,k)^k*x^k/k)+x*O(x^n)),n)}
    for(n=0,20,print1(a(n),", "))

Formula

Logarithmic derivative yields A224733.
a(n) ~ exp(-1/8) * 2^(2*n^2) / (Pi^(n/2) * n^(1 + n/2)). - Vaclav Kotesovec, Jan 26 2015
a(n) ~ (binomial(2*n,n))^n / n. - Vaclav Kotesovec, Jan 26 2015

A350595 a(n) = Sum_{k=0..2*n} (-1)^(n+k) * binomial(2*n,k)^n.

Original entry on oeis.org

1, 0, 6, 1680, 5562130, 248832363780, 157933807781230404, 1459371378373349655614400, 199540356506880704345957381087490, 408278793226256174470138460253382778465100, 12605249484391847030759523774663513363531264829120276
Offset: 0

Views

Author

Seiichi Manyama, Jan 08 2022

Keywords

Crossrefs

Main diagonal of A350594.
Cf. A224733.

Programs

  • Mathematica
    Table[Sum[(-1)^(n+k)*Binomial[2*n,k]^n,{k,0,2n}],{n,0,10}] (* Stefano Spezia, Jan 08 2022 *)
  • PARI
    a(n) = sum(k=0, 2*n, (-1)^(n+k)*binomial(2*n, k)^n);
    
  • Python
    from math import comb
    def A350595(n): return sum((-1 if (n+k) % 2 else 1)*comb(2*n,k)**n for k in range(2*n+1)) # Chai Wah Wu, Jan 08 2022

Formula

a(n) ~ c * A224733 = c * binomial(2*n, n)^n, where c = 0.30062580086898437298921168710510240913792796183034926496082316066508397... - Vaclav Kotesovec, Jan 15 2022

A254157 a(n) = binomial(3*n,n)^n.

Original entry on oeis.org

1, 3, 225, 592704, 60037250625, 244217432431215243, 40928832685064366701940736, 287432029715751041166252933120000000, 85609985515193235253656684862285741981771256961, 1091210761769150876962680951989752349788052377750396728515625
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 26 2015

Keywords

Comments

Generally, for p > 1 is
binomial(p*n,n) ~ (p^p/(p-1)^(p-1))^n * sqrt(p/(2*Pi*n*(p-1))) * (1 - (p^2-p+1)/(12*n*p*(p-1))).
binomial(p*n,n)^n ~ exp(-(p^2-p+1)/(12*p*(p-1))) * (p^p/(p-1)^(p-1))^(n^2) * (p/(2*Pi*n*(p-1)))^(n/2).

Crossrefs

Programs

  • Mathematica
    Table[Binomial[3n,n]^n,{n,0,10}]

Formula

a(n) ~ exp(-7/72) * 3^(3*n^2 + n/2) / (2^(2*n^2 + n) * Pi^(n/2) * n^(n/2)).
Showing 1-3 of 3 results.