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.

A266481 E.g.f.: Limit_{N->oo} [ Sum_{n>=0} (N + n)^(2*n) * (x/N)^n/n! ]^(1/N).

Original entry on oeis.org

1, 1, 5, 55, 993, 25501, 857773, 35850795, 1795564865, 104972371417, 7022842421301, 529428563641759, 44421725002096225, 4106744812439019765, 414834196219620026333, 45462732300569936279251, 5373006006732947705188737, 681229881246574750274962225, 92237589983019368975021777125, 13283769418970268811752725081607, 2027649185923009220298941142143201, 326999803592314489529958494308640461, 55558592280735155060861740192416874125
Offset: 0

Views

Author

Paul D. Hanna, Dec 30 2015

Keywords

Comments

Compare to: Limit_{N->oo} [ Sum_{n>=0} (N + n)^n * x^n/n! ]^(1/N) = Sum_{n>=0} (n+1)^(n-1) * x^n/n!.
Conjecture: a(p*n) = 1 (mod p) for n>=0 and all prime p.

Examples

			E.g.f.: A(x) = 1 + x + 5*x^2/2! + 55*x^3/3! + 993*x^4/4! + 25501*x^5/5! + 857773*x^6/6! + 35850795*x^7/7! + 1795564865*x^8/8! + 104972371417*x^9/9! + 7022842421301*x^10/10! +...
where A(x) equals the limit, as N -> oo, of the series
[1 + (N+1)^2*(x/N) + (N+2)^4*(x/N)^2/2! + (N+3)^6*(x/N)^3/3! + (N+4)^8*(x/N)^4/4! + (N+5)^10*(x/N)^5/5! + (N+6)^12*(x/N)^6/6! +...]^(1/N).
RELATED SERIES.
The following limit exists:
G(x) = Limit_{N->oo} [ Sum_{n>=0} (N + n)^(2*n) * (x/N)^n/n! ] / A(x)^N
where
G(x) = 1 + 2*x + 22*x^2/2! + 432*x^3/3! + 12220*x^4/4! + 451480*x^5/5! + 20591784*x^6/6! + 1117635008*x^7/7! + 70348179472*x^8/8! + 5037843612960*x^9/9! + 404453425948000*x^10/10! +...+ A266522(n)*x^n/n! +...
Logarithm of the g.f. A(x) begins:
Log(A(x)) = x + 4*x^2/2! + 42*x^3/3! + 752*x^4/4! + 19360*x^5/5! + 654912*x^6/6! + 27546736*x^7/7! + 1388207872*x^8/8! + 81621893376*x^9/9! + 5488951731200*x^10/10! +...+ A266526(n)*x^n/n! +...
and forms a diagonal in the triangles A266521 and A266488.
		

Crossrefs

Programs

  • PARI
    {A266526(n) = n! * polcoeff( polcoeff( log( sum(m=0,n+1, (m + y)^(2*m) *x^m/m! ) +x*O(x^n) ),n,x), n+1,y)}
    {a(n) = n! * polcoeff( exp( sum(m=1,n+1, A266526(m)*x^m/m! ) +x*O(x^n)), n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    /* Informal listing of terms 0..30 */
    \p100
    P(n) = sum(k=0,31, (n+k)^(2*k) * x^k/k! +O(x^31))
    Vec(round( serlaplace( subst(P(10^100)^(1/10^100),x,x/10^100) )*1.) )

Formula

E.g.f. exp( Sum_{n>=0} A266526(n)*x^n/n! ), where A266526(n) = [x^n*y^(n+1)/n!] log( Sum_{n>=0} (n + y)^(2*n) * x^n/n! ).
a(n) ~ c * d^n * n^(n-2), where d = 2*(1 + sqrt(2)) * exp(1 - sqrt(2)) = 3.19091339076710837219515616759285808414857..., c = sqrt(1 - 1/sqrt(2)) * exp(3 - 2*sqrt(2)) = 0.642492128663019850313957348436... . - Vaclav Kotesovec, Jan 01 2016, updated Mar 17 2024

A266526 a(n) = coefficient of x^n*y^(n+1)/n! in Log( Sum_{n>=0} (n + y)^(2*n) * x^n/n! ), for n>=1.

Original entry on oeis.org

1, 4, 42, 752, 19360, 654912, 27546736, 1388207872, 81621893376, 5488951731200, 415721105434624, 35026876903256064, 3250356630453317632, 329437813126362185728, 36214170617862339840000, 4291812357982293898231808, 545518054282041342531076096, 74032137722410904128877494272, 10684317262536125210489796296704, 1634019721630446295055397683200000
Offset: 1

Views

Author

Paul D. Hanna, Jan 01 2016

Keywords

Comments

Equals the logarithm of the e.g.f. of A266481.
Equals the right border of triangle A266521.

Examples

			E.g.f: A(x) = x + 4*x^2/2! + 42*x^3/3! + 752*x^4/4! + 19360*x^5/5! + 654912*x^6/6! + 27546736*x^7/7! + 1388207872*x^8/8! + 81621893376*x^9/9! + 5488951731200*x^10/10! +...
where exponentiation yields the e.g.f. of A266481:
exp(A(x)) = 1 + x + 5*x^2/2! + 55*x^3/3! + 993*x^4/4! + 25501*x^5/5! + 857773*x^6/6! + 35850795*x^7/7! + 1795564865*x^8/8! + 104972371417*x^9/9! +...+ A266481(n)*x^n/n! +...
which equals
Limit_{N->oo} [ Sum_{n>=0} (N + n)^(2*n) * (x/N)^n/n! ]^(1/N).
		

Crossrefs

Programs

  • PARI
    {a(n) = n! * polcoeff( polcoeff( log( sum(m=0,n+1, (m + y)^(2*m) *x^m/m! ) +x*O(x^n) ),n,x), n+1,y)}
    for(n=1,30, print1(a(n),", "))

Formula

a(n) ~ 2^(n - 1/4) * (1 + sqrt(2))^(n - 1/2) * exp((1 - sqrt(2))*n) * n^(n-2). - Vaclav Kotesovec, Mar 20 2024

A266522 E.g.f.: Limit_{N->oo} [ Sum_{n>=0} (N + n)^(2*n) * (x/N)^n/n! ] / F(x)^N, where F(x) = Limit_{N->oo} [ Sum_{n>=0} (N + n)^(2*n) * (x/N)^n/n! ]^(1/N).

Original entry on oeis.org

1, 2, 22, 432, 12220, 451480, 20591784, 1117635008, 70348179472, 5037843612960, 404453425948000, 35977638091065088, 3512312454013520832, 373346162796913784192, 42922941487808176036480, 5307003951337894697856000, 702183042248318469458657536, 98997224309112273722486891008, 14815674464782854979394204308992, 2345767767928443601985964232355840, 391750020994050554579656281189760000, 68820978855281989513379320801711429632
Offset: 0

Views

Author

Paul D. Hanna, Dec 30 2015

Keywords

Comments

The e.g.f. A(x) of this sequence also satisfies:
A(x*y) = Limit_{N->oo} [ Sum_{n>=0} (N + n*y)^(2*n) * (x/N)^n/n! ] / G(x,y)^N
where
G(x,y) = Limit_{N->oo} [ Sum_{n>=0} (N + n*y)^(2*n) * (x/N)^n/n! ]^(1/N)
for all real y.

Examples

			E.g.f.: A(x) = 1 + 2*x + 22*x^2/2! + 432*x^3/3! + 12220*x^4/4! + 451480*x^5/5! + 20591784*x^6/6! + 1117635008*x^7/7! + 70348179472*x^8/8! + 5037843612960*x^9/9! + 404453425948000*x^10/10! + ...
such that
A(x) = Limit_{N->oo} [ Sum_{n>=0} (N + n)^(2*n) * (x/N)^n/n! ] / F(x)^N
where
F(x) = Limit_{N->oo} [ Sum_{n>=0} (N + n)^(2*n) * (x/N)^n/n! ]^(1/N)
and
F(x) = 1 + x + 5*x^2/2! + 55*x^3/3! + 993*x^4/4! + 25501*x^5/5! + 857773*x^6/6! + 35850795*x^7/7! + 1795564865*x^8/8! + 104972371417*x^9/9! + 7022842421301*x^10/10! +...+ A266481(n)*x^n/n! + ...
RELATED SERIES.
log(A(x)) = 2*x + 18*x^2/2! + 316*x^3/3! + 8272*x^4/4! + 288048*x^5/5! + 12523584*x^6/6! + 652959872*x^7/7! + 39701769216*x^8/8! + 2758053332736*x^9/9! + ... + A266521(n,n)*x^n/n! + ...
		

Crossrefs

Formula

E.g.f.: exp( Sum_{n>=1} A266521(n,n)*x^n/n! ), where the e.g.f. of triangle A266521 is Log(Sum_{n>=0} (n + y)^(2*n) * x^n/n!). - Paul D. Hanna, Sep 30 2018
Showing 1-3 of 3 results.