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.

A249938 E.g.f.: Sum_{n>=0} exp(n^2*x) / 2^(n+1).

Original entry on oeis.org

1, 3, 75, 4683, 545835, 102247563, 28091567595, 10641342970443, 5315654681981355, 3385534663256845323, 2677687796244384203115, 2574844419803190384544203, 2958279121074145472650648875, 4002225759844168492486127539083, 6297562064950066033518373935334635, 11403568794011880483742464196184901963
Offset: 0

Views

Author

Paul D. Hanna, Nov 20 2014

Keywords

Comments

a(n) == 3 (mod 72) for n>0.
Conjectures from Federico Provvedi, Nov 07 2020: (Start)
For n>1, a(n+1) - a(n) == 0 (mod m) if and only if m divides 288.
This sequence is a periodic sequence modulo m, and if m is the k-th prime, the periods of {a(n)} over k-th prime is the sequence of the number of nonzero quadratic residues modulo k-th prime, for all k>0.
Example: k=9, m = prime(9) = 23, for n>0, {a(n) mod 23} generates a period of 11 elements {3, 6, 14, 22, 5, 3, 10, 2, 4, 5, 0}, hence A130290(9) = 11
(End)

Examples

			E.g.f.: A(x) = 1 + 3*x + 75*x^2/2! + 4683*x^3/3! + 545835*x^4/4! +...
where the e.g.f. equals the infinite series:
A(x) = 1/2 + exp(x)/2^2 + exp(4*x)/2^3 + exp(9*x)/2^4 + exp(16*x)/2^5 + exp(25*x)/2^6 + exp(36*x)/2^7 + exp(49*x)/2^8 +...
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[k! * StirlingS2[2*n, k],{k,0,2*n}],{n,0,20}] (* Vaclav Kotesovec, May 04 2015 *)
    Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*(i+r)^(n-r)/(i!*(k-i-r)!), {i, 0, k-r}], {k, r, n}]; a[n_] := Fubini[2n, 1]; a[0] = 1; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Mar 30 2016 *)
    Table[-PolyLog[-2*n, 2] / 2, {n, 0, 48}] (* Federico Provvedi, Nov 07 2020 *)
    HurwitzLerchPhi[1/2, -2*Range[0,48], 0] / 2 (* Federico Provvedi, Nov 11 2020 *)
    -HurwitzLerchPhi[2, -2*Range[0, 48], 1] (*Federico Provvedi,Nov 11 2020*)
  • PARI
    /* E.g.f.: Sum_{n>=0} exp(n^2*x)/2^(n+1) */
    \p100 \\ set precision
    {a(n) = round( n!*polcoeff(sum(m=0, 600, exp(m^2*x +x*O(x^n))/2^(m+1)*1.), n) )}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* E.g.f.: (2 - cosh(x)) / (5 - 4*cosh(x)): */
    {a(n) = local(X=x+O(x^(2*n+1))); (2*n)!*polcoeff( (2 - cosh(X)) / (5 - 4*cosh(X)) , 2*n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* Formula for a(n): */
    {Stirling2(n, k)=n!*polcoeff(((exp(x+x*O(x^n))-1)^k)/k!, n)}
    {a(n) = sum(k=0, 2*n, k! * Stirling2(2*n, k) )}
    for(n=0, 20, print1(a(n), ", "))

Formula

E.g.f.: (2 - cosh(x)) / (5 - 4*cosh(x)) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!.
a(n) = Sum_{k=0..2*n} k! * Stirling2(2*n, k) for n>=0.
a(n) = A000670(2*n), where A000670 is the Fubini numbers.
a(n) ~ (2*n)! / (2 * (log(2))^(2*n+1)). - Vaclav Kotesovec, May 04 2015
a(n) = Sum_{p=1..k, q=1..k} Stirling2(k,p)*Stirling2(k,q)*p!*q!*A008288(p, q) for n>1, where A008288 are the Delannoy numbers. See Chen link. - Michel Marcus, Apr 20 2017
a(n) = Sum_{k>=0} k^(2*n) / 2^(k + 1). - Ilya Gutkovskiy, Dec 19 2019
a(n) = -Polylog(-2*n, 2) / 2. - Federico Provvedi, Nov 07 2020
a(n) = Phi(1/2, -2*n, 0), where Phi(z,s,a) is the Hurwitz-Lerch Zeta transcendental function. - Federico Provvedi, Nov 11 2020

A068942 a(n) = Bo(n^2), n=0,1..., where Bo(n) are the ordered Bell numbers, A000670.

Original entry on oeis.org

1, 1, 75, 7087261, 5315654681981355, 106697365438475775825583498141, 144199280951655469628360978109406917583513090155, 27656793065414932606012896651489726461435178241015434306518713649426461
Offset: 0

Views

Author

Karol A. Penson, Mar 09 2002

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := PolyLog[-n^2, 1/2]/2; a[0] = 1; Table[a[n], {n, 0, 7}] (* Jean-François Alcover, Mar 30 2016 *)
    Table[Sum[k!*StirlingS2[n^2, k], {k, 0, n^2}], {n, 0, 10}] (* Vaclav Kotesovec, Jun 08 2021 *)
  • PARI
    a(n) = sum(k=0, n^2, k!*stirling(n^2, k, 2)); \\ Seiichi Manyama, Jan 17 2022

Formula

a(n) = Sum_{k>=1} (k^(n^2))/2^(k+1); this is the analog of the Dobinski formula.
Integral representation as n-th moment of a positive function on a positive half-axis, in Maple notation: a(n)=int(x^n*(sum(exp(-ln(x)^2/(4*ln(k))) / (2^k*sqrt(ln(k))), k=2..infinity)/(4*sqrt(Pi)*x)+Dirac(x-1)/4), x=0..infinity).
a(n) ~ (n^2)! / (2 * log(2)^(n^2 + 1)). - Vaclav Kotesovec, Jun 08 2021

A249940 E.g.f.: 1 + Sum_{n>=1} 2*exp(n^2*x) / 2^n.

Original entry on oeis.org

3, 12, 300, 18732, 2183340, 408990252, 112366270380, 42565371881772, 21262618727925420, 13542138653027381292, 10710751184977536812460, 10299377679212761538176812, 11833116484296581890602595500, 16008903039376673969944510156332, 25190248259800264134073495741338540
Offset: 0

Views

Author

Paul D. Hanna, Nov 19 2014

Keywords

Examples

			E.g.f.: A(x) = 3 + 12*x + 300*x^2/2! + 18732*x^3/3! + 2183340*x^4/4! +...
where the e.g.f. equals the infinite series:
A(x) = 1 + 2*exp(x)/2 + 2*exp(4*x)/2^2 + 2*exp(9*x)/2^3 + 2*exp(16*x)/2^4 + 2*exp(25*x)/2^5 + 2*exp(36*x)/2^6 +...
We also have the following series expansion:
3/(5 - 4*cosh(x)) = 3 + 12*x^2/2! + 300*x^4/4! + 18732*x^6/6! + 2183340*x^8/8! + 408990252*x^10/10! +...
		

Crossrefs

Programs

  • Mathematica
    nmax=20; Table[(CoefficientList[Series[3/(5-4*Cosh[x]), {x, 0, 2*nmax}], x] * Range[0, 2*nmax]!)[[n]],{n,1,2*nmax+2,2}] (* Vaclav Kotesovec, Nov 29 2014 *)
  • PARI
    /* E.g.f.: 3/(5 - 4*cosh(x)): */
    {a(n) = local(X=x+O(x^(2*n+1))); (2*n)!*polcoeff( 3/(5 - 4*cosh(X)), 2*n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    /* E.g.f.: 1 + Sum_{n>=1} 2*exp(n^2*x)/2^n */
    \p100 \\ set precision
    {a(n) = round( n!*polcoeff(1+2*sum(m=1,500,exp(m^2*x +x*O(x^n))/2^m*1.), n))}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    /* Formula for a(n): */
    {Stirling2(n, k)=n!*polcoeff(((exp(x+x*O(x^n))-1)^k)/k!, n)}
    {a(n) = if(n==0,3, sum(k=0, 2*n, 4*k! * Stirling2(2*n, k) ))}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* Formula for a(n): */
    {Stirling2(n, k)=n!*polcoeff(((exp(x+x*O(x^n))-1)^k)/k!, n)}
    {a(n) = if(n==0,3, 2*sum(k=1,(2*n+1)\3, (3*k)! * Stirling2(2*n+1,3*k) / k))}
    for(n=0,20,print1(a(n),", "))

Formula

E.g.f.: 3/(5 - 4*cosh(x)) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!.
a(n) = Sum_{k=0..2*n} 4*k! * Stirling2(2*n, k) for n>0 with a(0)=3.
a(n) = Sum_{k=1..[(2*n+1)/3]} 2*(3*k)! * Stirling2(2*n+1,3*k) / k for n>0 with a(0)=3, after Vladimir Kruchinin in A242858.
a(n) ~ 2 * (2*n)! / (log(2))^(2*n+1). - Vaclav Kotesovec, Nov 29 2014

A384408 Expansion of Product_{k>=1} 1/(1 - k^3 * x)^((1/2)^(k+1)).

Original entry on oeis.org

1, 13, 2426, 2393226, 7056543721, 46153703519501, 564874416706639304, 11596724623199364432312, 369937054535706501459633546, 17326810763609633232550088712162, 1140582994940898154002780391375267884, 101920298764725526200442366857326292990348
Offset: 0

Views

Author

Seiichi Manyama, May 28 2025

Keywords

Crossrefs

Programs

  • PARI
    a000670(n) = sum(k=0, n, k!*stirling(n, k, 2));
    my(N=20, x='x+O('x^N)); Vec(exp(sum(k=1, N, a000670(3*k)*x^k/k)))

Formula

G.f.: exp(Sum_{k>=1} A000670(3*k) * x^k/k).
a(n) ~ sqrt(Pi) * 3^(3*n + 1/2) * n^(3*n - 1/2) / (sqrt(2) * exp(3*n) * log(2)^(3*n+1)). - Vaclav Kotesovec, May 29 2025
Showing 1-4 of 4 results.