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.

A249939 E.g.f.: 1/(5 - 4*cosh(x)).

Original entry on oeis.org

1, 4, 100, 6244, 727780, 136330084, 37455423460, 14188457293924, 7087539575975140, 4514046217675793764, 3570250394992512270820, 3433125893070920512725604, 3944372161432193963534198500, 5336301013125557989981503385444, 8396749419933421378024498580446180
Offset: 0

Views

Author

Paul D. Hanna, Nov 19 2014

Keywords

Comments

a(n) = 4*A242858(2*n) for n>0.
a(n) = A249940(n)/3.
a(n) == 4 (mod 96) for n>0.

Examples

			E.g.f.: E(x) = 1 + 4*x^2/2! + 100*x^4/4! + 6244*x^6/6! + 727780*x^8/8! +...
where E(x) = 1/(5 - 4*cosh(x)) = -exp(x) / (2 - 5*exp(x) + 2*exp(2*x)).
ALTERNATE GENERATING FUNCTION.
E.g.f.: A(x) = 1 + 4*x + 100*x^2/2! + 6244*x^3/3! + 727780*x^4/4! +...
where 3*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 + 2*exp(49*x)/2^7 +...
		

Crossrefs

Programs

  • PARI
    /* E.g.f.: 1/(5 - 4*cosh(x)) */
    {a(n) = local(X=x+O(x^(2*n+1))); (2*n)!*polcoeff( 1/(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) = if(n==0, 1, sum(k=1, (2*n+1)\3, 2*(3*k-1)! * Stirling2(2*n+1, 3*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, 1, (4/3)*sum(k=0, 2*n, k! * Stirling2(2*n, k) ))}
    for(n=0, 20, print1(a(n), ", "))

Formula

E.g.f.: 1/3 + (2/3)*Sum_{n>=1} exp(n^2*x) / 2^n = Sum_{n>=0} a(n)*x^n/n!.
a(n) = (4/3) * Sum_{k=0..2*n} k! * Stirling2(2*n, k) for n>0 with a(0)=1.
a(n) = Sum_{k=1..[(2*n+1)/3]} 2 * (3*k-1)! * Stirling2(2*n+1, 3*k) for n>0 with a(0)=3, after Vladimir Kruchinin in A242858.

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

A247082 E.g.f.: (8 - 7*cosh(x)) / (13 - 12*cosh(x)).

Original entry on oeis.org

1, 5, 365, 66605, 22687565, 12420052205, 9972186170765, 11039636939221805, 16116066766061589965, 29996702068513925975405, 69334618695849722499185165, 194843145588759580915489113005, 654210085817395711127396030796365, 2586566313303319454399746941903834605, 11894287668430209899882926599828701863565
Offset: 0

Views

Author

Paul D. Hanna, Nov 28 2014

Keywords

Comments

The number of 3-level labeled linear rooted trees with 2*n leaves.
A bisection of A050351.
a(n) == 5 (mod 360) for n>0.

Examples

			E.g.f.: E(x) = 1 + 5*x^2/2! + 365*x^4/4! + 66605*x^6/6! + 22687565*x^8/8! +...
where E(x) = (8 - 7*cosh(x)) / (13 - 12*cosh(x)), or, equivalently,
E(x) = (7 - 16*exp(x) + 7*exp(2*x)) / (12 - 26*exp(x) + 12*exp(2*x)).
ALTERNATE GENERATING FUNCTION.
E.g.f.: A(x) = 1 + 5*x + 365*x^2/2! + 66605*x^3/3! + 22687565*x^4/4! +...
where
6*A(x) = 4 + exp(x)*(2/3) + exp(4*x)*(2/3)^2 + exp(9*x)*(2/3)^3 + exp(16*x)*(2/3)^4 + exp(25*x)*(2/3)^5 + exp(36*x)*(2/3)^6 + exp(49*x)*(2/3)^7 +...
		

Crossrefs

Programs

  • Mathematica
    nmax=20; Table[(CoefficientList[Series[(8-7*Cosh[x]) / (13-12*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.: (8 - 7*cosh(x)) / (13 - 12*cosh(x)): */
    {a(n) = local(X=x+O(x^(2*n+1))); (2*n)!*polcoeff( (8 - 7*cosh(X)) / (13 - 12*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) = if(n==0, 1, sum(k=0, 2*n, 2^(k-1) * k! * Stirling2(2*n, k) ))}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* As the Sum of an Infinite Series: */
    \p60 \\ set precision
    Vec(serlaplace(1/2+1/6*sum(n=0,2000,exp(n^2*x)*(2/3)^n*1.)))

Formula

E.g.f.: 1/2 + (1/6)*Sum_{n>=0} exp(n^2*x) * (2/3)^n = Sum_{n>=0} a(n)*x^n/n!.
a(n) = Sum_{k=0..2*n} 2^(k-1) * k! * Stirling2(2*n, k) for n>0 with a(0)=1. [After Vladeta Jovovic in A050351]
a(n) ~ (2*n)! / (6 * (log(3/2))^(2*n+1)). - Vaclav Kotesovec, Nov 29 2014

A250915 E.g.f.: (32 - 31*cosh(x)) / (41 - 40*cosh(x)).

Original entry on oeis.org

1, 9, 2169, 1306809, 1469709369, 2656472295609, 7042235448544569, 25740278881968596409, 124066865052334175027769, 762445058190042799428289209, 5818666543923901596429593478969, 53987940899344324456042542132654009, 598504142090716188282023260396781018169
Offset: 0

Views

Author

Paul D. Hanna, Nov 28 2014

Keywords

Comments

The number of 5-level labeled linear rooted trees with 2*n leaves.
A bisection of A050353.
a(n) == 9 (mod 2160) for n>0.

Examples

			E.g.f.: E(x) = 1 + 9*x^2/2! + 2169*x^4/4! + 1306809*x^6/6! + 1469709369*x^8/8! +...
where E(x) = (32 - 31*cosh(x)) / (41 - 40*cosh(x)).
ALTERNATE GENERATING FUNCTION.
E.g.f.: A(x) = 1 + 9*x + 2169*x^2/2! + 1306809*x^3/3! + 1469709369*x^4/4! +...
where
20*A(x) = 16 + exp(x)*(4/5) + exp(4*x)*(4/5)^2 + exp(9*x)*(4/5)^3 + exp(16*x)*(4/5)^4 + exp(25*x)*(4/5)^5 + exp(36*x)*(4/5)^6 +...
		

Crossrefs

Programs

  • Mathematica
    nmax=20; Table[(CoefficientList[Series[(32-31*Cosh[x]) / (41-40*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.: (32 - 31*cosh(x)) / (41 - 40*cosh(x)): */
    {a(n) = local(X=x+O(x^(2*n+1))); (2*n)!*polcoeff( (32 - 31*cosh(X)) / (41 - 40*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) = if(n==0, 1, sum(k=0, 2*n, 4^(k-1) * k! * Stirling2(2*n, k) ))}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* As the Sum of an Infinite Series: */
    \p100 \\ set precision
    Vec(serlaplace(3/4 + 1/20*sum(n=0,3000,exp(n^2*x)*(4/5)^n*1.)))

Formula

E.g.f.: 3/4 + (1/20)*Sum_{n>=0} exp(n^2*x) * (4/5)^n = Sum_{n>=0} a(n)*x^n/n!.
a(n) = Sum_{k=0..2*n} 4^(k-1) * k! * Stirling2(2*n, k) for n>0 with a(0)=1.
a(n) ~ (2*n)! / (20 * (log(5/4))^(2*n+1)). - Vaclav Kotesovec, Nov 29 2014
Showing 1-4 of 4 results.