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.

A136580 Row sums of triangle A136579.

Original entry on oeis.org

1, 1, 3, 7, 27, 127, 747, 5167, 41067, 368047, 3669867, 40284847, 482671467, 6267305647, 87660962667, 1313941673647, 21010450850667, 357001369769647, 6423384156578667, 122002101778601647, 2439325392333218667
Offset: 0

Views

Author

Gary W. Adamson, Jan 09 2008

Keywords

Examples

			a(4) = 27 = sum of row 4 terms, triangle A136579: (1 + 0 + 2 + 0 + 24) = 0! + 2! + 4!.
a(5) = 127 = sum of row 5 terms, triangle A136579: (0 + 1 + 0 + 6 + 0 + 120) = 1! + 3! + 5!
G.f. = 1 + x + 3*x^2 + 7*x^3 + 27*x^4 + 127*x^5 + 747*x^6 + 5167*x^7 + 41067*x^8 + ...
		

Crossrefs

Programs

  • Maple
    A136580 := proc(n)
        add( (n-2*i)!,i=0..floor(n/2) ) ;
    end proc: # R. J. Mathar, Jun 04 2021
  • Mathematica
    a[0] = 1; a[1] = 1; a[2] = 3; a[n_] := a[n] = n a[n-1] + a[n-2] - n a[n-3]; Table[a[n], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 29 2015 *)

Formula

G.f.: 2/(1-x^2)/G(0), where G(k)= 1 + 1/(1 - 1/(1 - 1/(2*x*(k+1)) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 29 2013
G.f.: Q(0)/(1-x^2), where Q(k) = 1 - x*(k+1)/( x*(k+1) - 1/(1 - x*(k+1)/( x*(k+1) - 1/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2013
From Vladimir Reshetnikov, Oct 29 2015: (Start):
a(n) = (-1)^n*exp(1)*Gamma(0, 1)/2 - Re(Gamma(0, -1))*exp(-1)/2 + (n+2)!*((-1)^n*Re(Gamma(-n-2, -1))*exp(-1)-Gamma(-n-2, 1)*exp(1))/2, where Gamma(a, x) is the upper incomplete Gamma function.
D-finite with recurrence: a(0) = 1, a(1) = 1, a(2) = 3, a(n) = n*a(n-1) + a(n-2) - n*a(n-3).
E.g.f.: 1/(1-x) + (exp(x-1)*(Ei(1)-Ei(1-x)) + exp(1-x)*(Ei(x-1)-Ei(-1)))/2, where Ei(x) is the exponential integral.
a(n+1)-a(n) = A153229(n+2) = (-1)^(n+1)*A058006(n+1).
(End)
0 = a(n)*(+a(n+1) - a(n+2) - a(n+3) + a(n+4)) + a(n+1)*(+a(n+1) - a(n+2) - 2*a(n+3)) + a(n+2)*(+a(n+2) + a(n+3) - a(n+4)) + a(n+3)*(+a(n+3)) for all n>=0. - Michael Somos, Oct 29 2015