OFFSET
0,2
FORMULA
a(n)=sum(A201201(n,k),k=0..n), n>=0.
Apparently a(n)+(2*n+1)*a(n-1)+n*(n+1)*a(n-2)=0, a(-1)=0, a(1)=1. - R. J. Mathar, Dec 07 2011
From Wolfdieter Lang, Dec 12 2011: (Start)
E.g.f. from A201201 with x=1, z->x: g(x)=(1+2*x)*exp(-1/(1+x))*(exp(1)-((Ei(1,-1/(1+x)) - Ei(1,-1))))/(1+x)^4 - x/(1+x)^3, with the exponential integral Ei. In order to obtain the series use first Ei(1,-y/(1+x)) - Ei(1,-y) and put y=1 afterwards.
This e.g.f. satisfies the homogeneous second-order differential equation: (1+x)^2*(d^2/dx^2)g(x) + (5+6*x)*(d/dx)g(x) + 6*g(x) = 0, with g(0)=1 and (d/dx)g(x)|_{x=0} = -3. This is equivalent to the recurrence conjectured by R. J. Mathar, which is thus proved.
(End)
MAPLE
A201202 := proc(n)
add(A201201(n, k), k=0..n) ;
end proc:
seq(A201202(n), n=0..20) ; # R. J. Mathar, Dec 07 2011
MATHEMATICA
a[n_, k_] := (-1)^(n-k)*((n+1)*(n+1)!/((k+1)*(k+1)!))*Binomial[n, k]*HypergeometricPFQ[{-(n-k), k, 1}, {-(n+1), k+2}, 1]; Table[Sum[a[n, k], {k, 0, n}], {n, 0, 20}] (* Jean-François Alcover, Jun 21 2013 *)
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Wolfdieter Lang, Dec 06 2011
STATUS
approved