OFFSET
0,2
FORMULA
a(n) = Sum_{k=0..n} binomial(2*n-k,n-k)*(n^2+n*k-k^2-k)/((2*n-k)*(2*n-k-1)), for n>=2.
G.f.: (1-5*x+4*x^2-(1-5*x)*sqrt(1-4x))/(2*x*(1-4*x))
a(n) = Sum_{k=1..n} (3-k)*binomial(2*n-k-1,n-1), n>0, a(0)=1. - Vladimir Kruchinin, Oct 18 2011
From Gary W. Adamson, Nov 14 2011: (Start)
a(n) is the sum of top row terms in M^n, M = an infinite square production matrix as follows, with the Fibonacci sequence as the left border:
1, 1, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
2, 1, 1, 1, 0, 0, ...
3, 1, 1, 1, 1, 0, ...
5, 1, 1, 1, 1, 1, ...
which means the top row of M^n is the n-th row in A189175. (End)
Conjecture: (n+1)*a(n) + 3*(1-3*n)*a(n-1) + 10*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 15 2011
a(n) = Sum_{k=0..n} (k+1) * A090181(n,k). - Alois P. Heinz, Apr 04 2024
EXAMPLE
a(3) = 15 since the top row of M^3 = (6, 5, 3, 1, 0, 0, 0, ...)
MATHEMATICA
T[n_, k_] := If[n==k, 1, Binomial[2n-k, n-k](n^2+n k-k^2-k)/((2n-k)(2n-k-1))]; Table[Sum[T[n, k], {k, 0, n}], {n, 0, 22}]
PROG
(Maxima) T(n, k):=if n=k then 1 else binomial(2*n-k, n-k)*(n^2+n*k-k^2-k)/((2*n-k)*(2*n-k-1));
makelist(sum(T(n, k), k, 0, n), n, 0, 22);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Emanuele Munarini, Apr 18 2011
STATUS
approved