OFFSET
0,2
COMMENTS
Sum of squares of terms in odd-indexed rows of triangle A104967 equals twice this sequence.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
FORMULA
G.f. A(x) satisfies: 2*(1+12*x)*A(x) - (1-16*x^2)*deriv(A(x), x) + 4 = 0.
a(n) = 2^(2*n-1)*(2 + Sum_{k=0..n-1} (-1)^k*binomial(2*k+1,k+1)/2^(2*k)). - Vaclav Kotesovec, Oct 28 2012
MATHEMATICA
Flatten[{1, Table[2^(2*n-1)*(2+Sum[(-1)^k*Binomial[2*k+1, k+1]/2^(2*k), {k, 0, n-1}]), {n, 1, 20}]}] (* Vaclav Kotesovec, Oct 28 2012 *)
PROG
(PARI) {a(n)=local(X=x+x*O(x^(2*n))); sum(k=0, 2*n, polcoeff(polcoeff((1-2*X)/(1-X-X*y*(1-2*X)), 2*n, x), k, y)^2)}
(Magma)
A104970:= func< n | n eq 0 select 1 else 4^n + (&+[(-1)^j*2^(2*n-2*j-1)*Binomial(2*j+1, j+1): j in [0..n-1]]) >;
[A104970(n): n in [0..40]]; // G. C. Greubel, Jun 09 2021
(Sage)
@cached_function
def A104967(n, k): return sum( (-2)^j*binomial(k+1, j)*binomial(n-j, k) for j in (0..n-k))
[A104970(n) for n in (0..50)] # G. C. Greubel, Jun 09 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 30 2005
STATUS
approved