OFFSET
0,3
COMMENTS
We have A051168(n,k) = (1/n) * Sum_{d | gcd(n,k)} mu(d) * binomial(n/d, k/d) for 0 <= k <= n with n > 0. If n is odd, gcd(2*n + 4, n) = 1. If n is even, gcd(2*n + 4, n) = 2 or 4, but mu(2) = -1 and mu(4) = 0. From these facts, we can prove the formula below. - Petros Hadjicostas, Jul 27 2020
FORMULA
a(n) = binomial(2*n + 4, n)/(2*n + 4), if n is odd, and a(n) = (binomial(2*n + 4, n) - binomial(n + 2, n/2))/(2*n + 4), if n is even. - Petros Hadjicostas, Jul 27 2020
D-finite with recurrence -(n+4) *(n+3) *(11*n-8)*a(n) +10 *(n+3) *(7*n^2+6*n-10) *a(n-1) -60 *n *(n^2-n-5)*a(n-2) -40 *n *(7*n^2+6*n-10) *a(n-3) +16*(n-1) *(13*n+4) *(2*n-3) *a(n-4)=0. - R. J. Mathar, Oct 28 2021
MAPLE
A050182 := proc(n)
binomial(2*n+4, n) ;
if type(n, 'even') then
%-binomial(n+2, n/2) ;
end if;
%/(2*n+4) ;
end proc:
seq(A050182(n), n=0..40) ; # R. J. Mathar, Oct 28 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved