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%I #22 Oct 28 2021 10:53:17
%S 0,1,3,12,40,143,497,1768,6288,22610,81686,297160,1086384,3991995,
%T 14732005,54587280,202995232,757398510,2834502346,10637507400,
%U 40023606896,150946230006,570534474698,2160865067312,8199711007200
%N a(n) = T(2*n+4, n), array T as in A051168 (a count of Lyndon words).
%C 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
%H <a href="/index/Lu#Lyndon">Index entries for sequences related to Lyndon words</a>
%F 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
%F 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
%p A050182 := proc(n)
%p binomial(2*n+4,n) ;
%p if type(n,'even') then
%p %-binomial(n+2,n/2) ;
%p end if;
%p %/(2*n+4) ;
%p end proc:
%p seq(A050182(n),n=0..40) ; # _R. J. Mathar_, Oct 28 2021
%Y A diagonal of the square array described in A051168.
%K nonn
%O 0,3
%A _Clark Kimberling_
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