A050182 a(n) = T(2*n+4, n), array T as in A051168 (a count of Lyndon words).

0, 1, 3, 12, 40, 143, 497, 1768, 6288, 22610, 81686, 297160, 1086384, 3991995, 14732005, 54587280, 202995232, 757398510, 2834502346, 10637507400, 40023606896, 150946230006, 570534474698, 2160865067312, 8199711007200

Offset: 0

Author

Clark Kimberling

Keywords

• nonn

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

Links

• Index entries for sequences related to Lyndon words

Crossrefs

A diagonal of the square array described in A051168.

Programs

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

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