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

0,3

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

Links

Table of n, a(n) for n=0..24.

Index entries for sequences related to Lyndon words

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

A diagonal of the square array described in A051168.

Sequence in context: A289652 A026071 A102839 * A222610 A162970 A126725

Adjacent sequences: A050179 A050180 A050181 * A050183 A050184 A050185

Keyword

nonn

Author

Clark Kimberling

Status

approved