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A370259

a(n) = (T(n,n+1) - 1)/n^3 for n >= 1, where T(n,x) is the n-th Chebyshev polynomial of the first kind.

1, 2, 9, 75, 961, 16900, 380689, 10498005, 343323841, 13007560326, 560789801881, 27125634729375, 1455389462287489, 85805768251305992, 5515372218107327521, 383931652351786775721, 28778117694539885440129, 2311202255914842794592010, 198009919900727928789497641, 18027589454633803742596931571

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OFFSET

1,2

COMMENTS

It appears that a(2*n+1) is always a square, while a(2*n) = (n + 1) * a square. See A370260 and A370261.

LINKS

Paolo Xausa, Table of n, a(n) for n = 1..350

FORMULA

a(n) = Sum_{k = 1..n} (2^k)*n^(k-2)*binomial(n+k, 2*k)/(n + k) (shows that a(n) is an integer).

a(n) = (cos(n*arccos(n+1)) - 1)/n^3.

a(n) = (A342205(n) - 1)/n^3.

a(n) = ( (n + 1 + sqrt(n*(n+2)))^n + (n + 1 - sqrt(n*(n+2)))^n - 2 )/(2*n^3).