OFFSET
0,2
COMMENTS
The sequence is conjectured to be integral [added 03 Mar 2024: now confirmed - see the Formula section].
LINKS
Paolo Xausa, Table of n, a(n) for n = 0..300
FORMULA
a(n) = sqrt( (T(2*n+1, 2*n+2) - 1)/(2*n+1)^3 ), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = sqrt( Sum_{k = 1..2*n+1} (2^k)*(2*n + 1)^(k-2)*binomial(2*n + k + 1, 2*k)/(2*n + k + 1) ).
a(n) = Sum_{k = 0..n} binomial(n+k, n-k)/(2*k + 1) * (4*n + 2)^k (shows the sequence to be integral) = R(n,2), where R(n, x) is the n-th row polynomial of A370262. - Peter Bala, Apr 03 2024
MAPLE
MATHEMATICA
Table[Sqrt[(ChebyshevT[k, k + 1] - 1)/k^3], {k, 1, 40, 2}] (* Paolo Xausa, Jul 24 2024 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Mar 11 2024
STATUS
approved