A370260 - OEIS

%I #17 Jul 24 2024 17:33:55

%S 1,3,31,617,18529,748859,38149567,2348482961,169641143873,

%T 14071599763379,1318414335714015,137720427724123513,

%U 15871136311527376801,2000355821099358166891,273735526097742996298111,40419227378551955037029921,6405616571975691389276400257

%N a(n) = sqrt(A370259(2*n+1)).

%C The sequence is conjectured to be integral [added 03 Mar 2024: now confirmed - see the Formula section].

%H Paolo Xausa, <a href="/A370260/b370260.txt">Table of n, a(n) for n = 0..300</a>

%F 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.

%F 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) ).

%F 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

%p A370259 := n -> simplify( (ChebyshevT(n, n+1) - 1)/n^3 ):

%p seq(sqrt(A370259(2*n+1)), n = 0..20);

%t Table[Sqrt[(ChebyshevT[k, k + 1] - 1)/k^3], {k, 1, 40, 2}] (* _Paolo Xausa_, Jul 24 2024 *)

%Y Cf. A008310, A342205, A370259, A370261, A370262.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Mar 11 2024