A115066 Chebyshev polynomial of the first kind T(n,x), evaluated at x=n.
1, 1, 7, 99, 1921, 47525, 1431431, 50843527, 2081028097, 96450076809, 4993116004999, 285573847759211, 17882714781360001, 1216895030905226413, 89415396036432386311, 7055673735003659189775, 595077930963909484707841, 53421565080956452077519377
Offset: 0
Examples
a(3) = 99 because T[3, x] = 4x^3 - 3x and T[3, 3] = 4*3^3 - 3*3 = 99.
References
- G. Freud, Orthogonal Polynomials, Pergamon Press, Oxford, 1966, p. 35.
- M. Rosenblum and J. Rovnyak, Hardy Classes and Operator Theory, Dover, New York, 1985, page 18.
- G. Szego, Orthogonal polynomials, Amer. Math. Soc., Providence, 1939, p. 29.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..351
Crossrefs
Programs
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Maple
with(orthopoly): seq(T(n,n),n=0..17);
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Mathematica
Table[ChebyshevT[n, n], {n, 0, 17}] (* Arkadiusz Wesolowski, Nov 17 2012 *) -
PARI
A115066(n)=cos(n*acos(n)) \\ M. F. Hasler, Apr 06 2012
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PARI
a(n) = polchebyshev(n, 1, n); \\ Seiichi Manyama, Dec 28 2018
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PARI
a(n) = if(n==0, 1, n*sum(k=0, n, (2*n-2)^k*binomial(n+k, 2*k)/(n+k))); \\ Seiichi Manyama, Mar 05 2021
Formula
a(n) = cos(n*arccos(n)).
a(n) ~ 2^(n-1) * n^n. - Vaclav Kotesovec, Jan 19 2019
a(n) = n * Sum_{k=0..n} (2*n-2)^k * binomial(n+k,2*k)/(n+k) for n > 0. - Seiichi Manyama, Mar 05 2021
It appears that a(2*n+1) == 0 (mod (2*n+1)^2) and 2*a(4*n+2) == -2 (mod (4*n+2)^4), while for k > 1, 2*a(2^k*(2*n+1)) == 2 (mod (2^k*(2*n+1))^4). - Peter Bala, Feb 01 2022
Extensions
Edited by N. J. A. Sloane, Apr 05 2006
Comments