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%I #15 Nov 10 2019 20:26:22
%S 1,1,1,1,4,3,1,6,15,10,1,8,28,56,35,1,10,45,120,210,126,1,12,66,220,
%T 495,792,462,1,14,91,364,1001,2002,3003,1716,1,16,120,560,1820,4368,
%U 8008,11440,6435,1,18,153,816,3060,8568,18564,31824,43758,24310,1,20,190
%N One half of even powers of 2*x in terms of Chebyshev's T-polynomials.
%C See A122366 for one half of odd powers of 2*x in terms of Chebyshev's T-polynomials.
%C This is, for n >= 1, the left half of Pascal's triangle for even rows with the central coefficients divided by 2.
%C The signed version of this triangle, b(n,k) := a(n,k)*(-1)^(n-k), appears in the formula (1/2)*(2*sin(phi))^(2*n) = (Sum_{k=0..n-1} b(n,k)*cos(2*(n-k)*phi)) + a(n,n).
%C Correspondingly, (1/2)*(4*(1-x^2))^n = (Sum_{k=0..n-1} b(n,k)*T(2*(n-k),x)) + a(n,n).
%C The proofs follow from Euler's formula 2*x = 2*cos(phi) = exp(i*phi) + exp(-i*phi) or 2*sqrt(1-x^2) = 2*sin(phi) = (exp(i*phi) - exp(-i*phi))/i and the binomial formula.
%D Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990. pp. 54-55, Ex. 1.5.31.
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 795.
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F a(n,k) = binomial(2*n,k), k=0..n-1 and a(n,n) = binomial(2*n,n)/2, n >= 1. Instead of a(0,0)=1 one should take 1/2.
%e [1/2];
%e [ 1, 1];
%e [ 1, 4, 3];
%e [ 1, 6, 15, 10];
%e [ 1, 8, 28, 56, 35];
%e ...
%e Row n=3: [1, 6, 15, 20/2 = 10] appears in ((2*x)^6)/2 = 1*T(6,x) + 6*T(4,x) + 15*T(2,x) + 10.
%e Row n=3: [1, 6, 15, 20/2 = 10] appears in ((2*cos(phi))^6)/2 = 1*cos(6*phi) + 6*cos(4*phi) + 15*cos(2*phi) + 10.
%e The signed row n=3, [-1, 6, -15, +20/2 = 10], appears in ((4*(1-x^2))^3)/2 = -1*T(6,x) + 6*T(4,x) - 15*T(2,x) + 10).
%e The signed row n=3, [-1, 6, -15, +20/2 = 10], appears in ((2*sin(phi))^6)/2 = -1*cos(6*phi) + 6*cos(4*phi) - 15*cos(2*phi) + 10.
%Y Cf. A122366, A008314, A008311.
%K nonn,tabl,easy
%O 0,5
%A _Wolfdieter Lang_, Mar 07 2007
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