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 A127673 One half of even powers of 2*x in terms of Chebyshev's T-polynomials. 4
 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, 495, 792, 462, 1, 14, 91, 364, 1001, 2002, 3003, 1716, 1, 16, 120, 560, 1820, 4368, 8008, 11440, 6435, 1, 18, 153, 816, 3060, 8568, 18564, 31824, 43758, 24310, 1, 20, 190 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS See A122366 for one half of odd powers of 2*x in terms of Chebyshev's T-polynomials. This is, for n >= 1, the left half of Pascal's triangle for even rows with the central coefficients divided by 2. 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). Correspondingly, (1/2)*(4*(1-x^2))^n = (Sum_{k=0..n-1} b(n,k)*T(2*(n-k),x)) + a(n,n). 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. REFERENCES 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. LINKS M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy]. M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 795. FORMULA 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. EXAMPLE [1/2]; [ 1, 1]; [ 1, 4,  3]; [ 1, 6, 15, 10]; [ 1, 8, 28, 56, 35]; ... 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. 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. 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). 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. CROSSREFS Cf. A122366, A008314, A008311. Sequence in context: A283299 A190157 A103552 * A016698 A200115 A038763 Adjacent sequences:  A127670 A127671 A127672 * A127674 A127675 A127676 KEYWORD nonn,tabl,easy AUTHOR Wolfdieter Lang, Mar 07 2007 STATUS approved

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Last modified December 11 12:33 EST 2019. Contains 329916 sequences. (Running on oeis4.)