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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

Table of n, a(n) for n=0..57.

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.

Index entries for sequences related to Chebyshev polynomials.

Index entries for triangles and arrays related to Pascal's triangle

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.)