A220670 Coefficient triangle for powers of x^2 of polynomials appearing in a generalized Melham conjecture on alternating sums of third powers of Chebyshev's S polynomials with odd indices. Coefficients in powers of x^2 of 2 + (-1)^n*S(2*n,x).
3, 3, -1, 3, -3, 1, 3, -6, 5, -1, 3, -10, 15, -7, 1, 3, -15, 35, -28, 9, -1, 3, -21, 70, -84, 45, -11, 1, 3, -28, 126, -210, 165, -66, 13, -1, 3, -36, 210, -462, 495, -286, 91, -15, 1, 3, -45, 330, -924, 1287, -1001, 455, -120, 17, -1, 3, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1
Offset: 0
Examples
The triangle a(n,p) begins: n\p 0 1 2 3 4 5 6 7 8 9 10 ... 0: 3 1: 3 -1 2: 3 -3 1 3: 3 -6 5 -1 4: 3 -10 15 -7 1 5: 3 -15 35 -28 9 -1 6: 3 -21 70 -84 45 -11 1 7: 3 -28 126 -210 165 -66 13 -1 8: 3 -36 210 -462 495 -286 91 -15 1 9: 3 -45 330 -924 1287 -1001 455 -120 17 -1 10: 3 -55 495 -1716 3003 -3003 1820 -680 153 -19 1 ... Row n=2: H(1,2,x^2) := (-3+x^2)*(0 - (S(1,x)/x)^3 + (S(3,x)/x)^3)/((1 - S(4,x))/x^2)^2 = 3 - 3*x^2 + x^4 = 2 + S(4,x). Row n=3: H(1,3,x^2) := (-3+x^2)*(0 - (S(1,x)/x)^3 + (S(3,x)/x)^3 - (S(5,x)/x)^3 )/((1 + S(6,x))/x^2)^2 = 3-6*x^2+5*x^4-x^6 = 2 - S(6,x).
Links
- R. S. Melham, Some conjectures concerning sums of odd powers of Fibonacci and Lucas numbers, The Fibonacci Quart. 46/47 (2008/2009), no. 4, 312-315.
- K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
- H. Prodinger, On a sum of Melham and its variants, The Fibonacci Quart. 46/47 (2008/2009), no. 3, 207-215.
- T. Wang and W. Zhang, Some identities involving Fibonacci, Lucas polynomials and their applications, Bull. Math. Soc. Sci. Math. Roumanie, Tome 55(103), No.1, (2012) 95-103.
Formula
a(n,p) = [x^(2*p)] H(1,n,x^2), with H(1,n,x^2) := (-3+x^2)*sum(((-1)^k)*(S(2*k-1,x)/x)^3,k=0..n)/((1 - (-1)^n*S(2*n,x))/x^2)^2, n >= 1, p = 0..n, and a(0,0):=3.
a(n,p) = [x^(2*p)] (2 + (-1)^n*S(2*n,x)), n >= 0, p = 0..n.
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