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A220670
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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).
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3
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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
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OFFSET
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0,1
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COMMENTS
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For the original Melham conjecture on sums of odd powers of even-indexed Fibonacci numbers see the references given in A217475. See especially the Wang and Zhang reference given there.
An analog conjecture stated for Chebyshev's S polynomials (see A049310) is product(tau(j,x),j=0..m)*sum(((-1)^k)*(S(2*k-1,x)/x)^(2*m+1),k=0..n)/(P(n,x^2)^2) = H(m,n,x^2), with P(n,x^2) := (1 - (-1)^n*S(2*n,x))/x^2 and certain integer polynomials H with degree (2*m-1)*n + binomial(m-1,2) in x^2. The coefficients of powers of x^2 of the monic integer polynomials tau(n,x):= 2*T(2*n+1,x/2)/x, with Chebyshev's T polynomials, are given by the signed A111125 triangle (see a comment there from Oct 23 2012). The coefficients of the powers of x^(2*j) of the polynomials P(n,x^2) are found in (-1)^(n-1)*A109954(n-1,j).
Here the conjecture is considered for m=1 (third powers): H(1,n,x^2) = sum(a(n,p)*x^(2*p),p=0..n), n >= 1. It is conjectured that in fact H(1,n,x^2) = 2 + (-1)^n*S(2*n,x). This has been checked by Maple for n=1..100. Therefore we have added a(0,0) = 3 (in the conjecture above this would be the undetermined 0/0).
The original Melham conjecture for m=1 (third powers), appears by putting x = i (the imaginary unit): 1*4*sum(F(2*k)^3)/(1-F(2*n+1))^2 = sum(a(n,p)*(-1)^p) = 2 + F(2*n+1) (the unsigned row sums of the present triangle). This m=1 identity is, of course, proved.
The row sums of this triangle are given by 2 + (-1)^n*S(2*n,1) = 2 + (-1)^n*(2/sqrt(3*sin((2*n+1)*Pi/3)) producing the period 6 sequence periodic(3, 2, 1, 1, 2, 3).
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LINKS
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K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
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FORMULA
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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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EXAMPLE
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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).
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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