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

0,1

Comments

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).

Links

Table of n, a(n) for n=0..65.

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.

Example

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).

Crossrefs

Cf, A049310, A111125 (signed), A109954 (signed), A217475, A220671 (fifth powers).

Sequence in context: A144437 A169609 A353527 * A264526 A138071 A206400

Adjacent sequences: A220667 A220668 A220669 * A220671 A220672 A220673

Keyword

sign,tabl

Author

Wolfdieter Lang, Jan 07 2013

Status

approved