A217475 Coefficients of polynomials in a Melham conjecture.

2, 1, -14, -3, 8, 4, 278, 3, -272, -92, 88, 44, -15016, 2188, 19392, 3932, -11528, -4488, 2552, 1276, 2172632, -589732, -3352096, -288860, 2774376, 809160, -1156056, -481052, 193952, 96976, -835765304, 313775572, 1463316448, -23403160, -1510122768, -308310816, 893501136, 303807944, -285885248, -123644400, 38596448, 19298224

Offset: 1

Author

Wolfdieter Lang, Oct 13 2012

Keywords

• sign
• tabf

Comments

The row length sequence for this array is [2,4,6,8,...] = 2*A000027.

A conjecture by Melham (see the reference, eq. 2.7) is:

sum(L(2*i+1),i=0..m)*sum(F(2*k)^(2*m+1),k=0..n) = (F(2*n+1)-1)^2*P(2*m-1,F(2*n+1)), where F=A000045 (Fibonacci), L=A000032 (Lucas) and P is an integer polynomial of degree 2*m-1 in x=F(2*n+1), for m >= 1 and n >= 0.

The table a(m,l) lists the coefficients of these polynomials for m=1..6. Thus the conjecture is certainly true for m=1..6.

P(2*m-1,x) = sum(a(m,l)*x^l,l=0..2*m-1), m>=1, where x= F(2*n+1), n>=0.

The absolute terms a(m,0), the first column entries, are given by A217474(m), m>=1.

See also the Wang and Zhang reference, Theorem 2. (D) and the Corollaries 2 and 3. Corollary 3 proves

sum(L(2*i+1),i=0..m)*sum(F(2*k)^(2*m+1),k=0..n) = (F(2*n+1)-1)*H(2*m,F(2*n+1)), with an integer polynomial of degree 2*n. (Thanks go to B. Cloitre for pointing out this paper). - Wolfdieter Lang, Oct 18 2012

Examples

			The array a(m,l) starts:
m\l     0      1        2      3      4     5     6      7 ...
1:      2      1
2:    -14     -3        8      4
3:    278      3     -272    -92     88    44
4: -15016   2188    19392   3932 -11528 -4488  2552   1276
...
Row 5: 2172632 -589732 -3352096 -288860 2774376 809160 -1156056 -481052 193952 96976.
Row 6: -835765304  313775572  1463316448  -23403160  -1510122768 -308310816,893501136 303807944 -285885248 -123644400  38596448  19298224.
Row 7: 851104689248 -394334131664 -1639772952576 174968334112 1989709620800 248542106736 -1492625407328 -403454346592 685716714144 253835649760 -178045414624 -78968332608 20108749408 10054374704.  Thus conjecture is true for m=7 as well.
m=1: 1*4*sum(F(2*k)^3,k=0..n) = 4*A163198(n) = (x-1)^2*(2 + x)  = 2-3*x+x^3 with x=F(2*n+1).  See also A217472, the example for m=1.
m=2: 1*4*11*sum(F(2*k)^5,k=0..n) = 44*A217471(n) = (x-1)^2* (-14 - 3*x + 8*x^2 + 4*x^3) = -14 + 25*x - 15*x^3 + 4*x^5 with x=F(2*n+1). See also A217472, the example for m=2.
		

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.
• 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.

Crossrefs

Cf. A217472, A217474.

Formula

a(m,l) = [x^l]P(2*m-1,x), m>-1, l=0..2*m-1, with the polynomial P appearing in the Melham conjecture stated in the comment section.