login
A217475
Coefficients of polynomials in a Melham conjecture.
5
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
1,1
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
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.
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.
EXAMPLE
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.
CROSSREFS
Sequence in context: A280412 A370826 A155729 * A288298 A288762 A187920
KEYWORD
sign,tabf
AUTHOR
Wolfdieter Lang, Oct 13 2012
STATUS
approved