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A217475 Coefficients of polynomials in a Melham conjecture.

%I A217475 #25 Jan 05 2025 19:51:39
%S A217475 2,1,-14,-3,8,4,278,3,-272,-92,88,44,-15016,2188,19392,3932,-11528,
%T A217475 -4488,2552,1276,2172632,-589732,-3352096,-288860,2774376,809160,
%U A217475 -1156056,-481052,193952,96976,-835765304,313775572,1463316448,-23403160,-1510122768,-308310816,893501136,303807944,-285885248,-123644400,38596448,19298224
%N A217475 Coefficients of polynomials in a Melham conjecture.
%C A217475 The row length sequence for this array is [2,4,6,8,...] = 2*A000027.
%C A217475 A conjecture by Melham (see the reference, eq. 2.7) is:
%C A217475   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.
%C A217475 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.
%C A217475   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.
%C A217475 The absolute terms a(m,0), the first column entries, are given by A217474(m), m>=1.
%C A217475 See also the Wang and Zhang reference, Theorem 2. (D) and the Corollaries 2 and 3. Corollary 3 proves
%C A217475   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
%H A217475 R. S. Melham, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/46_47-4/Melham.pdf">Some conjectures concerning sums of odd powers of Fibonacci and Lucas numbers</a>, The Fibonacci Quart. 46/47 (2008/2009), no. 4, 312-315.
%H A217475 T. Wang and W. Zhang, <a href="http://rms.unibuc.ro/bulletin/volumes/55-1/node9.html">Some identities involving Fibonacci, Lucas polynomials and their applications</a>, Bull. Math. Soc. Sci. Math. Roumanie, Tome 55(103), No.1, (2012) 95-103.
%F A217475 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.
%e A217475 The array a(m,l) starts:
%e A217475 m\l     0      1        2      3      4     5     6      7 ...
%e A217475 1:      2      1
%e A217475 2:    -14     -3        8      4
%e A217475 3:    278      3     -272    -92     88    44
%e A217475 4: -15016   2188    19392   3932 -11528 -4488  2552   1276
%e A217475 ...
%e A217475 Row 5: 2172632 -589732 -3352096 -288860 2774376 809160 -1156056 -481052 193952 96976.
%e A217475 Row 6: -835765304  313775572  1463316448  -23403160  -1510122768 -308310816,893501136 303807944 -285885248 -123644400  38596448  19298224.
%e A217475 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.
%e A217475 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.
%e A217475 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.
%Y A217475 Cf. A217472, A217474.
%K A217475 sign,tabf
%O A217475 1,1
%A A217475 _Wolfdieter Lang_, Oct 13 2012