|
%I #20 Dec 10 2019 12:14:48
%S 1,-3,1,25,-15,4,-553,455,-224,44,32220,-32664,22500,-8316,1276,
%T -4934996,5825600,-5028452,2640220,-771980,96976,1985306180,
%U -2636260484,2688531560,-1791505144,751934040,-181539072,19298224,-2096543510160,3060180107600,-3555908800752,2830338574800,-1521052125120,530958146400,-109131456720,10054374704
%N Coefficient table for polynomials used for the formula of partial sums of odd powers of even-indexed Fibonacci numbers.
%C The following formula is due to Ozeki (see the reference, Theorem 2, p. 109) and also to Prodinger (see the reference, p. 207). Here the version of Prodinger is given which coincides with the one of Ozeki (up to a misprint P instead of 1 in the latter).
%C sum(F(2*k)^(2*m+1),k=0..n) = sum(lambda(m,l)*F(2*n+1)^(2*l+1),l=0..m) + C(m), m>=0, n>= 0, with F=A000045 (Fibonacci), L=A000032 (Lucas),
%C lambda(m,l) = (-5)^(l-m)* sum(binomial(2*m+1,j)*binomial(m-j+l,m-j-l)*
%C (2*(m-j)+1)/L(2*(m-j)+1) ,j=0..m-l)/(2*l+1) and
%C C(m) = (1/5^m)*sum((-1)^(j-1)* binomial(2*m+1,j)*F(2*(m-j)+1)/L(2*(m-j)+1),j=0..m).
%C In order to have an integer triangle T(m,l) instead of the rational lambda(m,l) one uses the sequence pL(m) = product(L(2*i+1),i=0..m), m >= 0, given in A217473, with T(m,l) = pL(m)*lambda(m,l). Similarly, c(m) = pL(m)*C(m) gives the integer sequence A217474 = [-1, 2, -14, 278, -15016, 2172632, -835765304, 851104689248, ...].
%C Thus, pL(m)*sum(F(2*k)^(2*m+1),k=0..n) = sum(T(m,l)*F(2*n+1)^(2*l+1),l=0..m) + c(m), m >= 0, n >= 0.
%C For Melham's conjecture on pL(m)*sum(F(2*k)^(2*m+1),k=0..n) see A217475 where also the reference is given.
%H K. Ozeki, <a href="http://www.fq.math.ca/Papers1/46_47-2/Ozeki.pdf">On Melham's sum</a>, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
%H H. Prodinger, <a href="http://www.fq.math.ca/Papers1/46_47-3/Prodinger.pdf">On a sum of Melham and its variants</a>, The Fibonacci Quart. 46/47 2008/2009), no. 3, 207-215.
%F T(m,l) = pL(m)*lambda(m,l), m >= 0, l = 0..m, with pL(m) = A217473(m) and lambda(m,l) given in a comment above.
%e The triangle T(m,l) begins:
%e m\l 0 1 2 3 4 5 ...
%e 0: 1
%e 1: -3 1
%e 2: 25 -15 4
%e 3: -553 455 -224 44
%e 4: 32220 -32664 22500 -8316 1276
%e 5: -4934996 5825600 -5028452 2640220 -771980 96976
%e ...
%e row 6: 1985306180 -2636260484 2688531560 -1791505144 751934040 -181539072 19298224.
%e row 7: -2096543510160 3060180107600 -3555908800752 2830338574800 -1521052125120 530958146400 -109131456720 10054374704.
%e m=0: 1*sum(F(2*k)^1,k=0..n) = 1*F(2*n+1)^1 - 1, the last term comes from c(0) = A217474 = -1. See A027941.
%e m=1: 1*4*sum(F(2*k)^3,k=0..n) = -3*F(2*n+1)^1 +1*F(2*n+1)^3 + 2. See 4*A163198.
%e m=2: 1*4*11*sum(F(2*k)^5,k=0..n) = 25*F(2*n+1)^1 - 15*F(2*n+1)^3 + 4*F(2*n+1)^5 - 14. See 44*A217471.
%Y Cf. A217474, A217475.
%K sign,easy,tabl
%O 0,2
%A _Wolfdieter Lang_, Oct 12 2012
|
