A217472 Coefficient table for polynomials used for the formula of partial sums of odd powers of even-indexed Fibonacci numbers.

1, -3, 1, 25, -15, 4, -553, 455, -224, 44, 32220, -32664, 22500, -8316, 1276, -4934996, 5825600, -5028452, 2640220, -771980, 96976, 1985306180, -2636260484, 2688531560, -1791505144, 751934040, -181539072, 19298224, -2096543510160, 3060180107600, -3555908800752, 2830338574800, -1521052125120, 530958146400, -109131456720, 10054374704

Offset: 0

Author

Wolfdieter Lang, Oct 12 2012

Keywords

• sign
• easy
• tabl

Comments

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).

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),

lambda(m,l) = (-5)^(l-m)* sum(binomial(2*m+1,j)*binomial(m-j+l,m-j-l)*

(2*(m-j)+1)/L(2*(m-j)+1) ,j=0..m-l)/(2*l+1) and

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).

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, ...].

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.

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.

Examples

			The triangle T(m,l) begins:
m\l        0        1         2        3        4      5  ...
0:         1
1:        -3        1
2:        25      -15         4
3:      -553      455      -224       44
4:     32220   -32664     22500    -8316     1276
5:  -4934996  5825600  -5028452  2640220  -771980  96976
...
row 6:  1985306180   -2636260484   2688531560   -1791505144   751934040   -181539072    19298224.
row 7: -2096543510160  3060180107600 -3555908800752 2830338574800  -1521052125120  530958146400  -109131456720 10054374704.
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.
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.
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.
		

Links

• 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.

Crossrefs

Cf. A217474, A217475.

Formula

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.