A217471 -id:A217471 - cp's OEIS frontend

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

Wolfdieter Lang, Oct 13 2012

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

			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.

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.

Cf. A217472, A217474.

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.

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

Original entry on oeis.org

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

Wolfdieter Lang, Oct 12 2012

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.

			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.

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.

Cf. A217474, A217475.

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.

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A217475 Coefficients of polynomials in a Melham conjecture.

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