A217472 -id:A217472 - 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.

A217471 Partial sum of fifth power of the even-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 244, 33012, 4117113, 507401488, 62424765712, 7678070811369, 944346243245076, 116147016764564500, 14285140634333292625, 1756956185432949082176, 216091326285380812359744, 26577476188001703626949937

Offset: 0

Wolfdieter Lang, Oct 11 2012

For the o.g.f. for general powers of Fibonacci numbers F=A000045 see A056588 (row polynomials as numerators) and A055870 (row polynomials as denominator). The even part of the bisection leads to the o.g.f. for powers of F(2*n), and the partial sums of these powers are then given by dividing this o.g.f. by (1-x). For the o.g.f.s for F(n)^5 and F(2*n)^5 see A056572 and A215044, respectively.

The tables of the coefficient of the polynomials which appear in Ozeki's formula and in Melham's conjecture are found in A217472 and A217475, respectively (see References).

			a(2) = 244 = 2*(8-3)/5 - 610/20 + (832040-6765)/55^2 - 7/22.
a(2) = 244 = (1/11)*5^5 - (15/44)*5^3 + (25/44)*5 - 7/22.
a(2) = 244 = (5-1)^2*(4*5^3 + 8*5^2 - 3*5 - 14)/44
           = (4*5^3 + 8*5^2 - 3*5 - 14)*(4/11).

G. C. Greubel, Table of n, a(n) for n = 0..475
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.
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. A163198 (third powers).

Table[Sum[Fibonacci[2*k]^5, {k, 0, n}], {n, 0, 50}] (* G. C. Greubel, Apr 12 2017 *)
Accumulate[Fibonacci[Range[0,30,2]]^5] (* Harvey P. Dale, Jun 30 2025 *)

a(n) = sum(k=1, n, fibonacci(2*k)^5); \\ Michel Marcus, Feb 29 2016

a(n) = Sum_{k=0..n} F(2*k)^5, n>=0.
O.g.f.: x*(1+99*x+416*x^2+99*x^3+x^4)/((1-3*x+x^2)*(1-18*x+x^2)*(1-123*x+x^2)*(1-x)).
a(n) = 2*(F(2*(n+1)) - F(2*n))/5 - F(3*(2*n+1))/20 +
  (F(10*(n+1)) - F(10*n))/F(10)^2 - 7/22 (from the partial fraction decomposition of the o.g.f.).
a(n) = (1/11)*F(2*n+1)^5 - (15/44)*F(2*n+1)^3 + (25/44)*F(2*n+1) - 7/22 (from Ozeki reference, Theorem 2, p. 109 --- with a misprint -- and from Prodinger reference, p. 207).
a(n) =(F(2*n+1)-1)^2*(4*F(2*n+1)^3 + 8*F(2*n+1)^2 - 3*F(2*n+1) - 14)/44 (an example for Melham's conjecture, see the reference, eq. (2.7) for m=2).

A217474 Sequence used for the formula for partial sums of odd powers of even-indexed Fibonacci numbers.

Original entry on oeis.org

-1, 2, -14, 278, -15016, 2172632, -835765304, 851104689248, -2288258540319136, 16212819419809777952, -302332135138133434911104, 14824259801049378686209605248, -1909922987705772492088576593195136, 646210649409632730922299328304587407872

Offset: 0

Wolfdieter Lang, Oct 12 2012

This is the sequence c(m) used in the formula of Ozeki and Prodinger (see the references in A217472) for sum(F(2*k)^(2*m+1),k=1..n), m>=0, m>=0, given in A217472.

			a(2) = (1*4*11)*(-(1/25)*F(5)/L(5) + (1/5)*F(3)/(3) - (2/5)*F(1)/L(1)) = (1*4*11)*(-7/22) = -14.

a(n) = pL(n)*C(n), with pL(n)=A217473(n) and C(n) = (1/5^n)*sum((-1)^(j-1)*binomial(2*n+1,j)*F(2*(n-j)+1)/L(2*(n-j)+1),j=0..n), n>=0, with F=A000045 and L=A000032.

A217473 Product of the first n+1 odd-indexed Lucas numbers A000032.

Original entry on oeis.org

1, 4, 44, 1276, 96976, 19298224, 10054374704, 13714167096256, 48973290700730176, 457851294761126415424, 11206368290573330143917824, 718092873691648422292110244096, 120467978583384630972146706659789056

Offset: 0

Wolfdieter Lang, Oct 12 2012

This sequence is used in A217472 to produce the integer instead of rational Ozeki-Prodinger triangle, and the corresponding sequence A217474.

Cf. A000032, A001622 (phi), 2*A070825 (product of the first n+1 Lucas numbers).

Cf. A217472, A217474.

FoldList[Times, LucasL[Range[1, 26, 2]]] (* Amiram Eldar, Jul 09 2025 *)

a(n) = Product_{i=0..n} L(2*i+1), n >= 0, with the Lucas numbers L=A000032.

a(n) ~ c * phi^(n*(n+2)), where c = Product_{k>=1} (1 - 1/phi^(4*k+2)) = 0.93529381211738925891... . - Amiram Eldar, Jul 09 2025

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A217475 Coefficients of polynomials in a Melham conjecture.

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A217471 Partial sum of fifth power of the even-indexed Fibonacci numbers.

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A217474 Sequence used for the formula for partial sums of odd powers of even-indexed Fibonacci numbers.

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A217473 Product of the first n+1 odd-indexed Lucas numbers A000032.

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