A217475 -id:A217475 - cp's OEIS frontend

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

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.

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).

A220670 Coefficient triangle for powers of x^2 of polynomials appearing in a generalized Melham conjecture on alternating sums of third powers of Chebyshev's S polynomials with odd indices. Coefficients in powers of x^2 of 2 + (-1)^nS(2n,x).

Original entry on oeis.org

3, 3, -1, 3, -3, 1, 3, -6, 5, -1, 3, -10, 15, -7, 1, 3, -15, 35, -28, 9, -1, 3, -21, 70, -84, 45, -11, 1, 3, -28, 126, -210, 165, -66, 13, -1, 3, -36, 210, -462, 495, -286, 91, -15, 1, 3, -45, 330, -924, 1287, -1001, 455, -120, 17, -1, 3, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1

Offset: 0

Wolfdieter Lang, Jan 07 2013

For the original Melham conjecture on sums of odd powers of even-indexed Fibonacci numbers see the references given in A217475. See especially the Wang and Zhang reference given there.
An analog conjecture stated for Chebyshev's S polynomials (see A049310) is product(tau(j,x),j=0..m)*sum(((-1)^k)*(S(2*k-1,x)/x)^(2*m+1),k=0..n)/(P(n,x^2)^2) = H(m,n,x^2), with P(n,x^2) := (1 - (-1)^n*S(2*n,x))/x^2 and certain integer polynomials H with degree (2*m-1)*n + binomial(m-1,2) in x^2. The coefficients of powers of x^2 of the monic integer polynomials tau(n,x):= 2*T(2*n+1,x/2)/x, with Chebyshev's T polynomials, are given by the signed A111125 triangle (see a comment there from Oct 23 2012). The coefficients of the powers of x^(2*j) of the polynomials P(n,x^2) are found in (-1)^(n-1)*A109954(n-1,j).
Here the conjecture is considered for m=1 (third powers): H(1,n,x^2) = sum(a(n,p)*x^(2*p),p=0..n), n >= 1. It is conjectured that in fact H(1,n,x^2) = 2 + (-1)^n*S(2*n,x). This has been checked by Maple for n=1..100.  Therefore we have added a(0,0) = 3 (in the conjecture above this would be the undetermined 0/0).
The original Melham conjecture for m=1 (third powers), appears by putting x = i (the imaginary unit): 1*4*sum(F(2*k)^3)/(1-F(2*n+1))^2 = sum(a(n,p)*(-1)^p) = 2 + F(2*n+1) (the unsigned row sums of the present triangle). This m=1 identity is, of course, proved.
The row sums of this triangle are given by 2 + (-1)^n*S(2*n,1) = 2 + (-1)^n*((2/sqrt(3))*sin((2*n+1)*Pi/3)) producing the period 6 sequence periodic (3, 2, 1, 1, 2, 3).

			The triangle a(n,p) begins:
n\p 0    1    2     3    4     5    6    7   8   9 10 ...
0:  3
1:  3   -1
2:  3   -3    1
3:  3   -6    5    -1
4:  3  -10   15    -7    1
5:  3  -15   35   -28    9    -1
6:  3  -21   70   -84   45   -11    1
7:  3  -28  126  -210  165   -66   13  -1
8:  3  -36  210  -462  495  -286   91  -15   1
9:  3  -45  330  -924 1287 -1001  455 -120  17  -1
10: 3  -55  495 -1716 3003 -3003 1820 -680 153 -19  1
...
Row n=2: H(1,2,x^2) := (-3+x^2)*(0 - (S(1,x)/x)^3 + (S(3,x)/x)^3)/((1 - S(4,x))/x^2)^2 = 3 - 3*x^2 + x^4 =
  2 + S(4,x).
Row n=3:  H(1,3,x^2) := (-3+x^2)*(0 - (S(1,x)/x)^3 + (S(3,x)/x)^3 - (S(5,x)/x)^3 )/((1 + S(6,x))/x^2)^2 =  3-6*x^2+5*x^4-x^6 = 2 - S(6,x).

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.
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, A049310, A111125 (signed), A109954 (signed), A217475, A220671 (fifth powers).

a(n,p) = [x^(2*p)] H(1,n,x^2), with H(1,n,x^2) := (-3+x^2)*sum(((-1)^k)*(S(2*k-1,x)/x)^3,k=0..n)/((1 - (-1)^n*S(2*n,x))/x^2)^2, n >= 1, p = 0..n, and a(0,0):=3.

a(n,p) = [x^(2*p)] (2 + (-1)^n*S(2*n,x)), n >= 0, p = 0..n.

A220671 Coefficient array for powers of x^2 of polynomials appearing in a generalized Melham conjecture on alternating sums of fifth powers of Chebyshev S polynomials with odd indices.

Original entry on oeis.org

-14, 15, -20, 8, -1, 55, -170, 221, -153, 59, -12, 1, 115, -670, 1773, -2696, 2549, -1538, 589, -138, 18, -1, 195, -1850, 8215, -21530, 36330, -41110, 31865, -17080, 6314, -1579, 255, -24, 1, 295, -4150, 27735, -110795, 289540, -518290, 654595, -595805, 396316, -193906, 69641, -18129, 3327, -408, 30, -1

Offset: 1

Wolfdieter Lang, Jan 11 2013

The row lengths sequence is 3*n + 1 = A016777(n).
For the generalized Melham conjecture and links to the references concerned with the Melham conjecture on sums of fifth powers of even-indexed Fibonacci numbers see a comment under A220670.
Here the conjecture is considered for m=2 (fifth powers): H(2,n,x^2):= product(tau(j,x), j=0..2) * sum(((-1)^k)*(S(2*k-1,x)/x)^5, k=0..n) / (P(n,x^2)^2), with P(n,x^2):= (1 - (-1)^n*S(2*n,x))/x^2. For tau(j,x):= 2*T(2*j+1,x/2)/x, with Chebyshev's T polynomials see a Oct 23 2012 comment on A111125. For the polynomials P see signed A109954. The conjecture is that H(2,n,x^2) is an integer polynomial of degree 3*n:  H(2,n,x^2) = sum(a(n,p)*x^(2*p), p=0..3*n), n >= 1.
If one puts x = i (the imaginary unit) one obtains the original Melham conjecture for Fibonacci numbers F = A000045.
H(2,n,-1) = +44*sum(F(2*k)^5,k=0..n)/(1+F(2*n+1))^2, n>=1, which is conjectured to be -14 - 3*y(n) + 8*y(n)^2 + 4*y(n)^3, with y(n):=F(2*n+1) (see row m=2 of A217475).
It is conjectured that H(2,n,x^2) = h(2,n,x^2) - 3*z(n) + 8*z(n)^2 + 4*z(n)^3, with z(n):= ((-1)^n)*S(2*n,x), with h an integer polynomial of degree 3*n. See A220672 for the coefficients of h(2,n,x^2) for n = 1..5. Because h(2,n,-1) = -14 by the usual Melham conjecture, we put h(2,0,x^2) = -14.

			The array a(n,p) begins:
  n\p   0     1     2      3      4      5     6      7   8     9   10 11 12
  0:  -14
  1:   15   -20     8     -1
  2:   55  -170   221   -153     59    -12     1
  3:  115  -670  1773  -2696   2549  -1538   589   -138   18    -1
  4:  195 -1850  8215 -21530  36330 -41110 31865 -17080 6314 -1579 255 -24 1
...
Row n=5: [295, -4150, 27735, -110795, 289540, -518290, 654595, -595805, 396316, -193906, 69641, -18129, 3327, -408, 30, -1],
Row n=6: [415, -8120, 76118, -429531, 1599441, -4125672, 7621983, -10350335, 10539787, -8164410, 4853792, -2222153, 781514, -209172, 41823, -6047, 597, -36, 1].

Cf. A220670, A217475, A220672.

a(n,p) = [x^(2*p)] H(2,n,x^2), n>=1, with H(2,n,x^2) defined in a comment above. a(0,0) has been put to -14 ad hoc.

A220672 Coefficients of powers of x^2 of polynomials, called h(2,n,x^2), appearing in a conjecture on alternating sums of fifth powers of odd-indexed Chebyshev S polynomials stated in A220671.

Original entry on oeis.org

-14, 6, 5, -12, 3, 46, -95, 16, 75, -69, 24, -3, 106, -520, 928, -607, -351, 894, -651, 234, -42, 3, 186, -1600, 5840, -11355, 11005, -1110, -9615, 11580, -6906, 2433, -513, 60, -3, 286, -3775, 22360, -75595, 153515, -177565, 77115, 84495, -171324, 145302, -75831, 26235, -6057, 900, -78, 3

Offset: 0

Wolfdieter Lang, Jan 14 2013

The row lengths sequence for this irregular triangle is 3*n+1 = A016777(n).
A generalized Melham conjecture involving fifth powers (m=2) of odd-indexed Chebyshev S polynomials (see A049310) is H(2,n,x^2):= (x^2-3)*(x^4-5*x^2+5)*sum(((-1)^k)*(S(2*k-1,x)/x)^(2*m+1), k=0..n)/((1 - (-1)^n*S(2*n,x))/x^2)^2 = h(2,n,x^2) - 3*z(n) + 8*z(n)^2 + 4*z(n)^3, with z(n):= ((-1)^n)*S(2*n,x), and h an integer polynomial of degree 3*n.
  The present array a(n,p) appears as h(2,n,x^2) = sum(a(n,p)*x^(2*p),p=0..3*n), n >= 1. The entry a(0,0) := -14 has been used because, in accordance with the original Melham conjecture (see a comment on A220671), h(2,n,i^2), with the imaginary unit i, is conjectured to be -14, for all n >= 1.
  [-14, -3, 8, 4] is row m=2 of A217475.

			The array a(n,p) begins:
n\p   0     1    2     3     4    5     6    7    8  9
0:  -14
1:    6     5  -12     3
2:   46   -95   16    75   -69   24    -3
3:  106  -520  928  -607  -351  894  -651  234  -42  3
...
Row n=4: [186, -1600, 5840, -11355, 11005, -1110, -9615, 11580, -6906, 2433, -513, 60, -3];
Row n=5: [286, -3775, 22360, -75595, 153515, -177565, 77115, 84495, -171324, 145302, -75831, 26235, -6057, 900, -78, 3].
Thus the conjecture is true at least for n=1..5.

Cf. A220670, A220671.

a(n,p) = [x^(2p)] h(0,2,n,x^2), with the polynomial h defined above in a comment. The conjecture is that h is an integer polynomial of degree 3n in x^2.

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A220670 Coefficient triangle for powers of x^2 of polynomials appearing in a generalized Melham conjecture on alternating sums of third powers of Chebyshev's S polynomials with odd indices. Coefficients in powers of x^2 of 2 + (-1)^n*S(2*n,x).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A220671 Coefficient array for powers of x^2 of polynomials appearing in a generalized Melham conjecture on alternating sums of fifth powers of Chebyshev S polynomials with odd indices.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A220672 Coefficients of powers of x^2 of polynomials, called h(2,n,x^2), appearing in a conjecture on alternating sums of fifth powers of odd-indexed Chebyshev S polynomials stated in A220671.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A220670 Coefficient triangle for powers of x^2 of polynomials appearing in a generalized Melham conjecture on alternating sums of third powers of Chebyshev's S polynomials with odd indices. Coefficients in powers of x^2 of 2 + (-1)^nS(2n,x).