A215039 -id:A215039 - cp's OEIS frontend

A163198 Sum of the cubes of the first n even-indexed Fibonacci numbers.

0, 1, 28, 540, 9801, 176176, 3162160, 56744793, 1018249596, 18271762300, 327873509425, 5883451505856, 105574253853888, 1894453118539345, 33994581881622076, 610008020755286076, 10946149791725643705, 196420688230338021808, 3524626238354441796016, 63246851602149831726825

Offset: 0

Stuart Clary, Jul 24 2009

Natural bilateral extension (brackets mark index 0): ..., 9801, 540, 28, 1, 0, [0], 1, 28, 540, 9801, 176176, ... This is A163198-reversed followed by A163198. That is, A163198(-n) = A163198(n-1).

G. C. Greubel, Table of n, a(n) for n = 0..500
Stuart Clary and Paul D. Hemenway, On sums of cubes of Fibonacci numbers, Applications of Fibonacci Numbers, Vol. 5 (St. Andrews, 1992), 123-136, Kluwer Acad. Publ., 1993. See equations (3), (46), (47), and (49).
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.
K. Subba Rao, Some properties of Fibonacci numbers, Amer. Math. Monthly, 60(10):680-684, Dec. 1953. See page 682.
Index entries for linear recurrences with constant coefficients, signature (22,-77,77,-22,1).

Cf. A163194, A163195, A163196, A163197, A163199, A163200, A163201, A163202, A005968, A215039 (first differences).

a[n_Integer] := If[ n >= 0, Sum[ Fibonacci[2k]^3, {k, 1, n} ], -Sum[ Fibonacci[-2k]^3, {k, 1, -n - 1} ] ]
LinearRecurrence[{22, -77, 77, -22, 1}, {0, 1, 28, 540, 9801}, 50] (* G. C. Greubel, Dec 09 2016 *)
Accumulate[Fibonacci[Range[0,40,2]]^3] (* Harvey P. Dale, Nov 15 2023 *)

a(n) = sum(k=1, n, fibonacci(2*k)^3); \\ Michel Marcus, Feb 29 2016

concat([0], Vec(x*(1 + 6*x + x^2)/((1 - x)*(1 - 3*x + x^2 )*(1 - 18*x + x^2)) + O(x^50))) \\ G. C. Greubel, Dec 09 2016

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
a(n) = Sum_{k=1..n} F(2k)^3.
a(n) = A163199(n) + 1.
a(n) = (1/20)*(F(6n+3) - 12*F(2n+1) + 10).
a(n) = (1/4)*(F(2n+1)^3 - 3*F(2n+1) + 2). (K. Subba Rao)
a(n) = (1/4)*F(n)^2*L(n+1)^2*F(n-1)*L(n+2) = A163195(n) if n is even.
a(n) = (1/4)*L(n)^2*F(n+1)^2*L(n-1)*F(n+2) = A163197(n) if n is odd.
a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = 8.
a(n) - 22 a(n-1) + 77 a(n-2) - 77 a(n-3) + 22 a(n-4) - a(n-5) = 0.
G.f.: (x + 6*x^2 + x^3)/(1 - 22*x + 77*x^2 - 77*x^3 + 22*x^4 - x^5) = x*(1 + 6*x + x^2)/((1 - x)*(1 - 3*x + x^2 )*(1 - 18*x + x^2)).
a(n) = (F(2*n+1)-1)^2*(F(2*n+1) + 2)/4, n>=0. See the Melham reference for a general conjecture. - Wolfdieter Lang, Aug 10 2012

Melham and Ozeki references from Wolfdieter Lang, Aug 10 2012. Also Prodinger reference added, Oct 11 2012.

A215042 a(n) = F(8n)/L(2n) with n >= 0, F = A000045 (Fibonacci numbers) and L = A000032 (Lucas numbers).

Original entry on oeis.org

0, 7, 141, 2576, 46347, 831985, 14930208, 267913919, 4807525989, 86267568688, 1548008749155, 27777890017577, 498454011832896, 8944394323670071, 160500643816049277, 2880067194369984080, 51680708854856144763, 927372692193073296289

Offset: 0

Wolfdieter Lang, Aug 31 2012

This provides the second example for the Riordan transition matrix R mentioned in a comment to A078812 (here the column called there n=1 is relevant).

Index entries for linear recurrences with constant coefficients, signature (21,-56,21,-1).

Cf. A001906 (for F(4*n)/L(2*n) = F(2*n)), 24*A215043 (for F(12*n)/L(2*n)).

Cf. A000032, A000045.

[Fibonacci(8*n)/Lucas(2*n): n in [0..17]]; // Bruno Berselli, Aug 31 2012

Table[Fibonacci[8 n]/LucasL[2 n], {n, 0, 17}] (* Bruno Berselli, Aug 31 2012 *)
LinearRecurrence[{21,-56,21,-1},{0,7,141,2576},20] (* Harvey P. Dale, Jul 18 2021 *)

a(n) = 2*F(2*n) + 1*5*F(2*n)^3, n >= 0 (for the coefficients 2, 1,  see the second row of the Riordan matrix R = A078812 (with offset [0,0])).
a(n) = F(6*n) - F(2*n), n >= 0, (from the preceding line and a 5*F(2*n)^3 formula given in a comment on the signed triangle A111418, with l->2*n, n->1; see also 5*A215039).
O.g.f.: x*(7-6*x+7*x^2)/((1-3*x+x^2)*(1-18*x+x^2)). The partial fraction decomposition and recurrences lead to the preceding formula.

A215043 a(n) = F(12n)/(24L(2*n)), n >= 0, with F = A000045 (Fibonacci) and L = A000032 (Lucas).

Original entry on oeis.org

0, 2, 276, 34561, 4261992, 524393210, 64499742738, 7933009283134, 975696814205904, 120002796170968643, 14759368609635548580, 1815282342961539780022, 223264968937188026209956, 27459775899111901985784506

Offset: 0

Wolfdieter Lang, Aug 31 2012

24*a(n) is the third example for the Riordan transition matrix introduced in a comment on A078812 (with offset [0,0]). Take there l -> n, n -> 2. See the second formula below.

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (144,-2640,6930,-2640,144,-1).

Cf. A215042 (for F(8*n)/L(2*n)).

[Fibonacci(12*n)/(24*Lucas(2*n)): n in [0..15]]; // Vincenzo Librandi, Sep 02 2012

Table[Fibonacci[12*n]/(24*LucasL[2*n]), {n,0,15}] (* G. C. Greubel, Jun 30 2019 *)

lucas(n) = fibonacci(n+1) + fibonacci(n-1);
vector(15, n, n--; fibonacci(12*n)/(24*lucas(2*n))) \\ G. C. Greubel, Jun 30 2019

[fibonacci(12*n)/(24*lucas_number2(2*n,1,-1)) for n in (0..15)] # G. C. Greubel, Jun 30 2019

a(n) = F(12*n)/(24*L(2*n)), n >= 0, with F = A000045 (Fibonacci) and L = A000032 (Lucas).
a(n) = 3*F(2*n) + 20*F(2*n)^3 + 25*F(2*n)^5, n >= 0 (see the comment above).
O.g.f.: x*(2 - 12*x + 97*x^2 - 12*x^3 + 2*x^4)/((1 - 3*x + x^2)*(1 - 18*x + x^2)*(1 - 123*x + x^2)). From the o.g.f.s for the sequences appearing in the preceding formula, see A001906, A215039 and A215044.
a(n) = (L(8*n) + 1)*F(2*n)/24. - Ehren Metcalfe, Jun 04 2019

A215040 a(n) = F(2*n+1)^3, n>=0, with F = A000045 (Fibonacci).

Original entry on oeis.org

1, 8, 125, 2197, 39304, 704969, 12649337, 226981000, 4073003173, 73087061741, 1311494070536, 23533806109393, 422297015640625, 7577812474746632, 135978327528030989, 2440032083025183109, 43784599166913148552, 785682752921379769625, 14098504953417839657513

Offset: 0

Wolfdieter Lang, Aug 10 2012

Bisection (odd part) of A056570. From this follows the o.g.f., and its partial fraction decomposition leads to the explicit formula given below. For the computation see A215039 for a comment on Chebyshev's S-polynomials.

Harvey P. Dale, Table of n, a(n) for n = 0..797
Kenny B. Davenport, proposer, Problem B-1353, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 259.

Cf. A000045, A056570, A163200 (partial sums).

Fibonacci[2*Range[0,20]+1]^3 (* Harvey P. Dale, Jan 24 2013 *)

a(n) = F(2*n+1)^3, n>=0, with F=A000045.
O.g.f.: (1-x)*(1-12*x+x^2)/((1-3*x+x^2)*(1-18*x+x^2)) (from the bisection (odd part) of A056570).
a(n) = (12*F(2*n+1) + F(6*(n+1)) - F(6*n))/20, n>=0.
a(n) = 1 + (4*Sum_{k=1..n} F(6*k) + 3*Sum_{k=1..n} F(2*k)) / 5 (Davenport, 2024). - Amiram Eldar, Aug 29 2024

cp's OEIS Frontend

A163198 Sum of the cubes of the first n even-indexed Fibonacci numbers.

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A215042 a(n) = F(8*n)/L(2*n) with n >= 0, F = A000045 (Fibonacci numbers) and L = A000032 (Lucas numbers).

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A215043 a(n) = F(12*n)/(24*L(2*n)), n >= 0, with F = A000045 (Fibonacci) and L = A000032 (Lucas).

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A215040 a(n) = F(2*n+1)^3, n>=0, with F = A000045 (Fibonacci).

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A215042 a(n) = F(8n)/L(2n) with n >= 0, F = A000045 (Fibonacci numbers) and L = A000032 (Lucas numbers).

A215043 a(n) = F(12n)/(24L(2*n)), n >= 0, with F = A000045 (Fibonacci) and L = A000032 (Lucas).