A005968 -id:A005968 - cp's OEIS frontend

A128697 Sum of the eighth powers of the first n Fibonacci numbers.

0, 1, 2, 258, 6819, 397444, 17174660, 832905381, 38655764742, 1824449669638, 85558387560263, 4022147193262344, 188906406088298760, 8875457294194960201, 416941824416535235082, 19587673124144635235082, 920198619736386114829803, 43229838526402491973562764, 2030880577900713476799525260, 95408186647695095521364177901, 4482153365649947417785489568526

Offset: 0

Stuart Clary, Mar 23 2007

Natural bilateral extension (brackets mark index 0): ..., -17174660, -397444, -6819, -258, -2, -1, 0, [0], 1, 2, 258, 6819, 397444, 17174660, ... This is (-A128697)-reversed followed by A128697.

G. C. Greubel, Table of n, a(n) for n = 0..595
Index entries for linear recurrences with constant coefficients, signature (35,680,-5355,-7735,24752,-7735,-5355,680,35,-1).

Cf. A128698 (alternating sum).

Sums of other powers: A000071, A001654, A005968, A005969, A098531, A098532, A098533.

[(&+[Fibonacci(k)^8: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018

a[ n_Integer ] := If[ n >= 0, Sum[ Fibonacci[ k ]^8, {k, 1, n} ], Sum[ -Fibonacci[ -k ]^8, {k, 1, -n - 1} ] ]
Accumulate[Fibonacci[Range[0,20]]^8] (* Harvey P. Dale, Oct 26 2011 *)

a(n) = sum(k=1, n, fibonacci(k)^8); \\ Michel Marcus, Dec 10 2016

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} F(k)^8.
Closed form: a(n) = F(8n+4)/1875 - (-1)^n 4 F(6n+3)/625 + 28 F(4n+2)/625 - (-1)^n 56 F(2n+1)/625 + 7(2 n + 1)/125.
Recurrence: a(n) - 35 a(n-1) - 680 a(n-2) + 5355 a(n-3) + 7735 a(n-4) - 24752 a(n-5) + 7735 a(n-6) + 5355 a(n-7) - 680 a(n-8) - 35 a(n-9) + a(n-10) = 0.
G.f.: A(x) = (x - 33 x^2 - 492 x^3 + 1784 x^4 + 1784 x^5 - 492 x^6 - 33 x^7 + x^8)/(1 - 35 x - 680 x^2 + 5355 x^3 + 7735 x^4 - 24752 x^5 + 7735 x^6 + 5355 x^7 - 680 x^8 - 35 x^9 + x^10) = x*(1 + x)*(1 - 34 x - 458 x^2 + 2242 x^3 - 458 x^4 - 34 x^5 + x^6)/((1 - x)^2*(1 + 3 x + x^2)*(1 - 7 x + x^2)*(1 + 18 x + x^2)*(1 - 47 x + x^2)).

A163200 Sum of the cubes of the first n odd-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 9, 134, 2331, 41635, 746604, 13395941, 240376941, 4313380114, 77400441855, 1388894512391, 24922700621784, 447219716262409, 8025032191009041, 144003359719040030, 2584035442744223139, 46368634609657371691, 832051387531037141316, 14930556340948876798829

Offset: 0

Stuart Clary, Jul 24 2009

Natural bilateral extension (brackets mark index 0): ..., -41635, -2331, -134, -9, -1, [0], 1, 9, 134, 2331, 41635, ... This is (-A163200)-reversed followed by A163200, without repeating the 0. That is, a(-n) = -a(n). Thus a(n) is an odd function of n.

G. C. Greubel, Table of n, a(n) for n = 0..500
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 (21,-56,21,-1).

Cf. A005968, A163198, A163201, A163202.

[(1/4)*Fibonacci(2*n)*(Fibonacci(2*n)^2+3): n in [0..20]]; // Vincenzo Librandi, Dec 10 2016

a[n_Integer] := If[ n >= 0, Sum[ Fibonacci[2k-1]^3, {k, 1, n} ], -Sum[ Fibonacci[-2k+1]^3, {k, 1, -n} ] ]
LinearRecurrence[{21,-56,21,-1}, {0,1,9,134}, 50] (* or *) Table[(1/20)*(Fibonacci[6*n] + 12*Fibonacci[2*n]),{n,0,25}] (* G. C. Greubel, Dec 09 2016 *)
Join[{0},Accumulate[Fibonacci[Range[1,41,2]]^3]] (* Harvey P. Dale, Jul 20 2021 *)

concat([0],Vec(x*(1 - 12*x + x^2)/((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-1)^3.
a(n) = (1/20)*(F(6*n) + 12*F(2*n)).
a(n) = (1/4)*(F(2n)^3 + 3*F(2n)). (K. Subba Rao)
a(n) = (1/20)*F(2n)*(L(4n) + 13).
a(n) = (1/4)*F(2n)*(F(2n)^2 + 3).
a(n) - 21*a(n-1) + 56*a(n-2) - 21*a(n-3) + a(n-4) = 0.
G.f.: (x - 12*x^2 + x^3)/(1 - 21*x + 56*x^2 - 21*x^3 + x^4) = x*(1 - 12*x + x^2)/((1 - 3*x + x^2 )*(1 - 18*x + x^2)).

A363753 a(n) = Sum_{k=0..n} (-1)^kF(k-1)F(k)*F(k+1)/2, where F(n) is the Fibonacci number A000045(n).

Original entry on oeis.org

0, 0, 1, -2, 13, -47, 213, -879, 3762, -15873, 67342, -285098, 1207966, -5116586, 21674919, -91815276, 388937619, -1647563169, 6979194475, -29564334305, 125236542640, -530510487155, 2247278519916, -9519624520452, 40325776676748, -170822731106052, 723616701297373

Offset: 0

Hans J. H. Tuenter, Jun 19 2023

Alternating sum of the product of three consecutive Fibonacci numbers, divided by two.
Can also be seen as the alternating sum of the Fibonomial coefficients (n+1,3), A001655.
This sequence is part of a suite of sums over triple products of Fibonacci numbers. Subba Rao (1953) gives closed-form expressions for several Fibonacci sums of this type.

K. Subba Rao, Some properties of Fibonacci numbers, The American Mathematical Monthly, 60(10):680-684, December 1953.
Index entries for linear recurrences with constant coefficients, signature (-2,9,-3,-4,1).

Cf. A000045, A001655.
Other sequences with the product of three Fibonacci numbers as a summand (the sequence may have a shifted [and scaled] version of the summand given here).
A005968: F(k)^3,  A119284: (-1)^k*F(k)^3,  A215037: F(k-1)*F(k)*F(k+1),
A363753: (-1)^k*F(k-1)*F(k)*F(k+1),  A163198: F(2k)^3,  A163200: F(2k+1)^3,
A256178: F(2k)*F(2k+1)*F(2k+2),  this sequence: (-1)^k*F(k-1)*F(k)*F(k+1),
A363754: F(2k-1)*F(2k)*F(2k+1).

LinearRecurrence[{-2, 9, -3, -4, 1}, {0, 0, 1, -2, 13}, 27]

a(n) = ((-1)^n*(F(n+1)^3 - F(n)^3) + F(n+2) - 2)/8.
a(n) = ((-1)^n*F(3*n+1) + 4*F(n+2) - 5)/20.
a(n) = -2*a(n-1) + 9*a(n-2) - 3*a(n-3) - 4*a(n-4) + a(n-5).
a(-n) = A215037(n-3).
G.f.: x^2/((1 - x)*(1 + 4*x - x^2)*(1 - x - x^2)).
20*a(n) = (-1)^n*A033887(n) + 4*A000045(n+2) - 5. - R. J. Mathar, Jun 27 2023

A292278 a(n) = (Fibonacci(3*n-1) + 1)/2 for n >= 1.

Original entry on oeis.org

1, 3, 11, 45, 189, 799, 3383, 14329, 60697, 257115, 1089155, 4613733, 19544085, 82790071, 350704367, 1485607537, 6293134513, 26658145587, 112925716859, 478361013021, 2026369768941, 8583840088783, 36361730124071, 154030760585065, 652484772464329

Offset: 1

Vincenzo Librandi, Sep 13 2017

Problem B-1211 proposed by Hideyuki Ohtsuka (see Links section): For n >= 1, prove that Fibonacci(n-1)^3 + Sum_{k=1..n} Fibonacci(k)^3 = (Fibonacci(3*n-1) + 1)/2.

Proof. Let F(n-1)^3 = (F(3*n-3) + 3*(-1)^n*F(n-1))/5 (see Ralf Stephan's formula in A056570) and Sum_{k=1..n} F(k)^3 = (F(3*n+2) - 6*(-1)^(n)*F(n-1) + 5)/10 (see Benjamin & Timothy's formula in A005968), where F=A000045, n>0. Therefore, (F(3*n-3) + 3*(-1)^n*F(n-1))/5 + (F(3*n+2) - 6*(-1)^(n)*F(n-1) + 5)/10 = (2*F(3*n-3) + F(3*n+2) + 5)/10 = (2*(F(3*n-1) - F(3*n-2)) + (3*F(3*n-1) + 2*F(3*n-2)) + 5)/10 = (5*F(3*n-1) + 5)/10 = a(n). - Bruno Berselli, Sep 14 2017

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Hideyuki Ohtsuka, Problem B-1211, The Fibonacci Quarterly, Volume 55, Number 3 (August 2017), p. 276 (see Comments section).
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Cf. A000045; A005968, A056570.

[(Fibonacci(3*n-1)+1)/2: n in [1..30]];

Table[(Fibonacci[3 n - 1] + 1) / 2, {n, 40}]
LinearRecurrence[{5,-3,-1},{1,3,11},30] (* Harvey P. Dale, Mar 06 2024 *)

a(n) = (fibonacci(3*n-1)+1)/2; \\ Altug Alkan, Sep 13 2017

G.f.: x*(1 - 2*x - x^2)/((1 - x)*(1 - 4*x -x^2)).

a(n) = 5*a(n-1) - 3*a(n-2) - a(n-3).

Edited by Bruno Berselli, Sep 14 2017

A079716 a(n) = 2*(F(1)^3+F(2)^3+F(3)^3+...+F(p)^3)/(F(1)+F(2)+F(3)+...+F(p)) where p is the n-th prime and F(k) denotes the k-th Fibonacci number.

Original entry on oeis.org

2, 5, 27, 174, 7955, 54378, 2551019, 17482358, 821234595, 264431660859, 1812440734590, 583600131432954, 27416783156825867, 187917427075527110, 8828119011157298499, 2842626904464482727003, 915317035112361317429843, 6273676290103919245031562

Offset: 1

Benoit Cloitre, Feb 16 2003

Cf. A000040, A000045, A000071, A005968.

a(n) = 2*A005968(prime(n)) / A000071(prime(n)+2) [corrected by Sean A. Irvine, Aug 25 2025].

cp's OEIS Frontend

A128697 Sum of the eighth powers of the first n Fibonacci numbers.

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A163200 Sum of the cubes of the first n odd-indexed Fibonacci numbers.

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A363753 a(n) = Sum_{k=0..n} (-1)^k*F(k-1)*F(k)*F(k+1)/2, where F(n) is the Fibonacci number A000045(n).

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A292278 a(n) = (Fibonacci(3*n-1) + 1)/2 for n >= 1.

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A079716 a(n) = 2*(F(1)^3+F(2)^3+F(3)^3+...+F(p)^3)/(F(1)+F(2)+F(3)+...+F(p)) where p is the n-th prime and F(k) denotes the k-th Fibonacci number.

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Formula

A363753 a(n) = Sum_{k=0..n} (-1)^kF(k-1)F(k)*F(k+1)/2, where F(n) is the Fibonacci number A000045(n).