A119284 -id:A119284 - cp's OEIS frontend

A005968 Sum of cubes of first n Fibonacci numbers.

0, 1, 2, 10, 37, 162, 674, 2871, 12132, 51436, 217811, 922780, 3908764, 16558101, 70140734, 297121734, 1258626537, 5331629710, 22585142414, 95672204155, 405273951280, 1716768021816, 7272346018247, 30806152127640, 130496954475672, 552793970116297, 2341672834801754

Offset: 0

N. J. A. Sloane

From Alexander Adamchuk, Aug 07 2006: (Start)
The only two prime terms are a(2) = 2 and a(4) = 37.
The prime p divides a(p-1) iff p is in A045468.
The prime p divides a((p-1)/2) iff p is in A047650.
3^4 divides a(p) iff p is in A003628.
3^5 divides a(p) for p = {37,53,109,181,197,269,397,431,541,...}.
3^6 divides a(p) for p = {109,541,...}.
3^7 divides a(p) for p = {557,...}. (End)

Art Benjamin, Timothy A. Carnes, and Benoit Cloitre, Recounting the Sums of Cubes of Fibonacci Numbers, Congressus Numerantium, Proceedings of the Eleventh International Conference on Fibonacci Numbers and their Applications, (William Webb, ed.), Vol 194, pp. 45-51, 2009.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 14.
A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 18.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Art Benjamin, Timothy A. Carnes, and Benoit Cloitre, Counting the Sums of Cubes of Fibonacci Numbers.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
David Treeby, Hidden Formulas in Fibonacci Tilings, Fibonacci Quart. 54 (2016), no. 1, 23-30.
Index entries for linear recurrences with constant coefficients, signature (4,3,-9,2,1).

Partial sums of A056570.  Cf. A119284 (alternating sum).
Cf. A045468, A047650, A003628.
Sums of other powers: A000071, A001654, A005969, A098531, A098532, A098533, A128697.

[(1/10)*( Fibonacci(3*n+2)-(-1)^(n)*6*Fibonacci(n-1)+5 ): n in [0..30]]; // G. C. Greubel, Jan 17 2018

with(combinat): l[0] := 0: for i from 1 to 50 do l[i] := l[i-1]+fibonacci(i)^3; printf(`%d,`,l[i]) od: # James Sellers, May 29 2000
A005968:=(-1+2*z+z**2)/(z-1)/(z**2+4*z-1)/(z**2-z-1); # conjectured by Simon Plouffe in his 1992 dissertation

f[n_]:=(Fibonacci[n]*Fibonacci[n+1]^2+(-1)^(n-1)*Fibonacci[n-1]+1)/2;Table[f[n],{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Nov 22 2010 *)
Accumulate[Fibonacci[Range[0,20]]^3]
CoefficientList[Series[x*(1-2*x-x^2)/((1-x)*(1+x-x^2)*(1-4*x-x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)

a(n)=(fibonacci(n)*fibonacci(n+1)^2+(-1)^(n-1)*fibonacci(n-1)+1)/2

a(n)=(fibonacci(3*n+2)-(-1)^(n)*6*fibonacci(n-1)+5)/10

PARI
```
a(n)=sum(i=1,n,fibonacci(i)^3)
```

G.f.: x*(1-2*x-x^2)/((1-x)*(1+x-x^2)*(1-4*x-x^2)). - Ralf Stephan, Apr 23 2004
a(n) = (1/2)*(F(n)*F(n+1)^2 + (-1)^(n-1)*F(n-1) + 1). - Benoit Cloitre, Aug 06 2004
a(n) = Sum_{i=1..n} A000045(i)^3.
a(n) = (1/10)*(F(3*n+2) - (-1)^(n)*6*F(n-1) + 5). - Art Benjamin and Timothy A. Carnes
a(n+5) = 4*a(n+4) + 3*a(n+3) - 9*a(n+2) + 2*a(n+1) + a(n). - Benoit Cloitre, Sep 12 2004

More terms from James Sellers, May 29 2000

A119282 Alternating sum of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -2, 1, -4, 4, -9, 12, -22, 33, -56, 88, -145, 232, -378, 609, -988, 1596, -2585, 4180, -6766, 10945, -17712, 28656, -46369, 75024, -121394, 196417, -317812, 514228, -832041, 1346268, -2178310, 3524577, -5702888, 9227464, -14930353, 24157816, -39088170, 63245985, -102334156, 165580140, -267914297, 433494436, -701408734, 1134903169, -1836311904, 2971215072, -4807526977, 7778742048

Offset: 0

Stuart Clary, May 13 2006

Apart from signs, same as A008346.
Natural bilateral extension (brackets mark index 0): ..., 88, 54, 33, 20, 12, 7, 4, 2, 1, 0, [0], -1, 0, -2, 1, -4, 4, -9, 12, -22, 3, ... This is A000071-reversed followed by A119282.
Alternating sums of rows of the triangle in A141169. - Reinhard Zumkeller, Mar 22 2011

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Kyu-Hwan Lee, Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
Index entries for linear recurrences with constant coefficients, signature (0,2,-1)

Cf. A000071, A008346, A119283, A119284, A119285, A119286, A119287.

Cf. A128696, A128698.

[0] cat [(&+[(-1)^k*Fibonacci(k):k in [1..n]]): n in [1..30]]; // G. C. Greubel, Jan 17 2018

FoldList[#1 - Fibonacci@ #2 &, -Range@ 50] (* Michael De Vlieger, Jan 27 2016 *)
Accumulate[Table[(-1)^n Fibonacci[n], {n, 0, 49}]] (* Alonso del Arte, Apr 25 2017 *)

a(n) = sum(k=1, n, (-1)^k*fibonacci(k)); \\ Michel Marcus, Jan 27 2016

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k = 1..n} (-1)^k F(k).
Closed form: a(n) = (-1)^n F(n-1) - 1 = (-1)^n A008346(n-1).
Recurrence: a(n) - 2 a(n-2) + a(n-3)= 0.
G.f.: A(x) = -x/(1 - 2 x^2 + x^3) = -x/((1 - x)(1 + x - x^2)).
Another recurrence: a(n) = a(n-2) - a(n-1) - 1. - Rick L. Shepherd, Aug 12 2009

A119283 Alternating sum of the squares of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -4, 5, -20, 44, -125, 316, -840, 2185, -5736, 15000, -39289, 102840, -269260, 704909, -1845500, 4831556, -12649205, 33116020, -86698896, 226980625, -594243024, 1555748400, -4073002225, 10663258224, -27916772500, 73087059221, -191344405220, 500946156380, -1311494063981, 3433536035500, -8989114042584, 23533806092185, -61612304234040

Offset: 0

Stuart Clary, May 13 2006

Natural bilateral extension (brackets mark index 0): ..., 840, -316, 125, -44, 20, -5, 4, 0, 1, 0, [0], -1, 0, -4, 5, -20, 44, -125, 316, -840, 2185, ... This is (-A119283)-reversed followed by A119283.

			The first few squares of Fibonacci numbers are: 0, 1, 1, 4, 9, 25, 64, 169.
a(0) = 0.
a(1) = 0 - 1 = -1.
a(2) = 0 - 1 + 1 = 0.
a(3) = 0 - 1 + 1 - 4 = -4.
a(4) = 0 - 1 + 1 - 4 + 9 = 5.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,4,-1,-1)

Cf. A001654, A119282, A119284, A119285, A119286, A119287, A128696, A128698.

altFiboSqSum[n_Integer] := If[n >= 0, Sum[(-1)^k Fibonacci[k]^2, {k, n}], Sum[-(-1)^k Fibonacci[-k]^2, {k, 1, -n - 1}]]; altFiboSqSum[Range[0, 39]] (* Clary *)
Accumulate[Table[(-1)^n Fibonacci[n]^2, {n, 0, 39}]] (* Alonso del Arte, Apr 25 2017 *)
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,40]]^2,{1,-1},{2,-1,2}],2]] (* Harvey P. Dale, Jul 29 2024 *)

concat(0, Vec(-x*(1 + x) / ((1 - x)^2*(1 + 3*x + x^2)) + O(x^40))) \\ Colin Barker, Apr 25 2017

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k = 1..n} (-1)^k F(k)^2.
Closed form: a(n) = (-1)^n F(2n+1)/5 - (2 n + 1)/5.
Recurrence: a(n) + a(n-1) - 4 a(n-2) + a(n-3) + a(n-4) = 0.
G.f.: A(x) = (-x - x^2)/(1 + x - 4 x^2 + x^3 + x^4) = -x(1 + x)/((1 - x)^2 (1 + 3 x + x^2)).
a(n) = ( -1-2*n+(-1)^n*( A001906(n+1)-A001906(n) ))/5. - R. J. Mathar, Nov 16 2007
a(n) = (2^(-1-n)*(-5*2^(1+n)+5*(-3+sqrt(5))^n - sqrt(5)*(-3+sqrt(5))^n + (-3-sqrt(5))^n*(5+sqrt(5)) - 5*2^(2+n)*n)) / 25. - Colin Barker, Apr 25 2017

A119285 Alternating sum of the fourth powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -16, 65, -560, 3536, -25025, 169456, -1166880, 7983745, -54758496, 375223200, -2572072321, 17628580320, -120829829680, 828175410881, -5676410656400, 38906666170736, -266670338968385, 1827785480332240, -12527828615754816, 85867013279034625, -588541268397840576, 4033921854875707200, -27648911743562183425

Offset: 0

Stuart Clary, May 13 2006

Natural bilateral extension (brackets mark index 0): ..., -3536, 560, -65, 16, 0, 1, 0, [0], -1, 0, -16, 65, -560, 3536, -25025, ... This is (-A119285)-reversed followed by A119285.

Kunle Adegoke, Sums of fourth powers of Fibonacci and Lucas numbers, arXiv:1706.00407 [math.NT], 2017.
Index entries for linear recurrences with constant coefficients, signature (-5,15,15,-5,-1).

Cf. A005969, A119282, A119283, A119284, A119286, A119287, A128696, A128698.

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^4, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^4, {k, 1, -n - 1} ] ]
LinearRecurrence[{-5,15,15,-5,-1},{0,-1,0,-16,65},30] (* Harvey P. Dale, Apr 02 2018 *)

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} (-1)^k F(k)^4.
Closed form: a(n) = (-1)^n L(4n+2)/75 - (4/25) L(2n+1) + (-1)^n 3/25.
Factored closed form: a(n) = (-1)^n (1/3) F(n-2) F(n) F(n+1) F(n+3).
Recurrence: a(n) + 5 a(n-1) - 15 a(n-2) - 15 a(n-3) + 5 a(n-4) + a(n-5) = 0.
G.f.: A(x) = (-x - 5 x^2 - x^3)/(1 + 5 x - 15 x^2 - 15 x^3 + 5 x^4 + x^5) = -x(1 + 5 x + x^2)/((1 + x)(1 - 3 x + x^2)(1 + 7 x + x^2)).

A119286 Alternating sum of the fifth powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -32, 211, -2914, 29854, -341439, 3742662, -41692762, 461591613, -5122467836, 56794896388, -629924960005, 6985721085652, -77473909014348, 859194263419359, -9528629686028398, 105674040835291026, -1171943417651373875, 12997050199917354250, -144139501695851560726, 1598531543102764228825, -17727986584911448406232, 196606383515036414871336, -2180398207207766329269289

Offset: 0

Stuart Clary, May 13 2006

Natural bilateral extension (brackets mark index 0): ..., 3402, 277, 34, 2, 1, 0, [0], -1, 0, -32, 211, -2914, 29854, ... This is A098531-reversed followed by A119286.

Index entries for linear recurrences with constant coefficients, signature (-7,48,20,-100,32,9,-1)

Cf. A098531, A119282, A119283, A119284, A119285, A119287, A128696, A128698.

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^5, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^5, {k, 1, -n - 1} ] ]
LinearRecurrence[{-7,48,20,-100,32,9,-1},{0,-1,0,-32,211,-2914,29854},30] (* Harvey P. Dale, Jun 24 2018 *)

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} (-1)^k F(k)^5.
Closed form: a(n) = (-1)^n (1/275)(F(5n+1) + 2 F(5n+3)) - (1/10) F(3n+2) + (-1)^n (2/5) F(n-1) - 7/22; here F(5n+1) + 2 F(5n+3) = A001060(5n+1) = A013655(5n+2).
Recurrence: a(n) + 7 a(n-1) - 48 a(n-2) - 20 a(n-3) + 100 a(n-4) - 32 a(n-5) - 9 a(n-6) + a(n-7) = 0.
G.f.: A(x) = (-x - 7 x^2 + 16 x^3 + 7 x^4 - x^5)/(1 + 7 x - 48 x^2 - 20 x^3 + 100 x^4 - 32 x^5 - 9 x^6 + x^7) = -x(1 + 7 x - 16 x^2 - 7 x^3 + x^4)/((1 - x)(1 + x - x^2)(1 - 4 x - x^2)(1 + 11 x - x^2)).

A119287 Alternating sum of the sixth powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -64, 665, -14960, 247184, -4579625, 81186496, -1463617920, 26217022705, -470764268256, 8445336180000, -151560390359569, 2719538168853120, -48800836192146880, 875690649999921929, -15713664197268146000, 281970036429821245616, -5059748557502924705465, 90793493265349521060160, -1629223203785737022267136, 29235223670642547226470625

Offset: 0

Stuart Clary, May 13 2006

Natural bilateral extension (brackets mark index 0): ..., 14960, -665, 64, 0, 1, 0, [0], -1, 0, -64, 665, -14960, 247184, ... This is (-A119287)-reversed followed by A119287.

Index entries for linear recurrences with constant coefficients, signature (-12,117,156,-520,156,117,-12,-1).

Cf. A098532, A119282, A119283, A119284, A119285, A119286, A128696, A128698.

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^6, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^6, {k, 1, -n - 1} ] ]
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,30]]^6,{1,-1},{2,-1,2}],2]] (* Harvey P. Dale, Jul 23 2013 *)

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} (-1)^k F(k)^6.
a(n) = (-1)^n (1/250) F(6n+3) - (6/125) F(4n+2) + (-1)^n (3/25) F(2n+1) - (2/25)(2 n + 1).
Recurrence: a(n) + 12 a(n-1) - 117 a(n-2) - 156 a(n-3) + 520 a(n-4) - 156 a(n-5) - 117 a(n-6) + 12 a(n-7) + a(n-8) = 0.
G.f.: A(x) = (-x - 12 x^2 + 53 x^3 + 53 x^4 - 12 x^5 - x^6)/(1 + 12 x - 117 x^2 - 156 x^3 + 520 x^4 - 156 x^5 - 117 x^6 + 12 x^7 + x^8) = -x(1 + x)(1 + 11 x - 64 x^2 + 11 x^3 + x^4)/((1 - x)^2 (1 + 3 x + x^2)(1 - 7 x + x^2)(1 + 18 x + x^2)).

A128696 Alternating sum of the seventh powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -128, 2059, -76066, 2021086, -60727431, 1740361110, -50782989034, 1471652245341, -42759682650188, 1241158781898676, -36040175501820901, 1046363981321362852, -30381064378888637148, 882092032492683277335, -25611107658594421205278, 743603574761804566730466, -21590121866471006254739195, 626857059065125789349713930

Offset: 0

Stuart Clary, Mar 23 2007

Natural bilateral extension (brackets mark index 0): ..., 2177594, 80442, 2317, 130, 2, 1, 0, [0], -1, 0, -128, 2059, -76066, 2021086 ... This is A098533-reversed followed by A128696.

G. C. Greubel, Table of n, a(n) for n = 0..680
Index entries for linear recurrences with constant coefficients, signature (-20,294,819,-2912,728,1365,-252,-22,1).

Cf. A098533, A119282, A119283, A119284, A119285, A119286, A119287, A128698.

[(&+[(-1)^k*Fibonacci(k)^7: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018

a[ n_Integer ] := If[ n >= 0, Sum[ (-1)^k Fibonacci[ k ]^7, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k ]^7, {k, 1, -n - 1} ] ]
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,30]]^7,{1,-1}], 2]] (* Harvey P. Dale, May 11 2012 *)

a(n) = sum(k=1, n, (-1)^k*fibonacci(k)^7); \\ Michel Marcus, Dec 10 2016

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} (-1)^k F(k)^7.
Closed form: a(n) = (-1)^n (F(7n+7) - F(7n))/3625 + 7(F(5n+1) - 2 F(5n+4))/1375 + (-1)^n 21 F(3n+1)/250 - 7 F(n+2)/25 + 139/638.
Recurrence: a(n) + 20 a(n-1) - 294 a(n-2) - 819 a(n-3) + 2912 a(n-4) - 728 a(n-5) - 1365 a(n-6) + 252 a(n-7) + 22 a(n-8) - a(n-9) = 0.
G.f.: A(x) = (-x - 20 x^2 + 166 x^3 + 318 x^4 - 166 x^5 - 20 x^6 + x^7)/(1 + 20 x - 294 x^2 - 819 x^3 + 2912 x^4 - 728 x^5 - 1365 x^6 + 252 x^7 + 22 x^8 - x^9) = -x*(1 + 20 x - 166 x^2 - 318 x^3 + 166 x^4 + 20 x^5 - x^6)/ ((1 - x)*(1 - x - x^2)*(1 + 4 x - x^2)*(1 - 11 x - x^2)*(1 + 29 x - x^2)).

A128698 Alternating sum of the eighth powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -256, 6305, -384320, 16392896, -799337825, 37023521536, -1748770383360, 81985167507265, -3854603638194816, 181029655256841600, -8505521232849819841, 399560845889490455040, -18771170453838609544960, 881839776158402870049761, -41427800130507702988683200, 1946222939243803281837279296, -91431083130550578762727373345, 4295314095871701743501398017280

Offset: 0

Stuart Clary, Mar 23 2007

Natural bilateral extension (brackets mark index 0): ..., -16392896, 384320, -6305, 256, 0, 1, 0, [0], -1, 0, -256, 6305, -384320, 16392896, ... This is (-A128698)-reversed followed by A128698.

G. C. Greubel, Table of n, a(n) for n = 0..595

Cf. A128697, A119282, A119283, A119284, A119285, A119286, A119287, A128696.

[(&+[(-1)^k*Fibonacci(k)^8: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018

a[ n_Integer ] := If[ n >= 0, Sum[ (-1)^k Fibonacci[ k ]^8, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k ]^8, {k, 1, -n - 1} ] ]
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,20]]^8,{1,-1},{2,-1,2}],2]] (* Harvey P. Dale, May 04 2016 *)

a(n) = sum(k=1, n, (-1)^k*fibonacci(k)^8); \\ Michel Marcus, Dec 10 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} (-1)^k F(k)^8.
Closed form: a(n) = (-1)^n L(8n+4)/4375 - 2 L(6n+3)/625 + (-1)^n 28 L(4n+2)/1875 - 56 L(2n+1)/625 + (-1)^n 7/125.
Factored closed form: a(n) = (-1)^n (1/21) F(n-2) F(n) F(n+1) F(n+3) (3 F(n)^2 F(n+1)^2 + 4).
Recurrence: a(n) + 34 a(n-1) - 714 a(n-2) - 4641 a(n-3) + 12376 a(n-4) + 12376 a(n-5) - 4641 a(n-6) - 714 a(n-7) + 34 a(n-8) + a(n-9) = 0.
G.f.: A(x) = (-x - 34 x^2 + 458 x^3 + 2242 x^4 + 458 x^5 - 34 x^6 - x^7)/(1 + 34 x - 714 x^2 - 4641 x^3 + 12376 x^4 + 12376 x^5 - 4641 x^6 - 714 x^7 + 34 x^8 + x^9) = -x(1 + 34 x - 458 x^2 - 2242 x^3 - 458 x^4 + 34 x^5 + x^6)/((1 + x)(1 - 3 x + x^2)(1 + 7 x + x^2)(1 - 18 x + x^2)(1 + 47 x + x^2)).

A163201 Alternating sum of the cubes of the first n even-indexed Fibonacci numbers.

Original entry on oeis.org

0, -1, 26, -486, 8775, -157600, 2828384, -50754249, 910750554, -16342762150, 293258984975, -5262319011456, 94428483336576, -1694450381348881, 30405678381733850, -545607760491930150, 9790534010478427479, -175684004428133950624, 3152521545695969823584, -56569703818099420107225

Offset: 0

Stuart Clary, Jul 24 2009

Natural bilateral extension (brackets mark index 0): ..., 8775, -486, 26, -1, 0, [0], -1, 26, -486, 8775, -157600, ... This is A163201-reversed followed by A163201. That is, a(-n) = a(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. For the factored closed form, let alpha equal the imaginary unit in Equation (21).
Index entries for linear recurrences with constant coefficients, signature (-20,-35,35,20,1).

Cf. A119284, A163198, A163200, A163202.

[(-1)^n*(1/50)*(Lucas(6*n+3)-6*Lucas(2*n+1)+2*(-1)^n): n in [0..20]]; // Vincenzo Librandi, Dec 10 2016

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[2k]^3, {k, 1, n} ], -Sum[ (-1)^k Fibonacci[-2k]^3, {k, 1, -n - 1} ] ]
LinearRecurrence[{-20, -35, 35, 20, 1}, {0, -1, 26, -486, 8775}, 50] (* or *) Table[(-1)^n*(1/50)*(LucasL[6 n + 3] - 6 LucasL[2 n + 1] + 2*(-1)^n), {n, 0, 25}] (* G. C. Greubel, Dec 10 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 10 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} (-1)^k F(2k)^3.
a(n) = (1/50)*(L(6n+3) - 6 L(2n+1) + 2) if n is even.
a(n) = -(1/50)*(L(6n+3) - 6 L(2n+1) - 2) if n is odd.
a(n) = (1/2) * F(n)^2 * F(n+1)^2 * (L(2n+1) + 2) if n is even.
a(n) = -(1/2) * F(n)^2 * F(n+1)^2 * (L(2n+1) - 2) if n is odd.
a(n) + 21*a(n-1) + 56*a(n-2) + 21*a(n-3) + a(n-4) = 4.
a(n) + 20*a(n-1) + 35*a(n-2) - 35*a(n-3) - 20*a(n-4) - a(n-5) = 0.
G.f.: (-x + 6*x^2 - x^3)/(1 + 20*x + 35*x^2 - 35*x^3 - 20*x^4 - x^5) = -x*(1 - 6*x + x^2)/((1 - x)*(1 + 3*x + x^2)*(1 + 18*x + x^2)).

A163202 Alternating sum of the cubes of the first n odd-indexed Fibonacci numbers.

Original entry on oeis.org

0, -1, 7, -118, 2079, -37225, 667744, -11981593, 214999407, -3858003766, 69229057975, -1242265012561, 22291541096832, -400005474543793, 7177807000202839, -128800520527828150, 2311231562497354959, -41473367604415793593, 744209385316963976032, -13354295568100875681481

Offset: 0

Stuart Clary, Jul 24 2009

Natural bilateral extension (brackets mark index 0): ..., -37225, 2079, -118, 7, -1, [0], -1, 7, -118, 2079, -37225, ... This is A163202-reversed followed by A163202, without repeating the 0. That is, a(-n) = a(n). Thus a(n) is an even function of n.

			-x + 7*x^2 - 118*x^3 + 2079*x^4 - 37225*x^5 + 667744*x^6 - 11981593*x^7 + ... - _Michael Somos_, Aug 11 2009

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (-20, -35, 35, 20, 1).

Cf. A163198, A163200, A163201, A119284.

[((-1)^n*(Fibonacci(6*n)/2+Fibonacci(6*n-1)+ 3*Fibonacci(2*n-1)+3*Fibonacci(2*n+1))-7)/25: n in [0..20]]; // Vincenzo Librandi, Dec 19 2016

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[2k-1]^3, {k, 1, n} ], Sum[ (-1)^k Fibonacci[-2k+1]^3, {k, 1, -n} ] ]
Join[{0},Accumulate[Times@@@Partition[Riffle[Take[Fibonacci[Range[41]],{1,-1,2}]^3,{-1,1}],2]]] (* or *) LinearRecurrence[{-20,-35,35,20,1},{0,-1,7,-118,2079},20] (* Harvey P. Dale, Feb 19 2012 *)
Table[(-1)^n*(1/50)*(LucasL[6 n] + 6 LucasL[2 n] - 14*(-1)^n), {n,0,50}] (* G. C. Greubel, Dec 10 2016 *)

{a(n) = ((-1)^n * (fibonacci(6*n) / 2 + fibonacci(6*n - 1) + 3*fibonacci(2*n - 1) + 3*fibonacci(2*n + 1)) - 7) / 25} /* Michael Somos, Aug 11 2009 */

concat([0], Vec(-x*(1 + x)*(1 + 12*x +x^2)/((1 - x)*(1 + 3*x + x^2)*(1 + 18*x + x^2)) + O(x^50))) \\ G. C. Greubel, Dec 10 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} (-1)^k F(2k-1)^3.
a(n) = (1/50)*(L(6n) + 6 L(2n) - 14) if n is even.
a(n) = -(1/50)*(L(6n) + 6 L(2n) + 14) if n is odd.
a(n) = (1/10)*F(n)^2*(L(4 n) + 2 L(2n) + 9) if n is even.
a(n) = -(1/10)*F(n)^2*(L(4 n) - 2 L(2n) + 9) if n is odd.
a(n) + 21*a(n-1) + 56*a(n-2) + 21*a(n-3) + a(n-4) = -28.
a(n) + 20*a(n-1) + 35*a(n-2) - 35*a(n-3) - 20*a(n-4) - a(n-5) = 0.
G.f.: (-x - 13*x^2 - 13*x^3 - x^4)/(1 + 20*x + 35*x^2 - 35*x^3 - 20*x^4 - x^5) = -x*(1 + x)*(1 + 12*x +x^2)/((1 - x)*(1 + 3*x + x^2)*(1 + 18*x + x^2)).
a(-n) = a(n). - Michael Somos, Aug 11 2009

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A005968 Sum of cubes of first n Fibonacci numbers.

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A119283 Alternating sum of the squares of the first n Fibonacci numbers.

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A128696 Alternating sum of the seventh powers of the first n Fibonacci numbers.

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