A163201 -id:A163201 - 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.

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)).

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

A203171 Alternating sum of the fourth powers of the first n even-indexed Fibonacci numbers.

Original entry on oeis.org

0, -1, 80, -4016, 190465, -8960160, 421021536, -19779631105, 929225609456, -43653851217680, 2050801968082945, -96344039926706496, 4526119083346841280, -212631252937414840321, 9989142769386670981520, -469277078911056723578480

Offset: 0

Stuart Clary, Dec 30 2011

Natural bilateral extension (brackets mark index 0): ..., 8960160, -190465, 4016, -80, 1, 0, [0], -1, 80, -4016, 190465, -8960160, ... That is, a(-n) = -a(n-1).

G. C. Greubel, Table of n, a(n) for n = 0..500
R. S. Melham, Alternating sums of fourth powers of Fibonacci and Lucas numbers, The Fibonacci Quarterly, 38(3):254-259, June-July 2000.
Index entries for linear recurrences with constant coefficients, signature (-55,-385,-385,-55,-1)

Cf. A203169, A203170, A203172.

Cf. A156088, A163201.

with(combinat): A203171:=n->(-1)^n*(1/21)*fibonacci(2*n)*fibonacci(2*n+2)*(3*fibonacci(2*n+1)^2 - 5): seq(A203171(n), n=0..20); # Wesley Ivan Hurt, Jan 16 2017

a[n_Integer] := (-1)^n (1/525)(3*LucasL[8n+4] - 28*LucasL[4n+2] + 63); Table[a[n], {n, 0, 20}]

a(n) = sum(k=1, n, (-1)^k*fibonacci(2*k)^4); \\ Michel Marcus, Apr 16 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(2*k)^4.
Closed form: a(n) = (-1)^n (1/525)(3 L(8n+4) - 28 L(4n+2) + 63).
Factored closed form: a(n) = (-1)^n (1/21) F(2n) F(2n+2) (3 F(2n+1)^2 - 5).
Alternate factored closed form: a(n) = (-1)^n (1/21) F(2n) F(2n+2) (3 F(2n) F(2n+2) - 2).
Recurrence: a(n) + 55 a(n-1) + 385 a(n-2) + 385 a(n-3) + 55 a(n-4) + a(n-5) = 0.
G.f.: A(x) = (-x + 25 x^2 - x^3)/(1 + 55 x + 385 x^2 + 385 x^3 + 55 x^4 + x^5) = -x(1 - 25 x + x^2)/((1 + x)(1 + 7 x + x^2)(1 + 47 x + x^2)).

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

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

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

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A203171 Alternating sum of the fourth powers of the first n even-indexed Fibonacci numbers.

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