A163202 -id:A163202 - 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)).

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)).

A203172 Alternating sum of the fourth powers of the first n odd-indexed Fibonacci numbers.

Original entry on oeis.org

0, -1, 15, -610, 27951, -1308385, 61433856, -2885861665, 135572548335, -6369013518946, 299207991620175, -14056406104466881, 660351875572408320, -31022481722865482305, 1457396288941918481871, -68466603097469928960610

Offset: 0

Stuart Clary, Dec 30 2011

Natural bilateral extension (brackets mark index 0): ..., -1308385, 27951, -610, 15, -1, [0], -1, 15, -610, 27951, -1308385, ... That is, a(-n) = a(n).

Index entries for linear recurrences with constant coefficients, signature (-54,-330,0,330,54,1)

Cf. A203169, A203170, A203171.

Cf. A156089, A163202.

a[n_Integer] := (-1)^n (1/525)(3*LucasL[8n] + 28*LucasL[4n] + 63 - (-1)^n 125); Table[a[n], {n, 0, 20}]
LinearRecurrence[{-54,-330,0,330,54,1},{0,-1,15,-610,27951,-1308385},20] (* Harvey P. Dale, Jun 04 2025 *)

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)^4.
Closed form: a(n) = (-1)^n (1/525)(3 L(8n) + 28 L(4n) + 63 - (-1)^n 125).
Alternate closed form: a(n) = (1/21) F(2n)^2 (3 F(2n)^2 + 8) if n is even, a(n) = -(1/21)(3 F(2n)^4 + 8 F(2n)^2 + 10) if n is odd.
Recurrence: a(n) + 54 a(n-1) + 330 a(n-2) - 330 a(n-4) - 54 a(n-5) - a(n-6) = 0.
G.f.: A(x) = -(x + 39 x^2 + 130 x^3 + 39 x^4 + x^5)/(1 + 54 x + 330 x^2 - 330 x^4 - 54 x^5 - x^6) = -x(1 + 39 x + 130 x^2 + 39 x^3 + x^4)/((1 - x)(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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A163201 Alternating sum of the cubes of the first n even-indexed Fibonacci numbers.

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A203172 Alternating sum of the fourth powers of the first n odd-indexed Fibonacci numbers.

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