A163200 -id:A163200 - 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.

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

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

A203170 Sum of the fourth powers of the first n odd-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 17, 642, 29203, 1365539, 64107780, 3011403301, 141469813301, 6646055880582, 312223061019703, 14667837157106759, 689076118833981960, 32371909717271872585, 1520790680382055836761, 71444790066793903279242

Offset: 0

Stuart Clary, Dec 30 2011

Natural bilateral extension (brackets mark index 0): ..., -1365539, -29203, -642, -17, -1, [0], 1, 17, 642, 29203, 1365539, ... That is, a(-n) = -a(n).

Index entries for linear recurrences with constant coefficients, signature (56,-440,770,-440,56,-1).

Cf. A203169, A203171, A203172.

Cf. A001906, A103433, A163200.

a[n_Integer] := (1/75)(Fibonacci[8n] + 12*Fibonacci[4n] + 18 n); Table[a[n], {n, 0, 20}]

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} F(2k-1)^4.
Closed form: a(n) = (1/75)(F(8n) + 12 F(4n) + 18 n).
Recurrence: a(n) - 56 a(n-1) + 440 a(n-2) - 770 a(n-3) + 440 a(n-4) - 56 a(n-5) + a(n-6) = 0.
G.f.: A(x) = (x - 39 x^2 + 130 x^3 - 39 x^4 + x^5)/(1 - 56 x + 440 x^2 - 770 x^3 + 440 x^4 - 56 x^5 + x^6) = x(1 - 39 x + 130 x^2 - 39 x^3 + x^4)/((1 - x)^2 (1 - 7 x + x^2)(1 - 47 x + x^2)).

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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

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).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A203170 Sum of the fourth powers of the first n odd-indexed Fibonacci numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

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).