A163197 -id:A163197 - 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.

A163194 a(n) = F(n)^2 * L(n+1)^2 * F(n-1) * L(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.

Original entry on oeis.org

0, 0, 112, 2156, 39204, 704700, 12648640, 226979168, 4072998384, 73087049196, 1311494037700, 23533806023420, 422297015415552, 7577812474157376, 135978327526488304, 2440032083021144300, 43784599166902574820, 785682752921352087228, 14098504953417767184064, 252987406408599326907296

Offset: 0

Stuart Clary, Jul 24 2009

Natural bilateral extension (brackets mark index 0): ..., 39200, 2160, 108, 4, -4, [0], 0, 112, 2156, 39204, 704700, ... This is A163196-reversed followed by A163194. That is, A163194(-n) = A163196(n-1).

			G.f. = 112*x^2 + 2156*x^3 + 39204*x^4 + 704700*x^5 + 12648640*x^6 + ...

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 equation (3).

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. A163195, A163196, A163197, A163198, A163199.

[(Fibonacci(n)*Lucas(n+1))^2*(Fibonacci(n-1)*Lucas(n+2)): n in [0..30]]; // G. C. Greubel, Dec 21 2017

a[n_Integer] := Fibonacci[n]^2 LucasL[n+1]^2 Fibonacci[n-1] LucasL[n+2]
LinearRecurrence[{20, -35, -35, 20, -1}, {0, 0, 112, 2156, 39204}, 50] (* or *) Table[(1/5)*(Fibonacci[6n+3] - 12*Fibonacci[2n+1] + 10*(-1)^n), {n,0,25}] (* G. C. Greubel, Dec 09 2016 *)

for(n=0,30, print1((1/5)*(fibonacci(6*n+3) - 12*fibonacci(2*n+1) + 10*(-1)^n), ", ")) \\ G. C. Greubel, Dec 21 2017

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
a(n) = F(n)^2 * L(n+1)^2 * F(n-1) * L(n+2).
Unfactored form: a(n) = (1/5)*(F(6n+3) - 12*F(2n+1) + 10*(-1)^n).
Unfactored form: a(n) = F(2n+1)^3 - 3*F(2n+1) + 2*(-1)^n.
Summation form: a(n) = 4*Sum_{k=1..n} F(2k)^3 = 4 A163198(n) if n is even; a(n) = 4*Sum_{k=2..n} F(2k)^3 = 4 A163199(n) if n is odd. a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = 200*(-1)^n.
a(n) - 20 a(n-1) + 35 a(n-2) + 35 a(n-3) - 20 a(n-4) + a(n-5) = 0.
G.f.: (112*x^2 - 84*x^3 + 4*x^4)/(1 - 20*x + 35*x^2 + 35*x^3 - 20*x^4 + x^5) = 4*x^2*(28 - 21*x + x^2)/((1 + x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)).
a(n) - A163196(n) = 4*(-1)^n.

A163195 a(n) = (1/4)F(n)^2 L(n+1)^2 * F(n-1) * L(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.

Original entry on oeis.org

0, 0, 28, 539, 9801, 176175, 3162160, 56744792, 1018249596, 18271762299, 327873509425, 5883451505855, 105574253853888, 1894453118539344, 33994581881622076, 610008020755286075, 10946149791725643705, 196420688230338021807, 3524626238354441796016, 63246851602149831726824

Offset: 0

Stuart Clary, Jul 24 2009

Natural bilateral extension (brackets mark index 0): ..., 9800, 540, 27, 1, -1, [0], 0, 28, 539, 9801, 176175, ... This is A163197-reversed followed by A163195. That is, A163195(-n) = A163197(n-1).

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 equation (3).

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. A163194, A163196, A163197, A163198, A163199.

[(1/4)*(Fibonacci(n)*Lucas(n+1))^2*(Fibonacci(n-1)*Lucas(n+2)): n in [0..30]]; // G. C. Greubel, Dec 21 2017

a[n_Integer] := (1/4)*Fibonacci[n]^2 * LucasL[n+1]^2 * Fibonacci[n-1] * LucasL[n+2]
LinearRecurrence[{20,-35,-35,20,-1}, {0,0,28,539,9801}, 50] (* or *) Table[(Fibonacci[6n+3] - 12*Fibonacci[2n+1] + 10*(-1)^n)/20, {n,1,25}] (* G. C. Greubel, Dec 09 2016 *)

for(n=0,30, print1((fibonacci(6*n+3) - 12*fibonacci(2*n+1) + 10*(-1)^n)/20, ", ")) \\ G. C. Greubel, Dec 21 2017

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
a(n) = (1/4)*F(n)^2 * L(n+1)^2 * F(n-1) * L(n+2).
a(n) = (1/20)*(F(6n+3) - 12*F(2n+1) + 10*(-1)^n).
a(n) = (1/4)(F(2n+1)^3 - 3*F(2n+1) + 2*(-1)^n).
a(n) = Sum_{k=1..n} F(2k)^3 = A163198(n) if n is even.
a(n) = Sum_{k=2..n} F(2k)^3 = A163199(n) if n is odd.
a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = 50*(-1)^n.
a(n) - 20 a(n-1) + 35 a(n-2) + 35 a(n-3) - 20 a(n-4) + a(n-5) = 0.
G.f.: (28*x^2 - 21*x^3 + x^4)/(1 - 20*x + 35*x^2 + 35*x^3 - 20*x^4 + x^5) = x^2*(28 - 21*x + x^2)/((1 + x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)).
a(n) - A163197(n) = (-1)^n.

A163196 a(n) = L(n)^2 * F(n+1)^2 * L(n-1) * F(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.

Original entry on oeis.org

-4, 4, 108, 2160, 39200, 704704, 12648636, 226979172, 4072998380, 73087049200, 1311494037696, 23533806023424, 422297015415548, 7577812474157380, 135978327526488300, 2440032083021144304, 43784599166902574816, 785682752921352087232, 14098504953417767184060, 252987406408599326907300

Offset: 0

Stuart Clary, Jul 24 2009

Natural bilateral extension (brackets mark index 0): ..., 39204, 2156, 112, 0, 0, [-4], 4, 108, 2160, 39200, 704704, ... This is A163194-reversed followed by A163196. That is, A163196(-n) = A163194(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 equation (3).
Index entries for linear recurrences with constant coefficients, signature (20,-35,-35,20,-1).

Cf. A163194, A163195, A163197, A163198, A163199.

[(Lucas(n)*Fibonacci(n+1))^2*(Lucas(n-1)*Fibonacci(n+2)): n in [0..30]]; // G. C. Greubel, Dec 21 2017

a[n_Integer] := LucasL[n]^2*Fibonacci[n+1]^2*LucasL[n-1] *Fibonacci[n+2]
LinearRecurrence[{20, -35, -35, 20, -1}, {-4, 4, 108, 2160, 39200}, 50] (* or *) Table[(1/5)*(Fibonacci[6*n+3] - 12*Fibonacci[2*n+1] - 10*(-1)^n),{n,0,25}] (* G. C. Greubel, Dec 09 2016 *)

Vec( -4*(1 - 21*x + 28*x^2)/((1 + x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)) + O(x^50)) \\ G. C. Greubel, Dec 09 2016

for(n=0, 30, print1((1/5)*(fibonacci(6*n+3) - 12*fibonacci(2*n+1) - 10*(-1)^n), ", ")) \\ G. C. Greubel, Dec 21 2017

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
a(n) = L(n)^2 * F(n+1)^2 * L(n-1) * F(n+2).
a(n) = (1/5)*(F(6n+3) - 12*F(2n+1) - 10*(-1)^n).
a(n) = F(2n+1)^3 - 3*F(2n+1) - 2*(-1)^n.
a(n) = 4*Sum_{k=2..n} F(2k)^3 = 4*A163199(n) if n is even.
a(n) = 4*Sum_{k=1..n} F(2k)^3 = 4*A163198(n) if n is odd.
a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = - 200*(-1)^n.
a(n) - 20 a(n-1) + 35 a(n-2) + 35 a(n-3) - 20 a(n-4) + a(n-5) = 0.
G.f.: (-4 + 84*x - 112*x^2)/(1 - 20*x + 35*x^2 + 35*x^3 - 20*x^4 + x^5) = -4*(1 - 21*x + 28*x^2)/((1 + x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)).
A163194(n) - a(n) = 4*(-1)^n.

A163199 Sum of the cubes of the first n even-indexed Fibonacci numbers, minus 1.

Original entry on oeis.org

-1, 0, 27, 539, 9800, 176175, 3162159, 56744792, 1018249595, 18271762299, 327873509424, 5883451505855, 105574253853887, 1894453118539344, 33994581881622075, 610008020755286075, 10946149791725643704, 196420688230338021807, 3524626238354441796015, 63246851602149831726824

Offset: 0

Stuart Clary, Jul 24 2009

Natural bilateral extension (brackets mark index 0): ..., 9800, 539, 27, 0, -1, [-1], 0, 27, 539, 9800, 176175, ... This is A163199-reversed followed by A163199. That is, A163199(-n) = A163199(n-1).

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (22,-77,77,-22,1).

Cf. A163194, A163195, A163196, A163197, A163198.

a[n_Integer] := If[ n >= 0, Sum[ Fibonacci[2k]^3, {k, 1, n} ] - 1, -Sum[ Fibonacci[-2k]^3, {k, 1, -n - 1} ] - 1 ]
Accumulate[Fibonacci[Range[0,40,2]]^3]-1 (* Harvey P. Dale, Jan 03 2016 *)
LinearRecurrence[{22,-77,77,-22,1},{-1,0,27,539,9800},50] (* or *) Table[(1/20)*(F(6n+3) - 12*F(2n+1) - 10),{n,0,25}] (* G. C. Greubel, Dec 09 2016 *)

Vec(-(1 - 22*x + 50*x^2 - 22*x^3 + x^4)/((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 - 1.
a(n) = Sum_{k=2..n} F(2k)^3 for n > 0.
a(n) = A163198(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).
a(n) = (1/4)*L(n)^2*F(n+1)^2*L(n-1)*F(n+2) = A163197(n) if n is even.
a(n) = (1/4)*F(n)^2*L(n+1)^2*F(n-1)*L(n+2) = A163195(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.: (-1 + 22*x - 50*x^2 + 22*x^3 - x^4)/(1 - 22*x + 77*x^2 - 77*x^3 + 22*x^4 - x^5) = -(1 - 22*x + 50*x^2 - 22*x^3 + x^4)/((1 - x)*(1 - 3*x + x^2 )*(1 - 18*x + x^2)).

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

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A163194 a(n) = F(n)^2 * L(n+1)^2 * F(n-1) * L(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.

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A163195 a(n) = (1/4)*F(n)^2 * L(n+1)^2 * F(n-1) * L(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.

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A163196 a(n) = L(n)^2 * F(n+1)^2 * L(n-1) * F(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.

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A163199 Sum of the cubes of the first n even-indexed Fibonacci numbers, minus 1.

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A163195 a(n) = (1/4)F(n)^2 L(n+1)^2 * F(n-1) * L(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.