A156091 -id:A156091 - cp's OEIS frontend

A156085 One fourth of the sum of the squares of the first n Fibonacci numbers with index divisible by 3.

0, 1, 17, 306, 5490, 98515, 1767779, 31721508, 569219364, 10214227045, 183286867445, 3288949386966, 59017802097942, 1059031488375991, 19003548988669895, 341004850307682120, 6119083756549608264, 109802502767585266633, 1970325966059985191129

Offset: 0

Stuart Clary, Feb 04 2009

Natural bilateral extension (brackets mark index 0): ..., -5490, -306, -17, -1, 0, [0], 1, 17, 306, 5490, 98515, ... This is (-A156085)-reversed followed by A156085. That is, A156085(-n) = -A156085(n-1).

Harvey P. Dale, Table of n, a(n) for n = 0..798
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7
Index entries for linear recurrences with constant coefficients, signature (17, 17, -1).

Cf. A156084, A156090, A156091.

a[n_Integer] := If[ n >= 0, Sum[ (1/4) Fibonacci[3k]^2, {k, 1, n} ], -Sum[ (1/4) Fibonacci[-3k]^2, {k, 1, -n - 1} ] ]
Accumulate[Fibonacci[3*Range[0,20]]^2]/4 (* or *) LinearRecurrence[{17,17,-1},{0,1,17},30] (* Harvey P. Dale, Aug 17 2014 *)

concat(0, Vec(x/((1+x)*(1-18*x+x^2)) + O(x^25))) \\ Colin Barker, Mar 04 2016

a(n) = sum(k=1, n, fibonacci(3*k)^2)/4; \\ Michel Marcus, Mar 04 2016

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
a(n) = (1/4) sum_{k=1..n} F(3k)^2.
Closed form: a(n) = L(6n+3)/80 - (-1)^n/20.
Factored closed form: a(n) = (1/16) F(n) F(n+1) (L(n) - 1)(L(n) + 1)(L(2n+2) - 1) if n is even; a(n) = (1/16) F(n) F(n+1) (L(n+1) - 1)(L(n+1) + 1)(L(2n) - 1) if n is odd.
Recurrence: a(n) - 17 a(n-1) - 17 a(n-2) + a(n-3) = 0.
G.f.: A(x) = x/(1 - 17 x - 17 x^2 + x^3) = x/((1 + x)(1 - 18 x + x^2)).
a(n) = ((9+4*sqrt(5))^(-n)*(2-sqrt(5)-4*(-9-4*sqrt(5))^n+(2+sqrt(5))*(9+4*sqrt(5))^(2*n)))/80. - Colin Barker, Mar 04 2016

A156084 Sum of the squares of the first n Fibonacci numbers with index divisible by 3.

Original entry on oeis.org

0, 4, 68, 1224, 21960, 394060, 7071116, 126886032, 2276877456, 40856908180, 733147469780, 13155797547864, 236071208391768, 4236125953503964, 76014195954679580, 1364019401230728480, 24476335026198433056, 439210011070341066532

Offset: 0

Stuart Clary, Feb 04 2009

Natural bilateral extension (brackets mark index 0): ..., -21960, -1224, -68, -4, 0, [0], 4, 68, 1224, 21960, 394060, ... This is (-A156084)-reversed followed by A156084. That is, A156084(-n) = -A156084(n-1).

Index entries for linear recurrences with constant coefficients, signature (17,17,-1).

Partial sums of A014729.

Cf. A156085, A156090, A156091

a[n_Integer] := If[ n >= 0, Sum[ Fibonacci[3k]^2, {k, 1, n} ], -Sum[ Fibonacci[-3k]^2, {k, 1, -n - 1} ] ]

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(3k)^2.
Closed form: a(n) = L(6n+3)/20 - (-1)^n/5.
Factored closed form: a(n) = (1/4) F(n) F(n+1) (L(n) - 1)(L(n) + 1)(L(2n+2) - 1) if n is even; a(n) = (1/4) F(n) F(n+1) (L(n+1) - 1)(L(n+1) + 1)(L(2n) - 1) if n is odd.
Recurrence: a(n) - 17 a(n-1) - 17 a(n-2) + a(n-3) = 0.
G.f.: A(x) = 4 x/(1 - 17 x - 17 x^2 + x^3) = 4 x/((1 + x)(1 - 18 x + x^2)).
a(n) = 4*A156085(n). - R. J. Mathar, Aug 06 2019

A156090 Alternating sum of the squares of the first n Fibonacci numbers with index divisible by 3.

Original entry on oeis.org

0, -4, 60, -1096, 19640, -352460, 6324596, -113490320, 2036501104, -36543529620, 655747031980, -11766903046104, 211148507797800, -3788906237314396, 67989163763861220, -1220016041512187680, 21892299583455516896, -392841376460687116580

Offset: 0

Stuart Clary, Feb 04 2009

Natural bilateral extension (brackets mark index 0): ..., -19640, 1096, -60, 4, 0, [0], -4, 60, -1096, 19640, -352460, ... This is (-a(n))-reversed followed by a(n). That is, a(-n) = -a(n-1).

Muniru A Asiru, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (-16,34,-16,-1).

Cf. A000045, A156084, A156085, A156088, A156089, A156091.

a:=[0,-4,60,-1096];; for n in [5..20] do a[n]:=-16*a[n-1]+34*a[n-2]-16*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018

[(-1)^n*Fibonacci(6*n+3)/10 - (2*n + 1)/5: n in [0..20]]; // Vincenzo Librandi, Sep 12 2018

with(combinat,fibonacci): a:=n->add((-1)^k*fibonacci(3*k)^2,k=1..n): seq(a(n), n=0..20); # Muniru A Asiru, Sep 12 2018

a[n_]:= If[n >= 0, Sum[(-1)^k Fibonacci[3 k]^2, {k, 1, n}], Sum[-(-1)^k Fibonacci[-3 k]^2, {k, 1, -n-1}]];
LinearRecurrence[{-16,34,-16,-1}, {0,-4,60,-1096}, 30] (* Harvey P. Dale, Oct 24 2016 *)

def A156090(n): return ((-1)^n*fibonacci(6*n+3) -2*(2*n+1))//10 # G. C. Greubel, Jun 12 2025

Let F(n) be the n-th Fibonacci number, A000045(n), then: (Start)
a(n) = Sum_{k=1..n} (-1)^k*F(3*k)^2.
Closed form: a(n) = (-1)^n*F(6*n+3)/10 - (2*n + 1)/5.
Recurrence: a(n) + 17*a(n-1) - 17*a(n-2) - a(n-3) = -8.
Recurrence: a(n) + 16*a(n-1) - 34*a(n-2) + 16*a(n-3) + a(n-4) = 0.
G.f.: -4*x*(1 + x)/(1 + 16*x - 34*x^2 + 16*x^3 + x^4) = -4*x(1 + x)/((1 - x)^2*(1 + 18*x + x^2)). (End)
Limit_{n -> oo} a(n)/a(n-1) = -(9 + sqrt(80)). - A.H.M. Smeets, Sep 11 2018
E.g.f.: (-1/(5*sqrt(5)))*( exp(-9*x)*(2*sinh(p*x) - sqrt(5)*cosh(p*x)) + sqrt(5)*(1 + 2*x)*exp(x) ), where p = 4*sqrt(5). - G. C. Greubel, Jun 12 2025

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A156085 One fourth of the sum of the squares of the first n Fibonacci numbers with index divisible by 3.

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A156084 Sum of the squares of the first n Fibonacci numbers with index divisible by 3.

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A156090 Alternating sum of the squares of the first n Fibonacci numbers with index divisible by 3.

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