A156086 -id:A156086 - cp's OEIS frontend

A156093 One ninth of the alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4.

0, -1, 48, -2256, 105985, -4979040, 233908896, -10988739073, 516236827536, -24252142155120, 1139334444463105, -53524466747610816, 2514510602693245248, -118128473859834915841, 5549523760809547799280, -260709488284188911650320

Offset: 0

Stuart Clary, Feb 04 2009

Natural bilateral extension (brackets mark index 0): ..., -105985, 2256, -48, 1, 0, [0], -1, 48, -2256, 105985, -4979040, ... This is (-A156093)-reversed followed by A156093. That is, A156093(-n) = -A156093(n-1).

Index entries for linear recurrences with constant coefficients, signature (-48,-48,-1).

Cf. A156086, A156087, A156092.

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k (1/9) Fibonacci[4k]^2, {k, 1, n} ], Sum[ -(-1)^k (1/9) Fibonacci[-4k]^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) = (1/9) sum_{k=1..n} (-1)^k F(4k)^2.
Closed form: a(n) = (-1)^n (L(8n+4) - 7)/315.
Factored closed form: a(n) = (-1)^n F(4n) F(4n+4)/63.
Recurrence: a(n) + 47 a(n-1) + a(n-2) = (-1)^n.
Recurrence: a(n) + 48 a(n-1) + 48 a(n-2) + a(n-3) = 0.
G.f.: A(x) = -x/(1 + 48 x + 48 x^2 + x^3) = -x/((1 + x)(1 + 47 x + x^2)).

A156087 One ninth of the sum of the squares of the first n Fibonacci numbers with index divisible by 4.

Original entry on oeis.org

0, 1, 50, 2354, 110595, 5195620, 244083556, 11466731525, 538692298134, 25307071280790, 1188893657899015, 55852694849972936, 2623887764290829000, 123266872226818990089, 5790919106896201705210, 272049931151894661154810

Offset: 0

Stuart Clary, Feb 04 2009

Natural bilateral extension (brackets mark index 0): ..., -110595, -2354, -50, -1, 0, [0], 1, 50, 2354, 110595, 5195620, ... This is (-A156087)-reversed followed by A156087. That is, A156087(-n) = -A156087(n-1).

Cf. A156086, A156092, A156093.

a[n_Integer] := If[ n >= 0, Sum[ (1/9) Fibonacci[4k]^2, {k, 1, n} ], -Sum[ (1/9) Fibonacci[-4k]^2, {k, 1, -n - 1} ] ]
Accumulate[Fibonacci[4Range[0,20]]^2]/9 (* Harvey P. Dale, Sep 22 2011 *)

Let F(n) be the Fibonacci number A000045(n).
a(n) = sum_{k=1..n} F(4k)^2.
Closed form: a(n) = F(8n+4)/135 - (2n + 1)/45.
Recurrence: a(n) - 48 a(n-1) + 48 a(n-2) - a(n-3) = 2.
Recurrence: a(n) - 49 a(n-1) + 96 a(n-2) - 49 a(n-3) + a(n-4) = 0.
G.f.: A(x) = (x + x^2)/(1 - 49 x + 96 x^2 - 49 x^3 + x^4) = x (1 + x)/((1 - x)^2 (1 - 47 x + x^2)).

A156092 Alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4.

Original entry on oeis.org

0, -9, 432, -20304, 953865, -44811360, 2105180064, -98898651657, 4646131447824, -218269279396080, 10254010000167945, -481720200728497344, 22630595424239207232, -1063156264738514242569, 49945713847285930193520, -2346385394557700204852880

Offset: 0

Stuart Clary, Feb 04 2009

Natural bilateral extension (brackets mark index 0): ..., -953865, 20304, -432, 9, 0, [0], -9, 432, -20304, 953865, -44811360, ... This is (-A156092)-reversed followed by A156092. That is, A156092(-n) = -A156092(n-1).

Cf. A156086, A156087, A156093.

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[4k]^2, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[-4k]^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} (-1)^k F(4k)^2.
Closed form: a(n) = (-1)^n (L(8n+4) - 7)/35.
Factored closed form: a(n) = (-1)^n F(4n) F(4n+4)/7.
Recurrence: a(n) + 47 a(n-1) + a(n-2) = (-1)^n 9.
Recurrence: a(n) + 48 a(n-1) + 48 a(n-2) + a(n-3) = 0.
G.f.: A(x) = -9 x/(1 + 48 x + 48 x^2 + x^3) = -9 x/((1 + x)(1 + 47 x + x^2)).

cp's OEIS Frontend

A156093 One ninth of the alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4.

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A156087 One ninth of the sum of the squares of the first n Fibonacci numbers with index divisible by 4.

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Author

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Crossrefs

Programs

Mathematica

Formula

A156092 Alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4.

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Keywords

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Programs

Mathematica

Formula