A203169 Sum of the fourth powers of the first n even-indexed Fibonacci numbers.

0, 1, 82, 4178, 198659, 9349284, 439330980, 20639983621, 969645224182, 45552722051318, 2140008541351943, 100534850436141384, 4722997973709689160, 221880369994471370761, 10423654392318557192602, 489689876072761951752602

Offset: 0

Author

Stuart Clary, Dec 30 2011

Keywords

• nonn
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., -9349284, -198659, -4178, -82, -1, 0, [0], 1, 82, 4178, 198659, 9349284, ... That is, a(-n) = -a(n-1).

Links

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

Crossrefs

Cf. A203170, A203171, A203172.

Cf. A027941, A103434, A163198.

Programs

Mathematica

a[n_Integer] := (1/75)(Fibonacci[8n+4] - 12*Fibonacci[4n+2] + 9*(2*n+1)); Table[a[n], {n, 0, 20}]

Formula

Let F(n) be the Fibonacci number A000045(n).

a(n) = sum_{k=1..n} F(2k)^4.

Closed form: a(n) = (1/75)(F(8n+4) - 12 F(4n+2) + 9(2 n + 1)).

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 + 26 x^2 + 26 x^3 + x^4)/(1 - 56 x + 440 x^2 - 770 x^3 + 440 x^4 - 56 x^5 + x^6) = x(1 + x)(1 + 25 x + x^2)/((1 - x)^2 (1 - 7 x + x^2)(1 - 47 x + x^2)).