A119286 Alternating sum of the fifth powers of the first n Fibonacci numbers.

0, -1, 0, -32, 211, -2914, 29854, -341439, 3742662, -41692762, 461591613, -5122467836, 56794896388, -629924960005, 6985721085652, -77473909014348, 859194263419359, -9528629686028398, 105674040835291026, -1171943417651373875, 12997050199917354250, -144139501695851560726, 1598531543102764228825, -17727986584911448406232, 196606383515036414871336, -2180398207207766329269289

Offset

0,4

Comments

Natural bilateral extension (brackets mark index 0): ..., 3402, 277, 34, 2, 1, 0, [0], -1, 0, -32, 211, -2914, 29854, ... This is A098531-reversed followed by A119286.

Links

Table of n, a(n) for n=0..25.

Index entries for linear recurrences with constant coefficients, signature (-7,48,20,-100,32,9,-1)

Formula

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

a(n) = Sum_{k=1..n} (-1)^k F(k)^5.

Closed form: a(n) = (-1)^n (1/275)(F(5n+1) + 2 F(5n+3)) - (1/10) F(3n+2) + (-1)^n (2/5) F(n-1) - 7/22; here F(5n+1) + 2 F(5n+3) = A001060(5n+1) = A013655(5n+2).

Recurrence: a(n) + 7 a(n-1) - 48 a(n-2) - 20 a(n-3) + 100 a(n-4) - 32 a(n-5) - 9 a(n-6) + a(n-7) = 0.

G.f.: A(x) = (-x - 7 x^2 + 16 x^3 + 7 x^4 - x^5)/(1 + 7 x - 48 x^2 - 20 x^3 + 100 x^4 - 32 x^5 - 9 x^6 + x^7) = -x(1 + 7 x - 16 x^2 - 7 x^3 + x^4)/((1 - x)(1 + x - x^2)(1 - 4 x - x^2)(1 + 11 x - x^2)).

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^5, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^5, {k, 1, -n - 1} ] ]
LinearRecurrence[{-7, 48, 20, -100, 32, 9, -1}, {0, -1, 0, -32, 211, -2914, 29854}, 30] (* Harvey P. Dale, Jun 24 2018 *)

Crossrefs

Cf. A098531, A119282, A119283, A119284, A119285, A119287, A128696, A128698.

Sequence in context: A184020 A283336 A223023 * A125342 A126500 A160538

Adjacent sequences: A119283 A119284 A119285 * A119287 A119288 A119289

Keyword

sign,easy

Author

Stuart Clary, May 13 2006

Status

approved