|
Search: id:A119286
|
|
|
| A119286 |
|
Alternating sum of the fifth powers of the first n Fibonacci numbers. |
|
+0
9
|
|
| 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
(list; graph; listen)
|
|
|
OFFSET
|
0,4
|
|
|
COMMENT
|
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.
|
|
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} ] ]
|
|
CROSSREFS
|
Cf. A098531, A119282, A119283, A119284, A119285, A119287
Cf. A098531, A119282, A119283, A119284, A119285, A119287, A128696, A128698.
Sequence in context: A120046 A019560 A130811 this_sequence A125342 A126500 A070053
Adjacent sequences: A119283 A119284 A119285 this_sequence A119287 A119288 A119289
|
|
KEYWORD
|
sign,easy
|
|
AUTHOR
|
Stuart Clary (clary(AT)uakron.edu), May 13, 2006
|
|
|
Search completed in 0.002 seconds
|
