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A119282 Alternating sum of the first n Fibonacci numbers. 16
0, -1, 0, -2, 1, -4, 4, -9, 12, -22, 33, -56, 88, -145, 232, -378, 609, -988, 1596, -2585, 4180, -6766, 10945, -17712, 28656, -46369, 75024, -121394, 196417, -317812, 514228, -832041, 1346268, -2178310, 3524577, -5702888, 9227464, -14930353, 24157816, -39088170, 63245985, -102334156, 165580140, -267914297, 433494436, -701408734, 1134903169, -1836311904, 2971215072, -4807526977, 7778742048 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,4
COMMENTS
Apart from signs, same as A008346.
Natural bilateral extension (brackets mark index 0): ..., 88, 54, 33, 20, 12, 7, 4, 2, 1, 0, [0], -1, 0, -2, 1, -4, 4, -9, 12, -22, 3, ... This is A000071-reversed followed by A119282.
Alternating sums of rows of the triangle in A141169. - Reinhard Zumkeller, Mar 22 2011
LINKS
Kyu-Hwan Lee, Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
FORMULA
Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k = 1..n} (-1)^k F(k).
Closed form: a(n) = (-1)^n F(n-1) - 1 = (-1)^n A008346(n-1).
Recurrence: a(n) - 2 a(n-2) + a(n-3)= 0.
G.f.: A(x) = -x/(1 - 2 x^2 + x^3) = -x/((1 - x)(1 + x - x^2)).
Another recurrence: a(n) = a(n-2) - a(n-1) - 1. - Rick L. Shepherd, Aug 12 2009
MATHEMATICA
FoldList[#1 - Fibonacci@ #2 &, -Range@ 50] (* Michael De Vlieger, Jan 27 2016 *)
Accumulate[Table[(-1)^n Fibonacci[n], {n, 0, 49}]] (* Alonso del Arte, Apr 25 2017 *)
PROG
(PARI) a(n) = sum(k=1, n, (-1)^k*fibonacci(k)); \\ Michel Marcus, Jan 27 2016
(Magma) [0] cat [(&+[(-1)^k*Fibonacci(k):k in [1..n]]): n in [1..30]]; // G. C. Greubel, Jan 17 2018
CROSSREFS
Sequence in context: A175000 A355020 A008346 * A241513 A095293 A034409
KEYWORD
sign,easy
AUTHOR
Stuart Clary, May 13 2006
STATUS
approved

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Last modified March 29 05:43 EDT 2024. Contains 371264 sequences. (Running on oeis4.)