A127968 a(n) = F(n+1) + (1-(-1)^n)/2, where F() = Fibonacci numbers A000045.
1, 2, 2, 4, 5, 9, 13, 22, 34, 56, 89, 145, 233, 378, 610, 988, 1597, 2585, 4181, 6766, 10946, 17712, 28657, 46369, 75025, 121394, 196418, 317812, 514229, 832041, 1346269, 2178310, 3524578, 5702888, 9227465, 14930353, 24157817, 39088170, 63245986, 102334156
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1).
Programs
-
Magma
I:=[1,2,2,4]; [n le 4 select I[n] else Self(n-1) +2*Self(n-2) - Self(n-3) -Self(n-4): n in [1..30]]; // G. C. Greubel, May 04 2018
-
Mathematica
LinearRecurrence[{1,2,-1,-1},{1,2,2,4},40] (* Harvey P. Dale, Jun 19 2013 *)
-
PARI
Vec((1+x-2*x^2-x^3)/((1-x)*(1+x)*(1-x-x^2)) + O(x^50)) \\ Colin Barker, Jul 12 2017
Formula
G.f.: 1 / (1 - x - x^2) + x / (1 - x^2).
G.f.: (1 + x - 2*x^2 - x^3) / ((1 - x)*(1 + x)*(1 - x - x^2)).
From Colin Barker, Jul 12 2017: (Start)
a(n) = (5 - 5*(-1)^n + 2^(-n)*sqrt(5)*(-(1-sqrt(5))^(1+n) + (1+sqrt(5))^(1+n))) / 10.
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4) for n>3.
(End)
Comments