OFFSET
0,2
COMMENTS
A floretion-generated sequence relating to Fibonacci numbers and powers of 2. The sequence results from a particular transform of the sequence A000079*(-1)^n (powers of 2).
Floretion Algebra Multiplication Program, FAMP Code: 1jesforseq[ ( 5'i + .5i' + .5'ii' + .5e)*( + .5j' + .5'kk' + .5'ki' + .5e ) ], 1vesforseq = A000079(n+1)*(-1)^(n+1), ForType: 1A. Identity used: jesfor = jesrightfor + jesleftfor
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-2).
FORMULA
Suggestions made by Superseeker: a(n+2) - a(n+1) - a(n) = A042950(n+1).
Coefficients of g.f.*(1-x)/(1+x) match A099036.
Coefficients of g.f./(1+x) match A027934.
Coefficients of g.f./(1-x^2) match A008466;
a(n) = 3*2^n - Fibonacci(n+3) = A221719(n) + 1. - Ralf Stephan, May 20 2007, Hugo Pfoertner, Mar 06 2024
a(n) = (3*2^n - (2^(-n)*((1-sqrt(5))^n*(-2+sqrt(5)) + (1+sqrt(5))^n*(2+sqrt(5)))) / sqrt(5)). - Colin Barker, Aug 18 2017
MAPLE
with (combinat):a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=fibonacci(n-1)+2*a[n-1] od: seq(a[n], n=1..26); # Zerinvary Lajos, Mar 17 2008
MATHEMATICA
LinearRecurrence[{3, -1, -2}, {1, 3, 7}, 80] (* Vincenzo Librandi, Aug 18 2017 *)
CoefficientList[Series[(1-x)(1+x)/((2x-1)(x^2+x-1)), {x, 0, 40}], x] (* Harvey P. Dale, Oct 12 2024 *)
PROG
(Magma) [3*2^n-Fibonacci(n+3): n in [0..40]]; // Vincenzo Librandi, Aug 18 2017
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Creighton Dement, Feb 24 2005
STATUS
approved