A154691 Expansion of (1+x+x^2) / ((1-x)*(1-x-x^2)).
1, 3, 7, 13, 23, 39, 65, 107, 175, 285, 463, 751, 1217, 1971, 3191, 5165, 8359, 13527, 21889, 35419, 57311, 92733, 150047, 242783, 392833, 635619, 1028455, 1664077, 2692535, 4356615, 7049153, 11405771, 18454927, 29860701, 48315631, 78176335
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Tomislav Došlić and Biserka Kolarec, On Log-Definite Tempered Combinatorial Sequences, Mathematics (2025) Vol. 13, Iss. 7, 1179.
- Index entries for linear recurrences with constant coefficients, signature (2,0,-1).
Programs
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Haskell
a154691 n = a154691_list !! n a154691_list = 1 : zipWith (+) a154691_list (drop 2 $ map (* 2) a000045_list) -- Reinhard Zumkeller, Nov 17 2013
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Magma
A154691:= func< n | 2*Fibonacci(n+3) - 3 >; [A154691(n): n in [0..40]]; // G. C. Greubel, Jan 18 2025
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Maple
A154691 := proc(n) coeftayl( (1+x+x^2)/(1-x-x^2)/(1-x),x=0,n) ; end proc:
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Mathematica
Fibonacci[Range[3,60]]*2 -3 (* Vladimir Joseph Stephan Orlovsky, Mar 19 2010 *) CoefficientList[Series[(1 + x + x^2)/((1 - x - x^2)(1 - x)), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 18 2012 *)
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PARI
Vec((1+x+x^2) / ((1-x-x^2)*(1-x)) + O(x^60)) \\ Colin Barker, Feb 01 2017
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Python
def A154691(n): return 2*fibonacci(n+3) - 3 print([A154691(n) for n in range(41)]) # G. C. Greubel, Jan 18 2025
Formula
a(n+1) = a(n) + 2*A000045(n+2). - Reinhard Zumkeller, Nov 17 2013
From Colin Barker, Feb 01 2017: (Start)
a(n) = -3 + (2^(1-n)*((1-r)^n*(-2+r) + (1+r)^n*(2+r))) / r where r=sqrt(5).
a(n) = 2*a(n-1) - a(n-3) for n>2. (End)
a(n) = 2*Fibonacci(n+3) - 3. - Greg Dresden, Oct 10 2020
E.g.f.: 4*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 2*sqrt(5)*sinh(sqrt(5)*x/2))/5 - 3*exp(x). - Stefano Spezia, Apr 09 2025