A020956 a(n) = Sum_{k>=1} floor(tau^(n-k)) where tau is A001622.
1, 2, 4, 8, 14, 25, 42, 71, 117, 193, 315, 514, 835, 1356, 2198, 3562, 5768, 9339, 15116, 24465, 39591, 64067, 103669, 167748, 271429, 439190, 710632, 1149836, 1860482, 3010333, 4870830, 7881179, 12752025, 20633221, 33385263, 54018502, 87403783, 141422304
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Clark Kimberling, Problem 10520, Amer. Math. Mon. 103 (1996) p. 347.
- Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).
Programs
-
Magma
I:=[1,2,4,8,14]; [n le 5 select I[n] else 2*Self(n-1)+Self(n-2)-3*Self(n-3)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Nov 01 2016
-
Magma
[Lucas(n+1)-(2*n+5-(-1)^n)/4: n in [1..40]]; // G. C. Greubel, Apr 05 2024
-
Mathematica
LinearRecurrence[{2,1,-3,0,1}, {1,2,4,8,14}, 40] (* Vincenzo Librandi, Nov 01 2016 *)
-
PARI
Vec(x*(1-x^2+x^3)/((1-x-x^2)*(1+x)*(1-x)^2) + O(x^50)) \\ Michel Marcus, Nov 01 2016
-
Python
prpr = 0 prev = 1 for n in range(2,100): print(prev, end=", ") curr = prpr+prev + n//2 prpr = prev prev = curr # Alex Ratushnyak, Jul 30 2012
-
SageMath
[lucas_number2(n+1,1,-1) -(n+2+(n%2))//2 for n in range(1,41)] # G. C. Greubel, Apr 05 2024
Formula
G.f.: x*(1-x^2+x^3)/((1-x-x^2)*(1+x)*(1-x)^2). - Ralf Stephan, Apr 08 2004
a(n) = Lucas(n+1) - floor(n/2) - 1.
a(n) = Sum_{k=0..n-1} A014217(k).
a(n) = 2^(-2-n)*((-2)^n - 5*2^n + 2*(1-t)^(1+n) + 2*(1+t)^n + 2*t*(1+t)^n - 2^(1+n)*n) where t=sqrt(5). - Colin Barker, Feb 09 2017
From G. C. Greubel, Apr 05 2024: (Start)
a(n) = Lucas(n+1) - (1/4)*(2*n + 5 - (-1)^n).
E.g.f.: exp(x/2)*(cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2)) - (1/2)*((x+2)*cosh(x) + (x+3)*sinh(x)). (End)
Extensions
More terms from Vladeta Jovovic, Apr 04 2002