A047850 a(n) = (5^n + 3)/4.
1, 2, 7, 32, 157, 782, 3907, 19532, 97657, 488282, 2441407, 12207032, 61035157, 305175782, 1525878907, 7629394532, 38146972657, 190734863282, 953674316407, 4768371582032, 23841857910157, 119209289550782, 596046447753907
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (6,-5).
Programs
-
Magma
A047850:= func< n | (5^n + 3)/4 >; [A047850(n): n in [0..40]]; // G. C. Greubel, Jan 10 2025
-
Mathematica
(5^Range[0,40] +3)/4 (* G. C. Greubel, Jan 10 2025 *)
-
PARI
a(n)=(5^n+3)/4 \\ Charles R Greathouse IV, Oct 07 2015
-
Python
def A047850(n): return (5^n+3)//4 print([A047850(n) for n in range(41)]) # G. C. Greubel, Jan 10 2025
Formula
a(n) = A047848(2,n).
From Philippe Deléham, Oct 06 2009: (Start)
a(n) = 6*a(n-1) - 5*a(n-2) for n > 1.
G.f.: (1-4*x)/((1-x)*(1-5*x)). (End)
a(n) = 5*a(n-1) - 3 (with a(0)=1). - Vincenzo Librandi, Aug 06 2010
From Elmo R. Oliveira, Dec 10 2023: (Start)
a(n) = A132079(n)/2.
E.g.f.: (1/4)*(exp(5*x) + 3*exp(x)). (End)
Extensions
Better description from Michael Somos
Comments