A005059 a(n) = (5^n - 3^n)/2.
0, 1, 8, 49, 272, 1441, 7448, 37969, 192032, 966721, 4853288, 24325489, 121804592, 609554401, 3049366328, 15251614609, 76272421952, 381405156481, 1907154922568, 9536162033329, 47681972428112, 238413348924961, 1192077204978008, 5960417405949649, 29802181172927072
Offset: 0
Examples
For the fifth formula: a(4) = 1*125 + 3*25 + 9*5 + 27*1 = 272. [_Bruno Berselli_, Aug 07 2013]
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- M. A. Alekseyev and T. Berger, Solving the Tower of Hanoi with Random Moves. In: J. Beineke, J. Rosenhouse (eds.) The Mathematics of Various Entertaining Subjects: Research in Recreational Math, Princeton University Press, 2016, pp. 65-79. ISBN 978-0-691-16403-8
- Index entries for linear recurrences with constant coefficients, signature (8,-15).
Crossrefs
Programs
-
Magma
[(5^n - 3^n)/2: n in [0..30] ]; // Vincenzo Librandi, Aug 19 2011
-
Maple
A005059:=n->(5^n-3^n)/2: seq(A005059(n), n=0..30); # Wesley Ivan Hurt, Nov 18 2014
-
Mathematica
Join[{a=0,b=1},Table[c=8*b-15*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 19 2011 *) LinearRecurrence[{8, -15}, {0, 1}, 50] (* Sture Sjöstedt, Oct 17 2012 *) Table[(5^n - 3^n)/2, {n, 0, 23}] (* Michael De Vlieger, Jul 16 2017 *)
-
PARI
a(n)=(5^n-3^n)/2 \\ Charles R Greathouse IV, Jun 11 2013
-
Sage
[lucas_number1(n,8,15) for n in range(0, 21)] # Zerinvary Lajos, Apr 23 2009
Formula
From Paul Barry, Mar 03 2003: (Start)
a(n) = 8*a(n-1) - 15*a(n-2).
G.f.: x/((1-3*x)*(1-5*x)). (End)
a(n) = Sum_{k=1..n} 2^(k-1)*3^(n-k)*binomial(n,k). - Zerinvary Lajos, Sep 24 2006
a(n) = (r^n-s^n)/(r-s) with r=5 and s=3. - Sture Sjöstedt, Oct 17 2012
a(n) = Sum_{k=0..n-1} 3^k*5^(n-k-1) for n>0, a(0)=0. - Bruno Berselli, Aug 07 2013
E.g.f.: exp(3*x)*(exp(2*x) - 1)/2. - Stefano Spezia, Jul 23 2024
Comments