OFFSET
1,1
COMMENTS
It appears that if s(n) is a first-order rational sequence of the form s(0)=4, s(n) = (2*s(n-1)+1)/(s(n-1)+2), n > 0, then s(n) = a(n)/(a(n)-1), n > 0.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Nathan Fox, A Slow Relative of Hofstadter's Q-Sequence, arXiv:1611.08244 [math.NT], 2016.
Index entries for linear recurrences with constant coefficients, signature (4,-3).
FORMULA
a(n+1) = 3*a(n) - 1 for n > 1. - Reinhard Zumkeller, Jan 22 2011
G.f.: (5/2)*U(0) where U(k) = 1 + 2/(5*3^k + 5*3^k/(1 - 30*x*3^k/(15*x*3^k - 1/U(k+1)))); (continued fraction, 4-step). - Sergei N. Gladkovskii, Nov 01 2012
E.g.f.: (5/2)*U(0) where U(k) = 1 + 2/(5*3^k + 5*3^k/(1 - 30*x*3^k/(15*x*3^k - (k+1)/U(k+1)))); (continued fraction, 4-step). - Sergei N. Gladkovskii, Nov 01 2012
G.f.: x*(3-4*x) / ( (3*x-1)*(x-1) ). - R. J. Mathar, Jan 25 2015
E.g.f.: (5*exp(3*x) + 3*exp(x) - 8)/6. - Stefano Spezia, Aug 28 2023
EXAMPLE
G.f. = 3*x + 8*x^2 + 23*x^3 + 68*x^4 + 203*x^5 + 608*x^6 + 1823*x^7 + 5468*x^8 + ...
MATHEMATICA
Table[(5*3^(n-1) + 1)/2, {n, 30}] (* T. D. Noe, Oct 11 2012 *)
PROG
(PARI) a(n)=(5*3^(n-1)+1)/2 \\ Charles R Greathouse IV, Oct 11 2012
(Magma) [(5*3^(n-1) + 1)/2: n in [1..30]]; // Vincenzo Librandi, Oct 11 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Mallows and N. J. A. Sloane, Sep 16 2000
EXTENSIONS
Incorrect zeroth term removed by Jon Perry, Oct 11 2012
STATUS
approved