OFFSET
0,1
COMMENTS
Another renewal type of sequence: Let X, X(1),X(2),... denote independent random variables with pdf P(X=1) = P(X=2) = P(X=4) = 1/3. Let N(x) denote the first value of k such that X(1)*X(2)...*X(k) > x, and let H(x) = E(N(x)). The sequence a(n) is given by a(n) = 2^(n+1)*H(2^n).
LINKS
FORMULA
a(n) = n*2^(n+1) + (2^(n+3)+(-1)^n)/3.
a(n) = 3 * A045883(n+1).
G.f.: 3/((1 + x)*(1 - 2*x)^2). [Bruno Berselli, Oct 16 2014]
MATHEMATICA
Table[n 2^(n + 1) + (2^(n + 3) + (-1)^n)/3, {n, 0, 70}] (* Vincenzo Librandi, Oct 16 2014 *)
LinearRecurrence[{3, 0, -4}, {3, 9, 27}, 40] (* Harvey P. Dale, Feb 11 2024 *)
PROG
(PARI) a(n) = n*2^(n+1) + (2^(n+3)+(-1)^n)/3; \\ Michel Marcus, Oct 16 2014
(Magma) [n*2^(n+1)+(2^(n+3)+(-1)^n)/3: n in [0..30]]; // Vincenzo Librandi, Oct 16 2014
CROSSREFS
KEYWORD
nonn,easy,less
AUTHOR
Edward Omey, Jun 16 2011
EXTENSIONS
Formula corrected and more terms from Michel Marcus, Oct 16 2014
STATUS
approved