



2, 4, 4, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10
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OFFSET

1,1


COMMENTS

Arises in a renewal problem. Suppose X(1), X(2), ... are independent and identically distributed random variables with P(X = 1) = P(X = 2) = 0.5, and let N(n) denote the first value of k such that X(1)*X(2)*...*X(k) > x. Then a(n) gives the expected value of N(n), n = 1, 2, 3, ...


LINKS

Table of n, a(n) for n=1..31.
Edward Omey, A new type of renewal sequence.


FORMULA

a(n)= 2*floor(log(n)/log(2)).
a(n) = 2 + a(floor(n/2)).
G.f.: 2*(z + z^2 + z^4 + z^8 + z^16 + ...)/(1  z)
Other type of g.f.: H(z) = z^a(1) + z^a(2) + z^a(3) + ... = z^2/(1  (2*z)^2)


CROSSREFS

Sequence in context: A251546 A282894 A162550 * A113724 A161794 A111650
Adjacent sequences: A191679 A191680 A191681 * A191683 A191684 A191685


KEYWORD

nonn,easy


AUTHOR

Edward Omey, Jun 11 2011


STATUS

approved



