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A093608 Let b(0)=1; b(1)=1; b(n+2)=(e^g+1/e^g)*b(n+1)-b(n). a(n)=floor(b(n)). 5

%I #5 Oct 19 2017 03:14:29

%S 1,1,2,3,6,11,20,36,64,115,205,366,652,1162,2070,3687,6567,11696,

%T 20832,37103,66084,117701,209635,373375,665008,1184428,2109552,

%U 3757265,6691962,11918868,21228368,37809262,67341034,119939258,213620504

%N Let b(0)=1; b(1)=1; b(n+2)=(e^g+1/e^g)*b(n+1)-b(n). a(n)=floor(b(n)).

%C g is Euler's gamma, 0.5772156649...

%C a(n+1)/a(n) converges to e^g.

%C Young states, "It has been argued on probabilistic grounds that the expected number of primes p in the octave interval (x,2x) for which 2^p-1 is a prime is e^G where G is Euler's constant. Equivalently: If M(n) is the n-th Mersenne prime, then (log to base 2): log log M(n)/n ==> e^(-G)."

%D Robert M. Young, "Excursions in Calculus, An Interplay of the Continuous and the Discrete", MAA, 1992, p. 245.

%Y Cf. A090039, A090426, A090427, A093607.

%K nonn

%O 0,3

%A _Gary W. Adamson_, Apr 04 2004

%E Edited by _Don Reble_, Nov 14 2005

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Last modified April 24 22:17 EDT 2024. Contains 371964 sequences. (Running on oeis4.)