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A093608
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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)).
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5
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1, 1, 2, 3, 6, 11, 20, 36, 64, 115, 205, 366, 652, 1162, 2070, 3687, 6567, 11696, 20832, 37103, 66084, 117701, 209635, 373375, 665008, 1184428, 2109552, 3757265, 6691962, 11918868, 21228368, 37809262, 67341034, 119939258, 213620504
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OFFSET
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0,3
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COMMENTS
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g is Euler's gamma, 0.5772156649...
a(n+1)/a(n) converges to e^g.
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)."
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REFERENCES
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Robert M. Young, "Excursions in Calculus, An Interplay of the Continuous and the Discrete", MAA, 1992, p. 245.
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LINKS
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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