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A111497 Difference between successive terms of floor(10^n/Li(10^n) - 1). 0

%I #3 Oct 01 2013 17:58:21

%S 2,3,2,2,2,2,3,2,2,3,2,3,2,2,3,2,3,2,2,3,2,2,2,2,3,2,2,2,2,3,2,2,2,2,

%T 3,2,2,2,2,3,2,2,3,2,3,2,2,3,2,3,2,2,3,2,3,2,2,3,2,2,2,2,3,2,2,2,2,3,

%U 2,2,2,2,3,2,2,2,2,3,2,2,3,2,3,2,2,3,2,3,2,2,3,2,3,2,2,3,2,2,2,2

%N Difference between successive terms of floor(10^n/Li(10^n) - 1).

%C 10^n/Li(10^n) - 1) is the ratio of estimated composite numbers less than 10^n to the estimated prime numbers less than 10^n. Conjecture: 2 and 3 are the only numbers in this sequence.

%F Li(n) is the logarithmic integral which approximates the number of primes less than n. n Li(n) = Int dt/log(t) 2

%o (PARI) LiRatioDiff(m,n) = { local(x,p1,p2,a,b); forstep(x=m,n,2, p1=10.^x; p2=10^(x+1); a=floor(p1/Li(p1)-1); b=floor(p2/Li(p2)-1); print1(b-a,",") ) } Li(x) = \ Logarithmic integral { -eint1(log(1/x)) }

%K easy,nonn

%O 1,1

%A _Cino Hilliard_, Nov 16 2005

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