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A089680 Let g[n_]=(Prime[n]-n)/(n-PrimePi[n]). Then a(n) = Floor[Exp[((g[n]+g[n+1])-g[n*(n+1)]+1)/1.5]]. 0

%I #4 Mar 30 2012 17:34:14

%S 3,3,1,2,4,3,3,1,2,2,3,4,4,2,2,3,4,4,4,3,2,3,3,3,3,3,2,2,2,3,4,3,3,3,

%T 3,3,3,3,3,3,3,3,3,3,2,3,4,4,3,3,3,3,3,3,3,3,2,2,2,2,3,3,3,3,2,3,3,3,

%U 3,3,3,3,4,4,3,3,3,3,3,3,3,3,4,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,3

%N Let g[n_]=(Prime[n]-n)/(n-PrimePi[n]). Then a(n) = Floor[Exp[((g[n]+g[n+1])-g[n*(n+1)]+1)/1.5]].

%C Exponential version of log like weighted average function of primes and the prime distribution.

%t digits=200 g[n_]=(Prime[n]-n)/(n-PrimePi[n]) b=Table[Floor[Exp[((g[n]+g[n+1])-g[n*(n+1)]+1)/1.5]], {n, 1, digits}]

%K nonn

%O 1,1

%A _Roger L. Bagula_, Jan 04 2004

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