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Let b(0)=1/2, b(n) = (b(n-1)+Prime[n])/2; sequence gives 2^(n+1)*b(n).
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%I #4 Jul 14 2012 11:40:56

%S 1,5,17,57,169,521,1353,3529,8393,20169,49865,113353,264905,600777,

%T 1305289,2845385,6318793,14052041,30042825,65170121,139619017,

%U 292711113,624061129,1320315593,2813487817,6068267721,12846262985,26670688969

%N Let b(0)=1/2, b(n) = (b(n-1)+Prime[n])/2; sequence gives 2^(n+1)*b(n).

%C An integer sequence as a power of 2 representation of the running average of the primes (with 1 as the zeroth element).

%t a[0] = 1/2; a[n_] := a[n] = (a[n - 1] + Prime[n])/2; Table[2^(n + 1)*a[n], {n, 0, 27}]

%K nonn

%O 0,2

%A _Roger L. Bagula_, Nov 28 2005

%E More terms from _Robert G. Wilson v_, Nov 29 2005