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A112039 Let b(0)=1/2, b(n) = b(n-1) + Prime[n]/2; a(n)=b(2*n). 1

%I #5 Mar 30 2012 18:40:30

%S 3,9,21,39,65,99,141,191,251,320,396,482,581,686,797,926,1064,1214,

%T 1374,1544,1724,1916,2114,2331,2559,2795,3041,3301,3571,3850,4138,

%U 4447,4762,5096,5444,5800,6170,6551,6944,7349,7769,8201,8642,9095,9557,10030

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

%D H. L. Nelson, "Prime Sums", J. Rec. Math., 14 (1981), 205-206.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeSums.html">Prime Sums</a>.

%F Asymptotically b(n) ~ 2 * n^2 * (log 2 + log(n)). - _Jonathan Vos Post_, Nov 29 2005

%F b(n) = (1 + A007504(2*n))/2. - _Jonathan Vos Post_, Nov 29 2005

%t a[0] = 1/2; a[n_] := a[n] = a[n - 1] + Prime[n]/2 bb = Table[a[2*n], {n, 1, 200}]

%Y Cf. A000040, A007504.

%K nonn

%O 1,1

%A _Roger L. Bagula_, Nov 28 2005

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Last modified August 27 03:11 EDT 2024. Contains 375462 sequences. (Running on oeis4.)