Asymptotically b(n) ~ 2 * n^2 * (log 2 + log(n)). - _Jonathan Vos Post (jvospost3(AT)gmail.com), _, Nov 29 2005

b(n) = (1 + A007504(2*n))/2. - _Jonathan Vos Post (jvospost3(AT)gmail.com), _, Nov 29 2005

_Roger L. Bagula (rlbagulatftn(AT)yahoo.com), _, Nov 28 2005

Asymptotically b(n) ~ 2 * n^2 * (log 2 + log(n)). - Jonathan Vos Post (jvospost2jvospost3(AT)yahoogmail.com), Nov 29 2005

b(n) = (1 + A007504(2*n))/2. - Jonathan Vos Post (jvospost2jvospost3(AT)yahoogmail.com), Nov 29 2005

nonn,new

nonn

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

nonn,new

nonn

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

3, 9, 21, 39, 65, 99, 141, 191, 251, 320, 396, 482, 581, 686, 797, 926, 1064, 1214, 1374, 1544, 1724, 1916, 2114, 2331, 2559, 2795, 3041, 3301, 3571, 3850, 4138, 4447, 4762, 5096, 5444, 5800, 6170, 6551, 6944, 7349, 7769, 8201, 8642, 9095, 9557, 10030

1,1

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

Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PrimeSums.html">Prime Sums</a>.

Asymptotically b(n) ~ 2 * n^2 * (log 2 + log(n)). - Jonathan Vos Post (jvospost2(AT)yahoo.com), Nov 29 2005

b(n) = (1 + A007504(2*n))/2. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Nov 29 2005

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

Cf. A000040, A007504.

nonn

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Nov 28 2005

approved