A174989 Partial sums of A003602.

1, 2, 4, 5, 8, 10, 14, 15, 20, 23, 29, 31, 38, 42, 50, 51, 60, 65, 75, 78, 89, 95, 107, 109, 122, 129, 143, 147, 162, 170, 186, 187, 204, 213, 231, 236, 255, 265, 285, 288, 309, 320, 342, 348, 371, 383, 407, 409, 434, 447, 473, 480, 507, 521, 549, 553, 582, 597

Offset: 1

Author

Jonathan Vos Post, Apr 03 2010

Keywords

• nonn
• easy

Comments

I conjecture that infinitely many terms are prime. For n<=10^5, exactly 5115 terms are prime. For n<=10^7, there are 352704 prime terms. The largest prime for n<10^10 is at n=9999999983, a(n)=16666666618226308891. Below 10^100, n=(10^100)-345. Below 10^500, n=(10^500)-2414. - Griffin N. Macris, May 04 2016

Since (n^2+3n)/6 < a(n) < (n^2+5n+4)/6, the sum of reciprocals of this sequence converges to a value between 13/6 and 11/3, approximately 2.888. - Griffin N. Macris, May 07 2016

Crossrefs

Cf. A003602, A003603, A101279, A117303, A135013.

Programs

Mathematica

a[0]:=0;
a[n_]:=Ceiling[n/2](1+Ceiling[n/2])/2 + a[Floor[n/2]];
Array[a,50] (* Griffin N. Macris, May 04 2016 *)

Formula

a(n) = Sum{i=1..n} A003602(i) = Sum_{i=1..n} (A000265(i) + 1)/2.

From Griffin N. Macris, May 04 2016 (Start)

a(0) = 0; a(n) = A000217(ceiling(n/2)) + a(floor(n/2)).

Asymptotically, a(n) ~ (n^2+3n)/6. (End)

a(n) = (A135013(n) + n)/2. - Amiram Eldar, Dec 27 2022