



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
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OFFSET

1,2


COMMENTS

Partial sums of Kimberling's paraphrases: if n = (2k1)*2^m then a(n) = k. The subsequence of primes in this partial sum begins: 2, 5, 23, 29, 31, 89, 107, 109, 383, 409, 521.
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


LINKS

Table of n, a(n) for n=1..58.


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)


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 *)


CROSSREFS

Cf. A003602, A003603, A101279, A117303.
Sequence in context: A186077 A018498 A002048 * A190809 A067941 A259711
Adjacent sequences: A174986 A174987 A174988 * A174990 A174991 A174992


KEYWORD

nonn


AUTHOR

Jonathan Vos Post, Apr 03 2010


STATUS

approved



