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%I #12 Jan 18 2024 01:49:52
%S 1,2,2,3,3,3,3,4,4,4,3,5,3,5,5,5,3,6,3,6,5,5,3,7,5,5,6,6,3,7,3,6,6,5,
%T 7,8,3,5,6,8,3,8,3,7,8,5,3,9,5,7,6,7,3,9,6,8,6,5,3,11,3,5,9,7,7,8,3,7,
%U 6,10,3,11,3,5,9,7,7,8,3,10,8,5,3,11,7,5,6,9,3,12,6,7,6,5,7,11,3,8,10,10,3,9,3,9,11,5,3,12,3,9,6,10,3,9,7,7,10,5,7,14
%N Bisection of A137921.
%C Since the other bisection (A099774) is known - it is the bisection of A000005 - the obvious problem is to find a formula for this sequence.
%H N. J. A. Sloane, <a href="/A174199/b174199.txt">Table of n, a(n) for n = 1..5000</a>
%F From _Amiram Eldar_, Jan 18 2024: (Start)
%F a(n) = A137921(2*n).
%F Sum_{k=1..n} a(k) ~ (n/2) * (3*log(n) + 6*gamma - 7 + log(2)), where gamma is Euler's constant (A001620). (End)
%o (PARI) a(n) = sumdiv(2*n, d, ((2*n)%(d+1) != 0)); \\ _Amiram Eldar_, Jan 18 2024
%Y Cf. A000005, A001620, A099774, A137921.
%K nonn
%O 1,2
%A _N. J. A. Sloane_, Nov 26 2010
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