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A161841 Number of factors, with repetition, in all distinct pairs (a <= b) such that a*b = n.

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%I A161841 #18 Feb 05 2025 02:51:22
%S A161841 2,2,2,4,2,4,2,4,4,4,2,6,2,4,4,6,2,6,2,6,4,4,2,8,4,4,4,6,2,8,2,6,4,4,
%T A161841 4,10,2,4,4,8,2,8,2,6,6,4,2,10,4,6,4,6,2,8,4,8,4,4,2,12,2,4,6,8,4,8,2,
%U A161841 6,4,8,2,12,2,4,6,6,4,8,2,10,6,4,2,12,4,4,4,8,2,12,4,6,4,4,4,12,2,6,6,10
%N A161841 Number of factors, with repetition, in all distinct pairs (a <= b) such that a*b = n.
%H A161841 Robert Israel, <a href="/A161841/b161841.txt">Table of n, a(n) for n = 1..10000</a>
%F A161841 a(n) = A000005(n) + A010052(n) = A038548(n)*2.
%F A161841 Sum_{k=1..n} a(k) ~ (log(n) + 2*gamma - 1)*n + sqrt(n), where gamma is the Euler-Mascheroni constant A001620. - _Vaclav Kotesovec_, Jul 01 2021
%e A161841 a(16)=6 because there are three distinct pairs (a <= b) such that a*b = n: the pairs (1,16), (2,8) and (4,4). So the number of factors, with repetition, in all the pairs is equal to 6.
%p A161841 seq(numtheory:-tau(n) + `if`(issqr(n),1,0), n = 1 .. 200); # _Robert Israel_, Dec 23 2015
%t A161841 2*Ceiling[DivisorSigma[0, Range[100]]/2] (* _Paolo Xausa_, Feb 05 2025 *)
%Y A161841 Cf. A000005, A010052, A038548, A161842.
%K A161841 easy,nonn
%O A161841 1,1
%A A161841 _Omar E. Pol_, Jun 23 2009