%I #13 Nov 25 2023 22:09:42
%S 1,3,1,3,1,6,1,7,4,3,1,16,1,3,4,7,1,12,1,12,4,3,1,24,1,3,4,7,1,17,1,7,
%T 4,3,1,25,1,3,4,20,1,19,1,7,9,3,1,24,1,8,4,7,1,12,1,22,4,3,1,43,1,3,4,
%U 7,1,12,1,7,4,15,1,45,1,3,9,7,1,12,1,30,4,3,1,35,1
%N Sum of the divisors of n <= tau(n).
%C First differs from A126212(n) at a(25) = 1.
%F a(n) = Sum_{d|n, d<=tau(n)} d.
%e a(12) = 16. The sum of the divisors of 12 <= tau(12) = 6 are 1 + 2 + 3 + 4 + 6 = 16.
%t Table[Sum[k(1-Ceiling[n/k]+Floor[n/k]), {k, DivisorSigma[0, n]}], {n, 100}]
%o (PARI) a(n) = my(t=numdiv(n)); sumdiv(n, d, if (d <=t, d)); \\ _Michel Marcus_, Nov 25 2023
%Y Cf. A000005 (tau), A126212, A366979.
%K nonn
%O 1,2
%A _Wesley Ivan Hurt_, Nov 24 2023