%I
%S 1,1,2,3,4,5,5,7,8,9,7,11,9,13,14,15,10,17,11,19,20,20,13,23,18,22,23,
%T 27,16,29,17,31,28,28,28,35,19,30,30,39,20,41,21,41,42,34,22,47,26,46,
%U 39,48,23,52,41,54,42,42,25,59,26,45,55,56
%N a(1) = 1, a(n) counts m < n for which a(m) has at most as many divisors as n.
%H Peter Kagey, <a href="/A269524/b269524.txt">Table of n, a(n) for n = 1..10000</a>
%e a(1) = 1;
%e a(2) = 1 because a(1) has as many or fewer divisors as 2;
%e a(3) = 2 because a(1) and a(2) have as many or fewer divisors as 3;
%e a(4) = 3 because a(1), a(2), and a(3) have as many or fewer divisors than 4;
%e (...)
%e a(7) = 5 because a(1), a(2), a(3), a(4), and a(6) have as many or fewer divisors as 7.
%t a = {1}; Do[d = DivisorSigma[0, n]; AppendTo[a, Count[a, k_ /; DivisorSigma[0, k] <= d]], {n, 2, 64}]; a (* _Michael De Vlieger_, Feb 29 2016 *)
%o (Java)
%o int[] terms = new int[100];
%o terms[0] = 1;
%o for (int i = 1; i < 100; i++) {
%o int count = 0;
%o for (int j = 0; j < i; j++) {
%o if (divisors(terms[j]) <= divisors(i+1)) {
%o count = count + 1;
%o }
%o }
%o terms[i] = count;
%o }
%o (PARI) lista(nn) = {va = vector(nn); print1(va[1] = 1, ", "); for (n=2, nn, va[n] = sum(k=1, n1, numdiv(va[k]) <= numdiv(n)); print1(va[n], ", "););} \\ _Michel Marcus_, Feb 29 2016
%Y Cf. A032741.
%K easy,nonn
%O 1,3
%A _Alec Jones_, Feb 28 2016
