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%I #13 May 12 2023 21:36:44
%S 1,1,2,1,3,1,4,1,1,1,7,0,7,1,1,1,10,0,10,0,1,1,12,2,2,1,1,0,14,2,14,0,
%T 1,1,1,0,17,1,1,2,19,2,19,0,0,1,20,0,6,0,1,0,21,2,1,2,1,1,24,0,24,1,0,
%U 1,1,2,27,0,1,2,28,0,28,1,0,0,1,2,30,0,1,1,32,0,1,1,1,2,35,0,1,0,1,1,1,0,39
%N a(n) is number of earlier terms equal to number of proper divisors of n.
%C Similar to A125087, but instead of exponents, we use number of proper divisors.
%H Katarzyna Matylla, <a href="/A135591/b135591.txt">Table of n, a(n) for n = 1..1000</a>
%e a(12)=0 because 12 has 5 proper divisors (1, 2, 3, 4 and 6) and there is no 5 in a(1), a(2), ..., a(11).
%o (Maxima) max:1000; f:makelist(0,i,1,max); apr:makelist(0, i, 0, max); f[1]:1; apr[2]:1; for n:2 through max do block(f[n]:apr[divsum(n,0)], apr[f[n]+1]:apr[f[n]+1]+1);
%Y Cf. A125087.
%K nonn
%O 1,3
%A _Katarzyna Matylla_, Feb 25 2008
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