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A252056 a(n) is the least m such that m = A001065(j) = A001065(k) where j != k, A000005(j) = A000005(k) = n; or 0 if no such m exists.

%I A252056 #32 Jan 15 2015 12:41:41
%S A252056 0,1,0,13,0,73,0,106,9064940,4001,0,396,0
%N A252056 a(n) is the least m such that m = A001065(j) = A001065(k) where j != k, A000005(j) = A000005(k) = n; or 0 if no such m exists.
%C A252056 When n>2 and A001055(n)=1, then a(n)=0; because in that case, only a prime^n has n divisors, and then it is not possible to get twice the same value for sigma(x)-x. This happens for n=3, 5, 7, 11, 13, 17, 19, 23, 29, ... - _Michel Marcus_, Dec 16 2014
%C A252056 Note that for n=8, j and k do not have the same prime signature. - _Michel Marcus_, Dec 17 2014
%e A252056 For n=2, all primes have 2 divisors and satisfy sigma(x)-x=1, so a(2) = 1.
%e A252056 For n=4, 27 and 35 have 4 divisors and the sum of their proper divisors is 13 for both (1+3+9 and 1+5+7).
%e A252056 For n=6, 98 and 175 have 6 divisors and the sum of their proper divisors is 73 for both (1+2+7+14+49 and 1+5+7+25+35).
%e A252056 For n=8, 104 and 110 have 8 divisors and the sum of their proper divisors is 106 for both (1+2+4+8+13+26+52 and 1+2+5+10+11+22+55).
%e A252056 For n=9, 163^2*167^2 and 61^2*353^2 have 9 divisors and the sum of their proper divisors is 9064940 for both.
%e A252056 For n=10, 7203 and 7857 have 10 divisors and the sum of their proper divisors is 4001 for both.
%e A252056 For n=12, 276 and 306 have 12 divisors and the sum of their proper divisors is 396 for both.
%Y A252056 Cf. A000005 (number of divisors of n), A001065 (sum of proper divisors of n).
%Y A252056 Cf. A001055, A048138, A152454.
%K A252056 nonn,more
%O A252056 1,4
%A A252056 _Naohiro Nomoto_, Dec 13 2014
%E A252056 a(9)-a(13) from _Michel Marcus_, Dec 16 2014