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%I #18 Sep 22 2017 03:51:23
%S 0,1,1,2,1,0,1,3,2,0,1,3,1,0,0,4,1,3,1,3,0,0,1,4,2,0,3,3,1,0,1,5,0,0,
%T 0,0,1,0,0,4,1,0,1,3,3,0,1,5,2,3,0,3,1,4,0,4,0,0,1,2,1,0,3,6,0,0,1,3,
%U 0,0,1,5,1,0,3,3,0,0,1,5,4,0,1,2,0,0,0,4,1,2,0,3,0,0,0,6,1,3,3,0,1,0,1,4,0
%N a(n) = sum of the exponents occurring only once each in the prime-factorization of n.
%H Diana Mecum and Michael De Vlieger, <a href="/A136566/b136566.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from Diana Mecum)
%e 4200 = 2^3 * 3^1 * 5^2 * 7^1. The exponents of the prime factorization are therefore 3,1,2,1. The exponents occurring exactly once are 2 and 3. So a(4200) = 2+3 = 5.
%t Table[Total@ Flatten@ Select[Split[Sort[FactorInteger[n][[All, -1]]]], Length@ # == 1 &] - Boole[n == 1], {n, 105}] (* _Michael De Vlieger_, Sep 21 2017 *)
%o (PARI) a(n) = my(f=factor(n)[,2]); sum(k=1, #f, f[k]*(#select(x->(x==f[k]), f) == 1)); \\ _Michel Marcus_, Sep 22 2017
%Y Cf. A136565, A136567.
%K nonn
%O 1,4
%A _Leroy Quet_, Jan 07 2008
%E More terms from _Diana L. Mecum_, Jul 17 2008
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