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%I #11 Jun 07 2018 22:02:47
%S 0,3,2,1,0,0,2,1,0,0,0,0,0,1,0,1,0,0,0,2,1,0,0,0,0,0,0,0,0,0,1,0,3,2,
%T 1,0,0,2,1,0,0,0,0,2,1,0,0,0,0,2,1,0,0,4,3,2,1,0,0,0,0,1,0,0,0,0,0,1,
%U 0,0,0,0,0,3,2,1,0,0,0,0,0,0,0,0,3,2,1
%N Number of successors of n having the same value of A001221 as n.
%e For n = 2: 3, 4 and 5 have the same number of distinct prime divisors as 2, but 6 does not, so a(2) = 3.
%o (PARI) a(n) = my(k=n+1, i=0); while(omega(k)==omega(n), i++; k++); i
%Y Cf. A001221, A077655, A305235.
%K nonn
%O 1,2
%A _Felix Fröhlich_, May 28 2018
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