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%I #11 Mar 25 2020 15:59:09
%S 1,2,3,3,4,4,5,5,5,5,5,5,6,6,6,6,7,7,8,8,8,8,8,8,8,8,8,8,8,8,9,9,9,9,
%T 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,10
%N Number of solutions to Cototient(x) = A051953(x) = 2^n.
%C a(n) increases at A000043(n).
%C Since A051953(p) = 1 for p prime, and given that there are an infinite number of primes, we disregard a(0) = oo. - _Michael De Vlieger_, Mar 25 2020
%e InvCototient(2^0) has an infinite number of entries, so 2^0=1 is left out.
%e n=14: 2^14=16384, InvCototient(16384) = {24576,28672,31744,32512,32764,32768}, so a(14)=6;
%t Length /@ Most@ Split@ DeleteCases[Select[Array[# - EulerPhi[#] &, 10^6], IntegerQ@ Log2@ # &], 1] (* _Michael De Vlieger_, Mar 25 2020 *)
%Y Cf. A051953, A058764, A000043, A063740.
%K nonn
%O 1,2
%A _Labos Elemer_, Jun 13 2002
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