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%I #10 Dec 16 2021 10:38:44
%S 1,3,15,63,255,65535,4294967295
%N Numbers n such that 2^tau(n) = n + 1 (where tau(n) = number of divisors of n).
%C It is clear that n+1 must be a power of 2. Hence n=2^k-1 for some k. Found k=1, 2, 4, 6, 8, 16, 32. No other k<150. - _T. D. Noe_, Nov 04 2005
%F Note that this is different from the sequence A019434 - 2.
%t Select[Range[10^6], 2^DivisorSigma[0, # ] == # + 1 &]
%t 2^Select[Range[150], DivisorSigma[0, 2^#-1]==#&] - 1 (Noe)
%o (Python)
%o from sympy import divisor_count as tau
%o def afind(klimit, kstart=1):
%o for k in range(kstart, klimit+1):
%o m = 2**k - 1
%o if 2**tau(m) == m + 1: print(m, end=", ")
%o afind(klimit=100) # _Michael S. Branicky_, Dec 16 2021
%Y Cf. A046801 (number of divisors of 2^n-1), A019434.
%K nonn,hard,more
%O 1,2
%A _Joseph L. Pe_, Nov 02 2005
%E One more term from _T. D. Noe_, Nov 04 2005
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