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%I #35 Jul 25 2023 10:20:38
%S 1,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,
%T 97,101,103,107,109,113,121,127,131,137,139,149,151,157,163,167,173,
%U 179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269
%N A000120-deficient numbers.
%C For a more precise definition, see comment in A175522.
%C All odd primes (A065091) are in the sequence. Squares of the form (2^n+3)^2, n>=3, where 2^n+3 is prime (A057733), are also in the sequence. [Proof: (2^n+3)^2 = 2^(2*n)+2^(n+2)+2^(n+1)+2^3+1. Thus, since n>=3, A000120((2^n+3)^2)=5. Also, for primes of the form 2^n+3, (2^n+3)^2 has only two proper divisors, 1 and 2^n+3, so A000120(1)+A000120(2^n+3) = 4, and in conclusion, (2^n+3)^2 is deficient. QED]
%C It is natural to assume that there are infinitely many primes of the form 2^n+3 (by analogy with the Mersenne sequence 2^n-1). If this is true, the sequence contains infinitely many composite numbers, because it contains all of the form (2^n+3)^2.
%C a(n) = A006005(n) for n <= 30;
%H Reinhard Zumkeller, <a href="/A175524/b175524.txt">Table of n, a(n) for n = 1..1000</a>
%F A192895(a(n)) < 0. - _Reinhard Zumkeller_, Jul 12 2011
%t Select[Range[270], DivisorSum[#, DigitCount[#, 2, 1] &] < 2*DigitCount[#, 2, 1] &] (* _Amiram Eldar_, Jul 25 2023 *)
%o (Sage) is_A175524 = lambda x: sum(A000120(d) for d in divisors(x)) < 2*A000120(x)
%o A175524 = filter(is_A175524, IntegerRange(1, 10**4)) # _D. S. McNeil_, Dec 04 2010
%o (Haskell)
%o import Data.List (findIndices)
%o a175524 n = a175524_list !! (n-1)
%o a175524_list = map (+ 1) $ findIndices (< 0) a192895_list
%o -- _Reinhard Zumkeller_, Jul 12 2011
%o (PARI) is(n)=sumdiv(n,d,hammingweight(d))<2*hammingweight(n) \\ _Charles R Greathouse IV_, Jan 28 2016
%Y Cf. A175522 (perfect version), A175526 (abundant version), A000120, A005100, A005101, A006005, A192895.
%K nonn,base
%O 1,2
%A _Vladimir Shevelev_, Dec 03 2010
%E More terms from _Amiram Eldar_, Feb 18 2019
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