OFFSET
1,2
COMMENTS
For a more precise definition, see comment in A175522.
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]
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.
a(n) = A006005(n) for n <= 30;
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
FORMULA
A192895(a(n)) < 0. - Reinhard Zumkeller, Jul 12 2011
MATHEMATICA
Select[Range[270], DivisorSum[#, DigitCount[#, 2, 1] &] < 2*DigitCount[#, 2, 1] &] (* Amiram Eldar, Jul 25 2023 *)
PROG
(Haskell)
import Data.List (findIndices)
a175524 n = a175524_list !! (n-1)
a175524_list = map (+ 1) $ findIndices (< 0) a192895_list
-- Reinhard Zumkeller, Jul 12 2011
(PARI) is(n)=sumdiv(n, d, hammingweight(d))<2*hammingweight(n) \\ Charles R Greathouse IV, Jan 28 2016
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Vladimir Shevelev, Dec 03 2010
EXTENSIONS
More terms from Amiram Eldar, Feb 18 2019
STATUS
approved