

A175524


A000120deficient numbers.


14



1, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269
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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^n1). If this is true, the sequence contains infinitely many composite numbers, because it contains all of the form (2^n+3)^2.


LINKS



FORMULA



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 !! (n1)
a175524_list = map (+ 1) $ findIndices (< 0) a192895_list


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



