A175524 A000120-deficient numbers.

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

Offset: 1

Author

Vladimir Shevelev, Dec 03 2010

Keywords

• nonn
• base

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

Crossrefs

Cf. A175522 (perfect version), A175526 (abundant version), A000120, A005100, A005101, A006005, A192895.

Programs

Haskell

import Data.List (findIndices)
a175524 n = a175524_list !! (n-1)
a175524_list = map (+ 1) $ findIndices (< 0) a192895_list
-- Reinhard Zumkeller, Jul 12 2011

Mathematica

Select[Range[270], DivisorSum[#, DigitCount[#, 2, 1] &] < 2*DigitCount[#, 2, 1] &] (* Amiram Eldar, Jul 25 2023 *)

PARI

is(n)=sumdiv(n,d,hammingweight(d))<2*hammingweight(n) \\ Charles R Greathouse IV, Jan 28 2016

Sage

is_A175524 = lambda x: sum(A000120(d) for d in divisors(x)) < 2*A000120(x)
A175524 = filter(is_A175524, IntegerRange(1, 10**4)) # D. S. McNeil, Dec 04 2010

Formula

A192895(a(n)) < 0. - Reinhard Zumkeller, Jul 12 2011

Extensions

More terms from Amiram Eldar, Feb 18 2019