

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.
a(n) = A006005(n) for n <= 30;
A192895(a(n)) < 0. Reinhard Zumkeller, Jul 12 2011


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000


PROG

(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
(Haskell)
import Data.List (findIndices)
a175524 n = a175524_list !! (n1)
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

Cf. A175522 (perfect version), A175526 (abundant version), A000120, A005100, A005101.
Sequence in context: A056912 A075763 A074918 * A073579 A006005 A065091
Adjacent sequences: A175521 A175522 A175523 * A175525 A175526 A175527


KEYWORD

nonn


AUTHOR

Vladimir Shevelev, Dec 03 2010


EXTENSIONS

More terms from Amiram Eldar, Feb 18 2019


STATUS

approved



