A175522 A000120-perfect numbers.

2, 25, 95, 111, 119, 123, 125, 169, 187, 219, 221, 247, 289, 335, 365, 411, 415, 445, 485, 493, 505, 629, 655, 685, 695, 697, 731, 767, 815, 841, 871, 943, 949, 965, 985, 1003, 1139, 1207, 1241, 1261, 1263, 1273, 1343, 1387, 1465, 1469, 1507, 1513, 1529, 1563

Offset

1,1

Comments

Let A(n), n>=1, be an infinite positive sequence.

We call a number n:

A-deficient if Sum{d|n, d<n} A(d) < A(n),

A-abundant if Sum{d|n, d<n} A(d) > A(n),

and

A-perfect if Sum{d|n, d<n} A(d) = A(n),

depending on the sum over the proper divisors of n.

The definition generalizes the standard nomenclature of deficient (A005100), abundant (A005101) and perfect numbers (A000396), which is recovered by setting A(n) = n = A000027(n).

Conjecture: if there exist infinitely many A-deficient numbers and infinitely many A-abundant numbers, then there exist infinitely many A-perfect numbers.

Note that the sequence contains squares of all Fermat primes larger than 3 (see A019434). [This would also hold for squares of any hypothetical Fermat primes after the fifth one, 65537. Comment clarified by Antti Karttunen, May 14 2015]

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

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Reinhard Zumkeller)

Example

Proper divisors of 119 are 1,7,17. Since A000120(1)+A000120(7)+A000120(17)=A000120(119), then 119 is in the sequence.

Mathematica

binw[n_] := DigitCount[n, 2, 1]; Select[Range[1500], binw[#] == DivisorSum[#, binw[#1] &]/2 &] (* Amiram Eldar, Dec 14 2020 *)

Prog

(Sage) A000120 = lambda x: x.digits(base=2).count(1)
is_A175522 = lambda x: sum(A000120(d) for d in divisors(x)) == 2*A000120(x)
A175522 = filter(is_A175522, IntegerRange(1, 10**4))
# D. S. McNeil, Dec 04 2010
(Haskell)
import Data.List (elemIndices)
a175522 n = a175522_list !! (n-1)
a175522_list = map (+ 1) $ elemIndices 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
(Python)
from sympy import divisors
def A000120(n): return bin(n).count('1')
def aupto(limit):
  alst = []
  for m in range(1, limit+1):
    if A000120(m) == sum(A000120(d) for d in divisors(m)[:-1]): alst += [m]
  return alst
print(aupto(1563)) # Michael S. Branicky, Feb 25 2021

Crossrefs

Cf. A175524 (deficient version), A175526 (abundant version), A000120, A000396.

Subsequence of A257691 (non-abundant version).

Positions of zeros in A192895.

Cf. A019434, A253204.

Sequence in context: A269232 A181920 A123012 * A226418 A196011 A080515

Adjacent sequences: A175519 A175520 A175521 * A175523 A175524 A175525

Keyword

nonn

Author

Vladimir Shevelev, Dec 03 2010

Extensions

More terms from Amiram Eldar, Feb 18 2019

Status

approved