A175522 -id:A175522 - cp's OEIS frontend

A360639 Numbers k such that k and k+2 are both A000120-perfect numbers (A175522).

123, 219, 695, 1261, 1851, 1943, 3543, 5963, 7031, 7613, 7769, 7861, 10081, 11357, 11629, 12083, 13211, 13791, 14185, 15699, 15835, 15929, 16241, 18649, 20197, 20989, 22521, 23449, 23521, 23963, 24461, 27215, 27829, 28263, 28367, 29485, 29651, 30359, 30901, 31803

Offset: 1

Amiram Eldar, Feb 15 2023

The smallest gap between two consecutive A000120-perfect numbers is 2.

All terms of this sequence are odd.

			123 is a term since 123 and 125 are both in A175522: A093653(123)/A000120(123) = A093653(125)/A000120(125) = 12/6 = 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Subsequence of A175522.

Cf. A000120, A093653, A338452, A360640.

q[n_] := DivisorSum[n, DigitCount[#, 2, 1] &] == 2 * DigitCount[n, 2, 1]; seq[kmax_] := Module[{s = {}, k = 1, q1 = False, q2}, Do[q2 = q[k]; If[q1 && q2, AppendTo[s, k-2]]; q1 = q2, {k, 3, kmax, 2}]; s]; seq[32000]

lista(kmax) = {my(is1 = 0, is2); forstep(k=1, kmax, 2, is2 = (sumdiv(k, d, hammingweight(d)) == 2*hammingweight(k)); if(is1 && is2, print1(k-2, ", ")); is1 = is2); }

A360640 a(n) is the start of the least run of exactly n consecutive odd numbers that are A000120-perfect numbers (A175522).

Original entry on oeis.org

25, 123, 31803, 8019811, 130194395

Offset: 1

Amiram Eldar, Feb 15 2023

a(6) > 2*10^11, if it exists.

			Table of values of A000120 and A093653 for k = a(n), a(n)+2, ..., a(n)+2*(n-1):
  n |      a(n)              A000120(k)              A093653(k)
  --+----------------------------------------------------------
  1 |        25                       3                       6
  2 |       123                    6, 6.                 12, 12
  3 |     31803              10, 10, 11              20, 20, 22
  4 |   8019811          15, 15, 16, 15          30, 30, 32, 30
  5 | 130194395  17, 17, 18, 15, 16, 16  34, 34, 36, 30, 32, 32

Cf. A000120, A093653, A175522, A360639.

q[n_] := DivisorSum[n, DigitCount[#, 2, 1] &] == 2 * DigitCount[n, 2, 1]; seq[len_, nmax_] := Module[{s = Table[0, {len}], v = {}, n = 1, c = 0, m}, While[c <= len && n <= nmax, If[q[n], v = Join[v, {n}], m = Length[v]; v = {}; If[0 < m <= len && s[[m]] == 0, c++; s[[m]] = n - 2*m]]; n += 2]; s]; seq[3, 10^5]

lista(len, nmax) = {my(s = vector(len), v=[], n = 1, c = 0, m); while(c <= len && n <= nmax, if(sumdiv(n, d, hammingweight(d)) == 2 * hammingweight(n), v = concat(v, n), m =#v; v = []; if(0 < m && m <= len && s[m] == 0, c++; s[m] = n - 2*m)); n += 2); s};

A360643 a(n) is the least A000120-perfect number (A175522) whose binary weight (A000120) is n, or 0 if no such number exists.

Original entry on oeis.org

2, 0, 25, 169, 841, 95, 247, 943, 767, 5999, 6139, 16123, 30655, 90109, 122847, 245695, 522237, 1572591, 1966015, 3932095, 12582651, 28311519, 33423343, 100663023, 133693435, 402128831, 931135479, 1069547515, 1610612607, 11802771447, 12884901567, 25736249279

Offset: 1

Amiram Eldar, Feb 15 2023

Apparently, the sequence is increasing after n = 6.

			a(1) = 2 since A000120(2) = 1 and A093653(2)/A000120(2) = 4/2 = 2.
a(2) = 0 since there is no number m with binary weight 2 and with A093653(m) = 4.
a(3) = 25 since A000120(25) = 3 and A093653(25)/A000120(25) = 6/3 = 2, and 25 is the least number with this property.

Cf. A000120, A093653, A175522.

seq[len_, nmax_] := Module[{s = Table[-1, {len}], n = 3, c = 2, bw, dbw}, s[[1]] = 2; While[c < len && n <= nmax, bw = DigitCount[n, 2, 1]; If[bw <= len && s[[bw]] < 0, dbw = DivisorSum[n, DigitCount[#, 2, 1] &]; If[dbw == 2*bw, c++; s[[bw]] = n]]; n += 2]; s]; seq[16, 10^6]

lista(len, nmax) = {my(s = vector(len,i,-1), n = 3, c = 2, bw, dbw); s[1] = 2; while(c < len && n <= nmax, bw = hammingweight(n); if(bw <= len && s[bw] < 0, dbw = sumdiv(n, d, hammingweight(d)); if(dbw == 2*bw, c++; s[bw] = n)); n += 2); s};

A074206 Kalmár's [Kalmar's] problem: number of ordered factorizations of n.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 8, 1, 3, 3, 8, 1, 8, 1, 8, 3, 3, 1, 20, 2, 3, 4, 8, 1, 13, 1, 16, 3, 3, 3, 26, 1, 3, 3, 20, 1, 13, 1, 8, 8, 3, 1, 48, 2, 8, 3, 8, 1, 20, 3, 20, 3, 3, 1, 44, 1, 3, 8, 32, 3, 13, 1, 8, 3, 13, 1, 76, 1, 3, 8, 8, 3, 13, 1, 48, 8, 3, 1, 44, 3, 3, 3, 20, 1, 44, 3, 8, 3, 3, 3, 112

Offset: 0

N. J. A. Sloane, Apr 29 2003

a(0)=0, a(1)=1; thereafter a(n) is the number of ordered factorizations of n as a product of integers greater than 1.
Kalmár (1931) seems to be the earliest reference that mentions this sequence (as opposed to A002033). - N. J. A. Sloane, May 05 2016
a(n) is the permanent of the n-1 X n-1 matrix A with (i,j) entry = 1 if j|i+1 and = 0 otherwise. This is because ordered factorizations correspond to nonzero elementary products in the permanent. For example, with n=6, 3*2 -> 1,3,6 [partial products] -> 6,3,1 [reverse list] -> (6,3)(3,1) [partition into pairs with offset 1] -> (5,3)(2,1) [decrement first entry] -> (5,3)(2,1)(1,2)(3,4)(4,5) [append pairs (i,i+1) to get a permutation] -> elementary product A(1,2)A(2,1)A(3,4)A(4,5)A(5,3). - David Callan, Oct 19 2005
This sequence is important in describing the amount of energy in all wave structures in the Universe according to harmonics theory. - Ray Tomes (ray(AT)tomes.biz), Jul 22 2007
a(n) appears to be the number of permutation matrices contributing to the Moebius function. See A008683 for more information. Also a(n) appears to be the Moebius transform of A067824. Furthermore it appears that except for the first term a(n)=A067824(n)*(1/2). Are there other sequences such that when the Moebius transform is applied, the new sequence is also a constant factor times the starting sequence? - Mats Granvik, Jan 01 2009
Numbers divisible by n distinct primes appear to have ordered factorization values that can be found in an n-dimensional summatory Pascal triangle. For example, the ordered factorization values for numbers divisible by two distinct primes can be found in table A059576. - Mats Granvik, Sep 06 2009
Inverse Mobius transform of A174725 and also except for the first term, inverse Mobius transform of A174726. - Mats Granvik, Mar 28 2010
a(n) is a lower bound on the worst-case number of solutions to the probed partial digest problem for n fragments of DNA; see the Newberg & Naor reference, below. - Lee A. Newberg, Aug 02 2011
All integers greater than 1 are perfect numbers over this sequence (for definition of A-perfect numbers, see comment to A175522). - Vladimir Shevelev, Aug 03 2011
If n is squarefree, then a(n) = A000670(A001221(n)) = A000670(A001222(n)). - Vladimir Shevelev and Franklin T. Adams-Watters, Aug 05 2011
A034776 lists the values taken by this sequence. - Robert G. Wilson v, Jun 02 2012
From Gus Wiseman, Aug 25 2020: (Start)
Also the number of strict chains of divisors from n to 1. For example, the a(n) chains for n = 1, 2, 4, 6, 8, 12, 30 are:
  1  2/1  4/1    6/1    8/1      12/1      30/1
          4/2/1  6/2/1  8/2/1    12/2/1    30/2/1
                 6/3/1  8/4/1    12/3/1    30/3/1
                        8/4/2/1  12/4/1    30/5/1
                                 12/6/1    30/6/1
                                 12/4/2/1  30/10/1
                                 12/6/2/1  30/15/1
                                 12/6/3/1  30/6/2/1
                                           30/6/3/1
                                           30/10/2/1
                                           30/10/5/1
                                           30/15/3/1
                                           30/15/5/1
(End)
a(n) is also the number of ways to tile a strip of length log(n) with tiles having lengths {log(k) : k>=2}. - David Bevan, Jan 07 2025

			G.f. = x + x^2 + x^3 + 2*x^4 + x^5 + 3*x^6 + x^7 + 4*x^8 + 2*x^9 + 3*x^10 + ...
Number of ordered factorizations of 8 is 4: 8 = 2*4 = 4*2 = 2*2*2.

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 126, see #27.
R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 141.
Kalmár, Laszlo, A "factorisatio numerorum" problemajarol [Hungarian], Matemat. Fizik. Lapok, 38 (1931), 1-15.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 124.

N. J. A. Sloane, Table of n, a(n) for n = 0..20000 (first 10000 terms from T. D. Noe)
David Bevan and Julien Condé, Introducing irrational enumeration: analytic combinatorics for objects of irrational size, arXiv:2412.14682 [math.CO], 2024. See p. 11.
Peter Brown, Number of Ordered Factorizations.
Peter Brown, Number of Ordered Factorizations.
Benny Chor, Paul Lemke, and Ziv Mador, On the number of ordered factorizations of natural numbers, Discrete Math. 214 (2000), no. 1-3, 123-133. MR1743631 (2000m:11093).
Kristin DeVleming and Nikita Singh, Rational unicuspidal plane curves of low degree, arXiv:2311.15922 [math.AG], 2023. See p. 14.
T. M. A. Fink, Properties of the recursive divisor function and the number of ordered factorizations, arXiv:2307.09140 [math.NT], 2023.
E. Hille, A problem in factorisatio numerorum, Acta Arith., 2 (1936), 134-144.
E. Hille, The inversion problem of Möbius, Duke Math. J., 3 (1937), 549-568.
Shikao Ikehara, On Kalmar's Problem in "Factorisatio Numerorum", Proceedings of the Physico-Mathematical Society of Japan. 3rd Series, Vol. 21 (1939) pp. 208-219.
Shikao Ikehara, On Kalmar's Problem in "Factorisatio Numerorum" II, Proceedings of the Physico-Mathematical Society of Japan. 3rd Series, Vol. 23 (1941) pp. 767-774.
Laszlo Kalmár, Über die mittlere Anzahl der Produktdarstellungen der Zahlen. (Erste Mitteilung), Acta Litt. ac Scient. Szeged 5 (1931): 95-107.
M. Klazar and F. Luca, On the maximal order of numbers in the "factorisatio numerorum" problem, arXiv:math/0505352 [math.NT], 2005-2006.
Arnold Knopfmacher and Michael Mays, Ordered and Unordered Factorization of Integers, The Mathematica Journal, Volume 10, Issue 1 p. 72.
Arnau Mir, Francesc Rossello, and Lucia Rotger, Sound Colless-like balance indices for multifurcating trees, arXiv:1805.01329 [q-bio.PE], 2018.
Augustine O. Munagi, Labeled factorization of integers, INTEGERS: The Electronic Journal of Combinatorics 16:1 (2009), #R50.
L. A. Newberg and D. Naor, A lower bound on the number of solutions to the probed partial digest problem, Advances in Applied Mathematics, 14(2), 1993, 172-183. doi: 10.1006/aama.1993.1009.
Ray Tomes, The Maths and Physics of the Harmonics Theory.
Eric Weisstein's World of Mathematics, Perfect Partition.
Eric Weisstein's World of Mathematics, Ordered Factorization.
David W. Wilson, Comments on A074206 and related sequences.
David W. Wilson, Perl program for A074206.
Index entries for "core" sequences

Apart from initial term, same as A002033.
a(A002110) = A000670, row sums of A251683.
A173382 (and A025523) gives partial sums.
Cf. A008683, A050324, A027751, A001221, A034776, A000670, A050369.
A124433 has these as unsigned row sums.
A334996 has these as row sums.
A001055 counts factorizations.
A001222 counts prime factors with multiplicity.
A008480 counts ordered prime factorizations.
A067824 counts strict chains of divisors starting with n.
A122651 counts strict chains of divisors summing to n.
A253249 counts strict chains of divisors.
Cf. A000005, A025487, A046523, A059576, A122408, A124010, A153881 (Dirichlet inverse), A167865, A174725, A174726, A175522, A181819, A320390, A334997, A337105, A361665.

a074206 n | n <= 1 = n
| otherwise = 1 + (sum $ map (a074206 . (div n)) $
tail $ a027751_row n)
-- Reinhard Zumkeller, Oct 01 2012

a := array(1..150): for k from 1 to 150 do a[k] := 0 od: a[1] := 1: for j from 2 to 150 do for m from 1 to j-1 do if j mod m = 0 then a[j] := a[j]+a[m] fi: od: od: for k from 1 to 150 do printf(`%d,`,a[k]) od: # James Sellers, Dec 07 2000
A074206:= proc(n) option remember; if n > 1 then `+`(op(apply(A074206, numtheory[divisors](n)[1..-2]))) else n fi end: # M. F. Hasler, Oct 12 2018

a[0] = 0; a[1] = 1; a[n_] := a[n] = a /@ Most[Divisors[n]] // Total; a /@ Range[20000] (* N. J. A. Sloane, May 04 2016, based on program in A002033 *)
ccc[n_]:=Switch[n,0,{},1,{{1}},,Join@@Table[Prepend[#,n]&/@ccc[d],{d,Most[Divisors[n]]}]]; Table[Length[ccc[n]],{n,0,100}] (* _Gus Wiseman, Aug 25 2020 *)

A=vector(100);A[1]=1; for(n=2,#A,A[n]=1+sumdiv(n,d,A[d])); A/=2; A[1]=1; concat(0,A) \\ Charles R Greathouse IV, Nov 20 2012

{a(n) = if( n<2, n>0, my(A = divisors(n)); sum(k=1, #A-1, a(A[k])))}; /* Michael Somos, Dec 26 2016 */

A074206(n)=if(n>1, sumdiv(n, i, if(iA074206(i))),n) \\ M. F. Hasler, Oct 12 2018

A74206=[1]; A074206(n)={if(#A74206A074206(i)))} \\ Use memoization for computing many values. - M. F. Hasler, Oct 12 2018

first(n) = {my(res = vector(n, i, 1)); for(i = 2, n, for(j = 2, n \ i, res[i*j] += res[i])); concat(0, res)} \\ David A. Corneth, Oct 13 2018

first(n) = {my(res = vector(n, i, 1)); for(i = 2, n, d = divisors(i); res[i] += sum(j = 1, #d-1, res[d[j]])); concat(0, res)} \\ somewhat faster than progs above for finding first terms of n. \\ David A. Corneth, Oct 12 2018

a(n)={if(!n, 0, my(sig=factor(n)[,2], m=vecsum(sig)); sum(k=0, m, prod(i=1, #sig, binomial(sig[i]+k-1, k-1))*sum(r=k, m, binomial(r,k)*(-1)^(r-k))))} \\ Andrew Howroyd, Aug 30 2020

from math import prod
from functools import lru_cache
from sympy import divisors, factorint, prime
@lru_cache(maxsize=None)
def A074206(n): return sum(A074206(d) for d in divisors(prod(prime(i+1)**e for i,e in enumerate(sorted(factorint(n).values(),reverse=True))),generator=True,proper=True)) if n > 1 else n # Chai Wah Wu, Sep 16 2022

@cached_function
def minus_mu(n):
    if n < 2: return n
    return sum(minus_mu(d) for d in divisors(n)[:-1])
# Note that changing the sign of the sum gives the Möbius function A008683.
print([minus_mu(n) for n in (0..96)]) # Peter Luschny, Dec 26 2016

With different offset: a(n) = sum of all a(i) such that i divides n and i < n. - Clark Kimberling
a(p^k) = 2^(k-1) if k>0 and p is a prime.
Dirichlet g.f.: 1/(2-zeta(s)). - Herbert S. Wilf, Apr 29 2003
a(n) = A067824(n)/2 for n>1; a(A122408(n)) = A122408(n)/2. - Reinhard Zumkeller, Sep 03 2006
If p,q,r,... are distinct primes, then a(p*q)=3, a(p^2*q)=8, a(p*q*r)=13, a(p^3*q)=20, etc. - Vladimir Shevelev, Aug 03 2011 [corrected by Charles R Greathouse IV, Jun 02 2012]
a(0) = 0, a(1) = 1; a(n) = [x^n] Sum_{k=1..n-1} a(k)*x^k/(1 - x^k). - Ilya Gutkovskiy, Dec 11 2017
a(n) = a(A046523(n)); a(A025487(n)) = A050324(n): a(n) depends only on the nonzero exponents in the prime factorization of n, more precisely prime signature of n, cf. A124010 and A320390. - M. F. Hasler, Oct 12 2018
a(n) = A000670(A001221(n)) for squarefree n. In particular a(A002110(n)) = A000670(n). - Amiram Eldar, May 13 2019
a(n) = A050369(n)/n, for n>=1. - Ridouane Oudra, Aug 31 2019
a(n) = A361665(A181819(n)). - Pontus von Brömssen, Mar 25 2023
From Ridouane Oudra, Nov 02 2023: (Start)
If p,q are distinct primes, and n,m>0 then we have:
a(p^n*q^m) = Sum_{k=0..min(n,m)} 2^(n+m-k-1)*binomial(n,k)*binomial(m,k);
More generally: let tau[k](n) denote the number of ordered factorizations of n as a product of k terms, also named the k-th Piltz function (see A007425), then we have for n>1:
a(n) = Sum_{j=1..bigomega(n)} Sum_{k=1..j} (-1)^(j-k)*binomial(j,k)*tau[k](n), or
a(n) = Sum_{j=1..bigomega(n)} Sum_{k=0..j-1} (-1)^k*binomial(j,k)*tau[j-k](n). (End)

Originally this sequence was merged with A002033, the number of perfect partitions. Herbert S. Wilf suggested that it warrants an entry of its own.

A175526 A000120-abundant numbers.

Original entry on oeis.org

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 96, 98, 99, 100, 102, 104, 105, 106, 108, 110, 112, 114, 115, 116, 117, 118, 120

Offset: 1

Vladimir Shevelev, Dec 03 2010

nonn

Definition see in A175522. All even numbers > 2 are in the sequence.

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

Reinhard Zumkeller (first 1000 terms) & Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A175522 (perfect version), A175524 (deficient version), A257691 (complement, non-abundant version).
Cf. A000120, A192895.
Cf. also A005100, A005101.
a(n) differs from A091212(n) and from A205783(n+1) for the first time at n=37, where a(37) = 55, while 55 is missing from both A091212 and A205783.
Differs from A192506 for the first time at n=54, where a(54) = 77, while 77 is missing from A192506.

import Data.List (findIndices)
a175526 n = a175526_list !! (n-1)
a175526_list = map (+ 1) $ findIndices (> 0) a192895_list
-- Reinhard Zumkeller, Jul 12 2011

isA175526 := proc(n) s := 0 ; for d in (numtheory[divisors](n) minus {n}) do s := s+A000120(d) ; end do: evalb(s> A000120(n)) ; end proc:
for n from 1 to 120 do if isA175526(n) then printf("%d,",n); end if; end do: # R. J. Mathar, Jul 11 2011

okQ[n_] := DivisorSum[n, Total[IntegerDigits[#, 2]]*(-1)^Boole[#==n]&]>0; Select[Range[120], okQ] (* Jean-François Alcover, Dec 06 2015 *)

A192895(n) = sumdiv(n, d, hammingweight(d)*(-1)^(d==n)); \\ Charles R Greathouse IV, Feb 07 2013
isA175526(n) = (A192895(n) > 0);
n = 0; i = 0; while(i < 10000, n++; if(isA175526(n), i++; write("b175526.txt", i, " ", n)));
\\ Antti Karttunen, May 11 2015

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

is_A175526 = lambda x: sum(A000120(d) for d in divisors(x)) > 2*A000120(x)
A175526 = filter(is_A175526, IntegerRange(1, 10**4))
# D. S. McNeil, Dec 04 2010

A192895 A000120-deficiency of n.

Original entry on oeis.org

-1, 0, -1, 1, -1, 2, -2, 2, 1, 2, -2, 5, -2, 2, 1, 3, -1, 6, -2, 5, 3, 2, -3, 8, 0, 2, 1, 6, -3, 10, -4, 4, 4, 2, 3, 11, -2, 2, 2, 8, -2, 12, -3, 6, 7, 2, -4, 11, 1, 6, 1, 6, -3, 10, 1, 10, 2, 2, -4, 19, -4, 2, 5, 5, 4, 12, -2, 5, 4, 12, -3, 16, -2, 2, 8, 6

Offset: 1

Reinhard Zumkeller, Jul 12 2011

sign

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A000120, A033879, A093653, A175522, A175524, A175526, A292257.

Cf. A257691 (positions where a(n) <= 0), A294905 (and its char.fun).

a192895 n =
   sum (map a000120 $ filter ((== 0) . (mod n)) [1..n-1]) - a000120 n
a192895_list = map a192895 [1..]

a[n_] := DivisorSum[n, Total[IntegerDigits[#, 2]]*(-1)^Boole[# == n]&]; Array[a, 80] (* Jean-François Alcover, Dec 05 2015, adapted from PARI *)

a(n)=sumdiv(n,d,hammingweight(d)*(-1)^(d==n)) \\ Charles R Greathouse IV, Feb 07 2013

from sympy import divisors
def A192895(n): return sum((d.bit_count() if dChai Wah Wu, Jul 25 2023

a(n) = Sum(A000120(d): 1 <= d < n and n mod d = 0) - A000120(n); see A175522 for motivation and more information;
a(A175524(n)) < 0; a(A175522(n)) = 0; a(A175526(n)) > 0.
a(n) = A292257(n) - A000120(n). - Antti Karttunen, Nov 10 2017

A292257 a(n) is the total number of 1's in binary expansion of all proper divisors of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 4, 1, 3, 3, 4, 1, 7, 1, 5, 5, 4, 1, 8, 1, 7, 6, 5, 1, 10, 3, 5, 5, 9, 1, 14, 1, 5, 6, 4, 6, 13, 1, 5, 6, 10, 1, 15, 1, 9, 11, 6, 1, 13, 4, 9, 5, 9, 1, 14, 6, 13, 6, 6, 1, 23, 1, 7, 11, 6, 6, 14, 1, 7, 7, 15, 1, 18, 1, 5, 12, 9, 7, 16, 1, 13, 9, 5, 1, 24, 5, 6, 7, 13, 1, 26, 7, 11, 8, 7, 6, 16, 1, 11, 10, 15, 1, 14, 1, 13, 18

Offset: 1

Antti Karttunen, Oct 04 2017

If a(n) == A000120(n), then n is in A175522, if a(n) < A000120(n), then n is in A175524, and if a(n) > A000120(n), then n is in A175526.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Index entries for sequences related to binary expansion of n.

Cf. A000120, A093653, A175522, A175524, A175526, A192895, A290090, A293214.

a[n_] := DivisorSum[n, DigitCount[#, 2, 1] &, # < n &]; Array[a, 100] (* Amiram Eldar, Jul 20 2023 *)
Table[Total[Flatten[IntegerDigits[#,2]&/@Most[Divisors[n]]]],{n,120}] (* Harvey P. Dale, Oct 11 2024 *)

PARI
```
A292257(n) = sumdiv(n,d,(d
				
```

a(n) = Sum_{d|n, dA000120(d).
a(n) = A093653(n) - A000120(n).
a(n) = A192895(n) + A000120(n).
a(n) = A001222(A293214(n)).
A000035(a(n)) = A000035(A290090(n)). [Parity-wise equivalent with A290090.]

A175524 A000120-deficient numbers.

Original entry on oeis.org

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

Vladimir Shevelev, Dec 03 2010

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;

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

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

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

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

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

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

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

More terms from Amiram Eldar, Feb 18 2019

A257691 Numbers that are not A000120-abundant.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 95, 97, 101, 103, 107, 109, 111, 113, 119, 121, 123, 125, 127, 131, 137, 139, 149, 151, 157, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 211, 219, 221, 223, 227, 229, 233, 239, 241, 247, 251, 257

Offset: 1

Antti Karttunen, May 11 2015

nonn

A000120-nonabundant numbers: Numbers n for which A192895(n) <= 0.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000120, A192895.
Complement of A175526 (A000120-abundant numbers).
Disjoint union of A175522 (A000120-perfect numbers) and A175524 (A000120-deficient numbers).
Differs from A206074(n-1), A186891(n) and A257688(n) for the first time at n=19, where a(19) = 59, while A206074(18) = A186891(19) = A257688(19) = 55, a term missing from here.
Differs from A257689 for the first time at n=24, where a(24) = 79, while A257689(24) = 77, a term missing from here.

A192895[n_] := DivisorSum[n, Total[IntegerDigits[#, 2]]*(-1)^Boole[# == n]& ]; Reap[For[n=1, n<300, n++, If[A192895[n] <= 0, Sow[n]]]][[2, 1]] (* Jean-François Alcover, Dec 05 2015 *)

A192895(n) = sumdiv(n, d, hammingweight(d)*(-1)^(d==n));
isA257691(n) = (A192895(n) <= 0);
n=i=0; while(i < 10000, n++; if(isA257691(n), i++; write("b257691.txt", i, " ", n))); \\ Charles R Greathouse IV, Feb 07 2013

A175548 Binary weight of sigma(n).

Original entry on oeis.org

1, 2, 1, 3, 2, 2, 1, 4, 3, 2, 2, 3, 3, 2, 2, 5, 2, 4, 2, 3, 1, 2, 2, 4, 5, 3, 2, 3, 4, 2, 1, 6, 2, 4, 2, 5, 3, 4, 3, 4, 3, 2, 3, 3, 4, 2, 2, 5, 4, 5, 2, 3, 4, 4, 2, 4, 2, 4, 4, 3, 5, 2, 3, 7, 3, 2, 2, 6, 2, 2, 2, 4, 3, 4, 5, 3, 2, 3, 2, 5, 5, 6, 3, 3, 4, 2, 4, 4, 4, 5, 3, 3, 1, 2, 4, 6, 3, 5, 4, 5, 4, 4, 3, 4, 2

Offset: 1

Vladimir Shevelev, Dec 03 2010

The sequence is considered in connection with A175522, A175524, A175526.

a(n)=1 if n is in A046528. - Robert Israel, Nov 07 2017

			a(4) = 3 because the divisors of 4 add up to 7, a number which in binary is written as 3 ones.

Cf. A000203, A000120, A046528, A175522, A175524, A175526.

seq(convert(convert(numtheory:-sigma(n),base,2),`+`),n=1..100); # Robert Israel, Nov 07 2017

Table[Plus@@IntegerDigits[DivisorSigma[1, n], 2], {n, 80}] (* Alonso del Arte, Dec 03 2010 *)

a(n) = hammingweight(sigma(n)); \\ Michel Marcus, Feb 08 2016

a(n) = A000120(A000203(n)).

More terms from Antti Karttunen, Nov 07 2017

cp's OEIS Frontend

A360639 Numbers k such that k and k+2 are both A000120-perfect numbers (A175522).

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A360640 a(n) is the start of the least run of exactly n consecutive odd numbers that are A000120-perfect numbers (A175522).

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A360643 a(n) is the least A000120-perfect number (A175522) whose binary weight (A000120) is n, or 0 if no such number exists.

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A074206 Kalmár's [Kalmar's] problem: number of ordered factorizations of n.

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A175526 A000120-abundant numbers.

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A192895 A000120-deficiency of n.

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