A246544 Consider the aliquot parts, in ascending order, of a composite number. Take their sum and repeat the process deleting the minimum number and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves.

6, 21, 28, 85, 496, 2133, 8128, 19521, 77125, 97273, 176661, 615281, 4948133, 33550336, 68353213, 129127041, 8589869056

Offset: 1

Author

Paolo P. Lava, Aug 29 2014

Keywords

• nonn
• more

Comments

Similar to Keith numbers and Primonacci numbers, using proper divisors instead of digits or prime factors.

The perfect numbers A000396 are a subset.

The numbers of iterations are: 1, 2, 1, 3, 1, 2, 1, 2, 3, 4, 2, 5, ...; with 1 when the term is perfect. - Michel Marcus, Aug 30 2014

The 2-hyperperfect numbers A007593 are a subset, with number of iterations 2. - Michel Marcus, Sep 22 2014

Examples

			Aliquot parts of 85 are 1, 5 and 17:
  1 + 5 + 17 = 23;
  5 + 17 + 23 = 45;
  17 + 23 + 45 = 85.
Aliquot parts of 19521 are 1, 3, 9, 27, 81, 241, 723, 2169 and 6507:
  1 + 3 + 9 + 27 + 81 + 241 + 723 + 2169 + 6507 = 9761;
  3 + 9 + 27 + 81 + 241 + 723 + 2169 + 6507 + 9761 = 19521.
		

Crossrefs

Cf. A000396, A007629, A212875.

Programs

Maple

with(numtheory): P:=proc(q,h)
local a,b,k,n,t,v; v:=array(1..h);
for n from 2 to q do if not isprime(n) then
a:=sort([op(divisors(n))]); b:=nops(a)-1;
for k from 1 to b do v[k]:=a[k]; od;
t:=b+1; v[t]:=add(v[k],k=1..b);
while v[t]
				

Mathematica

A246544 = {};
For[n = 4, n <= 1000000, n++,
 If[PrimeQ[n], Continue[]];
 a = Most[Divisors[n]];
 sum = Total[a];
 While[sum < n, sum = Total[a = Join[Rest[a], {sum}]]];
 If[sum == n, AppendTo[A246544, n]];
]; A246544 (* Robert Price, Sep 08 2019 *)

PARI

lista(nn) = {forcomposite(n=1, nn, d = divisors(n); v = vector(#d-1, i, d[i]); vs = sum(i=1, #v, v[i]); ind = 1; while (vs < n, v = concat(v, vs); vs += vs - v[ind]; ind++;); if (vs == n, print1(n, ", ")););} \\ Michel Marcus, Aug 29 2014

Python

import math
def divs(n):
    large_divisors = []
    for i in range(1, int(math.sqrt(n) + 1)):
        if n % i == 0:
            yield i
            if i !=  n // i:
                large_divisors.insert(0, n / i)
    for divisor in large_divisors:
        yield divisor
a = 2
while a < 1000000000:
    q = list(divs(a))[:-1]
    r = sum(q)
    if r > a or len(q) == 1:
        pass
    elif r == a:
        print(a)
    else:
        c = 1
        while r < a:
            q.append(r)
            r = sum(q[c:])
            c += 1
        if r == a:
            print(a)
    a += 1
# David Consiglio, Jr., Sep 09 2014

Python

from sympy import divisors, isprime
A246544_list = []
for n in range(2,10**5):
    if not isprime(n):
        x = divisors(n)
        x.pop()
        y = sum(x)
        while y < n:
            x, y = x[1:]+[y], 2*y-x[0]
        if y == n:
            A246544_list.append(n)
# Chai Wah Wu, Nov 03 2014

Extensions

a(13)-a(15) from Michel Marcus, Aug 29 2014

a(16) from David Consiglio, Jr., Sep 06 2014

a(17) from Lars Blomberg, Oct 27 2014