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

1,1

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

Links

Table of n, a(n) for n=1..17.

Example

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.

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]<n do t:=t+1; v[t]:=add(v[k], k=t-b..t-1); od;
if v[t]=n then print(n); fi; fi; od; end: P(10^9, 1000);

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 *)

Prog

(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)
from itertools import count
from sympy import divisors
for a in count(2):
    q = divisors(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)
# 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

Crossrefs

Cf. A000396, A007629, A212875.

Sequence in context: A046467 A132184 A143322 * A034897 A347875 A383531

Adjacent sequences: A246541 A246542 A246543 * A246545 A246546 A246547

Keyword

nonn,more

Author

Paolo P. Lava, Aug 29 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

Status

approved