A363285 a(n) is the smallest multiple k of 2^n such that |sigma(k) - 2*k| = 2^n, where sigma = A000203.

1, 10, 12, 56, 752, 992, 12224, 16256, 654848, 7337984, 10483712, 12580864, 167763968, 67100672, 2684321792, 38654574592, 584115027968, 17179738112, 206158168064, 274877382656, 149533572988928, 123145293922304, 9288674164342784, 34902896977903616

Offset

0,2

Comments

I conjecture that this sequence is infinite.

If 2^n*(2^(j+1) +- 1) - 1 is an odd prime p then k = 2^(n+j)*p is a multiple of 2^n such that |sigma(k) - 2*k| = 2^n. It is not known whether these solutions always include a(n), but this is the case for n = 1..23. - Andrew Howroyd, May 26 2023

Links

Table of n, a(n) for n=0..23.

Example

2^3 = 8, and the proper divisors of 56 are 1, 2, 4, 7, 8, 14, 28, which add up to 64, which is 8 more than 56, and since 56 is also divisible by 8 (and since there is no smaller number for which these things are true), a(3) = 56.
2^4 = 16, and the proper divisors of 752 are 1, 2, 4, 8, 16, 47, 94, 188, 376, which add up to 736, which is 16 less than 752, and since 752 is also divisible by 16 (and since there is no smaller number for which these things are true), a(4) = 752.

Prog

(PARI) a(n)=for(j=1, oo, my(k=2^n*j); if(abs(sigma(k)-2*k) == 2^n, return(k))) \\ Andrew Howroyd, May 25 2023
(Python)
from itertools import count
from math import prod
from sympy import factorint
def A363285(n):
    m = 1<<n
    for r in count(1):
        s = prod((p**(e+1 if p&1 else e+n+1)-1)//(p-1) for p, e in factorint(r).items())*((m<<1)-1 if r&1 else 1)
        if abs(s-(r*m<<1))==m:
            return r*m # Chai Wah Wu, Jul 17 2023

Crossrefs

Cf. A000079, A000203, A292557, A292558.

Sequence in context: A266700 A242508 A367149 * A219917 A257039 A337076

Adjacent sequences: A363282 A363283 A363284 * A363286 A363287 A363288

Keyword

nonn

Author

Jesiah Darnell, May 25 2023

Extensions

a(10)-a(16) from Alois P. Heinz, May 25 2023

a(17)-a(23) from Andrew Howroyd, May 25 2023

Status

approved