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

%I A363285 #35 Jul 17 2023 12:39:28
%S A363285 1,10,12,56,752,992,12224,16256,654848,7337984,10483712,12580864,
%T A363285 167763968,67100672,2684321792,38654574592,584115027968,17179738112,
%U A363285 206158168064,274877382656,149533572988928,123145293922304,9288674164342784,34902896977903616
%N A363285 a(n) is the smallest multiple k of 2^n such that |sigma(k) - 2*k| = 2^n, where sigma = A000203.
%C A363285 I conjecture that this sequence is infinite.
%C A363285 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
%e A363285 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.
%e A363285 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.
%o A363285 (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
%o A363285 (Python)
%o A363285 from itertools import count
%o A363285 from math import prod
%o A363285 from sympy import factorint
%o A363285 def A363285(n):
%o A363285     m = 1<<n
%o A363285     for r in count(1):
%o A363285         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)
%o A363285         if abs(s-(r*m<<1))==m:
%o A363285             return r*m # _Chai Wah Wu_, Jul 17 2023
%Y A363285 Cf. A000079, A000203, A292557, A292558.
%K A363285 nonn
%O A363285 0,2
%A A363285 _Jesiah Darnell_, May 25 2023
%E A363285 a(10)-a(16) from _Alois P. Heinz_, May 25 2023
%E A363285 a(17)-a(23) from _Andrew Howroyd_, May 25 2023