OFFSET
1,1
COMMENTS
If p = 2^(1+t) + (1+2*t)*k - 1 is a prime, for some t > 0 and k even, then x = 2^t*p is in the sequence, where k is the value by which the sum of aliquot parts is increased. In this sequence k = -6; for t = 24 we get a(4) = 562945004142592. - Giovanni Resta, May 11 2018; updated by Max Alekseyev, Jun 18 2025
Terms using odd values of k seem very hard to find. Up to n = 10^12, only three such terms are known: 2, 98, and 8450, for k = 1, 5, and -7, respectively.
a(8) <= 42535295865117304469545626090002710528. - Max Alekseyev, Jul 30 2025
EXAMPLE
Aliquot parts of 104704 are 1, 2, 4, 8, 16, 32, 64, 128, 256, 409, 818, 1636, 3272, 6544, 13088, 26176, 52352 and (1-6) + (2-6) + (4-6) + (8-6) + (16-6) + (32-6) + (64-6) + (128-6) + (256-6) + (409-6) + (818-6) + (1636-6) + (3272-6) + (6544-6) + (13088-6) + (26176-6) + (52352-6) = 104704.
MAPLE
with(numtheory): P:=proc(q, k) local n;
for n from 1 to q do if 2*n=sigma(n)+k*(tau(n)-1) then print(n);
fi; od; end: P(10^12, -6);
MATHEMATICA
With[{k = -6}, Select[Range[10^6], DivisorSum[#, # + k &] - (# + k) == # &] ] (* Michael De Vlieger, May 14 2018 *)
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Paolo P. Lava, May 11 2018
EXTENSIONS
a(4) from Giovanni Resta confirmed and a(5)-a(7) added by Max Alekseyev, Jul 30 2025
STATUS
approved