OFFSET
1,1
COMMENTS
Searched up to n = 10^12.
From Giovanni Resta, May 11 2018: (Start)
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 = -8; for t = 23 we get 140734325850112, which is a term greater than 188403300352, but this does not exclude the existence of other intermediate terms following a different solution pattern.
In fact, there could be also sporadic solutions of the type x = 2^t*r*q, where r and q are prime and for which no closed form is known. E.g. for k = -8 we have x = 2^26*134442677*80216006459.
To find them, since d(n) = 4*(t+1) and sigma(n) = (2^(t+1)-1)*(1+r)*(1+q), the relation 2*n = sigma(n) + k*(d(n)-1) becomes 2^(t+1)*r*q = (2^(t+1)-1)*(1+r)*(1+q) + k*(4*t+3), which, for fixed t and k, is a quadratic Diophantine equation in r and q that could admit solutions with r and q prime.
(End)
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.
EXAMPLE
Aliquot parts of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and (1-8) + (2-8) + (3-8) + (4-8) + (5-8) + (6-8) + (8-8) + (10-8) + (12-8) + (15-8) + (20-8) + (24-8) + (30-8) + (40-8) + (60-8) = 120.
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, -8);
MATHEMATICA
With[{k = -8}, Select[Range[10^6], DivisorSum[#, # + k &] - (# + k) == # &] ] (* Michael De Vlieger, May 14 2018 *)
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Paolo P. Lava, Giovanni Resta, May 11 2018
STATUS
approved