OFFSET
1,1
COMMENTS
Searched up to n = 10^12.
a(24) > 10^18. - Hiroaki Yamanouchi, Aug 28 2018
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 = 20 we get 2199366139904, which is a term greater than 2544998588, but this does not exclude the existence of other intermediate terms following a different solution pattern.
In fact, there could also be 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 = 2^14*32771*268460149.
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 188 are 1, 2, 4, 47, 94; (1+8) + (2+8) + (4+8) + (47+8) + (94+8) = 188.
Aliquot parts of 370 are 1, 2, 5, 10, 37, 74, 185; (1+8) + (2+8) + (5+8) + (10+8) + (37+8) + (74+8) + (185+8) = 370.
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);
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Paolo P. Lava, Giovanni Resta, May 11 2018
EXTENSIONS
a(18)-a(23) from Hiroaki Yamanouchi, Aug 28 2018
STATUS
approved