OFFSET
1,1
COMMENTS
Searched up to n = 10^12.
a(16) > 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 = 4; for t = 64 we get 680564733841876936426822412823955505152, which is a term greater than 15381952750, 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 = 4 we have x = 2^17*500069*550959.
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 part of 5 is 1 and 1+4 = 5.
Aliquot parts of 182 are 1, 2, 7, 13, 14, 26, 91 and (1+4) + (2+4) + (7+4) + (13+4) + (14+4) + (26+4) + (91+4) = 182.
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, 4);
MATHEMATICA
With[{k = 4}, Select[Range[10^6], DivisorSum[#, # + k &] - (# + k) == # &] ] (* Michael De Vlieger, May 14 2018 *)
Select[Range[219*10^4], Total[Most[Divisors[#]]+4]==#&] (* The program generates the first 10 terms of the sequence. *) (* Harvey P. Dale, Aug 26 2020 *)
PROG
(PARI) isok(n) = sumdiv(n, d, if (d < n, d+4)) == n; \\ Michel Marcus, May 14 2018
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Paolo P. Lava, Giovanni Resta, May 11 2018
EXTENSIONS
a(15) from Hiroaki Yamanouchi, Aug 28 2018
STATUS
approved