

A304281


Numbers equal to the sum of their aliquot parts, each of them decreased by 2.


8




OFFSET

1,1


COMMENTS

Searched up to n = 10^12.
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 = 2; for t = 21 we get 8795910569984, which is a term greater than 8292352, but this does not exclude the existence of other intermediate terms following a different solution pattern. (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.


LINKS



EXAMPLE

Aliquot parts of 208 are 1 and 1, 2, 4, 8, 13, 16, 26, 52, 104 and (12) + (22) + (42) + (82) + (132) + (162) + (262) + (522) + (1042) = 208.


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, 2);


MATHEMATICA

With[{k = 2}, Select[Range[10^6], DivisorSum[#, # + k &]  (# + k) == # &] ] (* Michael De Vlieger, May 14 2018 *)


PROG

(PARI) isok(n) = sumdiv(n, d, if (d < n, d2)) == n; \\ Michel Marcus, May 14 2018


CROSSREFS

Cf. A000005, A000203, A000396, A304276, A304277, A304278, A304279, A304280, A304282, A304283, A304284.


KEYWORD

nonn,hard,more


AUTHOR



STATUS

approved



