

A304283


Numbers equal to the sum of their aliquot parts, each of them decreased by 6.


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 = 6; for t = 24 we get 562945004142592, which is a term greater than 8576892928, 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 104704 are 1, 2, 4, 8, 16, 32, 64, 128, 256, 409, 818, 1636, 3272, 6544, 13088, 26176, 52352 and (16) + (26) + (46) + (86) + (166) + (326) + (646) + (1286) + (2566) + (4096) + (8186) + (16366) + (32726) + (65446) + (130886) + (261766) + (523526) + = 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

Cf. A000005, A000203, A000396, A304276, A304277, A304278, A304279, A304280, A304281, A304282, A304284.


KEYWORD

nonn,hard,more


AUTHOR



STATUS

approved



