

A304284


Numbers equal to the sum of their aliquot parts, each of them decreased by 8.


8



120, 460, 1472, 57584, 69488, 76516, 93148, 231748, 600928, 1924096, 8009728, 8043652, 33626692, 1078034816, 2139324416, 2535523012, 8572567552, 188403300352
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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 = 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.


LINKS



EXAMPLE

Aliquot parts of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and (18) + (28) + (38) + (48) + (58) + (68) + (88) + (108) + (128) + (158) + (208) + (248) + (308) + (408) + (608) = 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

Cf. A000005, A000203, A000396, A304276, A304277, A304278, A304279, A304280, A304281, A304282, A304283.


KEYWORD

nonn,hard,more


AUTHOR



STATUS

approved



