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%I #24 May 19 2018 17:41:40
%S 208,945,1312,6464,8292352
%N Numbers equal to the sum of their aliquot parts, each of them decreased by 2.
%C Searched up to n = 10^12.
%C From _Giovanni Resta_, May 11 2018: (Start)
%C 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.
%C 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)
%C 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.
%e Aliquot parts of 208 are 1 and 1, 2, 4, 8, 13, 16, 26, 52, 104 and (1-2) + (2-2) + (4-2) + (8-2) + (13-2) + (16-2) + (26-2) + (52-2) + (104-2) = 208.
%p with(numtheory): P:=proc(q,k) local n;
%p for n from 1 to q do if 2*n=sigma(n)+k*(tau(n)-1) then print(n);
%p fi; od; end: P(10^12,-2);
%t With[{k = -2}, Select[Range[10^6], DivisorSum[#, # + k &] - (# + k) == # &] ] (* _Michael De Vlieger_, May 14 2018 *)
%o (PARI) isok(n) = sumdiv(n, d, if (d < n, d-2)) == n; \\ _Michel Marcus_, May 14 2018
%Y Cf. A000005, A000203, A000396, A304276, A304277, A304278, A304279, A304280, A304282, A304283, A304284.
%K nonn,hard,more
%O 1,1
%A _Paolo P. Lava_, _Giovanni Resta_, May 11 2018
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