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%I #22 May 23 2018 08:59:48
%S 104704,407715,8576892928
%N Numbers equal to the sum of their aliquot parts, each of them decreased by 6.
%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 = -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)
%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 104704 are 1, 2, 4, 8, 16, 32, 64, 128, 256, 409, 818, 1636, 3272, 6544, 13088, 26176, 52352 and (1-6) + (2-6) + (4-6) + (8-6) + (16-6) + (32-6) + (64-6) + (128-6) + (256-6) + (409-6) + (818-6) + (1636-6) + (3272-6) + (6544-6) + (13088-6) + (26176-6) + (52352-6) + = 104704.
%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,-6);
%t With[{k = -6}, Select[Range[10^6], DivisorSum[#, # + k &] - (# + k) == # &] ] (* _Michael De Vlieger_, May 14 2018 *)
%Y Cf. A000005, A000203, A000396, A304276, A304277, A304278, A304279, A304280, A304281, A304282, A304284.
%K nonn,hard,more
%O 1,1
%A _Paolo P. Lava_, May 11 2018
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