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In this sequence k = -4 and ; for t = 20 we get 2198850240512 that , which is another a term greater than 149775190016 , but this does not exclude the existence of other intermediate terms following a different solution pattern.

To find them, being 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) becames becomes 2^(t+1)*r*q = (2^(t+1)-1)*(1+r)*(1+q) + k*(4*t+3) that is, , which, for fixed t and k, is a quadratic Diophantine equation in r and q that could admit solutions with r and q prime.

It appears that the search for terms Terms using odd values of k is seem very hard: up to find. Up to n = 10^12 , only three such terms are known: 2 for k = 1, , 98 for k = 5 , and 8450 , for k = 1, 5, and -7 have been found, respectively.

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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.

To find them, being d(n) = 4*(t+1), sigma(n) = [(2^(t+1)-1])*(1+r)*(1+q), the relation 2*n = sigma(n) + k*(d(n)-1) becames 2^(t+1)*r*q = [(2^(t+1)-1])*(1+r)*(1+q) + k*(4*t+3) that is, fixed t and k, a quadratic Diophantine equation in r and q that could admit solutions with r and q prime.

Numbers equal to the sum of their aliquot parts , each of them decreased by 4.

It appears that values' the search for k terms using odd values of k is very hard: up to n = 10^12 only 2 for k = 1, 98 for k = 5 and 8450 for k = -7 have been found.

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With[{k = -4}, Select[Range[10^6], DivisorSum[#, # + k &] - (# + k) == # &] ] (* Michael De Vlieger, May 14 2018 *)

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Tested Searched up to n = 10^12.

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