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A304282
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Numbers equal to the sum of their aliquot parts, each of them decreased by 4.
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8
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OFFSET
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1,1
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COMMENTS
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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 = -4; for t = 20 we get 2198850240512, which is a term greater than 149775190016, 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 = -4 we have x = 2^31*4294967357*297528134446815421.
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.
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LINKS
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EXAMPLE
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Aliquot parts of 350 are 1, 2, 5, 7, 10, 14, 25, 35, 50, 70, 175 and (1-4) + (2-4) + (5-4) + (7-4) + (10-4) + (14-4) + (25-4) + (35-4) + (50-4) + (70-4) + (175-4) = 350.
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MAPLE
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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, -4);
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MATHEMATICA
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With[{k = -4}, Select[Range[10^6], DivisorSum[#, # + k &] - (# + k) == # &] ] (* Michael De Vlieger, May 14 2018 *)
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CROSSREFS
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Cf. A000005, A000203, A000396, A304276, A304277, A304278, A304279, A304280, A304281, A304283, A304284.
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KEYWORD
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nonn,hard,more
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AUTHOR
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STATUS
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approved
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