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A057709 Numbers k such that there is a unique m for which the sum of the aliquot parts of m (A001065) is k.

%I A057709 #28 Feb 16 2025 08:32:43
%S A057709 3,4,7,9,10,11,12,18,24,26,28,30,34,36,38,39,48,56,58,60,66,68,70,72,
%T A057709 78,80,82,84,86,94,98,102,112,116,118,122,126,128,132,138,142,144,158,
%U A057709 160,164,168,172,174,178,180,190,192,204,208,212,220,222,224,228,250
%N A057709 Numbers k such that there is a unique m for which the sum of the aliquot parts of m (A001065) is k.
%C A057709 Alanen (1972) used the term "hermit" for a number k such that x = k is the only solution to A001065(x) = k. These numbers are the perfect numbers (A000396) in this sequence. Of the first 4 perfect numbers, 6, 28, 496 and 8128, only 28 is a term. - _Amiram Eldar_, Mar 03 2021
%H A057709 Amiram Eldar, <a href="/A057709/b057709.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from T. D. Noe)
%H A057709 Jack David Alanen, <a href="https://ir.cwi.nl/pub/9143">Empirical study of aliquot series</a>, Ph.D Thesis, Yale University, 1972.
%H A057709 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RestrictedDivisorFunction.html">Restricted Divisor Function</a>.
%H A057709 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/AliquotSequence.html">Aliquot Sequence</a>.
%e A057709 12 is a member of the sequence because s(121)=12 (and because no other integer m satisfies s(m) = 12).
%e A057709 18 is included because the sum of aliquot parts of 289 = 1+17 = 18, this being the only number with this property. 6 is not included because the sum of aliquot parts of 6 = 1+2+3 = 6 and the sum of aliquot parts of 25 = 1+5 = 6.
%t A057709 seq[max_] := Module[{s = Table[0, {n, 1, max}], i}, Do[If[(i = DivisorSigma[1, n] - n) <= max, s[[i]]++], {n, 2, (max - 1)^2 }]; Position[s, 1] // Flatten]; seq[250] (* _Amiram Eldar_, Dec 26 2020 *)
%Y A057709 Cf. A000396, A001065, A005114, A048138, A057710.
%K A057709 nonn
%O A057709 1,1
%A A057709 _Jack Brennen_, Oct 24 2000
%E A057709 Removed 1 from the sequence. - _T. D. Noe_, Dec 02 2008