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%I #26 Jul 10 2020 03:52:38
%S 18,36,72,100,144,162,196,200,288,324,392,400,450,576,648,738,748,774,
%T 784,800,846,882,900,954,968,1062,1098,1152,1296,1352,1458,1568,1600,
%U 1764,1800,1936,2178,2500,2592,2704,2916,3042,3136,3200,3528,3600,3872
%N Abundant numbers (A005101) with no subset of their divisors such that the complement has the same sum.
%C A083206(a(n)) = 0; subsequence of A083210.
%C All numbers with an odd sum of divisors (either a square or twice a square, A028982) must be terms because for these numbers the two subsets will be of opposite parity. - _Robert G. Wilson v_, Apr 01 2010
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AbundantNumber.html">Abundant Number.</a>
%H Reinhard Zumkeller, <a href="/A083206/a083206.txt">Illustration of initial terms</a>
%e Divisors of n=18: {1,2,3,6,9,18}; 18 is pseudo-perfect (A005835): 18=9+6+3, but there exist no two complementary subsets of divisors having the same sum, therefore 18 is a term.
%t (* first do *) Needs["Combinatorica`"] (* then *) abQ[n_] := DivisorSigma[1, n] > 2 n; sq2sQ[n_] := IntegerQ@ Sqrt@ n || IntegerQ@ Sqrt@(n/2); fQ[n_] := Block[{d = Divisors@n, lmt = 1 + 2^DivisorSigma[0, n]/2, k, s}, k = 1 + Length@d; s = Plus @@ d/2; While[k < lmt && Plus @@ NthSubset[k, d] != s, k++ ]; If[k == lmt, True, False]]; lst = {}; k = 1; While[k < 10^3, If[abQ@k && sq2sQ@k && fQ@k, AppendTo[lst, k]; Print@k]; k++ ]; lst (* _Robert G. Wilson v_, Apr 01 2010 *)
%Y Disjoint union of A156903 and A171641. - _Amiram Eldar_, Jun 20 2020
%Y Cf. A000203, A005101, A028982, A083206, A083210.
%K nonn
%O 1,1
%A _Reinhard Zumkeller_, Apr 22 2003
%E a(21)-a(46) from _Robert G. Wilson v_, Apr 01 2010
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