A386213 Integers t having at least one nonempty subset of the set of its proper divisors for which the equation sigma(t) + r = m*t (m is any integer > 1, r is the sum of elements of such subset) is true.
2, 4, 6, 8, 10, 12, 15, 16, 18, 20, 21, 24, 28, 30, 32, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56, 60, 63, 64, 66, 70, 72, 75, 78, 80, 84, 88, 90, 96, 99, 100, 102, 104, 105, 108, 112, 114, 117, 120, 126, 128, 130, 132, 135, 136, 138, 140, 144, 150, 152, 153, 154, 156, 160, 162, 165
Offset: 1
Examples
24 is a term because for 24 the set of proper divisors is {1, 2, 3, 4, 6, 8, 12} and it has exactly 6 subsets which sum up to r satisfying the equation sigma(24) + r = k*24:
(1) sigma(24) + d_7(24) = 60 + 12 = 72 and 72 = 3*24,
(2) sigma(24) + (d_4(24) + d_6(24)) = 60 + (4 + 8) = 72 and 72 = 3*24,
(3) sigma(24) + (d_2(24) + d_4(24) + d_5(24)) = 60 + (2 + 4 + 6) = 72 and 72 = 3*24,
(4) sigma(24) + (d_1(24) + d_3(24) + d_6(24)) = 60 + (1 + 3 + 8) = 72 and 72 = 3*24,
(5) sigma(24) + (d_1(24) + d_2(24) + d_3(24) + d_5(24)) = 60 + (1 + 2 + 3 + 6) = 72 and 72 = 3*24,
(6) sigma(24) + (d_1(24) + d_2(24) + d_3(24) + d_4(24) + d_5(24) + d_6(24) + d_7(24)) = 60 + (1 + 2 + 3 + 4 + 6 + 8 + 12) = 96 and 96 = 4*24.
So 24 is (1, 2, 3 (in 2 variants), 4)-deficient-3-perfect and 7-deficient-4-perfect number.
Programs
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Mathematica
n = 1;l={};Do[x = 1;s=DivisorSigma[1,t];A=Most[Divisors[t]];B=Subsets[A]; Do[r=Total[B[[i]]];If[Mod[s+r,t]==0,x=x+1],{i,2,2^Length[A]}]; If[x>1,AppendTo[l,t];n=n+1],{t,1,165}];l (* James C. McMahon, Aug 25 2025 *) -
Maxima
(n:1, for t:1 thru 300 do (x:1, s:divsum(t), A:delete(t, divisors(t)), B:args(powerset(A)), for i:2 thru 2^(length(args(A))) do (r:apply("+", args(B[i])), if mod(s+r, t)=0 then (x:x+1)), if x>1 then (print(n, "", t), n:n+1)));
Comments