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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.
1
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
1,1
COMMENTS
The following table lists sequences which give k-deficient-m-perfect numbers:
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k/m | any m | 2 | 3 |
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any k | this sequence | A331627 \ {1} | - |
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1 | A385462 | A271816 \ {1} | A364977 \ A000396 |
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2 | - | A331628 | - |
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3 | - | A331629 | - |
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This sequence contains all, and only, (any k)-deficient-m-perfect numbers (m = 2,3,4,...), equivalently it contains all, and only, k-deficient-(any m)-perfect numbers (k = 1,2,3,...).
EXAMPLE
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.
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 *)
PROG
(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)));
CROSSREFS
KEYWORD
nonn
AUTHOR
Lechoslaw Ratajczak, Aug 12 2025
STATUS
approved