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A354591
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Numbers k that can be written as the sum of 4 divisors of k (not necessarily distinct).
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11
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4, 6, 8, 10, 12, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 44, 48, 50, 52, 54, 56, 60, 64, 66, 68, 70, 72, 76, 78, 80, 84, 88, 90, 92, 96, 100, 102, 104, 108, 110, 112, 114, 116, 120, 124, 126, 128, 130, 132, 136, 138, 140, 144, 148, 150, 152, 156, 160, 162, 164, 168, 170, 172
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OFFSET
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1,1
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COMMENTS
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This is true. In other words, k is in the sequence if and only if k is even and divisible by 3, 4 or 5. Proof: the positive integer solutions of 1/a + 1/b + 1/c + 1/d = 1 can be enumerated explicitly, and each contains at least one even number and at least one divisible by 3, 4 or 5. Of course k = k/a + k/b + k/c + k/d if and only if 1 = 1/a + 1/b + 1/c + 1/d, and this writes k as the sum of 4 divisors of k if k is divisible by a,b,c, and d. If k is even and divisible by 3, we can use 1 = 1/3 + 1/3 + 1/6 + 1/6; if divisible by 4, 1 = 1/4 + 1/4 + 1/4 + 1/4; if even and divisible by 5, 1 = 1/2 + 1/5 + 1/5 + 1/10. - Robert Israel, Sep 01 2022
The asymptotic density of this sequence is 11/30. - Amiram Eldar, Aug 08 2023
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-1).
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EXAMPLE
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20 is in the sequence since 20 = 10+5+4+1 = 5+5+5+5 where each summand divides 20.
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MAPLE
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F:= proc(x, S, j) option remember;
local s, k;
if j = 0 then return(x = 0) fi;
if S = [] or x > j*S[-1] then return false fi;
s:= S[-1];
for k from 0 to min(j, floor(x/s)) do
if procname(x-k*s, S[1..-2], j-k) then return true fi
od;
false
end proc:
select(t -> F(t, sort(convert(numtheory:-divisors(t), list)), 4), [$1..200]); # Robert Israel, Aug 31 2022
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MATHEMATICA
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q[n_, k_] := AnyTrue[Tuples[Divisors[n], k], Total[#] == n &]; Select[Range[200], q[#, 4] &] (* Amiram Eldar, Aug 19 2022 *)
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PROG
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(PARI) isok(k) = my(d=divisors(k)); forpart(p=k, if (setintersect(d, Set(p)) == Set(p), return(1)), , [4, 4]); \\ Michel Marcus, Aug 19 2022
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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