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A273124
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Numbers n such that the sum of the residues (mod k) of their aliquot parts is equal to n, for some 1 <= k <= n.
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2
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6, 24, 28, 30, 48, 80, 84, 90, 96, 108, 120, 126, 132, 140, 150, 156, 160, 168, 192, 198, 200, 204, 210, 216, 220, 228, 240, 252, 260, 264, 270, 276, 300, 312, 320, 330, 336, 348, 372, 378, 384, 390, 396, 400, 408, 420, 432, 440, 444, 448, 450, 456, 462, 480
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OFFSET
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1,1
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COMMENTS
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The majority of the numbers of the sequence admit just one or two values of k. Anyway, there are numbers with more than two values. The first five numbers with 3 values of k are:
2940: 308, 336 and 462;
3276: 280, 455 and 520;
4560: 384, 480 and 720;
9120: 800, 1000 and 1500;
9180: 792, 990 and 1485.
Perfect numbers x have x/2 - 1 different values of k: (x/2 + 1) <= k <= x - 1.
Is there any number, apart from the perfect ones, with more than 3 different values of k?
The least such n is 111360, which has k = 9680, 12100, 13200, 18150. - Robert Israel, May 18 2016
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LINKS
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EXAMPLE
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Aliquot parts of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54. If we choose k = 32 we get: 1 mod 32 = 1, 2 mod 32 = 2, 3 mod 32 = 3, 4 mod 32 = 4, 6 mod 32 = 6, 9 mod 32 = 9, 12 mod 32 = 12, 18 mod 32 = 18, 27 mod 32 = 27, 36 mod 32 = 4, 54 mod 32 = 22 and 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 27 + 4 + 22 = 108.
Aliquot parts of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60. In this case we can choose two different values for k: 24 and 30. In fact the sum of divisors from 1 to 20 is 86. Then 24 mod 24 = 0, 30 mod 24 = 6, 40 mod 24 = 16, 60 mod 24 = 12 and 86 + 0 + 6 + 16 + 12 = 120. Again, 24 mod 30 = 24, 30 mod 30 = 0, 40 mod 30 = 10, 60 mod 30 = 0 and 86 + 24 + 0 + 10 + 0 = 120.
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MAPLE
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with(numtheory): P:= proc(q) local a, b, j, k, n;
for n from 1 to q do a:=sort([op(divisors(n))]);
for k from 1 to n do b:=0; for j from 1 to nops(a)-1 do
b:=b+(a[j] mod k); od; if b=n then print(n); break; fi; od; od; end: P(10^5);
# alternative:
filter:= proc(n) local D, Ks, d, k, kmin;
uses numtheory;
D:= divisors(n) minus {n};
kmin:= ceil(n/nops(D))+1;
Ks:= divisors(convert(D, `+`)-n);
if Ks = {} then return true fi;
Ks:= select(k -> (k >= kmin and k <= n), Ks);
for k in Ks do
if add(d mod k, d=D) = n then return true fi
od:
false
end proc:
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MATHEMATICA
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Select[Range@ 500, Function[d, Length@ Select[Range@ #, Function[k, Total@ Map[Mod[#, k] &, d] == #]] > 0]@ Most@ Divisors@ # &] (* Michael De Vlieger, May 17 2016 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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