login
A273124
Numbers n such that the sum of the residues (mod k) of their aliquot parts is equal to n, for some 1 <= k <= n.
2
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
OFFSET
1,1
COMMENTS
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
EXAMPLE
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.
MAPLE
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:
select(filter, [$2..10^5]); # Robert Israel, May 18 2016
MATHEMATICA
Select[Range@ 500, Function[d, Length@ Select[Range@ #, Function[k, Total@ Map[Mod[#, k] &, d] == #]] > 0]@ Most@ Divisors@ # &] (* Michael De Vlieger, May 17 2016 *)
CROSSREFS
Cf. A265646.
Sequence in context: A274557 A072710 A349688 * A349686 A069235 A175200
KEYWORD
nonn,easy
AUTHOR
Paolo P. Lava, May 16 2016
STATUS
approved