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A226378
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Number of distinct sums i+j+k with i,j,k >= 0, i*j*k = n.
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4
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1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 2, 4, 1, 4, 1, 4, 2, 2, 1, 6, 2, 2, 3, 4, 1, 5, 1, 5, 2, 2, 2, 7, 1, 2, 2, 5, 1, 5, 1, 4, 4, 2, 1, 9, 2, 4, 2, 4, 1, 6, 2, 6, 2, 2, 1, 10, 1, 2, 4, 7, 2, 5, 1, 4, 2, 5, 1, 11, 1, 2, 4, 4, 2, 5, 1, 9, 4, 2, 1, 10
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OFFSET
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0,5
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LINKS
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FORMULA
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EXAMPLE
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For n = 4 = 1*1*4 = 1*2*2, 1+1+4 = 6 and 1+2+2 = 5, so there are two distinct sums, and a(4) = 2.
For n = 6 = 1*1*6 = 1*2*3, 1+1+6 = 8 and 1+2+3 = 6, so there are two distinct sums, and a(6) = 2.
For n = 36, of its A034836(36) = 8 factorizations as x*y*z with 1 <= x <= y <= z: 1*1*36 = 1*2*18 = 1*3*12 = 1*4*9 = 1*6*6 = 2*2*9 = 2*3*6 = 3*3*4, sums 1+6+6 and 2+2+9 are both 13, while other triples yield unique sums, thus a(36) = 8-1 = 7. (End)
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MATHEMATICA
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f[n_] := Length[Complement[Union[Flatten[Table[If[i*j*k == n, {i + j + k}], {i, 0, n}, {j, 0, n}, {k, 0, n}], 2]], {Null}]]; Table[f[n], {n, 0, 100}]
(* Second program, more efficient: *)
{1}~Join~Table[With[{D = Divisors@ n}, Length@ Union@ Reap[Map[Function[a, Map[Function[b, Map[Function[c, If[a b c == n, Sow[a + b + c]]], Select[D, # <= n/a b &]]], Select[D, # <= n/a &]]], D]][[-1, 1]] ], {n, 100}] (* Michael De Vlieger, Aug 24 2017 *)
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PROG
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(PARI) A226378(n) = { my(sums=Set()); if(!n, 1, fordiv(n, i, for(j=i, (n/i), if(!(n%j), for(k=j, n/(i*j), if(i*j*k==n, sums = Set(concat(sums, (i+j+k)))))))); length(sums)); }; \\ Antti Karttunen, Aug 30 2017
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CROSSREFS
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Cf. A008578 (gives the positions of 1's after a(0)=1).
Differs from A034836 for the first time at n=36.
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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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