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%I #49 Jan 19 2019 04:14:59
%S 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,
%T 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,
%U 4,2,5,1,11,1,2,4,4,2,5,1,9,4,2,1,10
%N Number of distinct sums i+j+k with i,j,k >= 0, i*j*k = n.
%H Robert Price and Michael De Vlieger, <a href="/A226378/b226378.txt">Table of n, a(n) for n = 0..10000</a> (first 100 terms from Robert Price).
%H Michael De Vlieger, <a href="/A226378/a226378_1.txt">Records and first positions of records in A226378(n), with 0 <= n <= 10^6.</a>
%F For n >= 1, a(n) <= A034836(n). - _Antti Karttunen_, Aug 30 2017
%e From _Antti Karttunen_, Aug 30 2017: (Start)
%e 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.
%e 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.
%e 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)
%t 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}]
%t (* Second program, more efficient: *)
%t {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 *)
%o (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
%Y Cf. A226357, A226359, A222945, A222947, A223133, A223134, A223135.
%Y Cf. A008578 (gives the positions of 1's after a(0)=1).
%Y Differs from A034836 for the first time at n=36.
%K nonn
%O 0,5
%A _Robert Price_, Jun 12 2013
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