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A034836
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Number of ways to write n as n = x*y*z with 1 <= x <= y <= z.
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35
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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, 8, 1, 2, 2, 6, 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, 12, 1, 2, 4, 4, 2, 5, 1, 9, 4, 2, 1, 10, 2, 2, 2, 6, 1, 10, 2, 4, 2, 2, 2, 12, 1, 4, 4, 8
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OFFSET
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1,4
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COMMENTS
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Number of boxes with integer edge lengths and volume n.
Starts the same as, but is different from, A033273. First values of n such that a(n) differs from A033273(n) are 36,48,60,64,72,80,84,90,96,100. - Benoit Cloitre, Nov 25 2002
a(n) depends only on the signature of n; the sorted exponents of n. For instance, a(12) and a(18) are the same because both 12 and 18 have signature (1,2). - T. D. Noe, Nov 02 2011
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LINKS
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FORMULA
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Given a number n, let s(1),...,s(m) be the signature list of n, and a(n) the resulting number in the sequence.
Then np = Product_{k=1..m} binomial(2+s(k),2) is the total number of products solely based on the combination of exponents. The multiplicity of powers is not taken into account (e.g., all combinations of 1,2,4 (6 times) but (2,2,2) only once). See next formulas to compute corrections for 3rd and 2nd powers.
Let ntp = Product_{k=1..m} (floor((s(k) - s(k) mod(3))/s(k))) if the number is a 3rd power or not resulting in 1 or 0.
Let nsq = Product_{k=1..m} (floor(s(k)/2) + 1) is the number of squares.
Conjecture: a(n) = (np + 3*(nsq - ntp) + 5*ntp)/6 = (np + 3*nsq + 2*ntp)/6.
Example: n = 1728; s = [3,6]; np = 10*28 = 280; nsq = 2*4 = 8; ntp = 1 so a(1728)=51 (as in the b-file).
(End)
a(n) = 1 iff n = 1 or n is prime (A008578).
a(n) = 2 iff n is semiprime (A001358) (see examples). (End)
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EXAMPLE
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a(12) = 4 because we can write 12 = 1*1*12 = 1*2*6 = 1*3*4 = 2*2*3.
a(36) = 8 because we can write 36 = 1*1*36 = 1*2*18 = 1*3*12 = 1*4*9 = 1*6*6 = 2*2*9 = 2*3*6 = 3*3*4.
For n = p*q, p < q primes: a(n) = 2 because we can write n = 1*1*pq = 1*p*q.
For n = p^2, p prime: a(n) = 2 because we can write n = 1*1*p^2 = 1*p*p.
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MAPLE
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f:=proc(n) local t1, i, j, k; t1:=0; for i from 1 to n do for j from i to n do for k from j to n do if i*j*k = n then t1:=t1+1; fi; od: od: od: t1; end;
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MATHEMATICA
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Table[c=0; Do[If[i<=j<=k && i*j*k==n, c++], {i, t=Divisors[n]}, {j, t}, {k, t}]; c, {n, 100}] (* Jayanta Basu, May 23 2013 *)
(* Similar to the first Mathematica code but with fewer steps in Do[..] *)
b=0; d=Divisors[n]; r=Length[d];
Do[If[d[[h]] d[[i]] d[[j]]==n, b++], {h, r}, {i, h, r}, {j, i, r}]; b (* Manfred Boergens, Apr 06 2021 *)
a[1] = 1; a[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, If[IntegerQ[Surd[n, 3]], 1/3, 0] + (Times @@ ((e + 1)*(e + 2)/2))/6 + (Times @@ (Floor[e/2] + 1))/2]; Array[a, 100] (* Amiram Eldar, Apr 19 2024 *)
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PROG
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(PARI) a(n) = {my(e = factor(n)[, 2]); (2 * ispower(n, 3) + vecprod(apply(x -> (x+1)*(x+2)/2, e)) + 3 * vecprod(apply(x -> x\2 + 1, e))) / 6; } \\ Amiram Eldar, Apr 19 2024
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CROSSREFS
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See also: writing n = x*y (A038548, unordered, A000005, ordered), n = x*y*z (this sequence, unordered, A007425, ordered), n = w*x*y*z (A007426, ordered)
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KEYWORD
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nonn,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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