OFFSET
1,4
COMMENTS
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
Number of 3D grids of n congruent cubes, in a box, modulo rotation (cf. A007425 and A140773 for boxes instead of cubes; cf. A038548 for the 2D case). - Manfred Boergens, Apr 06 2021
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
Dorin Andrica and Eugen J. Ionascu, On the number of polynomials with coefficients in [n], An. Şt. Univ. Ovidius Constanţa, Vol. 22, No. 1 (2013), pp. 13-23; alternative link.
FORMULA
From Ton Biegstraaten, Jan 04 2016: (Start)
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) >= A226378(n) for all n >= 1. - Antti Karttunen, Aug 30 2017
From Bernard Schott, Dec 12 2021: (Start)
a(n) = 1 iff n = 1 or n is prime (A008578).
a(n) = 2 iff n is semiprime (A001358) (see examples). (End)
a(n) = (2 * A010057(n) + A007425(n) + 3 * A046951(n))/6 (Andrica and Ionascu, 2013, p. 19, eq. 11). - Amiram Eldar, Apr 19 2024
EXAMPLE
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.
MAPLE
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;
# second Maple program:
A034836:=proc(n)
local a, b, i;
a:=0;
b:=(l, x, h)->l<=x and x<=h;
for i in select(`<=`, NumberTheory:-Divisors(n), iroot(n, 3)) do
a:=a+nops(select[2](b, i, NumberTheory:-Divisors(n/i), isqrt(n/i)))
od;
return a
end proc;
seq(A034836(n), n=1..100); # Felix Huber, Oct 02 2024
MATHEMATICA
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 *)
PROG
a(n)=sumdiv(n, d, if(d^3<=n, A038548(n/d) - sumdiv(n/d, d0, d0<d))) \\ Rick L. Shepherd, Aug 27 2006
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Definition simplified by Jonathan Sondow, Oct 03 2013
STATUS
approved