Mathematica Asked by user57467 on November 30, 2020
As an example I want to find all Integers up to a given limit which only have certain prime factors.
I can do it efficiently for the case of 2, 3, and 5 by
g = 100;
t1 = Table[{2^x*3^y*5^z}, {z, 0, Log[g]/Log[5]}, {y, 0, Log[g/(5^z)]/Log[3]}, {x, 0, Log[g/(5^z*3^y)]/Log[2]}]
t2 = Sort[Flatten[t1]]
But I am looking for a more general solution where I might have a set of k different factors. I tried Nest
but didn’ t get far. And I don’t want to use a function like FactorInteger
and then Select
the good ones. I am looking for a solution combining depth of Table
and a (maybe) recursive function of the limits.
Perhaps you are looking for smooth numbers. The function pSmoothOuter
is quite fast, but requires memory as the upper bound m
grows larger. Input pmax
is the maximum prime allowed.
pSmoothOuter[pmax_Integer, m_] :=
Block[{s},
s = 2^Range[0, Log[2, m]];
Do[
s = Pick[s = Flatten[Outer[Times, s, p^Range[0, Log[p, m]]]], UnitStep[m - s], 1],
{p, Prime[Reverse[Range[2, PrimePi[pmax]]]]}];
Sort[s]]
For example,
pSmoothOuter[5,10^2]
{1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32,
36, 40, 45, 48, 50, 54, 60, 64, 72, 75, 80, 81, 90, 96, 100}
AbsoluteTiming[{Total[#], Length[#]} &[pSmoothOuter[30, 10^8]]]
{0.015348, {2364148327261, 88415}}
The following version allows an input list of primes p
, not necessarily contiguous.
pSmoothOuter[p_List, m_] :=
Block[{s},
s = Min[p]^Range[0, Log[Min[p], m]];
Do[
s = Pick[s=Flatten[Outer[Times, s, q^Range[0, Log[q, m]]]], UnitStep[m - s], 1],
{q, Most@Reverse[Sort[p]]}];
Sort[s]]
For example,
pSmoothOuter[{5, 13, 11}, 800]
{1, 5, 11, 13, 25, 55, 65, 121, 125, 143, 169, 275, 325, 605, 625, 715}
Correct answer by KennyColnago on November 30, 2020
This should work. The construction is unwieldy, and it can be stream-lined, but at least this works:
makePrimes[numPrimes_, bound_] := Module[
{powers = (Prime@Range[numPrimes])^Array[x, numPrimes] // Reverse},
Sort@Flatten@Table[
Times @@ powers // Evaluate,
Sequence @@ MapThread[{#1, 0, #2} &,
{Reverse@Array[x, numPrimes],
Log[bound/Prepend[Most@Exp@Accumulate@Log@powers, 1]]/Reverse@Log@Prime@Range[numPrimes]}
] // Evaluate
]
]
Then
makePrimes[3, 100]
(* {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60, 64, 72, 75, 80, 81, 90, 96, 100} *)
is the same list as in the OP's example.
Answered by march on November 30, 2020
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