Jordan-Pólya numbers
Jordan-Pólya numbers (or J-P numbers for short) are the numbers that can be obtained by multiplying together one or more (not necessarily distinct) factorials.
Jordan-Pólya numbers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
- Example
480 is a J-P number because 480 = 2! x 2! x 5!.
- Task
Find and show on this page the first 50 J-P numbers.
What is the largest J-P number less than 100 million?
- Bonus
Find and show on this page the 800th, 1,800th, 2,800th and 3,800th J-P numbers and also show their decomposition into factorials in highest to lowest order.
Hint: These J-P numbers are all less than 2^53.
- References
- Wikipedia article : Jordan-Pólya number
- OEIS sequence A001013: Jordan-Pólya numbers
Wren
This uses the recursive PARI/Python algorithm in the OEIS entry.
import "./set" for Set
import "./seq" for Lst
import "./fmt" for Fmt
var JordanPolya = Fn.new { |lim, mx|
if (lim < 2) return [1]
var v = Set.new()
v.add(1)
var t = 1
if (!mx) mx = lim
for (k in 2..mx) {
t = t * k
if (t > lim) break
for (rest in JordanPolya.call((lim/t).floor, t)) {
v.add(t * rest)
}
}
return v.toList.sort()
}
var factorials = List.filled(19, 1)
var cacheFactorials = Fn.new {
var fact = 1
for (i in 2..18) {
fact = fact * i
factorials[i] = fact
}
}
var Decompose = Fn.new { |n, start|
if (!start) start = 18
if (start < 2) return []
var m = n
var f = []
for (i in start..2) {
while (m % factorials[i] == 0) {
f.add(i)
m = m / factorials[i]
if (m == 1) return f
}
}
return Decompose.call(n, start-1)
}
cacheFactorials.call()
var v = JordanPolya.call(2.pow(53)-1, null)
System.print("First 50 Jordan–Pólya numbers:")
Fmt.tprint("$4d ", v[0..49], 10)
System.write("\nThe largest Jordan–Pólya number before 100 million: ")
for (i in 1...v.count) {
if (v[i] > 1e8) {
Fmt.print("$,d\n", v[i-1])
break
}
}
for (i in [800, 1800, 2800, 3800]) {
Fmt.print("The $,r Jordan-Pólya number is : $,d", i, v[i-1])
var g = Lst.individuals(Decompose.call(v[i-1], null))
var s = g.map { |l|
if (l[1] == 1) return "%(l[0])!"
return Fmt.swrite("($d!)$S", l[0], l[1])
}.join(" x ")
Fmt.print("= $s\n", s)
}- Output:
First 50 Jordan–Pólya numbers: 1 2 4 6 8 12 16 24 32 36 48 64 72 96 120 128 144 192 216 240 256 288 384 432 480 512 576 720 768 864 960 1024 1152 1296 1440 1536 1728 1920 2048 2304 2592 2880 3072 3456 3840 4096 4320 4608 5040 5184 The largest Jordan–Pólya number before 100 million: 99,532,800 The 800th Jordan-Pólya number is : 18,345,885,696 = (4!)⁷ x (2!)² The 1,800th Jordan-Pólya number is : 9,784,472,371,200 = (6!)² x (4!)² x (2!)¹⁵ The 2,800th Jordan-Pólya number is : 439,378,587,648,000 = 14! x 7! The 3,800th Jordan-Pólya number is : 7,213,895,789,838,336 = (4!)⁸ x (2!)¹⁶
XPL0
Simple-minded brute force. 20 seconds on Pi4. No bonus.
int Factorials(1+12);
func IsJPNum(N0);
int N0;
int N, Limit, I, Q;
[Limit:= 7;
N:= N0;
loop [I:= Limit;
loop [Q:= N / Factorials(I);
if rem(0) = 0 then
[if Q = 1 then return true;
N:= Q;
]
else I:= I-1;
if I = 1 then
[if Limit = 1 then return false;
Limit:= Limit-1;
N:= N0;
quit;
]
];
];
];
int F, N, C, SN;
[F:= 1;
for N:= 1 to 12 do
[F:= F*N;
Factorials(N):= F;
];
Text(0, "First 50 Jordan-Polya numbers:^m^j");
Format(5, 0);
RlOut(0, 1.); \handle odd number exception
C:= 1; N:= 2;
loop [if IsJPNum(N) then
[C:= C+1;
if C <= 50 then
[RlOut(0, float(N));
if rem(C/10) = 0 then CrLf(0);
];
SN:= N;
];
N:= N+2;
if N >= 100_000_000 then quit;
];
Text(0, "^m^jThe largest Jordan-Polya number before 100 million: ");
IntOut(0, SN); CrLf(0);
]- Output:
First 50 Jordan-Polya numbers:
1 2 4 6 8 12 16 24 32 36
48 64 72 96 120 128 144 192 216 240
256 288 384 432 480 512 576 720 768 864
960 1024 1152 1296 1440 1536 1728 1920 2048 2304
2592 2880 3072 3456 3840 4096 4320 4608 5040 5184
The largest Jordan-Polya number before 100 million: 99532800