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
jq
Also works with gojq, the Go implementation of jq
Adapted from Wren
### Generic functions
# For gojq
def _nwise($n):
def n: if length <= $n then . else .[0:$n] , (.[$n:] | n) end;
n;
def lpad($len): tostring | ($len - length) as $l | (" " * $l) + .;
# tabular print
def tprint(columns; wide):
reduce _nwise(columns) as $row ("";
. + ($row|map(lpad(wide)) | join(" ")) + "\n" );
# Input: an array
# Output: a stream of pairs [$x, $frequency]
# A two-level dictionary is used: .[type][tostring]
def frequencies:
if length == 0 then empty
else . as $in
| reduce range(0; length) as $i ({};
$in[$i] as $x
| .[$x|type][$x|tostring] as $pair
| if $pair
then .[$x|type][$x|tostring] |= (.[1] += 1)
else .[$x|type][$x|tostring] = [$x, 1]
end )
| .[][]
end ;
# Output: the items in the stream up to but excluding the first for which cond is truthy
def emit_until(cond; stream): label $out | stream | if cond then break $out else . end;
### Jordan-Pólya numbers
# input: {factorial}
# output: an array
def JordanPolya($lim; $mx):
if $lim < 2 then [1]
else . + {v: [1], t: 1, k: 2}
| .mx = ($mx // $lim)
| until(.k > .mx or .t > $lim;
.t *= .k
| if .t <= $lim
then reduce JordanPolya(($lim/.t)|floor; .t)[] as $rest (.;
.v += [.t * $rest] )
| .k += 1
else .
end)
| .v
| unique
end;
# Cache m! for m <= $n
def cacheFactorials($n):
{fact: 1, factorial: [1]}
| reduce range(1; $n + 1) as $i (.;
.fact *= $i
| .factorial[$i] = .fact );
# input: {factorial}
def Decompose($n; $start):
if $start and $start < 2 then []
else
{ factorial,
start: ($start // 18),
m: $n,
f: [] }
| label $out
| foreach range(.start; 1; -1) as $i (.;
.i = $i
| .emit = null
| until (.emit or (.m % .factorial[$i] != 0);
.f += [$i]
| .m = (.m / .factorial[$i])
| if .m == 1 then .emit = .f else . end)
| if .emit then ., break $out else . end)
| if .emit then .emit
elif .i == 2 then Decompose($n; .start-1)
else empty
end
end;
# Input: {factorial}
# $v should be an array of J-P numbers
def synopsis($v):
(100, 800, 1800, 2800, 3800) as $i
| if $v[$i-1] == null
then "\nThe \($i)th Jordan-Pólya number was not found." | error
else "\nThe \($i)th Jordan-Pólya number is \($v[$i-1] )",
([Decompose($v[$i-1]; null) | frequencies]
| map( if (.[1] == 1) then "\(.[0])!" else "(\(.[0])!)^\(.[1])" end)
| " i.e. " + join(" * ") )
end ;
def task:
cacheFactorials(18)
| JordanPolya(pow(2;53)-1; null) as $v
| "\($v|length) Jordan–Pólya numbers have been found. The first 50 are:",
( $v[:50] | tprint(10; 4)),
"\nThe largest Jordan–Pólya number before 100 million: " +
"\(if $v[-1] > 1e8 then last(emit_until(. >= 1e8; $v[])) else "not found" end)",
synopsis($v) ;
task- Output:
gojq and jq produce the same results except that gojq produces the factorizations in a different order. The output shown here corresponds to the invocation: jq -nr -f jordan-polya-numbers.jq
3887 Jordan–Pólya numbers have been found. The first 50 are: 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: 99532800 The 100th Jordan-Pólya number is 92160 i.e. 6! * (2!)^7 The 800th Jordan-Pólya number is 18345885696 i.e. (4!)^7 * (2!)^2 The 1800th Jordan-Pólya number is 9784472371200 i.e. (6!)^2 * (4!)^2 * (2!)^15 The 2800th Jordan-Pólya number is 439378587648000 i.e. 14! * 7! The 3800th Jordan-Pólya number is 7213895789838336 i.e. (4!)^8 * (2!)^16
Julia
""" The aupto() function is taken from the Python code at oeis.org/A001013 """
function aupto(lim::T, mx::T = zero(T)) where T <: Integer
lim < 2 && return [one(T)]
v, t = [one(T)], one(T)
mx == 0 && (mx = lim)
for k in 2:mx
t *= k
t > lim && break
append!(v, [t * rest for rest in aupto(lim ÷ t, t)])
end
return unique(sort!(v))
end
const factorials = map(factorial, 2:18)
function factor_as_factorials(n::T) where T <: Integer
fac_exp = Tuple{Int, Int}[]
for idx in length(factorials):-1:1
m = n
empty!(fac_exp)
for i in idx:-1:1
expo = 0
while m % factorials[i] == 0
expo += 1
m ÷= factorials[i]
end
if expo > 0
push!(fac_exp, (i + 1, expo))
end
end
m == 1 && break
end
return fac_exp
end
const superchars = ["\u2070", "\u00b9", "\u00b2", "\u00b3", "\u2074",
"\u2075", "\u2076", "\u2077", "\u2078", "\u2079"]
super(n::Integer) = prod([superchars[i + 1] for i in reverse(digits(n))])
arr = aupto(2^53)
println("First 50 Jordan–Pólya numbers:")
foreach(p -> print(rpad(p[2], 6), p[1] % 10 == 0 ? "\n" : ""), enumerate(arr[1:50]))
println("\nThe largest Jordan–Pólya number before 100 million: ", arr[findlast(<(100_000_000), arr)])
for n in [800, 1800, 2800, 3800]
print("\nThe $(n)th Jordan-Pólya number is: $(arr[n])\n= ")
println(join(map(t -> "($(t[1])!)$(super(t[2])))", factor_as_factorials(arr[n])), " x "))
end
- 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: 99532800 The 800th Jordan-Pólya number is: 18345885696 = (4!)⁷) x (2!)²) The 1800th Jordan-Pólya number is: 9784472371200 = (6!)²) x (4!)²) x (2!)¹⁵) The 2800th Jordan-Pólya number is: 439378587648000 = (14!)¹) x (7!)¹) The 3800th Jordan-Pólya number is: 7213895789838336 = (4!)⁸) x (2!)¹⁶)
Nim
import std/[algorithm, math, sequtils, strformat, strutils, tables]
const Max = if sizeof(int) == 8: 20 else: 12
type Decomposition = CountTable[int]
const Superscripts: array['0'..'9', string] = ["⁰", "¹", "²", "³", "⁴", "⁵", "⁶", "⁷", "⁸", "⁹"]
func superscript(n: Natural): string =
## Return the Unicode string to use to represent an exponent.
if n == 1:
return ""
for d in $n:
result.add Superscripts[d]
proc `$`(d: Decomposition): string =
## Return the representation of a decomposition.
for (value, count) in sorted(d.pairs.toSeq, Descending):
result.add &"({value}!){superscript(count)}"
# List of Jordan-Pólya numbers and their decomposition.
var jordanPolya = @[1]
var decomposition: Table[int, CountTable[int]] = {1: initCountTable[int]()}.toTable
# Build the list and the decompositions.
for m in 2..Max: # Loop on each factorial.
let f = fac(m)
for k in 0..jordanPolya.high: # Loop on existing elements.
var n = jordanPolya[k]
while n <= int.high div f: # Multiply by successive powers of n!
let p = n
n *= f
jordanPolya.add n
decomposition[n] = decomposition[p]
decomposition[n].inc(m)
# Sort the numbers and remove duplicates.
jordanPolya = sorted(jordanPolya).deduplicate(true)
echo "First 50 Jordan-Pólya numbers:"
for i in 0..<50:
stdout.write &"{jordanPolya[i]:>4}"
stdout.write if i mod 10 == 9: '\n' else: ' '
echo "\nLargest Jordan-Pólya number less than 100 million: ",
insertSep($jordanPolya[jordanPolya.upperBound(100_000_000) - 1])
for i in [800, 1800, 2800, 3800]:
let n = jordanPolya[i - 1]
var d = decomposition[n]
echo &"\nThe {i}th Jordan-Pólya number is:"
echo &"{insertSep($n)} = {d}"
- 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 Largest Jordan-Pólya number less than 100 million: 99_532_800 The 800th Jordan-Pólya number is: 18_345_885_696 = (4!)⁷(2!)² The 1800th Jordan-Pólya number is: 9_784_472_371_200 = (6!)²(4!)²(2!)¹⁵ The 2800th Jordan-Pólya number is: 439_378_587_648_000 = (14!)(7!) The 3800th Jordan-Pólya number is: 7_213_895_789_838_336 = (4!)⁸(2!)¹⁶
Phix
with javascript_semantics function factorials_le(atom limit) sequence res = {} while true do atom nf = factorial(length(res)+1) if nf>limit then exit end if res &= nf end while return res end function function jp(atom limit) sequence res = factorials_le(limit) integer k=2 while k<=length(res) do atom rk = res[k] for l=2 to length(res) do atom kl = res[l]*rk if kl>limit then exit end if do integer p = binary_search(kl,res) if p<0 then p = abs(p) res = res[1..p-1] & kl & res[p..$] end if kl *= rk until kl>limit end for k += 1 end while return res end function function decompose(atom jp) atom jp0 = jp sequence f = factorials_le(jp) while true do string res = "" for i=length(f) to 2 by -1 do atom fi = f[i] if remainder(jp,fi)=0 then jp /= fi if length(res) then res &= " * " end if res &= sprintf("%d!",i) integer fc = 0 while remainder(jp,fi)=0 do jp /= fi fc += 1 end while if fc then res &= sprintf("^%d",fc+1) end if end if if jp=1 then return res end if end for f = f[1..-2] jp = jp0 end while end function atom t0 = time() sequence r = jp(power(2,53)-1) printf(1,"%d Jordan-Polya numbers found, the first 50 are:\n%s\n", {length(r),join_by(r[1..50],1,10," ",fmt:="%4d")}) printf(1,"The largest under 100 million: %,d\n",r[abs(binary_search(1e8,r))-1]) for i in {100,800,1800,2800,3800} do printf(1,"The %d%s is %,d = %s\n",{i,ord(i),r[i],decompose(r[i])}) end for ?elapsed(time()-t0)
- Output:
3887 Jordan-Polya numbers found, the first 50 are: 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 under 100 million: 99,532,800 The 100th is 92,160 = 6! * 2!^7 The 800th is 18,345,885,696 = 4!^7 * 2!^2 The 1800th is 9,784,472,371,200 = 6!^2 * 4!^2 * 2!^15 The 2800th is 439,378,587,648,000 = 14! * 7! The 3800th is 7,213,895,789,838,336 = 4!^8 * 2!^16 "0.3s"
Actually slightly faster under pwa/p2js than it is on desktop/Phix.
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