Juggler sequence
You are encouraged to solve this task according to the task description, using any language you may know.
Background of the juggler sequence:
Juggler sequences were publicized by an American mathematician and author Clifford A. Pickover. The name of the sequence gets it's name from the similarity of the rising and falling nature of the numbers in the sequences, much like balls in the hands of a juggler.
- Description
A juggler sequence is an integer sequence that starts with a positive integer a[0], with each subsequent term in the sequence being defined by the recurrence relation:
a[k + 1] = floor(a[k] ^ 0.5) if k is even or
a[k + 1] = floor(a[k] ^ 1.5) if k is odd
If a juggler sequence reaches 1, then all subsequent terms are equal to 1. This is known to be the case for initial terms up to 1,000,000 but it is not known whether all juggler sequences after that will eventually reach 1.
- Task
Compute and show here the following statistics for juggler sequences with an initial term of a[n] where n is between 20 and 39 inclusive:
- l[n] - the number of terms needed to reach 1.
- h[n] - the maximum value reached in that sequence.
- i[n] - the index of the term (starting from 0) at which the maximum is (first) reached.
If your language supports big integers with an integer square root function, also compute and show here the same statistics for as many as you reasonably can of the following values for n:
113, 173, 193, 2183, 11229, 15065, 15845, 30817, 48443, 275485, 1267909, 2264915, 5812827
Those with fast languages and fast machines may also like to try their luck at n = 7110201.
However, as h[n] for most of these numbers is thousands or millions of digits long, show instead of h[n]:
- d[n] - the number of digits in h[n]
The results can be (partially) verified against the table here.
- Related tasks
- See also
- oeis:A007320 Number of steps needed for Juggler sequence started at n to reach 1
- oeis:A094716 Largest value in the Juggler sequence started at n
AppleScript
Core language
Keeping within AppleScript's usable number range: <lang applescript>on juggler(n)
script o
property sequence : {n}
end script
set i to 1
set {max, pos} to {n, i}
repeat until (n = 1)
set n to n ^ (n mod 2 + 0.5) div 1
set end of o's sequence to n
set i to i + 1
if (n > max) then set {max, pos} to {n, i}
end repeat
return {n:n, sequence:o's sequence, |length|:i, max:max, maxPos:pos}
end juggler
on intToText(n)
if (n < 2 ^ 29) then return n as integer as text
set lst to {n mod 10 as integer}
set n to n div 10
repeat until (n = 0)
set beginning of lst to n mod 10 as integer
set n to n div 10
end repeat
return join(lst, "")
end intToText
on join(lst, delim)
set astid to AppleScript's text item delimiters set AppleScript's text item delimiters to delim set txt to lst as text set AppleScript's text item delimiters to astid return txt
end join
on task()
set output to {}
repeat with n from 20 to 39
set {|length|:len, max:max, maxPos:pos} to juggler(n)
set end of output to join({n, ": l[n] = ", len - 1, ", h[n] = ", intToText(max), ", i[n] = ", pos - 1}, "")
end repeat
return join(output, linefeed)
end task
task()</lang>
- Output:
<lang applescript>"20: l[n] = 3, h[n] = 20, i[n] = 0 21: l[n] = 9, h[n] = 140, i[n] = 4 22: l[n] = 3, h[n] = 22, i[n] = 0 23: l[n] = 9, h[n] = 110, i[n] = 1 24: l[n] = 3, h[n] = 24, i[n] = 0 25: l[n] = 11, h[n] = 52214, i[n] = 3 26: l[n] = 6, h[n] = 36, i[n] = 3 27: l[n] = 6, h[n] = 140, i[n] = 1 28: l[n] = 6, h[n] = 36, i[n] = 3 29: l[n] = 9, h[n] = 156, i[n] = 1 30: l[n] = 6, h[n] = 36, i[n] = 3 31: l[n] = 6, h[n] = 172, i[n] = 1 32: l[n] = 6, h[n] = 36, i[n] = 3 33: l[n] = 8, h[n] = 2598, i[n] = 2 34: l[n] = 6, h[n] = 36, i[n] = 3 35: l[n] = 8, h[n] = 2978, i[n] = 2 36: l[n] = 3, h[n] = 36, i[n] = 0 37: l[n] = 17, h[n] = 24906114455136, i[n] = 8 38: l[n] = 3, h[n] = 38, i[n] = 0 39: l[n] = 14, h[n] = 233046, i[n] = 3"</lang>
Shell script
One of AppleScript's main roles is telling other software to do things. This includes Unix executables, many of which come with the system. In the following, the 'do shell script' command feeds a script to the Bash shell, which script itself contains code to be passed to and executed by the "bc" executable. It's essentially a script within a script within a script. The text returned from "bc", which can handle larger numbers than core AppleScript, contains lines which are just the zeros returned by the 'juggler' function, so these are stripped out using "sed". The 'do shell script' command is supplied by the StandardAdditions OSAX which comes with the system as a standard AppleScript extension. So ironically, there's not a single command from the core language in the following code. But it's legitimate AppleScript and the input and output are both AppleScript text objects.
<lang applescript>do shell script "echo ' define juggler(n) {
#auto temp,i,max,pos
#scale = 0;
temp = n; i = 0; max = n; pos = i;
while (temp > 1) {
i = i + 1; temp = sqrt(temp ^ (1 + (temp % 2 * 2)));
if (temp > max) { max = temp; pos = i; }
}
if (n < 40) {
print n,\": l[n] = \",i,\", h[n] = \",max, \", i[n] = \",pos,\"\\n\";
} else {
print n,\": l[n] = \",i,\", d[n] = \",length(max), \", i[n] = \",pos,\"\\n\";
}
return;
}
for (n = 20 ; n < 40 ; n++) { juggler(n); } print \"\\n\"; juggler(113); juggler(173); juggler(193); juggler(2183); juggler(11229); juggler(15065); juggler(15845); # 91 seconds so far. juggler(30817); # Another 191 to here.
- juggler(48443) produced no result after running all night.
' | bc | sed -n '/^0$/ !p;'"</lang>
- Output:
<lang applescript>"20: l[n] = 3, h[n] = 20, i[n] = 0 21: l[n] = 9, h[n] = 140, i[n] = 4 22: l[n] = 3, h[n] = 22, i[n] = 0 23: l[n] = 9, h[n] = 110, i[n] = 1 24: l[n] = 3, h[n] = 24, i[n] = 0 25: l[n] = 11, h[n] = 52214, i[n] = 3 26: l[n] = 6, h[n] = 36, i[n] = 3 27: l[n] = 6, h[n] = 140, i[n] = 1 28: l[n] = 6, h[n] = 36, i[n] = 3 29: l[n] = 9, h[n] = 156, i[n] = 1 30: l[n] = 6, h[n] = 36, i[n] = 3 31: l[n] = 6, h[n] = 172, i[n] = 1 32: l[n] = 6, h[n] = 36, i[n] = 3 33: l[n] = 8, h[n] = 2598, i[n] = 2 34: l[n] = 6, h[n] = 36, i[n] = 3 35: l[n] = 8, h[n] = 2978, i[n] = 2 36: l[n] = 3, h[n] = 36, i[n] = 0 37: l[n] = 17, h[n] = 24906114455136, i[n] = 8 38: l[n] = 3, h[n] = 38, i[n] = 0 39: l[n] = 14, h[n] = 233046, i[n] = 3
113: l[n] = 16, d[n] = 27, i[n] = 9 173: l[n] = 32, d[n] = 82, i[n] = 17 193: l[n] = 73, d[n] = 271, i[n] = 47 2183: l[n] = 72, d[n] = 5929, i[n] = 32 11229: l[n] = 101, d[n] = 8201, i[n] = 54 15065: l[n] = 66, d[n] = 11723, i[n] = 25 15845: l[n] = 139, d[n] = 23889, i[n] = 43 30817: l[n] = 93, d[n] = 45391, i[n] = 39"</lang>
BQN
<lang bqn>Juggle ← {
Step ← ⌊⊢⋆(0.5 + 2|⊢)
¯1‿0‿0 + 3↑{
n‿imax‿max‿term ← 𝕩
𝕊⍟(term≠1) ⟨n+1, (max<term)⊑imax‿n, max⌈term, Step term⟩
} 0‿0‿0‿𝕩
}
>⟨"NLIH"⟩ ∾ (⊢∾Juggle)¨ 20+↕20</lang>
- Output:
┌─
╵ 'N' 'L' 'I' 'H'
20 3 0 20
21 9 4 140
22 3 0 22
23 9 1 110
24 3 0 24
25 11 3 52214
26 6 3 36
27 6 1 140
28 6 3 36
29 9 1 156
30 6 3 36
31 6 1 172
32 6 3 36
33 8 2 2598
34 6 3 36
35 8 2 2978
36 3 0 36
37 17 8 24906114455136
38 3 0 38
39 14 3 233046
┘
C++
<lang cpp>#include <cassert>
- include <iomanip>
- include <iostream>
- include <string>
- include <gmpxx.h>
using big_int = mpz_class;
auto juggler(int n) {
assert(n >= 1);
int count = 0, max_count = 0;
big_int a = n, max = n;
while (a != 1) {
if (a % 2 == 0)
a = sqrt(a);
else
a = sqrt(big_int(a * a * a));
++count;
if (a > max) {
max = a;
max_count = count;
}
}
return std::make_tuple(count, max_count, max, max.get_str().size());
}
int main() {
std::cout.imbue(std::locale(""));
std::cout << "n l[n] i[n] h[n]\n";
std::cout << "--------------------------------\n";
for (int n = 20; n < 40; ++n) {
auto [count, max_count, max, digits] = juggler(n);
std::cout << std::setw(2) << n << " " << std::setw(2) << count
<< " " << std::setw(2) << max_count << " " << max
<< '\n';
}
std::cout << '\n';
std::cout << " n l[n] i[n] d[n]\n";
std::cout << "----------------------------------------\n";
for (int n : {113, 173, 193, 2183, 11229, 15065, 15845, 30817, 48443,
275485, 1267909, 2264915, 5812827, 7110201, 56261531,
92502777, 172376627, 604398963}) {
auto [count, max_count, max, digits] = juggler(n);
std::cout << std::setw(11) << n << " " << std::setw(3) << count
<< " " << std::setw(3) << max_count << " " << digits
<< '\n';
}
}</lang>
- Output:
n l[n] i[n] h[n]
--------------------------------
20 3 0 20
21 9 4 140
22 3 0 22
23 9 1 110
24 3 0 24
25 11 3 52214
26 6 3 36
27 6 1 140
28 6 3 36
29 9 1 156
30 6 3 36
31 6 1 172
32 6 3 36
33 8 2 2598
34 6 3 36
35 8 2 2978
36 3 0 36
37 17 8 24906114455136
38 3 0 38
39 14 3 233046
n l[n] i[n] d[n]
----------------------------------------
113 16 9 27
173 32 17 82
193 73 47 271
2,183 72 32 5,929
11,229 101 54 8,201
15,065 66 25 11,723
15,845 139 43 23,889
30,817 93 39 45,391
48,443 157 60 972,463
275,485 225 148 1,909,410
1,267,909 151 99 1,952,329
2,264,915 149 89 2,855,584
5,812,827 135 67 7,996,276
7,110,201 205 119 89,981,517
56,261,531 254 92 105,780,485
92,502,777 191 117 139,486,096
172,376,627 262 90 449,669,621
604,398,963 327 172 640,556,693
F#
This task uses Isqrt_(integer_square_root)_of_X#F.23 <lang fsharp> // Juggler sequence. Nigel Galloway: August 19th., 2021 let J n=Seq.unfold(fun(n,i,g,l)->if n=1I then None else let e=match n.IsEven with true->Isqrt n |_->Isqrt(n**3) in Some((i,g,l),if e>i then (e,e,l+1,l+1) else (e,i,g,l+1)))(n,n,0,0)|>Seq.last printfn " n l[n] i[n] h[n]\n___________________"; [20I..39I]|>Seq.iter(fun n->let i,g,l=J n in printfn $"%d{int n}%5d{l+1}%5d{g} %A{i}") printfn " n l[n] i[n] d[n]\n________________________"; [113I;173I;193I;2183I;11229I;15065I;15845I;30817I]|>Seq.iter(fun n->let i,g,l=J n in printfn $"%8d{int n}%5d{l+1}%5d{g} %d{(bigint.Log10>>int>>(+)1) i}") </lang>
- Output:
n l[n] i[n] h[n]
___________________
20 3 0 20
21 9 4 140
22 3 0 22
23 9 1 110
24 3 0 24
25 11 3 52214
26 6 3 36
27 6 1 140
28 6 3 36
29 9 1 156
30 6 3 36
31 6 1 172
32 6 3 36
33 8 2 2598
34 6 3 36
35 8 2 2978
36 3 0 36
37 17 8 24906114455136
38 3 0 38
39 14 3 233046
n l[n] i[n] d[n]
________________________
113 16 9 27
173 32 17 82
193 73 47 271
2183 72 32 5929
11229 101 54 8201
15065 66 25 11723
15845 139 43 23889
30817 93 39 45391
Factor
<lang factor>USING: combinators formatting generalizations io kernel math math.extras math.functions.integer-logs math.order math.ranges sequences strings tools.memory.private ;
- next ( m -- n )
dup odd? [ dup dup * * ] when integer-sqrt ;
- new-max ( l i h a -- l i h a )
[ drop dup ] 2dip nip dup ;
- (step) ( l i h a -- l i h a )
[ 1 + ] 3dip 2dup < [ new-max ] when next ;
- step ( l i h a -- l i h a )
dup 1 = [ (step) ] unless ;
- juggler ( n quot: ( h -- obj ) -- l i h )
[ 0 0 ] [ dup [ step ] to-fixed-point drop ] [ call ] tri* [ 1 [-] ] dip ; inline
CONSTANT: fmt "%-8s %-8s %-8s %s\n"
- row. ( n quot -- )
dupd juggler [ commas ] 4 napply fmt printf ; inline
- dashes. ( n -- )
CHAR: - <string> print ;
- header. ( str -- )
[ "n" "l[n]" "i[n]" ] dip fmt printf 45 dashes. ;
- juggler. ( seq quot str -- )
header. [ row. ] curry each ; inline
20 39 [a,b] [ ] "h[n]" juggler. nl
{ 113 173 193 2183 11229 15065 15845 30817 } [ integer-log10 1 + ] "d[n]" juggler.</lang>
- Output:
n l[n] i[n] h[n] --------------------------------------------- 20 3 0 20 21 9 4 140 22 3 0 22 23 9 1 110 24 3 0 24 25 11 3 52,214 26 6 3 36 27 6 1 140 28 6 3 36 29 9 1 156 30 6 3 36 31 6 1 172 32 6 3 36 33 8 2 2,598 34 6 3 36 35 8 2 2,978 36 3 0 36 37 17 8 24,906,114,455,136 38 3 0 38 39 14 3 233,046 n l[n] i[n] d[n] --------------------------------------------- 113 16 9 27 173 32 17 82 193 73 47 271 2,183 72 32 5,929 11,229 101 54 8,201 15,065 66 25 11,723 15,845 139 43 23,889 30,817 93 39 45,391
Go
This originally took about 13.5 minutes to reach n = 5,812,827 on my machine (Intel core i7-8565U) using Go's native 'math/big' package.
However, when I exchanged that for Go's GMP wrapper there was a massive speed-up (now only 6.4 seconds to reach n = 5,812,827) and even 7,110,201 became viable with an overall time of 1 minute 40 seconds.
The next four record holders for the largest term (see talk page), are also doable but increased the overall time to nearly 24 minutes on my machine. <lang go>package main
import (
"fmt" "log" //"math/big" big "github.com/ncw/gmp" "rcu"
)
var zero = new(big.Int) var one = big.NewInt(1) var two = big.NewInt(2)
func juggler(n int64) (int, int, *big.Int, int) {
if n < 1 {
log.Fatal("Starting value must be a positive integer.")
}
count := 0
maxCount := 0
a := big.NewInt(n)
max := big.NewInt(n)
tmp := new(big.Int)
for a.Cmp(one) != 0 {
if tmp.Rem(a, two).Cmp(zero) == 0 {
a.Sqrt(a)
} else {
tmp.Mul(a, a)
tmp.Mul(tmp, a)
a.Sqrt(tmp)
}
count++
if a.Cmp(max) > 0 {
max.Set(a)
maxCount = count
}
}
return count, maxCount, max, len(max.String())
}
func main() {
fmt.Println("n l[n] i[n] h[n]")
fmt.Println("-----------------------------------")
for n := int64(20); n < 40; n++ {
count, maxCount, max, _ := juggler(n)
cmax := rcu.Commatize(int(max.Int64()))
fmt.Printf("%2d %2d %2d %s\n", n, count, maxCount, cmax)
}
fmt.Println()
nums := []int64{
113, 173, 193, 2183, 11229, 15065, 15845, 30817, 48443, 275485, 1267909,
2264915, 5812827, 7110201, 56261531, 92502777, 172376627, 604398963,
}
fmt.Println(" n l[n] i[n] d[n]")
fmt.Println("-------------------------------------")
for _, n := range nums {
count, maxCount, _, digits := juggler(n)
cn := rcu.Commatize(int(n))
fmt.Printf("%11s %3d %3d %s\n", cn, count, maxCount, rcu.Commatize(digits))
}
}</lang>
- Output:
n l[n] i[n] h[n]
-----------------------------------
20 3 0 20
21 9 4 140
22 3 0 22
23 9 1 110
24 3 0 24
25 11 3 52,214
26 6 3 36
27 6 1 140
28 6 3 36
29 9 1 156
30 6 3 36
31 6 1 172
32 6 3 36
33 8 2 2,598
34 6 3 36
35 8 2 2,978
36 3 0 36
37 17 8 24,906,114,455,136
38 3 0 38
39 14 3 233,046
n l[n] i[n] d[n]
-------------------------------------
113 16 9 27
173 32 17 82
193 73 47 271
2,183 72 32 5,929
11,229 101 54 8,201
15,065 66 25 11,723
15,845 139 43 23,889
30,817 93 39 45,391
48,443 157 60 972,463
275,485 225 148 1,909,410
1,267,909 151 99 1,952,329
2,264,915 149 89 2,855,584
5,812,827 135 67 7,996,276
7,110,201 205 119 89,981,517
56,261,531 254 92 105,780,485
92,502,777 191 117 139,486,096
172,376,627 262 90 449,669,621
604,398,963 327 172 640,556,693
Julia
<lang julia>using Formatting
function juggler(k, countdig=true, maxiters=20000)
m, maxj, maxjpos = BigInt(k), BigInt(k), BigInt(0)
for i in 1:maxiters
m = iseven(m) ? isqrt(m) : isqrt(m*m*m)
if m >= maxj
maxj, maxjpos = m, i
end
if m == 1
println(lpad(k, 9), lpad(i, 6), lpad(maxjpos, 6), lpad(format(countdig ?
ndigits(maxj) : Int(maxj), commas=true), 20), countdig ? " digits" : "")
return i
end
end
error("Juggler series starting with $k did not converge in $maxiters iterations")
end
println(" n l(n) i(n) h(n) or d(n)\n------------------------------------------") foreach(k -> juggler(k, false), 20:39) @time foreach(juggler,
[113, 173, 193, 2183, 11229, 15065, 15845, 30817, 48443, 275485, 1267909,
2264915, 5812827])
@time juggler(7110201)
</lang>
- Output:
n l(n) i(n) h(n) or d(n)
------------------------------------------
20 3 0 20
21 9 4 140
22 3 0 22
23 9 1 110
24 3 0 24
25 11 3 52,214
26 6 3 36
27 6 1 140
28 6 3 36
29 9 1 156
30 6 3 36
31 6 1 172
32 6 3 36
33 8 2 2,598
34 6 3 36
35 8 2 2,978
36 3 0 36
37 17 8 24,906,114,455,136
38 3 0 38
39 14 3 233,046
113 16 9 27 digits
173 32 17 82 digits
193 73 47 271 digits
2183 72 32 5,929 digits
11229 101 54 8,201 digits
15065 66 25 11,723 digits
15845 139 43 23,889 digits
30817 93 39 45,391 digits
48443 157 60 972,463 digits
275485 225 148 1,909,410 digits
1267909 151 99 1,952,329 digits
2264915 149 89 2,855,584 digits
5812827 135 67 7,996,276 digits
4.969021 seconds (113.27 k allocations: 2.023 GiB, 0.29% gc time)
7110201 205 119 89,981,517 digits
89.493898 seconds (1.11 M allocations: 27.713 GiB, 1.19% gc time)
Nim
Using only standard library, so limited to values of n less than 40.
<lang Nim>import math, strformat
func juggler(n: Positive): tuple[count: int; max: uint64; maxIdx: int] =
var a = n.uint64
result = (0, a, 0)
while a != 1:
let f = float(a)
a = if (a and 1) == 0: sqrt(f).uint64
else: uint64(f * sqrt(f))
inc result.count
if a > result.max:
result.max = a
result.maxIdx = result.count
echo "n l[n] h[n] i[n]" echo "——————————————————————————————" for n in 20..39:
let (l, h, i) = juggler(n)
echo &"{n} {l:2} {h:14} {i}"</lang>
- Output:
n l[n] h[n] i[n] —————————————————————————————— 20 3 20 0 21 9 140 4 22 3 22 0 23 9 110 1 24 3 24 0 25 11 52214 3 26 6 36 3 27 6 140 1 28 6 36 3 29 9 156 1 30 6 36 3 31 6 172 1 32 6 36 3 33 8 2598 2 34 6 36 3 35 8 2978 2 36 3 36 0 37 17 24906114455136 8 38 3 38 0 39 14 233046 3
Perl
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Juggler_sequence use warnings; use Math::BigInt lib => 'GMP';
print " n l(n) i(n) h(n) or d(n)\n"; print " ------- ---- ---- ------------\n"; for my $i ( 20 .. 39,
113, 173, 193, 2183, 11229, 15065, 15845, 30817,
48443, 275485, 1267909, 2264915, 5812827,
7110201 # tried my luck, luck takes about 94 seconds
)
{
my $max = my $n = Math::BigInt->new($i);
my $at = my $count = 0;
while( $n > 1 )
{
$n = sqrt( $n & 1 ? $n ** 3 : $n );
$count++;
$n > $max and ($max, $at) = ($n, $count);
}
if( length $max < 27 )
{
printf "%8d %4d %3d %s\n", $i, $count, $at, $max;
}
else
{
printf "%8d %4d %3d d(n) = %d digits\n", $i, $count, $at, length $max;
}
}</lang>
- Output:
n l(n) i(n) h(n) or d(n)
------- ---- ---- ------------
20 3 0 20
21 9 4 140
22 3 0 22
23 9 1 110
24 3 0 24
25 11 3 52214
26 6 3 36
27 6 1 140
28 6 3 36
29 9 1 156
30 6 3 36
31 6 1 172
32 6 3 36
33 8 2 2598
34 6 3 36
35 8 2 2978
36 3 0 36
37 17 8 24906114455136
38 3 0 38
39 14 3 233046
113 16 9 d(n) = 27 digits
173 32 17 d(n) = 82 digits
193 73 47 d(n) = 271 digits
2183 72 32 d(n) = 5929 digits
11229 101 54 d(n) = 8201 digits
15065 66 25 d(n) = 11723 digits
15845 139 43 d(n) = 23889 digits
30817 93 39 d(n) = 45391 digits
48443 157 60 d(n) = 972463 digits
275485 225 148 d(n) = 1909410 digits
1267909 151 99 d(n) = 1952329 digits
2264915 149 89 d(n) = 2855584 digits
5812827 135 67 d(n) = 7996276 digits
7110201 205 119 d(n) = 89981517 digits
Phix
You can run this online here.
with javascript_semantics include mpfr.e function juggler(integer n, bool bDigits=false) atom t0 = time(), t1 = time()+1 assert(n>=1) mpz a = mpz_init(n), h = mpz_init(n) integer l = 0, hi = 0 while mpz_cmp_si(a,1)!=0 do if mpz_odd(a) then mpz_pow_ui(a,a,3) end if mpz_sqrt(a,a) l += 1 if mpz_cmp(a,h)>0 then mpz_set(h,a) hi = l end if if platform()!=JS and time()>t1 then progress("working (l=%d)\r",{l}) t1 = time()+1 end if end while atom hd = iff(bDigits?mpz_sizeinbase(h,10):mpz_get_atom(h)) t0 = time()-t0 string t = iff(t0>=1?" ("&elapsed(t0)&")":"") return {n, l, hd, hi, t} end function procedure main() atom t0 = time() printf(1," n l[n] h[n] i[n]\n") printf(1,"-----------------------------------\n") for n=20 to 39 do printf(1,"%2d %2d %,18d %2d %s\n", juggler(n)) end for sequence nums = {113, 173, 193, 2183, 11229, 15065, 15845, 30817} -- alas mpz_sqrt() throws a wobbly over the rest: -- 48443, 275485, 1267909, 2264915, 5812827, 7110201} printf(1,"\n n l[n] d[n] i[n]\n") printf(1,"-----------------------------\n") -- ... and pwa/p2js(/mpfr.js) only copes with the first 3 for i=1 to iff(platform()=JS?3:length(nums)) do printf(1,"%9d %3d %,6d %2d %s\n", juggler(nums[i],true)) end for ?elapsed(time()-t0) end procedure main()
- Output:
n l[n] h[n] i[n]
-----------------------------------
20 3 20 0
21 9 140 4
22 3 22 0
23 9 110 1
24 3 24 0
25 11 52,214 3
26 6 36 3
27 6 140 1
28 6 36 3
29 9 156 1
30 6 36 3
31 6 172 1
32 6 36 3
33 8 2,598 2
34 6 36 3
35 8 2,978 2
36 3 36 0
37 17 24,906,114,455,136 8
38 3 38 0
39 14 233,046 3
n l[n] d[n] i[n]
-----------------------------
113 16 27 9
173 32 82 17
193 73 271 47
2183 72 5,929 32
11229 101 8,201 54
15065 66 11,723 25
15845 139 23,889 43
30817 93 45,391 39
"0.1s"
Actually pwa/p2js will cope a bit further than that 193, but at a pretty glacial rate (your browser's JavaScript BigInt implementation/performance might differ from mine):
2183 72 5,929 32 (1 minute and 8s)
11229 101 8,201 54 (2 minutes and 33s)
15065 66 11,723 25 (7 minutes and 43s)
15845 139 23,889 43 (1 hour, 10 minutes and 32s)
"1 hour, 21 minutes and 47s"
(Everything prior was over and done in 0.1s) I think I can fairly safely predict that 30817 would be over 4 hours. Should someone provide a better implementation of mpz_sqrt() for either or both of desktop/Phix and pwa/p2js's mpfr.js (the latter is currently a trivial 10 liner), things might improve...
Python
Slowed to a crawl at n of 1267909, so did not run for larger n. <lang python>from math import isqrt
def juggler(k, countdig=True, maxiters=1000):
m, maxj, maxjpos = k, k, 0
for i in range(1, maxiters):
m = isqrt(m) if m % 2 == 0 else isqrt(m * m * m)
if m >= maxj:
maxj, maxjpos = m, i
if m == 1:
print(f"{k: 9}{i: 6,}{maxjpos: 6}{len(str(maxj)) if countdig else maxj: 20,}{' digits' if countdig else }")
return i
print("ERROR: Juggler series starting with $k did not converge in $maxiters iterations")
print(" n l(n) i(n) h(n) or d(n)\n-------------------------------------------")
for k in range(20, 40):
juggler(k, False)
for k in [113, 173, 193, 2183, 11229, 15065, 15845, 30817, 48443, 275485, 1267909]:
juggler(k)
</lang>
- Output:
n l(n) i(n) h(n) or d(n)
--------------------------------------------
20 3 0 20
21 9 4 140
22 3 0 22
23 9 1 110
24 3 0 24
25 11 3 52,214
26 6 3 36
27 6 1 140
28 6 3 36
29 9 1 156
30 6 3 36
31 6 1 172
32 6 3 36
33 8 2 2,598
34 6 3 36
35 8 2 2,978
36 3 0 36
37 17 8 24,906,114,455,136
38 3 0 38
39 14 3 233,046
113 16 9 27 digits
173 32 17 82 digits
193 73 47 271 digits
2183 72 32 5,929 digits
11229 101 54 8,201 digits
15065 66 25 11,723 digits
15845 139 43 23,889 digits
30817 93 39 45,391 digits
48443 157 60 972,463 digits
275485 225 148 1,909,410 digits
1267909 151 99 1,952,329 digits
Quackery
<lang Quackery> [ dip number$
over size - space swap of swap join echo$ ] is recho ( n n --> )
[ 1 & ] is odd ( n --> b )
[ dup dup * * ] is cubed ( n --> n )
[ dup 1
[ 2dup > while
+ 1 >>
2dup / again ]
drop nip ] is sqrt ( n --> n )
[ nested
[ dup -1 peek 1 != while
dup -1 peek
dup odd if cubed
sqrt join again ] ] is juggler ( n --> [ )
[ dup 4 recho
juggler
dup size 1 -
3 recho
0 swap behead swap
witheach
[ 2dup < if
[ rot drop
i^ 1+ unrot
swap ]
drop ]
15 recho 2 recho cr ] is stats ( n --> )
20 times [ i^ 20 + stats ]</lang>
- Output:
20 3 20 0 21 9 140 4 22 3 22 0 23 9 110 1 24 3 24 0 25 11 52214 3 26 6 36 3 27 6 140 1 28 6 36 3 29 9 156 1 30 6 36 3 31 6 172 1 32 6 36 3 33 8 2598 2 34 6 36 3 35 8 2978 2 36 3 36 0 37 17 24906114455136 8 38 3 38 0 39 14 233046 3
Raku
Reaches 30817 fairly quickly but later values suck up enough memory that it starts thrashing the disk cache and performance drops off a cliff (on my system). Killed it after 10 minutes and capped list at 30817. Could rewrite to not try to hold entire sequence in memory at once, but probably not worth it. If you want sheer numeric calculation performance, Raku is probably not where it's at.
<lang perl6>use Lingua::EN::Numbers; sub juggler (Int $n where * > 0) { $n, { $_ +& 1 ?? .³.&isqrt !! .&isqrt } … 1 }
sub isqrt ( \x ) { my ( $X, $q, $r, $t ) = x, 1, 0 ;
$q +<= 2 while $q ≤ $X ;
while $q > 1 {
$q +>= 2; $t = $X - $r - $q; $r +>= 1;
if $t ≥ 0 { $X = $t; $r += $q }
}
$r
}
say " n l[n] i[n] h[n]"; for 20..39 {
my @j = .&juggler; my $max = @j.max; printf "%2s %4d %4d %s\n", .&comma, +@j-1, @j.first(* == $max, :k), comma $max;
}
say "\n n l[n] i[n] d[n]"; ( 113, 173, 193, 2183, 11229, 15065, 15845, 30817 ).hyper(:1batch).map: {
my $start = now;
my @j = .&juggler;
my $max = @j.max;
printf "%10s %4d %4d %10s %6.2f seconds\n", .&comma, +@j-1, @j.first(* == $max, :k),
$max.chars.&comma, (now - $start);
}</lang>
- Output:
n l[n] i[n] h[n]
20 3 0 20
21 9 4 140
22 3 0 22
23 9 1 110
24 3 0 24
25 11 3 52,214
26 6 3 36
27 6 1 140
28 6 3 36
29 9 1 156
30 6 3 36
31 6 1 172
32 6 3 36
33 8 2 2,598
34 6 3 36
35 8 2 2,978
36 3 0 36
37 17 8 24,906,114,455,136
38 3 0 38
39 14 3 233,046
n l[n] i[n] d[n]
113 16 9 27 0.01 seconds
173 32 17 82 0.01 seconds
193 73 47 271 0.09 seconds
2,183 72 32 5,929 1.05 seconds
11,229 101 54 8,201 1.98 seconds
15,065 66 25 11,723 2.05 seconds
15,845 139 43 23,889 10.75 seconds
30,817 93 39 45,391 19.60 seconds
REXX
REXX doesn't have a native integer sqrt function, so one was utilized that was previously written.
Also, one optimization was to examine the last decimal digit for oddness (instead of getting the remainder when
dividing by two).
Another optimization was to reduce the number of digits after the sqrt was calculated. <lang rexx>/*REXX program calculates and displays the juggler sequence for any positive integer*/ numeric digits 20 /*define the number of decimal digits. */ parse arg LO HI list /*obtain optional arguments from the CL*/ if LO= | LO="," then do; LO= 20; HI= 39; end /*Not specified? Then use the default.*/ if HI= | HI="," then HI= LO /* " " " " " " */ w= length( commas(HI) ) /*find the max width of any number N. */ d= digits(); dd= d + d%3 + 1 /*get # numeric digits; another version*/ if LO>0 then say c('n',w) c("steps",7) c('maxIndex',8) c("biggest term" ,dd) if LO>0 then say c( ,w,.) c("" ,7,.) c( ,8,.) c("" ,dd,.)
do n=LO to HI while n>0; steps= juggler(n) /*invoke JUGGLER: find the juggler num.*/
nc= commas(n) /*maybe add commas to N. */
say right(nc, w) c(steps, 8) c(imx, 8) right( commas(mx), dd-1)
end /*n*/
/*biggest term isn't shown for list #s.*/
xtra= words(list) /*determine how many numbers in list. */ if xtra==0 then exit 0 /*no special ginormous juggler numbers?*/ w= max(9, length( commas( word(list, xtra) ) ) ) /*find the max width of any list number*/ say; say; say say c('n',w) c("steps",7) c('maxIndex',8) c("max number of digits",dd) say c( ,w,.) c("" ,7,.) c( ,8,.) c("" ,dd,.)
do p=1 for xtra; n= word(list, p) /*process each of the numbers in list. */
steps= juggler(n); nc= commas(n) /*get a number & pass it on to JUGGLER.*/
say right(nc, w) c(steps, 8) c(imx, 8) right( commas(length(mx)), dd-1)
end /*p*/
exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ c: parse arg c1,c2,c3; if c3== then c3= ' '; else c3= '─'; return center(c1,c2,c3) commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ? /*──────────────────────────────────────────────────────────────────────────────────────*/ Isqrt: procedure; parse arg x; q= 1; r= 0 /*obtain X; R will be the Isqrt(X). */
do until q>x; q= q * 4 /*multiply Q by four until > X. */
end /*until*/
do while q>1; q= q % 4 /*processing while Q>1; quarter Q. */
t= x - r - q; r= r % 2 /*set T ──► X-R-Q; halve R. */
if t>=0 then do; x= t; r= r + q /*T≥0? Then set X ──► T; add Q ──► R.*/
end
end /*while*/; return r /*return the integer square root of X.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ juggler: parse arg #; imx= 0; mx= # /*get N; set index of MX and MX ──► 0.*/
@.=0; @.1=1; @.3=1; @.5=1; @.7=1; @.9=1 /*set semaphores (≡1) for odd dec. digs*/
do j=1 until z==1 /*here is where we begin to juggle. */
parse var # -1 _ /*obtain the last decimal digit of #. */
if @._ then do; cube= #*#*# /*Odd? Then compute integer sqrt(#**3).*/
if pos(., cube)>0 then do /*need to increase decimal digits.*/
parse var cube with 'E' x
if x>=digits() then numeric digits x+2
end
z= Isqrt(#*#*#) /*compute the integer sqrt(#**3) */
end
else z= Isqrt(#) /*Even? Then compute integer sqrt(#). */
L= length(z)
if L>=d then numeric digits L+1 /*reduce the number of numeric digits. */
if z>mx then do; mx= z; imx= j; end /*found a new max; set MX; set IMX. */
#= z
end /*j*/; return j</lang>
- output when using the inputs: , , 113 173 193 2183 11229 15065 30817 48443
n steps maxIndex biggest term
── ─────── ──────── ───────────────────────────
20 3 0 20
21 9 4 140
22 3 0 22
23 9 1 110
24 3 0 24
25 11 3 52,214
26 6 3 36
27 6 1 140
28 6 3 36
29 9 1 156
30 6 3 36
31 6 1 172
32 6 3 36
33 8 2 2,598
34 6 3 36
35 8 2 2,978
36 3 0 36
37 17 8 24,906,114,455,136
38 3 0 38
39 14 3 233,046
n steps maxIndex max number of digits
───────── ─────── ──────── ───────────────────────────
113 16 9 27
173 32 17 82
193 73 47 271
2,183 72 32 5,929
11,229 101 54 8,201
15,065 66 25 11,723
15,845 139 43 23,889
30,817 93 39 45,391
48,443 157 60 972,463
Ruby
<lang ruby>def juggler(k) = k.even? ? Integer.sqrt(k) : Integer.sqrt(k*k*k)
(20..39).chain([113, 173, 193, 2183, 11229, 15065, 15845, 30817, 48443, 275485, 1267909, 2264915]).each do |k|
k1 = k
l = h = i = 0
until k == 1 do
h, i = k, l if k > h
l += 1
k = juggler(k)
end
if k1 < 40 then
puts "#{k1}: l[n] = #{l}, h[n] = #{h}, i[n] = #{i}"
else
puts "#{k1}: l[n] = #{l}, d[n] = #{h.to_s.size}, i[n] = #{i}"
end
end </lang>
- Output:
20: l[n] = 3, h[n] = 20, i[n] = 0 21: l[n] = 9, h[n] = 140, i[n] = 4 22: l[n] = 3, h[n] = 22, i[n] = 0 23: l[n] = 9, h[n] = 110, i[n] = 1 24: l[n] = 3, h[n] = 24, i[n] = 0 25: l[n] = 11, h[n] = 52214, i[n] = 3 26: l[n] = 6, h[n] = 36, i[n] = 3 27: l[n] = 6, h[n] = 140, i[n] = 1 28: l[n] = 6, h[n] = 36, i[n] = 3 29: l[n] = 9, h[n] = 156, i[n] = 1 30: l[n] = 6, h[n] = 36, i[n] = 3 31: l[n] = 6, h[n] = 172, i[n] = 1 32: l[n] = 6, h[n] = 36, i[n] = 3 33: l[n] = 8, h[n] = 2598, i[n] = 2 34: l[n] = 6, h[n] = 36, i[n] = 3 35: l[n] = 8, h[n] = 2978, i[n] = 2 36: l[n] = 3, h[n] = 36, i[n] = 0 37: l[n] = 17, h[n] = 24906114455136, i[n] = 8 38: l[n] = 3, h[n] = 38, i[n] = 0 39: l[n] = 14, h[n] = 233046, i[n] = 3 113: l[n] = 16, d[n] = 27, i[n] = 9 173: l[n] = 32, d[n] = 82, i[n] = 17 193: l[n] = 73, d[n] = 271, i[n] = 47 2183: l[n] = 72, d[n] = 5929, i[n] = 32 11229: l[n] = 101, d[n] = 8201, i[n] = 54 15065: l[n] = 66, d[n] = 11723, i[n] = 25 15845: l[n] = 139, d[n] = 23889, i[n] = 43 30817: l[n] = 93, d[n] = 45391, i[n] = 39 48443: l[n] = 157, d[n] = 972463, i[n] = 60 275485: l[n] = 225, d[n] = 1909410, i[n] = 148 1267909: l[n] = 151, d[n] = 1952329, i[n] = 99 2264915: l[n] = 149, d[n] = 2855584, i[n] = 89
Wren
This took just over 17 minutes to reach n = 30,817 on my machine and I gave up after that. <lang ecmascript>import "/fmt" for Fmt import "/big" for BigInt
var zero = BigInt.zero var one = BigInt.one var two = BigInt.two
var juggler = Fn.new { |n|
if (n < 1) Fiber.abort("Starting value must be a positive integer.")
var a = BigInt.new(n)
var count = 0
var maxCount = 0
var max = a.copy()
while (a != one) {
if (a.isEven) {
a = a.isqrt
} else {
a = (a.square * a).isqrt
}
count = count + 1
if (a > max) {
max = a
maxCount = count
}
}
return [count, maxCount, max, max.toString.count]
}
System.print("n l[n] i[n] h[n]") System.print("-----------------------------------") for (n in 20..39) {
var res = juggler.call(n)
Fmt.print("$2d $2d $2d $,i", n, res[0], res[1], res[2])
} System.print() var nums = [113, 173, 193, 2183, 11229, 15065, 15845, 30817] System.print(" n l[n] i[n] d[n]") System.print("----------------------------") for (n in nums) {
var res = juggler.call(n)
Fmt.print("$,6d $3d $3d $,6i", n, res[0], res[1], res[3])
}</lang>
- Output:
n l[n] i[n] h[n] ----------------------------------- 20 3 0 20 21 9 4 140 22 3 0 22 23 9 1 110 24 3 0 24 25 11 3 52,214 26 6 3 36 27 6 1 140 28 6 3 36 29 9 1 156 30 6 3 36 31 6 1 172 32 6 3 36 33 8 2 2,598 34 6 3 36 35 8 2 2,978 36 3 0 36 37 17 8 24,906,114,455,136 38 3 0 38 39 14 3 233,046 n l[n] i[n] d[n] ---------------------------- 113 16 9 27 173 32 17 82 193 73 47 271 2,183 72 32 5,929 11,229 101 54 8,201 15,065 66 25 11,723 15,845 139 43 23,889 30,817 93 39 45,391