Juggler sequence
- 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 task
- 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
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. <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
}
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("%9s %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
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
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
Wren
This took about 14.5 minutes to reach n = 15,845 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 $,i", n, res[0], res[1], res[2])
} System.print() var nums = [113, 173, 193, 2183, 11229, 15065, 15845] System.print(" n l[n] i[n] d[n]") System.print("----------------------------") for (n in nums) {
var res = juggler.call(n)
Fmt.print("$,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