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Summarize and say sequence

From Rosetta Code
Task
Summarize and say sequence
You are encouraged to solve this task according to the task description, using any language you may know.

There are several ways to generate a self-referential sequence. One very common one (the Look-and-say sequence) is to start with a positive integer, then generate the next term by concatenating enumerated groups of adjacent alike digits:

       0, 10, 1110, 3110, 132110, 1113122110, 311311222110 ...

The terms generated grow in length geometrically and never converge.

Another way to generate a self-referential sequence is to summarize the previous term.

Count how many of each alike digit there is, then concatenate the sum and digit for each of the sorted enumerated digits. Note that the first five terms are the same as for the previous sequence.

       0, 10, 1110, 3110, 132110, 13123110, 23124110 ... 

Sort the digits largest to smallest. Do not include counts of digits that do not appear in the previous term.

Depending on the seed value, series generated this way always either converge to a stable value or to a short cyclical pattern. (For our purposes, I'll use converge to mean an element matches a previously seen element.) The sequence shown, with a seed value of 0, converges to a stable value of 1433223110 after 11 iterations. The seed value that converges most quickly is 22. It goes stable after the first element. (The next element is 22, which has been seen before.)


Task

Find all the positive integer seed values under 1000000, for the above convergent self-referential sequence, that takes the largest number of iterations before converging. Then print out the number of iterations and the sequence they return. Note that different permutations of the digits of the seed will yield the same sequence. For this task, assume leading zeros are not permitted.

Seed Value(s): 9009 9090 9900

Iterations: 21 

Sequence: (same for all three seeds except for first element)
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110
Related tasks


Other tasks related to string operations:
Metrics
Counting
Remove/replace
Anagrams/Derangements/shuffling
Find/Search/Determine
Formatting
Song lyrics/poems/Mad Libs/phrases
Tokenize
Sequences


Also see



11l[edit]

Translation of: C++
[String] result
V longest = 0
 
F make_sequence(n) -> N
DefaultDict[Char, Int] map
L(c) n
map[c]++
 
V z = ‘’
L(k) sorted(map.keys(), reverse' 1B)
z ‘’= Char(code' map[k] + ‘0’.code)
z ‘’= k
 
I :longest <= z.len
 :longest = z.len
I z !C :result
 :result [+]= z
make_sequence(z)
 
L(test) [‘9900’, ‘9090’, ‘9009’]
result.clear()
longest = 0
make_sequence(test)
print(‘[#.] Iterations: #.’.format(test, result.len + 1))
print(result.join("\n"))
print("\n")
Output:
[9900] Iterations: 21
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110


[9090] Iterations: 21
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110


[9009] Iterations: 21
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Ada[edit]

with Ada.Text_IO; use Ada.Text_IO;
with Ada.Containers.Vectors;
procedure SelfRef is
subtype Seed is Natural range 0 .. 1_000_000;
subtype Num is Natural range 0 .. 10;
type NumList is array (0 .. 10) of Num;
package IO is new Ada.Text_IO.Integer_IO (Natural);
package DVect is new Ada.Containers.Vectors (Positive, NumList);
 
function Init (innum : Seed) return NumList is
list : NumList := (others => 0);
number : Seed := innum; d : Num;
begin
loop
d := Num (number mod 10);
list (d) := list (d) + 1;
number := number / 10; exit when number = 0;
end loop; return list;
end Init;
 
procedure Next (inoutlist : in out NumList) is
list : NumList := (others => 0);
begin
for i in list'Range loop
if inoutlist (i) /= 0 then
list (i) := list (i) + 1;
list (inoutlist (i)) := list (inoutlist (i)) + 1;
end if;
end loop; inoutlist := list;
end Next;
 
procedure Show (list : NumList) is begin
for i in reverse list'Range loop
if list (i) > 0 then
IO.Put (list (i), Width => 1); IO.Put (i, Width => 1);
end if;
end loop; New_Line;
end Show;
 
function Iterate (theseed : Seed; p : Boolean) return Natural is
list : NumList := Init (theseed);
vect : DVect.Vector;
begin
vect.Append (list);
loop
if p then Show (list); end if;
Next (list); exit when vect.Contains (list); vect.Append (list);
end loop;
return Integer (DVect.Length (vect)) + 1;
end Iterate;
 
mseed : Seed;
len, maxlen : Natural := 0;
begin
for i in Seed'Range loop
len := Iterate (i, False);
if len > maxlen then mseed := i; maxlen := len; end if;
end loop;
IO.Put (maxlen, Width => 1); Put_Line (" Iterations:");
IO.Put (mseed, Width => 1); New_Line;
len := Iterate (mseed, True);
end SelfRef;
Output:
21 Iterations:
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Aime[edit]

Translation of: C
text
next(text s)
{
integer c, e, l;
index v;
data d;
 
l = ~s;
while (l) {
v[-s[l -= 1]] += 1;
}
 
for (c, e in v) {
b_form(d, "%d%c", e, -c);
}
 
return d;
}
 
integer
depth(text s, integer i, record r)
{
integer d;
 
d = 0;
r_j_integer(d, r, s);
if (d <= 0) {
i += 1;
d += d ? i : -i;
r[s] = d;
i = depth(next(s), i, r);
d = r[s];
if (d <= 0) {
r[s] = d = i + 1;
}
}
 
return d;
}
 
integer
main(void)
{
integer d, e, i;
record r;
list l;
 
d = 0;
i = 1000000;
while (i) {
i -= 1;
e = depth(itoa(i), 0, r);
if (e == d) {
lb_p_integer(l, i);
} elif (d < e) {
d = e;
l_clear(l);
lb_p_integer(l, i);
}
}
 
o_("longest length is ", d, "\n");
while (l_o_integer(i, l, 0)) {
text s;
 
o_("\n", i, "\n");
e = d - 1;
s = itoa(i);
while (e) {
o_(s = next(s), "\n");
e -= 1;
}
}
 
return 0;
}
Output:
longest length is 21

9900
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

9090
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

AutoHotkey[edit]

Not optimized in the slightest.

 
; The following directives and commands speed up execution:
#NoEnv
SetBatchlines -1
ListLines Off
Process, Priority,, high
 
iterations := 0, seed := "Seeds: "
 
Loop 1000000
If (newIterations := CountSubString(list := ListSequence(A_Index), "`n")) > iterations
iterations := newiterations
,final := "`nIterations: " iterations+1 "`nSequence:`n`n" A_Index "`n" list
,seed := A_Index " "
else if (newIterations = iterations)
seed .= A_Index " "
MsgBox % "Seeds: " . seed . final
ListSequence(seed){
While !InStr("`n" . out, "`n" (d:=Describe(seed)) "`n")
out .= d "`n", seed := d
return out
}
 
Describe(n){
Loop 10
If (t:=CountSubString(n, 10-A_Index))
out .= t . (10-A_Index)
return out
}
 
CountSubstring(fullstring, substring){
StringReplace, junk, fullstring, %substring%, , UseErrorLevel
return errorlevel
}
 

Output:

Seeds: 9009 9090 9900 
Iterations: 21
Sequence:

9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

BBC BASIC[edit]

      *FLOAT64
DIM list$(30)
maxiter% = 0
maxseed% = 0
FOR seed% = 0 TO 999999
list$(0) = STR$(seed%)
iter% = 0
REPEAT
list$(iter%+1) = FNseq(list$(iter%))
IF VALlist$(iter%+1) <= VALlist$(iter%) THEN
FOR try% = iter% TO 0 STEP -1
IF list$(iter%+1) = list$(try%) EXIT REPEAT
NEXT
ENDIF
iter% += 1
UNTIL FALSE
IF iter% >= maxiter% THEN
IF iter% > maxiter% CLS
maxiter% = iter%
maxseed% = seed%
PRINT "Seed " ;seed% " has "; iter% " iterations"
ENDIF
NEXT
PRINT '"Sequence:"
number$ = STR$(maxseed%)
FOR i% = 1 TO maxiter%
PRINT number$
number$ = FNseq(number$)
NEXT
END
 
DEF FNseq(n$)
LOCAL I%, o$, d%()
DIM d%(9)
FOR I% = 1 TO LEN(n$)
d%(ASCMID$(n$,I%)-&30) += 1
NEXT
FOR I% = 9 TO 0 STEP -1
IF d%(I%) o$ += STR$d%(I%) + STR$I%
NEXT
= o$

Output:

Seed 9009 has 21 iterations
Seed 9090 has 21 iterations
Seed 9900 has 21 iterations

Sequence:
9900
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Bracmat[edit]

( ( self-referential
= seq N next
. ( next
= R S d f
. 0:?S
& whl
' (@(!arg:%@?d ?arg)&(.!d)+!S:?S)
& :?R
& whl
' ( !S:#?f*(.?d)+?S
& !f !d !R:?R
)
& str$!R
)
& 1:?N
& !arg:?seq
& whl
' ( next$!arg:?arg
& ~(!seq:? !arg ?)
& !arg !seq:?seq
& 1+!N:?N
)
& (!seq.!N)
)
& ( Perm
= permutations S p
.  :?permutations
& ( perm
= prefix List original A Z p
.  !arg:(?prefix.)
& str$!prefix:?p
& (!S:?+(.!p)+?|(.!p)+!S:?S)
| !arg:(0 ?.?)&
|  !arg:(?prefix.?List:?original)
& whl
' ( @(!List:%?A ?Z)
& perm$(!prefix !A.!Z)
& str$(!Z !A):~!original:?List
)
)
& 0:?S
& perm$(.!arg)
& :?permutations
& whl
' ( !S:?*(.?p)+?S
& !p !permutations:?permutations
)
& !permutations
)
& -1:?i:?max
& :?seqs
& whl
' ( 1+!i:<1000000:?i
& ( @(!i:? %@?a >%@!a ?)
| self-referential$!i
 : ( ?seq
. ( >!max:?max&:?seqs
| !max
)
& ( "Seed Value(s):" Perm$!i
. "Sequence: (same for all three seeds except for first element)
"
 !seq
)
 !seqs
 : ?seqs
)
|
)
)
& out$("Iterations:" !max !seqs)
);

Output:

  Iterations:
  21
  ( Seed Value(s): 9900 9090 9009
  .   Sequence: (same for all three seeds except for first element)

      19182716152413228110
      19281716151413427110
      19281716151423228110
      29181716151413328110
      19182716151423129110
      19181716151413327110
      191726151423128110
      191716151413326110
      1916251423127110
      1916151413325110
      19251413226110
      19151423125110
      191433125110
      191413323110
      1923224110
      1923123110
      19323110
      19222110
      192210
      2920
      9900
  )

C[edit]

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
typedef struct rec_t rec_t;
struct rec_t {
int depth;
rec_t * p[10];
};
 
rec_t root = {0, {0}};
 
#define USE_POOL_ALLOC
#ifdef USE_POOL_ALLOC /* not all that big a deal */
rec_t *tail = 0, *head = 0;
#define POOL_SIZE (1 << 20)
inline rec_t *new_rec()
{
if (head == tail) {
head = calloc(sizeof(rec_t), POOL_SIZE);
tail = head + POOL_SIZE;
}
return head++;
}
#else
#define new_rec() calloc(sizeof(rec_t), 1)
#endif
 
rec_t *find_rec(char *s)
{
int i;
rec_t *r = &root;
while (*s) {
i = *s++ - '0';
if (!r->p[i]) r->p[i] = new_rec();
r = r->p[i];
}
return r;
}
 
/* speed up number to string conversion */
char number[100][4];
void init()
{
int i;
for (i = 0; i < 100; i++)
sprintf(number[i], "%d", i);
}
 
void count(char *buf)
{
int i, c[10] = {0};
char *s;
 
for (s = buf; *s; c[*s++ - '0']++);
 
for (i = 9; i >= 0; i--) {
if (!c[i]) continue;
s = number[c[i]];
 
*buf++ = s[0];
if ((*buf = s[1])) buf++;
 
*buf++ = i + '0';
}
 
*buf = '\0';
}
 
int depth(char *in, int d)
{
rec_t *r = find_rec(in);
 
if (r->depth > 0)
return r->depth;
 
d++;
if (!r->depth) r->depth = -d;
else r->depth += d;
 
count(in);
d = depth(in, d);
 
if (r->depth <= 0) r->depth = d + 1;
return r->depth;
}
 
int main(void)
{
char a[100];
int i, d, best_len = 0, n_best = 0;
int best_ints[32];
rec_t *r;
 
init();
 
for (i = 0; i < 1000000; i++) {
sprintf(a, "%d", i);
d = depth(a, 0);
 
if (d < best_len) continue;
if (d > best_len) {
n_best = 0;
best_len = d;
}
if (d == best_len)
best_ints[n_best++] = i;
}
 
printf("longest length: %d\n", best_len);
for (i = 0; i < n_best; i++) {
printf("%d\n", best_ints[i]);
sprintf(a, "%d", best_ints[i]);
for (d = 0; d <= best_len; d++) {
r = find_rec(a);
printf("%3d: %s\n", r->depth, a);
count(a);
}
putchar('\n');
}
 
return 0;
}
Output:
longest length: 21
9009
 21: 9009
 20: 2920
 19: 192210
 18: 19222110
 17: 19323110
 16: 1923123110
 15: 1923224110
 14: 191413323110
 13: 191433125110
 12: 19151423125110
 11: 19251413226110
 10: 1916151413325110
  9: 1916251423127110
  8: 191716151413326110
  7: 191726151423128110
  6: 19181716151413327110
  5: 19182716151423129110
  4: 29181716151413328110
  3: 19281716151423228110
  2: 19281716151413427110
  2: 19182716152413228110
  2: 19281716151413427110

9090
 21: 9090
 20: 2920
 19: 192210
 18: 19222110
 17: 19323110
 16: 1923123110
 15: 1923224110
 14: 191413323110
 13: 191433125110
 12: 19151423125110
 11: 19251413226110
 10: 1916151413325110
  9: 1916251423127110
  8: 191716151413326110
  7: 191726151423128110
  6: 19181716151413327110
  5: 19182716151423129110
  4: 29181716151413328110
  3: 19281716151423228110
  2: 19281716151413427110
  2: 19182716152413228110
  2: 19281716151413427110

9900
 21: 9900
 20: 2920
 19: 192210
 18: 19222110
 17: 19323110
 16: 1923123110
 15: 1923224110
 14: 191413323110
 13: 191433125110
 12: 19151423125110
 11: 19251413226110
 10: 1916151413325110
  9: 1916251423127110
  8: 191716151413326110
  7: 191726151423128110
  6: 19181716151413327110
  5: 19182716151423129110
  4: 29181716151413328110
  3: 19281716151423228110
  2: 19281716151413427110
  2: 19182716152413228110
  2: 19281716151413427110

C++[edit]

 
#include <iostream>
#include <string>
#include <map>
#include <vector>
#include <algorithm>
 
std::map<char, int> _map;
std::vector<std::string> _result;
size_t longest = 0;
 
void make_sequence( std::string n ) {
_map.clear();
for( std::string::iterator i = n.begin(); i != n.end(); i++ )
_map.insert( std::make_pair( *i, _map[*i]++ ) );
 
std::string z;
for( std::map<char, int>::reverse_iterator i = _map.rbegin(); i != _map.rend(); i++ ) {
char c = ( *i ).second + 48;
z.append( 1, c );
z.append( 1, i->first );
}
 
if( longest <= z.length() ) {
longest = z.length();
if( std::find( _result.begin(), _result.end(), z ) == _result.end() ) {
_result.push_back( z );
make_sequence( z );
}
}
}
int main( int argc, char* argv[] ) {
std::vector<std::string> tests;
tests.push_back( "9900" ); tests.push_back( "9090" ); tests.push_back( "9009" );
for( std::vector<std::string>::iterator i = tests.begin(); i != tests.end(); i++ ) {
make_sequence( *i );
std::cout << "[" << *i << "] Iterations: " << _result.size() + 1 << "\n";
for( std::vector<std::string>::iterator j = _result.begin(); j != _result.end(); j++ ) {
std::cout << *j << "\n";
}
std::cout << "\n\n";
}
return 0;
}
 
Output:
[9900] Iterations: 21
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

[9090] Iterations: 21
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

[9009] Iterations: 21
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110


Clojure[edit]

(defmacro reduce-with
"simplifies form of reduce calls"
[bindings & body]
(assert (and (vector? bindings) (= 4 (count bindings))))
(let [[acc init, item sequence] bindings]
`(reduce (fn [~acc ~item] [email protected]) ~init ~sequence)))
 
(defn digits
"maps e.g. 2345 => [2 3 4 5]"
[n] (->> n str seq (map #(- (int %) (int \0))) vec))
 
(defn dcount
"handles case (probably impossible in this range) of digit count > 9"
[ds] (let [c (count ds)] (if (< c 10) c (digits c))))
 
(defn summarize-prev
"produces the summary sequence for a digit sequence"
[ds]
(->> ds (sort >) (partition-by identity) (map (juxt dcount first)) flatten vec)
 
(defn convergent-sequence
"iterates summarize-prev until a duplicate is found; returns summary step sequence"
[ds]
(reduce-with [cur-seq [], ds (iterate summarize-prev ds)]
(if (some #{ds} cur-seq)
(reduced cur-seq)
(conj cur-seq ds))))
 
(defn candidate-seq
"only try an already sorted digit sequence, so we only try equivalent seeds once;
e.g. 23 => []; 32 => (convergent-sequence [3 2])"

[n]
(let [ds (digits n)]
(if (apply >= ds) (convergent-sequence ds) [])))
 
(defn find-longest
"the meat of the task; returns summary step sequence(s) of max length within the range"
[limit]
(reduce-with [max-seqs [[]], new-seq (map candidate-seq (range 1 limit))]
(let [cmp (compare (-> max-seqs first count) (count new-seq))]
(cond
(pos? cmp) max-seqs
(neg? cmp) [new-seq]
(zero? cmp) (conj max-seqs new-seq)))))
 
(def results (find-longest 1000000))

The above code saves a lot of time by only calculating summary step sequences for one of a set of equivalent seeds: e.g. it only calculates for 4321, not for all 24 digit permutations 1234, 1243, 1324,.... So for output this creates a some extra work to reconstitute the permuted digit sequences for the result seed(s). Clojure doesn't have a standard permutations function (though there's one in the contributed library clojure.math.combinations), but the one here will serve.

(defn perms
"produce all the permutations of a finite sequence"
[ds]
(if (empty? ds)
[]
(let [rotseq (for [n (range (count ds))] (concat (drop n ds) (take n ds)))]
(reduce-with [rs [], [[d & ds]] rotseq]
(concat rs (if (empty? ds) [[d]] (map #(cons d %) (perms ds))))))))
 
(doseq [result results]
(let [seed (first result)
seeds (->> seed perms (map vec) set sort (remove (comp zero? first)))]
(apply println "Seed value(s):" (map #(apply str %) seeds)))))
(println)
(println "Iterations:" (count result))
(println)
(println "Sequence:")
(doseq [ds result]
(println (apply str ds))))

CoffeeScript[edit]

This example is incomplete. This code only produces one of the seeds, not all of them. Please ensure that it meets all task requirements and remove this message.

This takes less than a second to run, even though the only real optimization is to exclude integers that don't have their digits descending.

 
sequence = (n) ->
cnts = {}
for c in n.toString()
d = parseInt(c)
incr cnts, d
 
seq = []
while true
s = ''
for i in [9..0]
s += "#{cnts[i]}#{i}" if cnts[i]
if s in seq
break
seq.push s
 
new_cnts = {}
for digit, cnt of cnts
incr new_cnts, cnt
incr new_cnts, digit
cnts = new_cnts
seq
 
incr = (h, k) ->
h[k] ?= 0
h[k] += 1
 
descending = (n) ->
return true if n < 10
tens = n / 10
return false if n % 10 > tens % 10
descending(tens)
 
max_len = 0
for i in [1..1000000]
if descending(i)
seq = sequence(i)
if seq.length > max_len
max_len = seq.length
max_seq = seq
max_i = i
 
console.log max_i, max_seq
 
 
 9900 ["2920", "192210", "19222110", "19323110", "1923123110", "1923224110", "191413323110", 
"191433125110", "19151423125110", "19251413226110", "1916151413325110", "1916251423127110", "1
91716151413326110", "191726151423128110", "19181716151413327110", "19182716151423129110", 
"29181716151413328110", "19281716151423228110", "19281716151413427110", "19182716152413228110"]

Common Lisp[edit]

Doesn't do cache, and takes forever.

(defun count-and-say (str)
(let* ((s (sort (map 'list #'identity str) #'char>))
(out (list (first s) 0)))
(loop for x in s do
(if (char= x (first out))
(incf (second out))
(setf out (nconc (list x 1) out))))
(format nil "~{~a~^~}" (nreverse out))))
 
(defun ref-seq-len (n &optional doprint)
(let ((s (format nil "~d" n)) hist)
(loop (push s hist)
(if doprint (format t "~a~%" s))
(setf s (count-and-say s))
(loop for item in hist
for i from 0 to 2 do
(if (string= s item) (return-from ref-seq-len (length hist)))))))
 
(defun find-longest (top)
(let (nums (len 0))
(dotimes (x top)
(let ((l (ref-seq-len x)))
(if (> l len) (setf len l nums nil))
(if (= l len) (push x nums))))
(list nums len)))
 
(let ((r (find-longest 1000000)))
(format t "Longest: ~a~%" r)
(ref-seq-len (first (first r)) t))
output
Longest: ((9900 9090 9009) 21)
9900
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

D[edit]

Slow High-level Version[edit]

Translation of: Ruby
import std.stdio, std.algorithm, std.conv;
 
string[] selfReferentialSeq(string n, string[] seen=[]) nothrow {
__gshared static string[][string] cache;
if (n in cache)
return cache[n];
if (seen.canFind(n))
return [];
 
int[10] digit_count;
foreach (immutable d; n)
digit_count[d - '0']++;
string term;
foreach_reverse (immutable d; 0 .. 10)
if (digit_count[d] > 0)
term ~= text(digit_count[d], d);
return cache[n] = [n] ~ selfReferentialSeq(term, [n] ~ seen);
}
 
void main() {
enum int limit = 1_000_000;
int max_len;
int[] max_vals;
 
foreach (immutable n; 1 .. limit) {
const seq = n.text().selfReferentialSeq();
if (seq.length > max_len) {
max_len = seq.length;
max_vals = [n];
} else if (seq.length == max_len)
max_vals ~= n;
}
 
writeln("values: ", max_vals);
writeln("iterations: ", max_len);
writeln("sequence:");
foreach (const idx, const val; max_vals[0].text.selfReferentialSeq)
writefln("%2d %s", idx + 1, val);
}
Output:
values: [9009, 9090, 9900]
iterations: 21
sequence:
 1 9009
 2 2920
 3 192210
 4 19222110
 5 19323110
 6 1923123110
 7 1923224110
 8 191413323110
 9 191433125110
10 19151423125110
11 19251413226110
12 1916151413325110
13 1916251423127110
14 191716151413326110
15 191726151423128110
16 19181716151413327110
17 19182716151423129110
18 29181716151413328110
19 19281716151423228110
20 19281716151413427110
21 19182716152413228110

More Efficient Version[edit]

Translation of: Python
import std.range, std.algorithm;
 
struct Permutations(bool doCopy=true, T) {
T[] items;
int r;
bool stopped;
int[] indices, cycles;
static if (!doCopy)
T[] result;
 
this(T)(T[] items, int r=-1) pure nothrow @safe {
this.items = items;
immutable int n = items.length;
if (r < 0)
r = n;
this.r = r;
immutable n_minus_r = n - r;
if (n_minus_r < 0) {
this.stopped = true;
} else {
this.stopped = false;
this.indices = n.iota.array;
//this.cycles = iota(n, n_minus_r, -1).array; // Not nothrow.
this.cycles = iota(n_minus_r + 1, n + 1).retro.array;
}
 
static if (!doCopy)
result = new T[r];
}
 
@property bool empty() const pure nothrow @safe @nogc {
return this.stopped;
}
 
static if (doCopy) {
@property T[] front() const pure nothrow @safe {
assert(!this.stopped);
auto result = new T[r];
foreach (immutable i, ref re; result)
re = items[indices[i]];
return result;
}
} else {
@property T[] front() pure nothrow @safe @nogc {
assert(!this.stopped);
foreach (immutable i, ref re; this.result)
re = items[indices[i]];
return this.result;
}
}
 
void popFront() pure nothrow /*@safe*/ @nogc {
assert(!this.stopped);
int i = r - 1;
while (i >= 0) {
immutable int j = cycles[i] - 1;
if (j > 0) {
cycles[i] = j;
indices[i].swap(indices[$ - j]);
return;
}
cycles[i] = indices.length - i;
immutable int n1 = indices.length - 1;
assert(n1 >= 0);
immutable int num = indices[i];
 
// copy isn't @safe.
indices[i + 1 .. n1 + 1].copy(indices[i .. n1]);
indices[n1] = num;
i--;
}
 
this.stopped = true;
}
}
 
Permutations!(doCopy, T) permutations(bool doCopy=true, T)
(T[] items, int r=-1)
pure nothrow @safe {
return Permutations!(doCopy, T)(items, r);
}
 
// ---------------------------------
 
import std.stdio, std.typecons, std.conv, std.algorithm, std.array,
std.exception, std.string;
 
enum maxIters = 1_000_000;
 
string A036058(in string ns) pure nothrow @safe {
return ns.representation.group.map!(t => t[1].text ~ char(t[0])).join;
}
 
int A036058_length(bool doPrint=false)(string numberString="0") {
int iterations = 1;
int queueIndex;
string[3] lastThree;
 
while (true) {
static if (doPrint)
writefln("  %2d %s", iterations, numberString);
 
numberString = numberString
.dup
.representation
.sort()
.release
.assumeUTF;
 
if (lastThree[].canFind(numberString))
break;
assert(iterations < maxIters);
lastThree[queueIndex] = numberString;
numberString = numberString.A036058;
iterations++;
queueIndex++;
queueIndex %= 3;
}
 
return iterations;
}
 
Tuple!(int,int[]) max_A036058_length(R)(R startRange = 11.iota) {
bool[string] alreadyDone;
auto max_len = tuple(-1, (int[]).init);
 
foreach (n; startRange) {
immutable sns = n
.to!(char[])
.representation
.sort()
.release
.assumeUTF;
 
if (sns !in alreadyDone) {
alreadyDone[sns] = true;
const size = sns.A036058_length;
if (size > max_len[0])
max_len = tuple(size, [n]);
else if (size == max_len[0])
max_len[1] ~= n;
}
}
return max_len;
}
 
void main() {
//const (lenMax, starts) = maxIters.iota.max_A036058_length;
const lenMax_starts = maxIters.iota.max_A036058_length;
immutable lenMax = lenMax_starts[0];
const starts = lenMax_starts[1];
 
// Expand:
int[] allStarts;
foreach (immutable n; starts) {
bool[string] set;
foreach (const k; permutations!false(n.to!(char[]), 4))
if (k[0] != '0')
set[k.idup] = true;
//allStarts ~= set.byKey.to!(int[]);
allStarts ~= set.byKey.map!(to!int).array;
}
 
allStarts = allStarts.sort().filter!(x => x < maxIters).array;
 
writefln("The longest length, followed by the number(s) with the
longest sequence length for starting sequence numbers below maxIters
are:
Iterations = %d and sequence-starts = %s."
, lenMax, allStarts);
 
writeln("Note that only the first of any sequences with the same
digits is printed below. (The others will differ only in their first
term)."
);
 
foreach (immutable n; starts) {
writeln;
A036058_length!true(n.text);
}
}

The output is similar to the Python entry.

Faster Low-level Version[edit]

Translation of: C

From the C version, with a memory pool for a faster tree allocation.

import core.stdc.stdio, core.stdc.stdlib;
 
struct MemoryPool(T, int MAX_BLOCK_BYTES = 1 << 17) {
static assert(!is(T == class),
"MemoryPool is designed for native data.");
static assert(MAX_BLOCK_BYTES >= 1,
"MemoryPool: MAX_BLOCK_BYTES must be >= 1 bytes.");
static assert(MAX_BLOCK_BYTES >= T.sizeof,
"MemoryPool: MAX_BLOCK_BYTES must be" ~
" bigger than a T.");
static if (T.sizeof * 5 > MAX_BLOCK_BYTES)
pragma(msg, "MemoryPool: Block is very small.");
 
alias Block = T[MAX_BLOCK_BYTES / T.sizeof];
static __gshared Block*[] blocks;
static __gshared T* nextFree, lastFree;
 
static T* newItem() nothrow {
if (nextFree >= lastFree) {
blocks ~= cast(Block*)calloc(1, Block.sizeof);
if (blocks[$ - 1] == null)
exit(1);
nextFree = blocks[$ - 1].ptr;
lastFree = nextFree + Block.length;
}
 
return nextFree++;
}
 
// static void freeAll() nothrow {
// foreach (blockPtr; blocks)
// free(blockPtr);
// blocks.length = 0;
// nextFree = null;
// lastFree = null;
// }
}
 
struct Rec { // Tree node
int length;
Rec*[10] p;
}
 
__gshared int nNodes;
__gshared Rec* rec_root;
__gshared MemoryPool!Rec recPool;
 
Rec* findRec(char* s, Rec* root) nothrow {
int c;
Rec* next;
 
while (true) {
c = *s;
s++;
if (!c)
break;
c -= '0';
next = root.p[c];
if (!next) {
nNodes++;
next = recPool.newItem;
root.p[c] = next;
}
root = next;
}
return root;
}
 
void nextNum(char* s) nothrow @nogc {
int[10] cnt;
for (int i = 0; s[i]; i++)
cnt[s[i] - '0']++;
 
foreach_reverse (i; 0 .. 10) {
if (!cnt[i])
continue;
s += sprintf(s, "%d%c", cnt[i], i + '0');
}
}
 
int getLen(char* s, int depth) nothrow {
auto r = findRec(s, rec_root);
if (r.length > 0)
return r.length;
 
depth++;
if (!r.length)
r.length = -depth;
else
r.length += depth;
 
nextNum(s);
depth = 1 + getLen(s, depth);
 
if (r.length <= 0)
r.length = depth;
return r.length;
}
 
void main() nothrow {
enum MAXN = 1_000_000;
 
int[100] longest;
int nLongest, ml;
char[32] buf;
rec_root = recPool.newItem();
 
foreach (immutable i; 0 .. MAXN) {
sprintf(buf.ptr, "%d", i);
int l = getLen(buf.ptr, 0);
if (l < ml)
continue;
if (l > ml) {
nLongest = 0;
ml = l;
}
longest[nLongest] = i;
nLongest++;
}
 
printf("seq leng: %d\n\n", ml);
foreach (immutable i; 0 .. nLongest) {
sprintf(buf.ptr, "%d", longest[i]);
// print len+1 so we know repeating starts from when
foreach (immutable l; 0 .. ml + 1) {
printf("%2d: %s\n", getLen(buf.ptr, 0), buf.ptr);
nextNum(buf.ptr);
}
printf("\n");
}
 
printf("Allocated %d Rec tree nodes.\n", nNodes);
//recPool.freeAll;
}

Faster than the C entry, run-time is about 1.16 seconds using the dmd compiler (about 1.5 without memory pool). Same output as the C entry.

EchoLisp[edit]

Extra credit: searching up to 1e+10 does not find a longer sequence.

 
(lib 'hash)
(lib 'list) ;; permutations
 
(define H (make-hash))
 
;; G R A P H
;; generate 'normalized' starter vectors D[i] = number of digits 'i' (0 <=i < 10)
;; reduce graph size : 9009, 9900 .. will be generated once : vector #(2 0 0 0 0 0 0 0 0 2)
 
(define (generate D dstart ndigits (sd 0))
(when (> ndigits 0)
(set! sd (vector-ref D dstart)) ;; save before recurse
(for ((i (in-range 0 (1+ ndigits))))
#:continue (and ( = i 0) (> dstart 0))
(vector-set! D dstart i)
(sequence D) ;; sequence length from D
(for ((j (in-range (1+ dstart) 10)))
(generate D j (- ndigits i))))
(vector-set! D dstart sd))) ;; restore
 
 
;; compute follower of D (at most 99 same digits)
(define (dnext D (dd 0))
(define N (make-vector 10))
(for ((d D) (i 10))
#:continue (zero? d)
(vector-set! N i (1+ (vector-ref N i)))
(if (< d 10)
(vector-set! N d (1+ (vector-ref N d))) ;; d < 9
(begin
(set! dd (modulo d 10))
(vector-set! N dd (1+ (vector-ref N dd)))
(set! dd (quotient d 10))
(vector-set! N dd (1+ (vector-ref N dd))))))
N)
 
 
;; update all nodes in same sequence
;; seq-length (next D) = 1 - seq-length(D)
(define (sequence D)
(define (dists D d)
(unless (hash-ref H D)
(hash-set H D d)
(dists (dnext D ) (1- d))))
 
(unless (hash-ref H D)
(dists D (sequence-length D))))
 
;; sequence length from D
;; stops on loop found (node N)
(define (sequence-length D )
(define (looper N looplg depth) ;; looper 2 : a b a
(when ( > depth 0)
(hash-set H N looplg)
(looper (dnext N) looplg (1- depth))))
 
(define followers (make-hash))
(define N (dnext D))
(define seqlg 0)
(define looplg 0)
 
(hash-set followers D 0)
 
(set! seqlg
(for ((lg (in-naturals 1 )))
#:break (hash-ref H N) => (+ lg (hash-ref H N)) ;; already known
#:break (hash-ref followers N) => lg ;; loop found
(hash-set followers N lg)
(set! N (dnext N))))
 
;; update nodes in loop : same seq-length
(when (hash-ref followers N) ;; loop found
(set! looplg ( - seqlg (hash-ref followers N)))
(looper N looplg looplg))
 
seqlg )
 
;; O U T P U T
;; backwards from D - normalized vector - to numbers (as strings)
(define (starters D)
(define (not-leading-zero list) (!zero? (first list)))
(map list->string
(filter not-leading-zero (make-set (permutations (for/fold (acc null) ((d D) (i 10))
#:continue (zero? d)
(append acc (for/list ((j d)) i))))))))
 
 
;; printing one node
(define (D-print D)
(set! D (reverse (vector->list D)))
(for/string ( (d D) (i (in-range 9 -1 -1)))
#:continue (zero? d)
(string-append d i)))
 
;; print graph contents
(define (print-sequence D)
(writeln 1 (starters D))
(writeln 2 (D-print D ))
(for ((i (in-range 1 (hash-ref H D))))
(writeln (+ i 2) (D-print (setv! D (dnext D))))))
 
 
;; TA S K
(define (task (n 6) (verbose #t))
(generate (make-vector 10) 0 n)
(define seqmax (apply max (hash-values H)))
(when verbose (for ((kv H))
#:continue (!= (rest kv ) seqmax)
(print-sequence (first kv))))
 
(writeln (expt 10 n) '--> 'max-sequence= (1+ seqmax) 'nodes= (length (hash-values H))))
 
 
Output:
 
(task 6)
1 (9009 9090 9900)
2 2920
3 192210
4 19222110
5 19323110
6 1923123110
7 1923224110
8 191413323110
9 191433125110
10 19151423125110
11 19251413226110
12 1916151413325110
13 1916251423127110
14 191716151413326110
15 191726151423128110
16 19181716151413327110
17 19182716151423129110
18 29181716151413328110
19 19281716151423228110
20 19281716151413427110
21 19182716152413228110
1000000 --> max-sequence= 21 nodes= 10926
 
(task 7 #f)
10000000 --> max-sequence= 21 nodes= 23432
(task 8 #f)
100000000 --> max-sequence= 21 nodes= 47359
(task 9 #f)
1000000000 --> max-sequence= 21 nodes= 97455
(task 10 #f)
10000000000 --> max-sequence= 21 nodes= 188493
 
 

Eiffel[edit]

Only checks numbers where digits are in ascending order. Digits with trailing zeros have to be treated as ascending numbers (special case). Calculates all the permutations in the end.

 
class
SELF_REFERENTIAL_SEQUENCE
 
create
make
 
feature
 
make
local
i: INTEGER
length, max: INTEGER_64
do
create seed_value.make
create sequence.make (25)
create permuted_values.make
from
i := 1
until
i > 1000000
loop
length := check_length (i.out)
if length > max then
max := length
seed_value.wipe_out
seed_value.extend (i)
elseif length = max then
seed_value.extend (i)
end
sequence.wipe_out
i := next_ascending (i).to_integer
end
io.put_string ("Maximal length: " + max.out)
io.put_string ("%NSeed Value: %N")
across
seed_value as s
loop
permute (s.item.out, 1)
end
across
permuted_values as p
loop
io.put_string (p.item + "%N")
end
io.put_string ("Sequence:%N")
max := check_length (seed_value [1].out)
across
sequence as s
loop
io.put_string (s.item)
io.new_line
end
end
 
next_ascending (n: INTEGER_64): STRING
-- Next number with ascending digits after 'n'.
-- Numbers with trailing zeros are treated as ascending numbers.
local
st: STRING
first, should_be, zero: STRING
i: INTEGER
do
create Result.make_empty
create zero.make_empty
st := (n + 1).out
from
until
st.count < 2
loop
first := st.at (1).out
if st [2] ~ '0' then
from
i := 3
until
i > st.count
loop
zero.append ("0")
i := i + 1
end
Result.append (first + first + zero)
st := ""
else
should_be := st.at (2).out
if first > should_be then
should_be := first
end
st.remove_head (2)
st.prepend (should_be)
Result.append (first)
end
end
if st.count > 0 then
Result.append (st [st.count].out)
end
end
 
feature {NONE}
 
seed_value: SORTED_TWO_WAY_LIST [INTEGER]
 
permuted_values: SORTED_TWO_WAY_LIST [STRING]
 
sequence: ARRAYED_LIST [STRING]
 
permute (a: STRING; k: INTEGER)
-- All permutations of 'a'.
require
count_positive: a.count > 0
k_valid_index: k > 0
local
t: CHARACTER
b: STRING
found: BOOLEAN
do
across
permuted_values as p
loop
if p.item ~ a then
found := True
end
end
if k = a.count and a [1] /= '0' and not found then
create b.make_empty
b.deep_copy (a)
permuted_values.extend (b)
else
across
k |..| a.count as c
loop
t := a [k]
a [k] := a [c.item]
a [c.item] := t
permute (a, k + 1)
t := a [k]
a [k] := a [c.item]
a [c.item] := t
end
end
end
 
check_length (i: STRING): INTEGER_64
-- Length of the self referential sequence starting with 'i'.
local
found: BOOLEAN
j: INTEGER
s: STRING
do
create s.make_from_string (i)
from
until
found
loop
sequence.extend (s)
s := next (s)
from
j := sequence.count - 1
until
j < 1
loop
if sequence [j] ~ s then
found := True
end
j := j - 1
end
end
Result := sequence.count
end
 
next (n: STRING): STRING
-- Next item after 'n' in a self referential sequence.
local
i, count: INTEGER
counter: ARRAY [INTEGER]
do
create counter.make_filled (0, 0, 9)
create Result.make_empty
from
i := 1
until
i > n.count
loop
count := n [i].out.to_integer
counter [count] := counter [count] + 1
i := i + 1
end
from
i := 9
until
i < 0
loop
if counter [i] > 0 then
Result.append (counter [i].out + i.out)
end
i := i - 1
end
end
 
end
 
Output:
Maximal length: 21
Seed Value:
9009
9090
9900 
Sequence:
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

F#[edit]

Takes ~0.4 sec. to filter numbers less than 1 million with digits in descending order, so don't know why all the emphasis on optimization. Doesn't use any strings which maybe is good.

 
// Summarize and say sequence . Nigel Galloway: April 23rd., 2021
let rec fN g=let n=let n,g=List.head g|>List.countBy id|>List.unzip in [email protected](g|>List.collect(fun g->if g<10 then [g] else [g/10;g%10]))
if List.contains n g then g.Tail|>List.rev else fN(n::g)
let rec fG n g=seq{yield! n; if g>1 then yield! fG(n|>Seq.collect(fun n->[for g in 0..List.head n->g::n]))(g-1)}
let n,g=seq{yield [0]; yield! fG(Seq.init 9 (fun n->[n+1])) 6}
|>Seq.fold(fun(n,l) g->let g=fN [g] in match g.Length with e when e<n->(n,l) |e when e>n->(e,[[g]]) |e->(n,[g]::l))(0,[])
printfn "maximum number of iterations is %d" (n+1)
for n in g do for n in n do
printf "Permutations of "; n.Head|>List.rev|>List.iter(printf "%d"); printfn " produce:"
for n in n do (for n,g in List.countBy id n|>List.sort|>List.rev do printf "%d%d" g n); printfn ""
 
Output:
maximum number of iterations is 21
Permutations of 9900 produce:
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Factor[edit]

Like the Eiffel example, this program saves time by considering only seed numbers whose digits are in increasing order (zeros are exempt). This ensures that extra permutations of a number are not searched, as they produce equivalent sequences (aside from the first element). For instance,   21   is the first number to be skipped because it's a permutation of   12.

USING: assocs grouping io kernel math math.combinatorics
math.functions math.ranges math.statistics math.text.utils
prettyprint sequences sets ;
IN: rosetta-code.self-referential-sequence
 
: next-term ( seq -- seq ) histogram >alist concat ;
 
! Output the self-referential sequence, given a seed value.
: srs ( seq -- seq n )
V{ } clone [ 2dup member? ] [ 2dup push [ next-term ] dip ]
until nip dup length ;
 
: digit-before? ( m n -- ? ) dup zero? [ 2drop t ] [ <= ] if ;
 
! The numbers from 1 rto n sans permutations.
: candidates ( n -- seq )
[1,b] [ 1 digit-groups reverse ] map
[ [ digit-before? ] monotonic? ] filter ;
 
: max-seed ( n -- seq ) candidates [ srs nip ] supremum-by ;
 
: max-seeds ( n -- seq )
max-seed <permutations> members [ first zero? ] reject ;
 
: digits>number ( seq -- n ) [ 10^ * ] map-index sum ;
 
: >numbers ( seq -- seq ) [ digits>number ] map ;
 
: main ( -- )
"Seed value(s): " write
1,000,000 max-seeds
[ [ reverse ] map >numbers . ]
[ first srs ] bi
"Iterations: " write .
"Sequence:" print >numbers . ;
 
MAIN: main
Output:
Seed value(s): V{ 9009 9090 9900 }
Iterations: 21
Sequence:
V{
    9009
    2920
    221910
    22192110
    19323110
    1923123110
    1923224110
    191413323110
    191433125110
    19151423125110
    19251413226110
    1916151413325110
    1916251423127110
    191716151413326110
    191726151423128110
    19181716151413327110
    19182716151423129110
    29181716151413328110
    19281716151423228110
    19281716151413427110
    19182716152413228110
}

Go[edit]

Brute force

package main
 
import (
"fmt"
"strconv"
)
 
func main() {
var maxLen int
var seqMaxLen [][]string
for n := 1; n < 1e6; n++ {
switch s := seq(n); {
case len(s) == maxLen:
seqMaxLen = append(seqMaxLen, s)
case len(s) > maxLen:
maxLen = len(s)
seqMaxLen = [][]string{s}
}
}
fmt.Println("Max sequence length:", maxLen)
fmt.Println("Sequences:", len(seqMaxLen))
for _, seq := range seqMaxLen {
fmt.Println("Sequence:")
for _, t := range seq {
fmt.Println(t)
}
}
}
 
func seq(n int) []string {
s := strconv.Itoa(n)
ss := []string{s}
 
for {
dSeq := sortD(s)
d := dSeq[0]
nd := 1
s = ""
for i := 1; ; i++ {
if i == len(dSeq) {
s = fmt.Sprintf("%s%d%c", s, nd, d)
break
}
if dSeq[i] == d {
nd++
} else {
s = fmt.Sprintf("%s%d%c", s, nd, d)
d = dSeq[i]
nd = 1
}
}
for _, s0 := range ss {
if s == s0 {
return ss
}
}
ss = append(ss, s)
}
panic("unreachable")
}
 
func sortD(s string) []rune {
r := make([]rune, len(s))
for i, d := range s {
j := 0
for ; j < i; j++ {
if d > r[j] {
copy(r[j+1:], r[j:i])
break
}
}
r[j] = d
}
return r
}

Output:

Max sequence length: 21
Sequences: 3
Sequence:
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110
Sequence:
9090
2920
...
19182716152413228110
Sequence:
9900
2920
...
19182716152413228110

Groovy[edit]

Solution:

Number.metaClass.getSelfReferentialSequence = {
def number = delegate as String; def sequence = []
 
while (!sequence.contains(number)) {
sequence << number
def encoded = new StringBuilder()
((number as List).sort().join('').reverse() =~ /(([0-9])\2*)/).each { matcher, text, digit ->
encoded.append(text.size()).append(digit)
}
number = encoded.toString()
}
sequence
}
 
def maxSeqSize = { List values ->
values.inject([seqSize: 0, seeds: []]) { max, n ->
if (n % 100000 == 99999) println 'HT'
else if (n % 10000 == 9999) print '.'
def seqSize = n.selfReferentialSequence.size()
switch (seqSize) {
case max.seqSize: max.seeds << n
case { it < max.seqSize }: return max
default: return [seqSize: seqSize, seeds: [n]]
}
}
}

Test:

def max = maxSeqSize(0..<1000000)
 
println "\nLargest sequence size among seeds < 1,000,000\n"
println "Seeds: ${max.seeds}\n"
println "Size: ${max.seqSize}\n"
println "Sample sequence:"
max.seeds[0].selfReferentialSequence.each { println it }

Output:

Largest sequence size among seeds < 1,000,000

Seeds: [9009, 9090, 9900]

Size: 21

Sample sequence:
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Haskell[edit]

Brute force and quite slow:

import Data.Set (Set, member, insert, empty)
import Data.List (group, sort)
 
step :: String -> String
step = concatMap (\list -> show (length list) ++ [head list]) . group . sort
 
findCycle :: (Ord a) => [a] -> [a]
findCycle = aux empty where
aux set (x : xs)
| x `member` set = []
| otherwise = x : aux (insert x set) xs
 
select :: [[a]] -> [[a]]
select = snd . foldl (\(len, acc) xs -> case len `compare` length xs of
LT -> (length xs, [xs])
EQ -> (len, xs : acc)
GT -> (len, acc)) (0, [])
 
main :: IO ()
main = mapM_ (mapM_ print) $ -- Print out all the numbers
select $ -- find the longest ones
map findCycle $ -- run the sequences until there is a repeat
map (iterate step) $ -- produce the sequence
map show -- turn the numbers into digits
[1..1000000] -- The input seeds
 

Icon and Unicon[edit]

link printf 
 
procedure main()
every L := !longestselfrefseq(1000000) do
every printf(" %i : %i\n",i := 1 to *L,L[i])
end
 
 
procedure longestselfrefseq(N) #: find longest sequences from 1 to N
 
mlen := 0
every L := selfrefseq(n := 1 to N) do {
if mlen <:= *L then
ML := [L]
else if mlen = *L then
put(ML,L)
}
 
return ML
end
 
procedure selfrefseq(n) #: return list of sequence oeis:A036058 for seed n
S := set()
L := []
every p := seq(1) do
if member(S,n) then return L # ends at a repeat
else {
insert(S,n)
put(L,n)
n := nextselfrefseq(n)
}
end
 
procedure nextselfrefseq(n) #: return next element of sequence oeis:A036058
every (Counts := table(0))[integer(!n)] +:= 1 # count digits
every (n := "") ||:= (0 < Counts[i := 9 to 0 by -1]) || i # assemble counts
return integer(n)
end

printf.icn provides printf, sprintf, fprintf, etc.

Sample of Output:
 1 : 9009
 2 : 2920
 3 : 192210
 4 : 19222110
 5 : 19323110
 6 : 1923123110
 7 : 1923224110
 8 : 191413323110
 9 : 191433125110
 10 : 19151423125110
 11 : 19251413226110
 12 : 1916151413325110
 13 : 1916251423127110
 14 : 191716151413326110
 15 : 191726151423128110
 16 : 19181716151413327110
 17 : 19182716151423129110
 18 : 29181716151413328110
 19 : 19281716151423228110
 20 : 19281716151413427110
 21 : 19182716152413228110
 1 : 9090
 2 : 2920
 ... (manually removed, same as above)
 21 : 19182716152413228110
 1 : 9900
 2 : 2920
 ... (manually removed, same as above)
 21 : 19182716152413228110

The following (admittedly overdense) version produces output matching the problem statement and avoids repeating sequences that arise from 'similar' seeds. It does not assume that only one equivalence class of similar seeds exists at the maximum sequence length. As with the first example, it works in both Icon and Unicon.

 link strings   # to get csort()
 
procedure main(A)
limit := A[1] | 1000000 # Allow alternate limit
mSq := 0
# May have multiple 'unique' sequence sets (unrelated seeds) so use table
every s := [n := 1 to limit, sequence(n)] do {
if mSq <:= *s[2] then mT := table() # new max, start over
if mSq == *s[2] then insert((/mT[n := csort(n)] := set()) | mT[n],s)
}
dumpSequences(mT)
end
 
procedure sequence(n) # produce sequence of SDS with seed n
every (repeats := [], iter := seq(), put(repeats, n)) do
if (n := nElem(n)) == !repeats then return repeats # Converged
end
 
procedure nElem(n) # given n, produce its self-description
every (n1 := "", c := !cset(n)) do
(every (d := 0) +:= (upto(c, n),1)) | (n1 := d||c||n1)
return n1
end
 
procedure dumpSequences(seqTab) # Show each 'unique' sequence in table
every writes("Seeds:" | (!!seqTab)[1], " ")
write("\n\nIterations: ",*(!!seqTab)[2])
every s := !seqTab do (write() & every write(!(!s\1)[2]))
end
 

Output with limit = 1000000:

Seeds: 9009 9090 9900 

Iterations: 21

9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

J[edit]

Given:

require'stats'
digits=: 10&#.inv"0 :. ([: "[email protected]; (<'x'),~":&.>)
summar=: (#/.~ ,@,. ~.)@\:~&.digits
sequen=: [email protected](, [email protected]{:)^:_
values=: ~. \:~&.digits i.1e6
allvar=: [:(#~(=&<.&(10&^.) >./))@~.({~ [email protected]#)&.(digits"1)

The values with the longest sequence are:

   ;allvar&.> values #~ (= >./) #@sequen"0 values
9900 9090 9009
# sequen 9900
21
,.sequen 9900
9900
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Notes:

digits is an invertible function that maps from a number to a sequence of digits and back where the inverse transform converts numbers to strings, concatenates them, and then back to a number.

   digits 321
3 2 1
digits inv 34 5
345

summar computes the summary successor.

   summar 0 1 2
10 11 12

sequen computes the complete non-repeating sequence of summary successors

The computation for values could have been made much more efficient. Instead, though, all one million integers have their digits sorted in decreasing order, and then the unique set of them is found.

Finally, allvar finds all variations of a number which would have the same summary sequence based on the permutations of that number's digits.

Java[edit]

Works with: Java version 8
import java.util.*;
import java.util.concurrent.ConcurrentHashMap;
import java.util.stream.IntStream;
 
public class SelfReferentialSequence {
 
static Map<String, Integer> cache = new ConcurrentHashMap<>(10_000);
 
public static void main(String[] args) {
Seeds res = IntStream.range(0, 1000_000)
.parallel()
.mapToObj(n -> summarize(n, false))
.collect(Seeds::new, Seeds::accept, Seeds::combine);
 
System.out.println("Seeds:");
res.seeds.forEach(e -> System.out.println(Arrays.toString(e)));
 
System.out.println("\nSequence:");
summarize(res.seeds.get(0)[0], true);
}
 
static int[] summarize(int seed, boolean display) {
String n = String.valueOf(seed);
 
String k = Arrays.toString(n.chars().sorted().toArray());
if (!display && cache.get(k) != null)
return new int[]{seed, cache.get(k)};
 
Set<String> seen = new HashSet<>();
StringBuilder sb = new StringBuilder();
 
int[] freq = new int[10];
 
while (!seen.contains(n)) {
seen.add(n);
 
int len = n.length();
for (int i = 0; i < len; i++)
freq[n.charAt(i) - '0']++;
 
sb.setLength(0);
for (int i = 9; i >= 0; i--) {
if (freq[i] != 0) {
sb.append(freq[i]).append(i);
freq[i] = 0;
}
}
if (display)
System.out.println(n);
n = sb.toString();
}
 
cache.put(k, seen.size());
 
return new int[]{seed, seen.size()};
}
 
static class Seeds {
int largest = Integer.MIN_VALUE;
List<int[]> seeds = new ArrayList<>();
 
void accept(int[] s) {
int size = s[1];
if (size >= largest) {
if (size > largest) {
largest = size;
seeds.clear();
}
seeds.add(s);
}
}
 
void combine(Seeds acc) {
acc.seeds.forEach(this::accept);
}
}
}
Seeds:
[9009, 21]
[9090, 21]
[9900, 21]

Sequence:
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

JavaScript[edit]

 
function selfReferential(n) {
n = n.toString();
let res = [n];
const makeNext = (n) => {
let matchs = {
'9':0,'8':0,'7':0,'6':0,'5':0,'4':0,'3':0,'2':0,'1':0,'0':0}, res = [];
for(let index=0;index<n.length;index++)
matchs[n[index].toString()]++;
for(let index=9;index>=0;index--)
if(matchs[index]>0)
res.push(matchs[index],index);
return res.join("").toString();
}
for(let i=0;i<10;i++)
res.push(makeNext(res[res.length-1]));
return [...new Set(res)];
}
 

jq[edit]

# Given any array, produce an array of [item, count] pairs for each run.
def runs:
reduce .[] as $item
( [];
if . == [] then [ [ $item, 1] ]
else .[length-1] as $last
| if $last[0] == $item
then (.[0:length-1] + [ [$item, $last[1] + 1] ] )
else . + [[$item, 1]]
end
end ) ;
 
# string to string
def next_self_referential:
def runs2integer: # input is an array as produced by runs,
# i.e. an array of [count, n] pairs, where count is an int,
# and n is an "exploded" digit
reduce .[] as $pair
(""; . + ($pair[1] | tostring) + ([$pair[0]]|implode) ) ;
 
explode | sort | reverse | runs | runs2integer;
 
# Given an integer as a string,
# compute the entire sequence (of strings) to convergence:
def sequence_of_self_referentials:
def seq:
. as $ary
| (.[length-1] | next_self_referential) as $next
| if ($ary|index($next)) then $ary
else $ary + [$next] | seq
end;
[.] | seq;
 
def maximals(n):
def interesting:
tostring | (. == (explode | sort | reverse | implode));
 
reduce range(0;n) as $i
([[], 0]; # maximalseeds, length
if ($i | interesting) then
($i|tostring|sequence_of_self_referentials|length) as $length
| if .[1] == $length then [ .[0] + [$i], $length]
elif .[1] < $length then [ [$i], $length]
else .
end
else .
end );
 
def task(n):
maximals(n) as $answer
| "The maximal length to convergence for seeds up to \(n) is \($answer[1]).",
"The corresponding seeds are the allowed permutations",
"of the representative number(s): \($answer[0][])",
"For each representative seed, the self-referential sequence is as follows:",
($answer[0][] | tostring
| ("Representative: \(.)",
"Self-referential sequence:",
(sequence_of_self_referentials | map(tonumber))))
 ;
 
task(1000000)
Output:
$ jq -n -r -f Self_referential_sequence.jq
The maximal length to convergence for seeds up to 1000000 is 21.
The corresponding seeds are the allowed permutations
of the representative number(s): 9900
For each representative seed, the self-referential sequence is as follows:
Representative: 9900
Self-referential sequence:
[
9900,
2920,
192210,
19222110,
19323110,
1923123110,
1923224110,
191413323110,
191433125110,
19151423125110,
19251413226110,
1916151413325110,
1916251423127110,
191716151413326100,
191726151423128100,
19181716151413326000,
19182716151423128000,
29181716151413330000,
19281716151423230000,
19281716151413430000,
19182716152413230000
]

Julia[edit]

const seen = Dict{Vector{Char}, Vector{Char}}()
 
function findnextterm(prevterm)
counts = Dict{Char, Int}()
reversed = Vector{Char}()
for c in prevterm
if !haskey(counts, c)
counts[c] = 0
end
counts[c] += 1
end
for c in sort(collect(keys(counts)))
if counts[c] > 0
push!(reversed, c)
if counts[c] == 10
push!(reversed, '0'); push!(reversed, '1')
else
push!(reversed, Char(UInt8(counts[c]) + UInt8('0')))
end
end
end
reverse(reversed)
end
 
function findsequence(seedterm)
term = seedterm
sequence = Vector{Vector{Char}}()
while !(term in sequence)
push!(sequence, term)
if !haskey(seen, term)
nextterm = findnextterm(term)
seen[term] = nextterm
end
term = seen[term]
end
return sequence
end
 
function selfseq(maxseed)
maxseqlen = -1
maxsequences = Vector{Pair{Int, Vector{Char}}}()
for i in 1:maxseed
seq = findsequence([s[1] for s in split(string(i), "")])
seqlen = length(seq)
if seqlen > maxseqlen
maxsequences = [Pair(i, seq)]
maxseqlen = seqlen
elseif seqlen == maxseqlen
push!(maxsequences, Pair(i, seq))
end
end
println("The longest sequence length is $maxseqlen.")
for p in maxsequences
println("\n Seed: $(p[1])")
for seq in p[2]
println(" ", join(seq, ""))
end
end
end
 
selfseq(1000000)
 
Output:

The longest sequence length is 21.

Seed: 9009
    9009
    2920
    192210
    19222110
    19323110
    1923123110
    1923224110
    191413323110
    191433125110
    19151423125110
    19251413226110
    1916151413325110
    1916251423127110
    191716151413326110
    191726151423128110
    19181716151413327110
    19182716151423129110
    29181716151413328110
    19281716151423228110
    19281716151413427110
    19182716152413228110
Seed: 9090
    9090
    2920
    192210
    19222110
    19323110
    1923123110
    1923224110
    191413323110
    191433125110
    19151423125110
    19251413226110
    1916151413325110
    1916251423127110
    191716151413326110
    191726151423128110
    19181716151413327110
    19182716151423129110
    29181716151413328110
    19281716151423228110
    19281716151413427110
    19182716152413228110
Seed: 9900
    9900
    2920
    192210
    19222110
    19323110
    1923123110
    1923224110
    191413323110
    191433125110
    19151423125110
    19251413226110
    1916151413325110
    1916251423127110
    191716151413326110
    191726151423128110
    19181716151413327110
    19182716151423129110
    29181716151413328110
    19281716151423228110
    19281716151413427110
    19182716152413228110

Kotlin[edit]

// version 1.1.2
 
const val LIMIT = 1_000_000
 
val sb = StringBuilder()
 
fun selfRefSeq(s: String): String {
sb.setLength(0) // faster than using a local StringBuilder object
for (d in '9' downTo '0') {
if (d !in s) continue
val count = s.count { it == d }
sb.append("$count$d")
}
return sb.toString()
}
 
fun permute(input: List<Char>): List<List<Char>> {
if (input.size == 1) return listOf(input)
val perms = mutableListOf<List<Char>>()
val toInsert = input[0]
for (perm in permute(input.drop(1))) {
for (i in 0..perm.size) {
val newPerm = perm.toMutableList()
newPerm.add(i, toInsert)
perms.add(newPerm)
}
}
return perms
}
 
fun main(args: Array<String>) {
val sieve = IntArray(LIMIT) // all zero by default
val elements = mutableListOf<String>()
for (n in 1 until LIMIT) {
if (sieve[n] > 0) continue
elements.clear()
var next = n.toString()
elements.add(next)
while (true) {
next = selfRefSeq(next)
if (next in elements) {
val size = elements.size
sieve[n] = size
if (n > 9) {
val perms = permute(n.toString().toList()).distinct()
for (perm in perms) {
if (perm[0] == '0') continue
val k = perm.joinToString("").toInt()
sieve[k] = size
}
}
break
}
elements.add(next)
}
}
val maxIterations = sieve.max()!!
for (n in 1 until LIMIT) {
if (sieve[n] < maxIterations) continue
println("$n -> Iterations = $maxIterations")
var next = n.toString()
for (i in 1..maxIterations) {
println(next)
next = selfRefSeq(next)
}
println()
}
}
Output:
9009 -> Iterations = 21
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

9090 -> Iterations = 21
9090
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

9900 -> Iterations = 21
9900
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Lua[edit]

Runs in about nine seconds under LuaJIT. Uses memoisation via the global table 'nextTerm'.

-- Return the next term in the self-referential sequence
function findNext (nStr)
local nTab, outStr, pos, count = {}, "", 1, 1
for i = 1, #nStr do nTab[i] = nStr:sub(i, i) end
table.sort(nTab, function (a, b) return a > b end)
while pos <= #nTab do
if nTab[pos] == nTab[pos + 1] then
count = count + 1
else
outStr = outStr .. count .. nTab[pos]
count = 1
end
pos = pos + 1
end
return outStr
end
 
-- Return boolean indicating whether table t contains string s
function contains (t, s)
for k, v in pairs(t) do
if v == s then return true end
end
return false
end
 
-- Return the sequence generated by the given seed term
function buildSeq (term)
local sequence = {}
repeat
table.insert(sequence, term)
if not nextTerm[term] then nextTerm[term] = findNext(term) end
term = nextTerm[term]
until contains(sequence, term)
return sequence
end
 
-- Main procedure
nextTerm = {}
local highest, seq, hiSeq = 0
for i = 1, 10^6 do
seq = buildSeq(tostring(i))
if #seq > highest then
highest = #seq
hiSeq = {seq}
elseif #seq == highest then
table.insert(hiSeq, seq)
end
end
io.write("Seed values: ")
for _, v in pairs(hiSeq) do io.write(v[1] .. " ") end
print("\n\nIterations: " .. highest)
print("\nSample sequence:")
for _, v in pairs(hiSeq[1]) do print(v) end
Output:
Seed values: 9009 9090 9900

Iterations: 21

Sample sequence:
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Mathematica[edit]

selfRefSequence[ x_ ] := [email protected]@[email protected][[email protected]{RotateRight[[email protected],1], Range[0,9]},Except[{0,_}]]
DisplaySequence[ x_ ] := NestWhileList[selfRefSequence,x,UnsameQ[##]&,4]
data= {#, [email protected][#]}&/@Range[1000000];
Print["Values: ", Select[data ,#[[2]] == [email protected][[;;,2]]&][[1,;;]]]
Print["Iterations: ", [email protected][#]&/@Select[data ,#[[2]] == [email protected][[;;,2]]&][[1,;;]]]
[email protected][data, #[[2]] == [email protected][[;;,2]]&][[1]]//Column
Values: {9009, 9090, 9900}
Iterations: 21
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110
19281716151413427110

Nim[edit]

A version which uses a cache to store the number of iterations and thus avoids to compute the sequence for each permutation. The version without cache runs in more than 9 seconds. This version runs in less than 300 ms.

import algorithm, sequtils, sets, strutils, tables
 
var cache: Table[seq[char], int] # Maps key -> number of iterations.
 
 
iterator sequence(seed: string): string =
## Yield the successive strings of a sequence.
 
var history: HashSet[string]
history.incl seed
var current = seed
yield current
 
while true:
var counts = current.toCountTable()
var next: string
for ch in sorted(toSeq(counts.keys), Descending):
next.add $counts[ch] & ch
if next in history: break
current = move(next)
history.incl current
yield current
 
 
proc seqLength(seed: string): int =
## Return the number of iterations for the given seed.
let key = sorted(seed, Descending)
if key in cache: return cache[key]
result = toSeq(sequence(seed)).len
cache[key] = result
 
 
var seeds: seq[int]
var itermax = 0
for seed in 0..<1_000_000:
let itercount = seqLength($seed)
if itercount > itermax:
itermax = itercount
seeds = @[seed]
elif itercount == itermax:
seeds.add seed
 
echo "Maximum iterations: ", itermax
echo "Seed values: ", seeds.join(", ")
echo "Sequence for $#:".format(seeds[0])
for s in sequence($seeds[0]): echo s
Output:
Maximum iterations: 21
Seed values: 9009, 9090, 9900
Sequence for 9009:
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Perl[edit]

sub next_num {
my @a;
$a[$_]++ for split '', shift;
join('', map(exists $a[$_] ? $a[$_].$_ : "", reverse 0 .. 9));
}
 
my %cache;
sub seq {
my $a = shift;
my (%seen, @s);
until ($seen{$a}) {
$seen{$a} = 1;
push(@s, $a);
last if !wantarray && $cache{$a};
$a = next_num($a);
}
return (@s) if wantarray;
 
my $l = $cache{$a};
if ($l) { $cache{$s[$_]} = $#s - $_ + $l for (0 .. $#s); }
else {
$l++ while ($s[-$l] != $a);
$cache{pop @s} = $l for (1 .. $l);
$cache{pop @s} = ++$l while @s;
}
$cache{$s[0]}
}
 
my (@mlist, $mlen);
for (1 .. 100_000) { # 1_000_000 takes very, very long
my $l = seq($_);
next if $l < $mlen;
 
if ($l > $mlen) { $mlen = $l; @mlist = (); }
push @mlist, $_;
}
 
print "longest ($mlen): @mlist\n";
print join("\n", seq($_)), "\n\n" for @mlist;

Phix[edit]

Optimisation idea taken from CoffeeScript, completes in under a second.

string n = "000000"
 
function incn()
for i=length(n) to 1 by -1 do
if n[i]='9' then
if i=1 then return false end if
n[i]='0'
else
n[i] += 1
exit
end if
end for
return true
end function
 
sequence res = {}, bestseen
integer maxcycle = 0
 
procedure srs()
sequence seen, this = n
integer cycle = 1
while length(this)>1 and this[1]='0' do
this = this[2..$]
end while
integer ch = this[1]
for i=2 to length(this) do
if this[i]>ch then return end if
ch = this[i]
end for
seen = {this}
while 1 do
sequence digits = repeat(0,10)
for i=1 to length(this) do
digits[this[i]-'0'+1] += 1
end for
string next = ""
for i=length(digits) to 1 by -1 do
if digits[i]!=0 then
next &= sprint(digits[i])
next &= i+'0'-1
end if
end for
if find(next,seen) then exit end if
seen = append(seen,next)
this = next
cycle += 1
end while
if cycle>maxcycle then
res = {seen[1]}
maxcycle = cycle
bestseen = seen
elsif cycle=maxcycle then
res = append(res,seen[1])
end if
end procedure
 
while 1 do
srs()
if not incn() then exit end if
end while
 
-- add non-leading-0 perms:
for i=length(res) to 1 by -1 do
string ri = res[i]
for p=1 to factorial(length(ri)) do
string pri = permute(p,ri)
if pri[1]!='0' and not find(pri,res) then
res = append(res,pri)
end if
end for
end for
?res
puts(1,"cycle length is ") ?maxcycle
pp(bestseen,{pp_Nest,1})
{"9900","9009","9090"}
cycle length is 21
{"9900",
 "2920",
 "192210",
 "19222110",
 "19323110",
 "1923123110",
 "1923224110",
 "191413323110",
 "191433125110",
 "19151423125110",
 "19251413226110",
 "1916151413325110",
 "1916251423127110",
 "191716151413326110",
 "191726151423128110",
 "19181716151413327110",
 "19182716151423129110",
 "29181716151413328110",
 "19281716151423228110",
 "19281716151413427110",
 "19182716152413228110"}

PicoLisp[edit]

Using 'las' from Look-and-say sequence#PicoLisp:

(de selfRefSequence (Seed)
(let L (mapcar format (chop Seed))
(make
(for (Cache NIL (not (idx 'Cache L T)))
(setq L
(las (flip (sort (copy (link L))))) ) ) ) ) )
 
(let Res NIL
(for Seed 1000000
(let N (length (selfRefSequence Seed))
(cond
((> N (car Res)) (setq Res (list N Seed)))
((= N (car Res)) (queue 'Res Seed)) ) ) )
(println 'Values: (cdr Res))
(println 'Iterations: (car Res))
(mapc prinl (selfRefSequence (cadr Res))) )

Output:

Values: (9009 9090 9900)
Iterations: 21
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Python[edit]

The number generation function follows that of Look-and-say with a sort. only the first of any set of numbers with the same digits has the length of its sequence calculated in function max_A036058_length, although no timings were taken to check if the optimisation was of value.

from itertools import groupby, permutations
 
def A036058(number):
return ''.join( str(len(list(g))) + k
for k,g in groupby(sorted(str(number), reverse=True)) )
 
def A036058_length(numberstring='0', printit=False):
iterations, last_three, queue_index = 1, ([None] * 3), 0
 
def A036058(number):
# rely on external reverse-sort of digits of number
return ''.join( str(len(list(g))) + k
for k,g in groupby(number) )
 
while True:
if printit:
print("  %2i %s" % (iterations, numberstring))
numberstring = ''.join(sorted(numberstring, reverse=True))
if numberstring in last_three:
break
assert iterations < 1000000
last_three[queue_index], numberstring = numberstring, A036058(numberstring)
iterations += 1
queue_index +=1
queue_index %=3
return iterations
 
def max_A036058_length( start_range=range(11) ):
already_done = set()
max_len = (-1, [])
for n in start_range:
sn = str(n)
sns = tuple(sorted(sn, reverse=True))
if sns not in already_done:
already_done.add(sns)
size = A036058_length(sns)
if size > max_len[0]:
max_len = (size, [n])
elif size == max_len[0]:
max_len[1].append(n)
return max_len
 
lenmax, starts = max_A036058_length( range(1000000) )
 
# Expand
allstarts = []
for n in starts:
allstarts += [int(''.join(x))
for x in set(k
for k in permutations(str(n), 4)
if k[0] != '0')]
allstarts = [x for x in sorted(allstarts) if x < 1000000]
 
print ( '''\
The longest length, followed by the number(s) with the longest sequence length
for starting sequence numbers below 1000000 are:
Iterations = %i and sequence-starts = %s.'''
 % (lenmax, allstarts) )
 
print ( '''
Note that only the first of any sequences with the same digits is printed below.
(The others will differ only in their first term)'''
)
 
for n in starts:
print()
A036058_length(str(n), printit=True)
Output
The longest length, followed by the number(s) with the longest sequence length
for starting sequence numbers below 1000000 are:
  Iterations = 21 and sequence-starts = [9009, 9090, 9900].

Note that only the first of any sequences with the same digits is printed below.
(The others will differ only in their first term)

   1 9009
   2 2920
   3 192210
   4 19222110
   5 19323110
   6 1923123110
   7 1923224110
   8 191413323110
   9 191433125110
  10 19151423125110
  11 19251413226110
  12 1916151413325110
  13 1916251423127110
  14 191716151413326110
  15 191726151423128110
  16 19181716151413327110
  17 19182716151423129110
  18 29181716151413328110
  19 19281716151423228110
  20 19281716151413427110
  21 19182716152413228110

q[edit]

ls:{raze(string 1_ deltas d,count x),'x d:where differ x}  / look & say
sumsay:ls desc@ / summarize & say
 
seeds:group desc each string til 1000000 / seeds for million integers
seq:(key seeds)!30 sumsay\'key seeds / sequences for unique seeds
top:max its:(count distinct@)each seq / count iterations
 
/ report results
rpt:{1 x,": ",y,"\n\n";}
rpt["Seeds"]" "sv string raze seeds where its=top / all forms of top seed/s
rpt["Iterations"]string top
rpt["Sequence"]"\n\n","\n"sv raze seq where its=top
Output:
Seeds: 9009 9090 9900

Iterations: 21

Sequence:

9900
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Racket[edit]

 
#lang racket
 
(define (next s)
(define v (make-vector 10 0))
(for ([c s])
(define d (- (char->integer #\9) (char->integer c)))
(vector-set! v d (add1 (vector-ref v d))))
(string-append* (for/list ([x v] [i (in-range 9 -1 -1)] #:when (> x 0))
(format "~a~a" x i))))
 
(define (seq-of s)
(reverse (let loop ([ns (list s)])
(define n (next (car ns)))
(if (member n ns) ns (loop (cons n ns))))))
 
(define (sort-string s)
(list->string (sort (string->list s) char>?)))
 
(define-values [len nums seq]
(for/fold ([*len #f] [*nums #f] [*seq #f])
([n (in-range 1000000 -1 -1)]) ; start at the high end
(define s (number->string n))
(define sorted (sort-string s))
(cond [(equal? s sorted)
(define seq (seq-of s))
(define len (length seq))
(cond [(or (not *len) (> len *len)) (values len (list s) seq)]
[(= len *len) (values len (cons s *nums) seq)]
[else (values *len *nums *seq)])]
 ;; not sorted: see if it's a permutation of the best
[else (values
*len
(if (and *nums (member sorted *nums)) (cons s *nums) *nums)
*seq)])))
(printf "Numbers: ~a\nLength: ~a\n" (string-join nums ", ") len)
(for ([n seq]) (printf " ~a\n" n))
 
Output:
Numbers: 9009, 9090, 9900
Length: 21
  9900
  2920
  192210
  19222110
  19323110
  1923123110
  1923224110
  191413323110
  191433125110
  19151423125110
  19251413226110
  1916151413325110
  1916251423127110
  191716151413326110
  191726151423128110
  19181716151413327110
  19182716151423129110
  29181716151413328110
  19281716151423228110
  19281716151413427110
  19182716152413228110

Raku[edit]

(formerly Perl 6)

Works with: rakudo version 2018.03
my @list;
my $longest = 0;
my %seen;
 
for 1 .. 1000000 -> $m {
next unless $m ~~ /0/; # seed must have a zero
my $j = join '', $m.comb.sort;
next if %seen{$j}:exists; # already tested a permutation
%seen{$j} = '';
my @seq = converging($m);
my %elems;
my $count;
for @seq[] -> $value { last if ++%elems{$value} == 2; $count++; };
if $longest == $count {
@list.push($m);
}
elsif $longest < $count {
$longest = $count;
@list = $m;
print "\b" x 20, "$count, $m"; # monitor progress
}
};
 
for @list -> $m {
say "\nSeed Value(s): ", my $seeds = ~permutations($m).unique.grep( { .substr(0,1) != 0 } );
my @seq = converging($m);
my %elems;
my $count;
for @seq[] -> $value { last if ++%elems{$value} == 2; $count++; };
say "\nIterations: ", $count;
say "\nSequence: (Only one shown per permutation group.)";
.say for |@seq[^$count], "\n";
}
 
sub converging ($seed) { return $seed, -> $l { join '', map { $_.value.elems~$_.key }, $l.comb.classify({$^b}).sort: {-$^c.key} } ... * }
 
sub permutations ($string, $sofar? = '' ) {
return $sofar unless $string.chars;
my @perms;
for ^$string.chars -> $idx {
my $this = $string.substr(0,$idx)~$string.substr($idx+1);
my $char = substr($string, $idx,1);
@perms.push( |permutations( $this, join '', $sofar, $char ) );
}
return @perms;
}
Output:
Seed Value(s): 9009 9090 9900

Iterations: 21

Sequence: (Only one shown per permutation group.)
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

REXX[edit]

The REXX language supports   sparse   (stemmed) arrays, so this program utilizes REXX's hashing of
array elements to speed up the checking to see if a sequence has been generated before.

/*REXX pgm generates a self─referential sequence and displays sequences with max length.*/
parse arg LO HI . /*obtain optional arguments from the CL*/
if LO=='' | LO=="," then LO= 1 /*Not specified? Then use the default.*/
if HI=='' | HI=="," then HI=1000000 - 1 /* " " " " " " */
max$=; seeds=; maxL=0 /*inialize some defaults and counters. */
 
do #=LO to HI; n=#; @.=0; @.#=1 /*loop thru seed; define some defaults.*/
$=n
do c=1 until x==n; x=n /*generate a self─referential sequence.*/
n=; do k=9 by -1 for 10 /*generate a new sequence (downwards). */
_=countstr(k, x) /*obtain the number of sequence counts.*/
if _\==0 then n=n || _ || k /*is count > zero? Then append it to N*/
end /*k*/
if @.n then leave /*has sequence been generated before ? */
$=$'-'n; @.n=1 /*add the number to sequence and roster*/
end /*c*/
 
if c==maxL then do; seeds=seeds # /*is the sequence equal to max so far ?*/
max$=max$ $ /*append this self─referential # to $ */
end
else if c>maxL then do; seeds=# /*use the new number as the new seed. */
maxL=c; max$=$ /*also, set the new maximum L; max seq.*/
end /* [↑] have we found a new best seq ? */
end /*#*/
 
say ' seeds that had the most iterations: ' seeds
say 'the maximum self─referential length: ' maxL
 
do j=1 for words(max$) ; say
say copies('─',30) "iteration sequence for: " word(seeds,j) ' ('maxL "iterations)"
q=translate( word( max$, j), ,'-')
do k=1 for words(q); say word(q, k)
end /*k*/
end /*j*/ /*stick a fork in it, we're all done. */
output   when using the default inputs:


(Shown at five-sixths size.)

 seeds that had the most iterations:  9009 9090 9900
the maximum self─referential length:  21

────────────────────────────── iteration sequence for:  9009   (21 iterations)
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

────────────────────────────── iteration sequence for:  9090   (21 iterations)
9090
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

────────────────────────────── iteration sequence for:  9900   (21 iterations)
9900
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Ruby[edit]

Cached for performance

$cache = {}
def selfReferentialSequence_cached(n, seen = [])
return $cache[n] if $cache.include? n
return [] if seen.include? n
 
digit_count = Array.new(10, 0)
n.to_s.chars.collect {|char| digit_count[char.to_i] += 1}
term = ''
9.downto(0).each do |d|
if digit_count[d] > 0
term += digit_count[d].to_s + d.to_s
end
end
term = term.to_i
$cache[n] = [n] + selfReferentialSequence_cached(term, [n] + seen)
end
 
limit = 1_000_000
max_len = 0
max_vals = []
 
1.upto(limit - 1) do |n|
seq = selfReferentialSequence_cached(n)
if seq.length > max_len
max_len = seq.length
max_vals = [n]
elsif seq.length == max_len
max_vals << n
end
end
 
puts "values: #{max_vals.inspect}"
puts "iterations: #{max_len}"
puts "sequence:"
selfReferentialSequence_cached(max_vals[0]).each_with_index do |val, idx|
puts "%2d %d" % [idx + 1, val]
end

output

values: [9009, 9090, 9900]
iterations: 21
sequence:
 1 9009
 2 2920
 3 192210
 4 19222110
 5 19323110
 6 1923123110
 7 1923224110
 8 191413323110
 9 191433125110
10 19151423125110
11 19251413226110
12 1916151413325110
13 1916251423127110
14 191716151413326110
15 191726151423128110
16 19181716151413327110
17 19182716151423129110
18 29181716151413328110
19 19281716151423228110
20 19281716151413427110
21 19182716152413228110

Scala[edit]

This example creates a ParVector, which is a collection type that inherently uses parallel processing, of all seeds within the range, maps each seed to a tuple containing the seed, the sequence, and the number of iterations, sorts the collection by decreasing sequence length, then shows the relevant information for the maximal sequences at the head of the collection.

import spire.math.SafeLong
 
import scala.annotation.tailrec
import scala.collection.parallel.immutable.ParVector
 
object SelfReferentialSequence {
def main(args: Array[String]): Unit = {
val nums = ParVector.range(1, 1000001).map(SafeLong(_))
val seqs = nums.map{ n => val seq = genSeq(n); (n, seq, seq.length) }.toVector.sortWith((a, b) => a._3 > b._3)
val maxes = seqs.takeWhile(t => t._3 == seqs.head._3)
 
println(s"Seeds: ${maxes.map(_._1).mkString(", ")}\nIterations: ${maxes.head._3}")
for(e <- maxes.distinctBy(a => nextTerm(a._1.toString))){
println(s"\nSeed: ${e._1}\n${e._2.mkString("\n")}")
}
}
 
def genSeq(seed: SafeLong): Vector[String] = {
@tailrec
def gTrec(seq: Vector[String], n: String): Vector[String] = {
if(seq.contains(n)) seq
else gTrec(seq :+ n, nextTerm(n))
}
gTrec(Vector[String](), seed.toString)
}
 
def nextTerm(num: String): String = {
@tailrec
def dTrec(digits: Vector[(Int, Int)], src: String): String = src.headOption match{
case Some(n) => dTrec(digits :+ ((n.asDigit, src.count(_ == n))), src.filter(_ != n))
case None => digits.sortWith((a, b) => a._1 > b._1).map(p => p._2.toString + p._1.toString).mkString
}
dTrec(Vector[(Int, Int)](), num)
}
}
Output:
Seeds: 9009, 9090, 9900
Iterations: 21

Seed: 9009
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

Tcl[edit]

proc nextterm n {
foreach c [split $n ""] {incr t($c)}
foreach c {9 8 7 6 5 4 3 2 1 0} {
if {[info exist t($c)]} {append r $t($c) $c}
}
return $r
}
# Local context of lambda term is just for speed
apply {limit {
# Build a digit cache; this adds quite a bit of speed
set done [lrepeat [set l2 [expr {$limit * 100}]] 0]
# Iterate over search space
set maxlen 0
set maxes {}
for {set i 0} {$i < $limit} {incr i} {
if {[lindex $done $i]} continue
# Compute the sequence length for this value (with help from cache)
set seq {}
for {set seed $i} {$seed ni $seq} {set seed [nextterm $seed]} {
if {$seed < $l2 && [lindex $done $seed]} {
set len [expr {[llength $seq] + [lindex $done $seed]}]
break
}
set len [llength [lappend seq $seed]]
}
# What are we going to do about it?
if {$len > $maxlen} {
set maxlen $len
set maxes [list $i]
} elseif {$len == $maxlen} {
lappend maxes $i
}
# Update the cache with what we have learned
foreach n $seq {
if {$n < $l2} {lset done $n $len}
incr len -1
}
}
# Output code
puts "max length: $maxlen"
foreach c $maxes {puts $c}
puts "Sample max-len sequence:"
set seq {}
# Rerun the sequence generator for printing; faster for large limits
for {set seed [lindex $c 0]} {$seed ni $seq} {set seed [nextterm $seed]} {
lappend seq $seed
puts "\t$seed"
}
}} 1000000

Output:

max length: 21
9009
9090
9900
Sample max-len sequence:
	9900
	2920
	192210
	19222110
	19323110
	1923123110
	1923224110
	191413323110
	191433125110
	19151423125110
	19251413226110
	1916151413325110
	1916251423127110
	191716151413326110
	191726151423128110
	19181716151413327110
	19182716151423129110
	29181716151413328110
	19281716151423228110
	19281716151413427110
	19182716152413228110

TXR[edit]

Translation of: Clojure
[edit]

This is a close, almost expression-by-expression transliteration of the Clojure version.

;; Syntactic sugar for calling reduce-left
(defmacro reduce-with ((acc init item sequence) . body)
^(reduce-left (lambda (,acc ,item) ,*body) ,sequence ,init))
 
 ;; Macro similar to clojure's ->> and ->
(defmacro opchain (val . ops)
^[[chain ,*[mapcar [iffi consp (op cons 'op)] ops]] ,val])
 
;; Reduce integer to a list of integers representing its decimal digits.
(defun digits (n)
(if (< n 10)
(list n)
(opchain n tostring list-str (mapcar (op - @1 #\0)))))
 
(defun dcount (ds)
(digits (length ds)))
 
;; Perform a look-say step like (1 2 2) --"one 1, two 2's"-> (1 1 2 2).
(defun summarize-prev (ds)
(opchain ds copy (sort @1 >) (partition-by identity)
(mapcar [juxt dcount first]) flatten))
 
;; Take a starting digit string and iterate the look-say steps,
;; to generate the whole sequence, which ends when convergence is reached.
(defun convergent-sequence (ds)
(reduce-with (cur-seq nil ds [giterate true summarize-prev ds])
(if (member ds cur-seq)
(return-from convergent-sequence cur-seq)
(nconc cur-seq (list ds)))))
 
;; A candidate sequence is one which begins with montonically
;; decreasing digits. We don't bother with (9 0 9 0) or (9 0 0 9);
;; which yield identical sequences to (9 9 0 0).
(defun candidate-seq (n)
(let ((ds (digits n)))
(if [apply >= ds]
(convergent-sequence ds))))
 
;; Discover the set of longest sequences.
(defun find-longest (limit)
(reduce-with (max-seqs nil new-seq [mapcar candidate-seq (range 1 limit)])
(let ((cmp (- (opchain max-seqs first length) (length new-seq))))
(cond ((> cmp 0) max-seqs)
((< cmp 0) (list new-seq))
(t (nconc max-seqs (list new-seq)))))))
 
(defvar *results* (find-longest 1000000))
 
(each ((result *results*))
(flet ((strfy (list) ;; (strfy '((1 2 3 4) (5 6 7 8))) -> ("1234" "5678")
(mapcar [chain (op mapcar tostring) cat-str] list)))
(let* ((seed (first result))
(seeds (opchain seed perm uniq (remove-if zerop @1 first))))
(put-line `Seed value(s): @(strfy seeds)`)
(put-line)
(put-line `Iterations: @(length result)`)
(put-line)
(put-line `Sequence: @(strfy result)`))))
Output:
$ txr self-ref-seq.tl

Seed value(s): 9900 9090 9009

Iterations: 21

Sequence: 9900 2920 192210 19222110 19323110 1923123110 1923224110 191413323110 191433125110 19151423125110 19251413226110 1916151413325110 1916251423127110 191716151413326110 191726151423128110 19181716151413327110 19182716151423129110 29181716151413328110 19281716151423228110 19281716151413427110 19182716152413228110

Translation of: Common Lisp
[edit]

Mostly the same logic. The count-and-say function is based on the same steps, but stays in the string domain instead of converting the input to a list, and then the output back to a string. It also avoids building the output backwards and reversing it, so out must be accessed on the right side inside the loop. This is easy due to Python-inspired array indexing semantics: -1 means last element, -2 second last and so on.

Like in Common Lisp, TXR's sort is destructive, so we take care to use copy-str.

(defun count-and-say (str)
(let* ((s [sort (copy-str str) <])
(out `@[s 0]0`))
(each ((x s))
(if (eql x [out -1])
(inc [out -2])
(set out `@{out}[email protected]`)))
out))
 
(defun ref-seq-len (n : doprint)
(let ((s (tostring n)) hist)
(while t
(push s hist)
(if doprint (pprinl s))
(set s (count-and-say s))
(each ((item hist)
(i (range 0 2)))
(when (equal s item)
(return-from ref-seq-len (length hist)))))))
 
(defun find-longest (top)
(let (nums (len 0))
(each ((x (range 0 top)))
(let ((l (ref-seq-len x)))
(when (> l len) (set len l) (set nums nil))
(when (= l len) (push x nums))))
(list nums len)))
Output:
Longest: ((9900 9090 9009 99) 21)
9900
2029
102219
10212219
10313219
1031122319
1041222319
103132131419
105112331419
10511223141519
10612213142519
1051321314151619
1071122314251619
106132131415161719
108112231415261719
10713213141516171819
10911223141516271819
10813213141516171829
10812223141516172819
10714213141516172819
10812213241516271819

Translation of: Racket
[edit]

;; Macro very similar to Racket's for/fold
(defmacro for-accum (accum-var-inits each-vars . body)
(let ((accum-vars [mapcar first accum-var-inits])
(block-sym (gensym))
(next-args [mapcar (ret (progn @rest (gensym))) accum-var-inits])
(nvars (length accum-var-inits)))
^(let ,accum-var-inits
(flet ((iter (,*next-args)
,*[mapcar (ret ^(set ,@1 ,@2)) accum-vars next-args]))
(each ,each-vars
,*body)
(list ,*accum-vars)))))
 
(defun next (s)
(let ((v (vector 10 0)))
(each ((c s))
(inc [v (- #\9 c)]))
(cat-str
(collect-each ((x v)
(i (range 9 0 -1)))
(when (> x 0)
`@[email protected]`)))))
 
(defun seq-of (s)
(for* ((ns ()))
((not (member s ns)) (reverse ns))
((push s ns) (set s (next s)))))
 
(defun sort-string (s)
[sort (copy s) >])
 
(tree-bind (len nums seq)
(for-accum ((*len nil) (*nums nil) (*seq nil))
((n (range 1000000 0 -1))) ;; start at the high end
(let* ((s (tostring n))
(sorted (sort-string s)))
(if (equal s sorted)
(let* ((seq (seq-of s))
(len (length seq)))
(cond ((or (not *len) (> len *len)) (iter len (list s) seq))
((= len *len) (iter len (cons s *nums) seq))))
(iter *len
(if (and *nums (member sorted *nums)) (cons s *nums) *nums)
*seq))))
(put-line `Numbers: @{nums ", "}\nLength: @len`)
(each ((n seq)) (put-line ` @n`)))
Output:
Numbers: 9009, 9090, 9900
Length: 21
  9900
  2920
  192210
  19222110
  19323110
  1923123110
  1923224110
  191413323110
  191433125110
  19151423125110
  19251413226110
  1916151413325110
  1916251423127110
  191716151413326110
  191726151423128110
  19181716151413327110
  19182716151423129110
  29181716151413328110
  19281716151423228110
  19281716151413427110
  19182716152413228110

Wren[edit]

Translation of: Kotlin
Library: Wren-seq
Library: Wren-math
import "/seq" for Lst
import "/math" for Nums
 
var limit = 1e6
 
var selfRefSeq = Fn.new { |s|
var sb = ""
for (d in "9876543210") {
if (s.contains(d)) {
var count = s.count { |c| c == d }
sb = sb + "%(count)%(d)"
}
}
return sb
}
 
var permute // recursive
permute = Fn.new { |input|
if (input.count == 1) return [input]
var perms = []
var toInsert = input[0]
for (perm in permute.call(input[1..-1])) {
for (i in 0..perm.count) {
var newPerm = perm.toList
newPerm.insert(i, toInsert)
perms.add(newPerm)
}
}
return perms
}
 
var sieve = List.filled(limit, 0)
var elements = []
for (n in 1...limit) {
if (sieve[n] == 0) {
elements.clear()
var next = n.toString
elements.add(next)
while (true) {
next = selfRefSeq.call(next)
if (elements.contains(next)) {
var size = elements.count
sieve[n] = size
if (n > 9) {
var perms = permute.call(n.toString.toList).map { |p| p.join("") }.toList
perms = Lst.distinct(perms)
for (perm in perms) {
if (perm[0] != "0") {
var k = Num.fromString(perm.join(""))
sieve[k] = size
}
}
}
break
}
elements.add(next)
}
}
}
var maxIterations = Nums.max(sieve)
for (n in 1...limit) {
if (sieve[n] >= maxIterations) {
System.print("%(n) -> Iterations = %(maxIterations)")
var next = n.toString
for (i in 1..maxIterations) {
System.print(next)
next = selfRefSeq.call(next)
}
System.print()
}
}
Output:
9009 -> Iterations = 21
9009
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

9090 -> Iterations = 21
9090
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

9900 -> Iterations = 21
9900
2920
192210
19222110
19323110
1923123110
1923224110
191413323110
191433125110
19151423125110
19251413226110
1916151413325110
1916251423127110
191716151413326110
191726151423128110
19181716151413327110
19182716151423129110
29181716151413328110
19281716151423228110
19281716151413427110
19182716152413228110

zkl[edit]

N:=0d1_000_001;
 
fcn lookAndJustSaying(seed){ // numeric String --> numeric String
"9876543210".pump(String,'wrap(n){
(s:=seed.inCommon(n)) and String(s.len(),n) or ""
});
}
fcn sequence(seed){ // numeric string --> sequence until it repeats
seq:=L();
while(not seq.holds(seed)){ seq.append(seed); seed=lookAndJustSaying(seed); }
seq
}
fcn decending(str) //--> True if digits are in descending (or equal) order
{ (not str.walker().zipWith('<,str[1,*]).filter1()) }
 
szs:=List.createLong(25); max:=0;
foreach seed in (N){
z:=seed.toString();
if(decending(z)){ // 321 generates same sequence as 312,132,123,213
len:=sequence(z).len();
if(len>max) szs.clear();
if(len>=max){ szs.append(seed.toString()); max=len; }
}
}
 
// List permutations of longest seeds
// ("9900"-->(((9,0,0,9),...))-->((9,0,0,9),...)-->("9009"...)
// -->remove numbers w/leading zeros-->remove dups
zs:=szs.apply(Utils.Helpers.permute).flatten().apply("concat")
.filter(fcn(s){ s[0]!="0" }) : Utils.Helpers.listUnique(_);
println(max," iterations for ",zs.concat(", "));
zs.pump(Console.println,sequence,T("concat",", "));

Ignoring permutations cut run time from 4 min to 9 sec.

Output:
21 iterations for 9900, 9090, 9009
9900, 2920, 192210, 19222110, 19323110, 1923123110, 1923224110, 191413323110, 191433125110, 19151423125110, 19251413226110, 1916151413325110, 1916251423127110, 191716151413326110, 191726151423128110, 19181716151413327110, 19182716151423129110, 29181716151413328110, 19281716151423228110, 19281716151413427110, 19182716152413228110
9090, 2920, 192210, 19222110, 19323110, 1923123110, 1923224110, 191413323110, 191433125110, 19151423125110, 19251413226110, 1916151413325110, 1916251423127110, 191716151413326110, 191726151423128110, 19181716151413327110, 19182716151423129110, 29181716151413328110, 19281716151423228110, 19281716151413427110, 19182716152413228110
9009, 2920, 192210, 19222110, 19323110, 1923123110, 1923224110, 191413323110, 191433125110, 19151423125110, 19251413226110, 1916151413325110, 1916251423127110, 191716151413326110, 191726151423128110, 19181716151413327110, 19182716151423129110, 29181716151413328110, 19281716151423228110, 19281716151413427110, 19182716152413228110