Distinct palindromes within decimal numbers

From Rosetta Code
Distinct palindromes within decimal numbers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Find all distinct palindromes contained as substrings in decimal representation of n. Note that for the purpose of the initial function, a single digit will be considered a palindrome.

Task
  1. Find all the palindromes including single digits in the integers from 100 to 125, inclusive.
  2. Determine which if any of the following decimal integers contain palindromes of 2 digits or more:
 9, 169, 12769, 1238769, 123498769, 12346098769, 1234572098769,
 123456832098769, 12345679432098769, 1234567905432098769, 123456790165432098769,
 83071934127905179083, 1320267947849490361205695
Also see
Related tasks

Palindrome_detection
Longest_palindromic_substrings
N_1's_followed_by_a_3


Factor

Works with: Factor version 0.99 2021-02-05
USING: formatting io kernel math math.ranges present prettyprint
sequences sequences.extras sets ;

: dpal ( n -- seq )
    present all-subseqs members [ dup reverse = ] filter ;

! task 1
"Number  Palindromes" print
100 125 [a..b] [ dup pprint bl bl bl bl bl dpal .  ] each nl

! task 2
"Number                    Has no >= 2 digit-palindromes?" print
{ 9 169 12769 1238769 123498769 12346098769 1234572098769
  123456832098769 12345679432098769 1234567905432098769
  123456790165432098769 83071934127905179083
  1320267947849490361205695 }
[ dup dpal [ length 2 < ] reject empty? "%-25d %u\n" printf ]
each
Output:
Number  Palindromes
100     { "1" "0" "00" }
101     { "1" "0" "101" }
102     { "1" "0" "2" }
103     { "1" "0" "3" }
104     { "1" "0" "4" }
105     { "1" "0" "5" }
106     { "1" "0" "6" }
107     { "1" "0" "7" }
108     { "1" "0" "8" }
109     { "1" "0" "9" }
110     { "1" "0" "11" }
111     { "1" "11" "111" }
112     { "1" "2" "11" }
113     { "1" "3" "11" }
114     { "1" "4" "11" }
115     { "1" "5" "11" }
116     { "1" "6" "11" }
117     { "1" "7" "11" }
118     { "1" "8" "11" }
119     { "1" "9" "11" }
120     { "1" "2" "0" }
121     { "1" "2" "121" }
122     { "1" "2" "22" }
123     { "1" "2" "3" }
124     { "1" "2" "4" }
125     { "1" "2" "5" }

Number                    Has no >= 2 digit-palindromes?
9                         t
169                       t
12769                     t
1238769                   t
123498769                 t
12346098769               t
1234572098769             t
123456832098769           t
12345679432098769         t
1234567905432098769       t
123456790165432098769     t
83071934127905179083      t
1320267947849490361205695 f

Go

Translation of: Wren
package main

import (
    "fmt"
    "sort"
    "strings"
)

func substrings(s string) []string {
    var ss []string
    n := len(s)
    for i := 0; i < n; i++ {
        for j := 1; j <= n-i; j++ {
            ss = append(ss, s[i:i+j])
        }
    }
    return ss
}

func reversed(s string) string {
    var sb strings.Builder
    for i := len(s) - 1; i >= 0; i-- {
        sb.WriteByte(s[i])
    }
    return sb.String()
}

func main() {
    fmt.Println("Number  Palindromes")
    for i := 100; i <= 125; i++ {
        var pals []string
        ss := substrings(fmt.Sprintf("%d", i))
        for _, s := range ss {
            if s == reversed(s) {
                pals = append(pals, s)
            }
        }
        m := make(map[string]bool)
        for _, pal := range pals {
            m[pal] = true
        }
        pals = pals[:0]
        for k := range m {
            pals = append(pals, k)
        }
        sort.Slice(pals, func(i, j int) bool {
            if len(pals[i]) == len(pals[j]) {
                return pals[i] < pals[j]
            }
            return len(pals[i]) < len(pals[j])
        })
        fmt.Printf("%d   %3s\n", i, pals)
    }
    nums := []string{
        "9", "169", "12769", "1238769", "123498769", "12346098769", "1234572098769",
        "123456832098769", "12345679432098769", "1234567905432098769", "123456790165432098769",
        "83071934127905179083", "1320267947849490361205695",
    }
    fmt.Println("\nNumber                    Has no >= 2 digit palindromes")
    for _, num := range nums {
        tmp := substrings(num)
        var ss []string
        for _, t := range tmp {
            if len(t) > 1 {
                ss = append(ss, t)
            }
        }
        none := true
        for _, s := range ss {
            if s == reversed(s) {
                none = false
                break
            }
        }
        fmt.Printf("%-25s %t\n", num, none)
    }
}
Output:
Number  Palindromes
100   [  0   1  00]
101   [  0   1 101]
102   [  0   1   2]
103   [  0   1   3]
104   [  0   1   4]
105   [  0   1   5]
106   [  0   1   6]
107   [  0   1   7]
108   [  0   1   8]
109   [  0   1   9]
110   [  0   1  11]
111   [  1  11 111]
112   [  1   2  11]
113   [  1   3  11]
114   [  1   4  11]
115   [  1   5  11]
116   [  1   6  11]
117   [  1   7  11]
118   [  1   8  11]
119   [  1   9  11]
120   [  0   1   2]
121   [  1   2 121]
122   [  1   2  22]
123   [  1   2   3]
124   [  1   2   4]
125   [  1   2   5]

Number                    Has no >= 2 digit palindromes
9                         true
169                       true
12769                     true
1238769                   true
123498769                 true
12346098769               true
1234572098769             true
123456832098769           true
12345679432098769         true
1234567905432098769       true
123456790165432098769     true
83071934127905179083      true
1320267947849490361205695 false

jq

Works with jq (but the second task requires a version with support for "big integer" literals)
Works with gojq, the Go implementation of jq (provided `keys_unsorted` is replaced by `keys`)

In this entry, except for 0 itself, palindromes involving leading 0s are ignored.

One feature of the implementation given here is that uniqueness is achieved without any sorting; however, if gojq is used, then `keys` would have to be used, thus entailing a sort after distinctness has been achieved.

Note that for the second task, and for very large integers (greater than 2^53) in general, accurate results require gojq, or a version of jq supporting "big integer" literals.

# input: an integer
# output a stream of distinct integers, each representing an admissible palindrome

def palindromes:
  # input: an array
  def is_palindrome: . == reverse;
  def all_palindromes:
    (tostring|explode)
    | range(0; length) as $i
    | range($i+1; length+1) as $j
    | .[$i:$j] # candidate
    | select(is_palindrome)
    | implode
    # trim leading 0s:
    | if length > 1 then sub("^00*"; "") else . end
    | select(length>0) ;

  INDEX(all_palindromes; .) | keys_unsorted[] | tonumber;

def task1:
  " i   palindromes",
   (range(100; 126)
    | "\(.)  \([palindromes]|join(" "))" );
   
def task2:
 (9, 169, 12769, 1238769, 123498769, 12346098769, 1234572098769,
 123456832098769, 12345679432098769, 1234567905432098769, 123456790165432098769,
 83071934127905179083, 1320267947849490361205695)
 | select( any(palindromes; . > 99) );

task1,
"\nThe integers amongst those in the problem statement that have 2 or more digits:",
task2
Output:
 i   palindromes
100  1 0
101  1 101 0
102  1 0 2
103  1 0 3
104  1 0 4
105  1 0 5
106  1 0 6
107  1 0 7
108  1 0 8
109  1 0 9
110  1 11 0
111  1 11 111
112  1 11 2
113  1 11 3
114  1 11 4
115  1 11 5
116  1 11 6
117  1 11 7
118  1 11 8
119  1 11 9
120  1 2 0
121  1 121 2
122  1 2 22
123  1 2 3
124  1 2 4
125  1 2 5

The integers amongst those in the problem statement that have 2 or more digits:
1320267947849490361205695

Julia

function allpalindromes(a::Vector{T}) where T
    pals = Vector{Vector{T}}([[x] for x in a])
    len = length(a)
    len < 2 && return pals
    a == reverse(a) && push!(pals, a)
    len == 2 && return pals
    for i in 2:len
        left = a[1:i]
        left == reverse(left) && push!(pals, left)
    end
    return unique(vcat(pals, allpalindromes(a[2:end])))
end

println("Number  Palindromes")
for n in 100:125
    println(" ", rpad(n, 7), sort(allpalindromes(digits(n))))
end

palindrome2plusfree(n) = (a = allpalindromes(digits(n)); all(x -> length(x) == 1, a))

println("\nNumber                    Has no >= 2 digit palindromes")
for n in [9, 169, 12769, 1238769, 123498769, 12346098769, 1234572098769, 123456832098769,
    12345679432098769, 1234567905432098769, 123456790165432098769, 83071934127905179083, 1320267947849490361205695]
    println(rpad(n, 26), palindrome2plusfree(n))
end
Output:
Number  Palindromes
 100    [[0], [0, 0], [1]]
 101    [[0], [1], [1, 0, 1]]
 102    [[0], [1], [2]]
 103    [[0], [1], [3]]
 104    [[0], [1], [4]]
 105    [[0], [1], [5]]
 106    [[0], [1], [6]]
 107    [[0], [1], [7]]
 108    [[0], [1], [8]]
 109    [[0], [1], [9]]
 110    [[0], [1], [1, 1]]
 111    [[1], [1, 1], [1, 1, 1]]
 112    [[1], [1, 1], [2]]
 113    [[1], [1, 1], [3]]
 114    [[1], [1, 1], [4]]
 115    [[1], [1, 1], [5]]
 116    [[1], [1, 1], [6]]
 117    [[1], [1, 1], [7]]
 118    [[1], [1, 1], [8]]
 119    [[1], [1, 1], [9]]
 120    [[0], [1], [2]]
 121    [[1], [1, 2, 1], [2]]
 122    [[1], [2], [2, 2]]
 123    [[1], [2], [3]]
 124    [[1], [2], [4]]
 125    [[1], [2], [5]]

Number                    Has no >= 2 digit palindromes
9                         true
169                       true
12769                     true
1238769                   true
123498769                 true
12346098769               true
1234572098769             true
123456832098769           true
12345679432098769         true
1234567905432098769       true
123456790165432098769     true
83071934127905179083      true
1320267947849490361205695 false

Mathematica/Wolfram Language

ClearAll[ContainsPalindromeQ]
ContainsPalindromeQ[n_Integer, minlength_ : 1, b_ : 10] := Select[DeleteDuplicates@Subsequences[IntegerDigits[n, b], {minlength, \[Infinity]}], PalindromeQ]
ContainsPalindromeQ /@ Range[100, 125] // Column
Length[ContainsPalindromeQ[#, 2]] > 0 & /@ {9, 169, 12769, 1238769, 
  123498769, 12346098769, 1234572098769, 123456832098769, 
  12345679432098769, 1234567905432098769, 123456790165432098769, 
  83071934127905179083, 1320267947849490361205695}
Output:
{{1},{0},{0,0}}
{{1},{0},{1,0,1}}
{{1},{0},{2}}
{{1},{0},{3}}
{{1},{0},{4}}
{{1},{0},{5}}
{{1},{0},{6}}
{{1},{0},{7}}
{{1},{0},{8}}
{{1},{0},{9}}
{{1},{0},{1,1}}
{{1},{1,1},{1,1,1}}
{{1},{2},{1,1}}
{{1},{3},{1,1}}
{{1},{4},{1,1}}
{{1},{5},{1,1}}
{{1},{6},{1,1}}
{{1},{7},{1,1}}
{{1},{8},{1,1}}
{{1},{9},{1,1}}
{{1},{2},{0}}
{{1},{2},{1,2,1}}
{{1},{2},{2,2}}
{{1},{2},{3}}
{{1},{2},{4}}
{{1},{2},{5}}

{False, False, False, False, False, False, False, False, False, False, False, False, True}

Nim

import algorithm, sequtils, sets, strutils

iterator substrings(s: string): string =
  ## Yield the substrings of the given string.
  for istart in 0..s.high:
    for istop in istart..s.high:
      yield s[istart..istop]

func isPalindrome(s: string): bool =
  ## Return true if "s" is a palindrome.
  result = true
  for i in 0..(s.high div 2):
    if s[i] != s[s.high - i]: return false

func cmpPal(s1, s2: string): int =
  ## Compare two palindromes (used for sort).
  result = cmp(s1.len, s2.len)
  if result == 0:
    result = cmp(s1[0], s2[0])

func palindromes(str: string): seq[string] =
  ## Return the sorted list of palindromes contained in "str".
  var palSet: HashSet[string]
  for s in substrings(str):
    if s.isPalindrome: palSet.incl s
  result = sorted(toSeq(palSet), cmpPal)


when isMainModule:

  for n in 100..125:
    echo n, ": ", palindromes($n).mapIt(it.align(3)).join(" ")

  echo()
  for s in ["9", "169", "12769", "1238769", "123498769", "12346098769",
            "1234572098769", "123456832098769", "12345679432098769",
            "1234567905432098769", "123456790165432098769",
            "83071934127905179083", "1320267947849490361205695"]:
    let pals2 = palindromes(s).filterIt(it.len >= 2)
    let verb = if pals2.len == 0: " doesn't contain palindromes "
              else: " contains at least one palindrome "
    echo s, verb, "of two digits or more"
Output:
100:   0   1  00
101:   0   1 101
102:   0   1   2
103:   0   1   3
104:   0   1   4
105:   0   1   5
106:   0   1   6
107:   0   1   7
108:   0   1   8
109:   0   1   9
110:   0   1  11
111:   1  11 111
112:   1   2  11
113:   1   3  11
114:   1   4  11
115:   1   5  11
116:   1   6  11
117:   1   7  11
118:   1   8  11
119:   1   9  11
120:   0   1   2
121:   1   2 121
122:   1   2  22
123:   1   2   3
124:   1   2   4
125:   1   2   5

9 doesn't contain palindromes of two digits or more
169 doesn't contain palindromes of two digits or more
12769 doesn't contain palindromes of two digits or more
1238769 doesn't contain palindromes of two digits or more
123498769 doesn't contain palindromes of two digits or more
12346098769 doesn't contain palindromes of two digits or more
1234572098769 doesn't contain palindromes of two digits or more
123456832098769 doesn't contain palindromes of two digits or more
12345679432098769 doesn't contain palindromes of two digits or more
1234567905432098769 doesn't contain palindromes of two digits or more
123456790165432098769 doesn't contain palindromes of two digits or more
83071934127905179083 doesn't contain palindromes of two digits or more
1320267947849490361205695 contains at least one palindrome of two digits or more

Perl

#!/usr/bin/perl

# https://rosettacode.org/wiki/Distinct_Palindromes_Within_Decimal_Numbers
use strict;
use warnings;

for ( 100 .. 125 )
  {
  my %found;
  /.+(?{$& eq reverse $& and $found{$&}++})(*FAIL)/;
  print "$_ => @{[ sort { $a <=> $b } keys %found ]}\n"
  }

/..+(??{$& == (reverse $&) ? '' : '(*FAIL)' })/ and
  print "$_ has a palindrome of 2 or more\n"
  for ' 9, 169, 12769, 1238769, 123498769, 12346098769, 1234572098769,
    123456832098769, 12345679432098769, 1234567905432098769,
    123456790165432098769, 83071934127905179083, 1320267947849490361205695'
    =~ /\d+/g;
Output:
100 => 00 0 1
101 => 0 1 101
102 => 0 1 2
103 => 0 1 3
104 => 0 1 4
105 => 0 1 5
106 => 0 1 6
107 => 0 1 7
108 => 0 1 8
109 => 0 1 9
110 => 0 1 11
111 => 1 11 111
112 => 1 2 11
113 => 1 3 11
114 => 1 4 11
115 => 1 5 11
116 => 1 6 11
117 => 1 7 11
118 => 1 8 11
119 => 1 9 11
120 => 0 1 2
121 => 1 2 121
122 => 1 2 22
123 => 1 2 3
124 => 1 2 4
125 => 1 2 5
1320267947849490361205695 has a palindrome of 2 or more

Phix

procedure show_all_palindromes(string s, integer longest=0)
    sequence res = {}
    for i=1 to length(s) do
        if longest=0 then
            res = append(res,{1,s[i..i]})
        end if
        for j=0 to iff(i>1 and s[i-1]=s[i]?2:1) do
            integer rev = j,
                    fwd = 1
            while rev<i and i+fwd<=length(s) and s[i-rev]=s[i+fwd] do
                if longest=0 then
                    res = append(res,{rev+fwd+1,s[i-rev..i+fwd]})
                end if
                rev += 1
                fwd += 1
            end while
            if longest>0 then
                longest = max(longest,rev+fwd-1)
            end if
        end for
    end for
    if longest then
        printf(1,"%-26s %t\n",{s,longest<3})
    else
        printf(1," %s    %s\n",{s,join(vslice(unique(res),2))})
    end if
end procedure

printf(1,"Number  Palindromes\n")
papply(apply(true,sprintf,{{"%d"},tagset(125,100)}),show_all_palindromes)

constant tests = split_any("""9, 169, 12769, 1238769, 123498769, 12346098769, 
      1234572098769, 123456832098769, 12345679432098769, 1234567905432098769, 
      123456790165432098769, 83071934127905179083, 1320267947849490361205695"""," ,\n")
printf(1,"\nNumber           Has no >2 digit palindromes\n")
papply(true,show_all_palindromes,{tests,1})
Output:
Number  Palindromes
 100    0 1 00
 101    0 1 101
 102    0 1 2
 103    0 1 3
 104    0 1 4
 105    0 1 5
 106    0 1 6
 107    0 1 7
 108    0 1 8
 109    0 1 9
 110    0 1 11
 111    1 11 111
 112    1 2 11
 113    1 3 11
 114    1 4 11
 115    1 5 11
 116    1 6 11
 117    1 7 11
 118    1 8 11
 119    1 9 11
 120    0 1 2
 121    1 2 121
 122    1 2 22
 123    1 2 3
 124    1 2 4
 125    1 2 5

Number           Has no >2 digit palindromes
9                          true
169                        true
12769                      true
1238769                    true
123498769                  true
12346098769                true
1234572098769              true
123456832098769            true
12345679432098769          true
1234567905432098769        true
123456790165432098769      true
83071934127905179083       true
1320267947849490361205695  false

Raku

A minor modification of the Longest palindromic substrings task. As such, works for any string, not just integers.

use Sort::Naturally;

sub getpal ($str) {
    my @chars = $str.comb;
    my @pal = flat @chars,
    (1 ..^ @chars).map: -> \idx {
        my @s;
        for 1, 2 {
           my int ($rev, $fwd) = $_, 1;
           loop {
                quietly last if ($rev > idx) || (@chars[idx - $rev] ne @chars[idx + $fwd]);
                $rev = $rev + 1;
                $fwd = $fwd + 1;
            }
            @s.push: @chars[idx - $rev ^..^ idx + $fwd].join if $rev + $fwd > 2;
            last if @chars[idx - 1] ne @chars[idx];
        }
        next unless +@s;
        @s
    }
    @pal.unique.sort({.chars, .&naturally});
}

say 'All palindromic substrings including (bizarrely enough) single characters:';
put "$_ => ", getpal $_ for 100..125;
put "\nDo these strings contain a minimum two character palindrome?";
printf "%25s => %s\n", $_, getpal($_).tail.chars > 1 for flat
    9, 169, 12769, 1238769, 123498769, 12346098769, 1234572098769,
    123456832098769, 12345679432098769, 1234567905432098769,
    123456790165432098769, 83071934127905179083, 1320267947849490361205695,
    <Do these strings contain a minimum two character palindrome?>
Output:
All palindromic substrings including (bizarrely enough) single characters:
100 => 0 1 00
101 => 0 1 101
102 => 0 1 2
103 => 0 1 3
104 => 0 1 4
105 => 0 1 5
106 => 0 1 6
107 => 0 1 7
108 => 0 1 8
109 => 0 1 9
110 => 0 1 11
111 => 1 11 111
112 => 1 2 11
113 => 1 3 11
114 => 1 4 11
115 => 1 5 11
116 => 1 6 11
117 => 1 7 11
118 => 1 8 11
119 => 1 9 11
120 => 0 1 2
121 => 1 2 121
122 => 1 2 22
123 => 1 2 3
124 => 1 2 4
125 => 1 2 5

Do these strings contain a minimum two character palindrome?
                        9 => False
                      169 => False
                    12769 => False
                  1238769 => False
                123498769 => False
              12346098769 => False
            1234572098769 => False
          123456832098769 => False
        12345679432098769 => False
      1234567905432098769 => False
    123456790165432098769 => False
     83071934127905179083 => False
1320267947849490361205695 => True
                       Do => False
                    these => True
                  strings => False
                  contain => False
                        a => False
                  minimum => True
                      two => False
                character => True
              palindrome? => False

REXX

This REXX version can handle strings or numbers.

/*REXX pgm finds distinct palindromes contained in substrings  (decimal #s or strings). */
parse arg LO HI mL $$                            /*obtain optional arguments from the CL*/
if LO='' | LO=","  then LO= 100                  /*Not specified?  Then use the default.*/
if HI='' | HI=","  then HI= 125                  /* "      "         "   "   "     "    */
if mL='' | mL=","  then mL=   2                  /* "      "         "   "   "     "    */
if $$='' | $$=","  then $$= 9 169 12769 1238769 12346098769 1234572098769 123456832098769,
                            12345679432098769 1234567905432098769 123456790165432098769  ,
                            83071934127905179083 1320267947849490361205695,
                           'Do these strings contain a minimum two character palindrome?',
                           'amanaplanacanalpanama'         /*a man a plan a canal panama*/ 
w= length(HI)                                    /*max width of LO ──► HI for alignment.*/

       do j=LO  to HI;  #= Dpal(j, 1)            /*get # distinct palindromes, minLen=1 */
       say right(j, w)  ' has '  #    " palindrome"s(#)': '    $
       end   /*j*/

#= words($$);      if #==0  then exit 0          /*No special words/numbers?  Then exit.*/

       do m=1  for #;     w= max(w, length(word($$, m)))   /*find max width string in $$*/
       end   /*m*/
say
       do j=1  for #;     z= word($$, j)         /*obtain a string in the  $$  list.    */
       #= Dpal(z, mL)                            /*get # distinct palindromes, minLen=mL*/
       _= left(':', #>0); @has= ' has ';                     @of='of length'
       say right(z, w)    @has   #   " palindrome"s(#,,' ')  @of  mL  "or more"_  space($)
       end   /*j*/
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
s:    if arg(1)==1  then return arg(3);    return word( arg(2) 's', 1)
/*──────────────────────────────────────────────────────────────────────────────────────*/
Dpal: procedure expose @. !. $ w;  parse arg x, mL;    $=;   !.= 0;   #= 0;   L= length(x)
         do   j=1  for L                         /*test for primality for all substrings*/
           do k=1  to  L-j+1                     /*search for substrings (including  X).*/
           y= strip( substr(x, j, k) )           /*extract a substring from the X string*/
           if length(y)<mL | y\==reverse(y)  then iterate   /*too short or ¬palindromic?*/
           if \!.y  then do;  $= $ right(y, w);  !.y= 1;  #= # + 1;  end
           end   /*k*/
         end     /*j*/
      return #
output   when using the default inputs:
100  has  3  palindromes:     1   0  00
101  has  3  palindromes:     1 101   0
102  has  3  palindromes:     1   0   2
103  has  3  palindromes:     1   0   3
104  has  3  palindromes:     1   0   4
105  has  3  palindromes:     1   0   5
106  has  3  palindromes:     1   0   6
107  has  3  palindromes:     1   0   7
108  has  3  palindromes:     1   0   8
109  has  3  palindromes:     1   0   9
110  has  3  palindromes:     1  11   0
111  has  3  palindromes:     1  11 111
112  has  3  palindromes:     1  11   2
113  has  3  palindromes:     1  11   3
114  has  3  palindromes:     1  11   4
115  has  3  palindromes:     1  11   5
116  has  3  palindromes:     1  11   6
117  has  3  palindromes:     1  11   7
118  has  3  palindromes:     1  11   8
119  has  3  palindromes:     1  11   9
120  has  3  palindromes:     1   2   0
121  has  3  palindromes:     1 121   2
122  has  3  palindromes:     1   2  22
123  has  3  palindromes:     1   2   3
124  has  3  palindromes:     1   2   4
125  has  3  palindromes:     1   2   5

                        9  has  0  palindromes of length 2 or more
                      169  has  0  palindromes of length 2 or more
                    12769  has  0  palindromes of length 2 or more
                  1238769  has  0  palindromes of length 2 or more
              12346098769  has  0  palindromes of length 2 or more
            1234572098769  has  0  palindromes of length 2 or more
          123456832098769  has  0  palindromes of length 2 or more
        12345679432098769  has  0  palindromes of length 2 or more
      1234567905432098769  has  0  palindromes of length 2 or more
    123456790165432098769  has  0  palindromes of length 2 or more
     83071934127905179083  has  0  palindromes of length 2 or more
1320267947849490361205695  has  3  palindromes of length 2 or more: 202 494 949
                       Do  has  0  palindromes of length 2 or more
                    these  has  1  palindrome  of length 2 or more: ese
                  strings  has  0  palindromes of length 2 or more
                  contain  has  0  palindromes of length 2 or more
                        a  has  0  palindromes of length 2 or more
                  minimum  has  3  palindromes of length 2 or more: minim ini mum
                      two  has  0  palindromes of length 2 or more
                character  has  1  palindrome  of length 2 or more: ara
              palindrome?  has  0  palindromes of length 2 or more
    amanaplanacanalpanama  has  12  palindromes of length 2 or more: ama amanaplanacanalpanama manaplanacanalpanam ana anaplanacanalpana naplanacanalpan aplanacanalpa planacanalp lanacanal anacana nacan aca

Sidef

func palindromes(arr) {
    gather {
        for a in (0..arr.end), b in (a .. arr.end) {
            var sublist = arr.items(a..b -> ...)
            take(sublist) if (sublist == sublist.flip)
        }
    }.uniq
}

for n in (100..125) {
    say "#{n} -> #{palindromes(n.digits).sort.map{.join}.sort_by{.len}.join(' ')}"
}

[9, 169, 12769, 1238769, 123498769, 12346098769, 1234572098769,
 123456832098769, 12345679432098769, 1234567905432098769, 123456790165432098769,
 83071934127905179083, 1320267947849490361205695, "amanaplanacanalpanama"].each {|n|
    var p = palindromes(n.kind_of(Number) ? n.digits : n.chars).grep { .len >= 2}
    say ("#{'%25s' % n} has #{'%2d' % p.len} palindromes of length 2 or more: ",
        p.sort.map{.join}.sort_by{.len}.join(' '))
}
Output:
100 -> 0 1 00
101 -> 0 1 101
102 -> 0 1 2
103 -> 0 1 3
104 -> 0 1 4
105 -> 0 1 5
106 -> 0 1 6
107 -> 0 1 7
108 -> 0 1 8
109 -> 0 1 9
110 -> 0 1 11
111 -> 1 11 111
112 -> 1 2 11
113 -> 1 3 11
114 -> 1 4 11
115 -> 1 5 11
116 -> 1 6 11
117 -> 1 7 11
118 -> 1 8 11
119 -> 1 9 11
120 -> 0 1 2
121 -> 1 2 121
122 -> 1 2 22
123 -> 1 2 3
124 -> 1 2 4
125 -> 1 2 5
                        9 has  0 palindromes of length 2 or more: 
                      169 has  0 palindromes of length 2 or more: 
                    12769 has  0 palindromes of length 2 or more: 
                  1238769 has  0 palindromes of length 2 or more: 
                123498769 has  0 palindromes of length 2 or more: 
              12346098769 has  0 palindromes of length 2 or more: 
            1234572098769 has  0 palindromes of length 2 or more: 
          123456832098769 has  0 palindromes of length 2 or more: 
        12345679432098769 has  0 palindromes of length 2 or more: 
      1234567905432098769 has  0 palindromes of length 2 or more: 
    123456790165432098769 has  0 palindromes of length 2 or more: 
     83071934127905179083 has  0 palindromes of length 2 or more: 
1320267947849490361205695 has  3 palindromes of length 2 or more: 202 494 949
    amanaplanacanalpanama has 12 palindromes of length 2 or more: aca ama ana nacan anacana lanacanal planacanalp aplanacanalpa naplanacanalpan anaplanacanalpana manaplanacanalpanam amanaplanacanalpanama

Wren

Library: Wren-seq
Library: Wren-fmt
Library: Wren-sort
import "./seq" for Lst
import "./fmt" for Fmt
import "./sort" for Sort

var substrings = Fn.new { |s|
    var ss = []
    var n = s.count
    for (i in 0...n) {
        for (j in 1..n-i) ss.add(s[i...i + j])
    }
    return ss
}

System.print("Number  Palindromes")
for (i in 100..125) {
    var pals = []
    var ss = substrings.call(i.toString)
    for (s in ss) {
        if (s == s[-1..0]) pals.add(s)
    }
    pals = Lst.distinct(pals)
    var cmp = Fn.new { |i, j| (i.count - j.count).sign }
    Sort.insertion(pals, cmp) // sort by length
    Fmt.print("$d   $3s", i, pals)
}

var nums = [
    "9", "169", "12769", "1238769", "123498769", "12346098769", "1234572098769",
    "123456832098769", "12345679432098769", "1234567905432098769", "123456790165432098769",
    "83071934127905179083", "1320267947849490361205695"
]
System.print("\nNumber                    Has no >= 2 digit palindromes")
for (num in nums) {
    var ss = substrings.call(num).where { |s| s.count > 1 }
    var none = !ss.any { |s| s == s[-1..0] }
    Fmt.print("$-25s $s", num, none)
}
Output:
Number  Palindromes
100     1   0  00
101     1   0 101
102     1   0   2
103     1   0   3
104     1   0   4
105     1   0   5
106     1   0   6
107     1   0   7
108     1   0   8
109     1   0   9
110     1   0  11
111     1  11 111
112     1   2  11
113     1   3  11
114     1   4  11
115     1   5  11
116     1   6  11
117     1   7  11
118     1   8  11
119     1   9  11
120     1   2   0
121     1   2 121
122     1   2  22
123     1   2   3
124     1   2   4
125     1   2   5

Number                    Has no >= 2 digit palindromes
9                         true
169                       true
12769                     true
1238769                   true
123498769                 true
12346098769               true
1234572098769             true
123456832098769           true
12345679432098769         true
1234567905432098769       true
123456790165432098769     true
83071934127905179083      true
1320267947849490361205695 false