Non-palindromic numbers

Question

A strictly non-palindromic number N is a number that isn't a palindrome in any base (in bases 2 to N-2). These numbers are listed on OEIS

For example, the number 19 in base 2,3,4,5,6,...17 is: 10011,201,103,34,31,...12. None of these representations is palindromic, so the number is strictly non-palindromic.

For this challenge, you need to return a truthy value if the number is non-palindromic, otherwise a falsy value.

• You may assume the number passed to you is greater than or equal to 0.
• Your program should work for values up to your languages' integer size.

Test cases:

Truthy:

0
1
2
3
4
6
11
19
47
53
79
103
389
997
1459

Falsy:

5
7
8
9
10
13
16
43
48
61
62
101
113
211
1361

This is a code-golf, so make your answers as short as possible!

Answer 1

C, 82 bytes

p(n,a,b,c,r){c=0;for(b=1;++b<n-2;c+=r==n)for(a=n,r=0;a>0;a/=b)r=r*b+a%b;return!c;}

Ideone it!

Explanation

This code reverses n in base b and stores in r:

for(a=n,r=0;a>0;a/=b)r=r*b+a%b;

The outer loop counts the number of bases from 2 to n-1 in which n is a palindrome.

If n is non-palindromic, the count would be 1 (n must be a palindrome in base n-1).

Answer 2

Python 2, 71 bytes

n=input();b=1
while b<n-2:
 m=n;r=0;b+=1
 while m/(r!=n):r=r*b+m%b;m/=b

Output is via exit code, where 0 is truthy and 1 is falsy. Test it on Ideone.

Answer 3

S.I.L.O.S, 206 bytes

GOTO e
lbld
c - 1
GOTO c
lble
readIO 
n = i
i - 3
b = i
b + 1
GOTO f
lbla
a = n
r = 0
lblb
m = a
m % b
r * b
r + m
a / b
if a b
r - n
r |
if r d
lblc
c + 1
i - 1
b - 1
lblf
if i a
c / c
c - 1
c |
printInt c

Try it online!

Port of my answer in C.

Answer 4

Haskell, 75 68 bytes

(a!c)b|a<1=c|x<-c*b+mod a b=div a b!x$b
f n=all((/=n).(n!0))[2..n-2]

Answer 5

Jelly, 9 bytes

bRµ⁼"US<3

Try it online! or verify all test cases.

How it works

bRµ⁼"US<3  Main link. Argument: n

 R         Range; yield [1, ..., n].
b          Convert n to all bases between 1 and n, yielding a 2D array A>
  µ        Begin a new, monadic chain. Argument: A
     U     Upend; reverse the 1D arrays in A.
   ⁼"      Zipwith equal; yield 1 for each array that matches its inverse.
      S    Sum; add the resulting Booleans.
           If n > 1, the sum will be 2 if n is strictly non-palindromic (it is only
           a palindrome in bases 1 and n - 1), and greater than 2 otherwise.
           For 0 and 1, the sum will be 0 (sum of the empty array) and 1 (only a
           palindrome in base 1); both are less than 2.
       <3  Compare the sum with 3, yielding the desired Boolean.

Answer 6

Pyth, 12 10 bytes

Saved two bytes with Dennis' trick.

>3sm_IjQdS

Try it online!

Explanation:

         S (Q)   Get all the bases we need by building a list from 1 to Q
   m               For all bases d in the bases list:
      jQd           cast Q to base d as a list
    _I              and check to see if the list is palindromic (invariant on reversal)
                  Compile all the results back into a list
  s                Sum the results (a shorter form of any), gives 3 or more for palindromics 
                    (2 is the usual because of bases 1 and Q-1)
>3                 And verify that the sum is greater than three to get non-palindromics

Answer 7

Mathematica, 58 43 bytes

!Or@@Table[#==#~IntegerReverse~i,{i,2,#-2}]&

TIL that #~IntegerReverse~i reverses the digits of the input when written in base i.

Answer 8

Brachylog, 14 bytes

¬{⟦₆bk∋;?ḃ₍.↔}

Try it online!

Outputs through predicate success or failure, which prints true. or false. if run as a program.

¬{           }    It cannot be shown that
        ?         the input
       ; ḃ₍       in a base
      ∋           which is an element of
  ⟦₆              the range from 1 to the input - 1
    b             without its first element
     k            or its last element
           .      can be unified with both the output variable
            ↔     and its reverse.

Answer 9

Perl6, 110 72 65

my &f={?all(map {{.reverse ne$_}(@(.polymod: $^a xx*))},2..$_-2)}

Couldn't use base since that's broken for any base above 36.

Previous attempts

my &a={$^a??flat($a%$^b,a($a div$b,$b))!!()};my &f=-> $n {?all(map {.reverse ne$_ given @(a($n,$_))},2..$n-2)}
my &f=->\n {?all(map {.reverse ne$_ given @(n.polymod: $_ xx*)},2..n-2)}

Answer 10

JavaScript (ES6), 83 bytes

f=(n,i=n-2,g=n=>n<i?[n]:[...g(n/i|0),n%i])=>i<2||`${a=g(n)}`!=a.reverse()&&f(n,i-1)

<input type=number oninput=o.textContent=f(this.value);><pre id=o>

Answer 11

PHP, 68 bytes

for($b=$argn;--$b;)strrev($c=base_convert($argn,10,$b))!=$c?:die(1);

takes input from STDIN, exits with 1 for falsy, 0 for truthy. Run with -R.

Answer 12

APL(NARS), chars 47, bytes 94

{⍵≤4:1⋄∼∨/{⍵≡⌽⍵}¨{⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}w}¨2..¯2+w←⍵}

where {(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵} would be the function conversion one positive omega in digits number base alpha, and {⍵≡⌽⍵} would be the function check palindrome... test:

  f←{⍵≤4:1⋄∼∨/{⍵≡⌽⍵}¨{⍵{(⍺⍴⍨⌊1+⍺⍟⍵)⊤⍵}w}¨2..¯2+w←⍵}
  f¨0 1 2 3 4 6 11 19 47 53 79 103 389 997 1459
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
  f¨5 7 8 9 10 13 16 43 48 61 62 101 113 211 1361
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

Answer 13

C, 129 bytes

f(n,b,k,j){int a[99];for(b=2;b+2<n;++b){for(j=0,k=n;a[j]=k%b,k/=b;++j);for(;k<j&&a[k]==a[j];++k,--j);if(k>=j)return 0;}return 1;}

Answer 14

C, 77 bytes

h(n,b,k,z){for(z=0,k=n;z+=k%b,k/=b;z*=b);return b+3>n?1:z==n?0:h(n,++b,0,0);}

recursive exercise...i change (b+2>=n) with (b+3>n) whithout debugging...

main()
{int  v[]={0,1,2,3, 4, 6,11,19,47,53,79,103,389,997,1459},
  n[]={5,7,8,9,10,13,16,43,48,61,62,101,113,211,1361}, m;
    // 0 1 2 3  4  5  6  7  8  9 10  11  12  13   14
 for(m=0; m<15; ++m)
    printf("%u=%u\n", v[m], h(v[m],2,0,0));
 for(m=0; m<15; ++m)
    printf("%u=%u\n", n[m], h(n[m],2,0,0));
}

/*
 77
 0=1
 1=1
 2=1
 3=1
 4=1
 6=1
 11=1
 19=1
 47=1
 53=1
 79=1
 103=1
 389=1
 997=1
 1459=1
 5=0
 7=0
 8=0
 9=0
 10=0
 13=0
 16=0
 43=0
 48=0
 61=0
 62=0
 101=0
 113=0
 211=0
 1361=0
*/