Non-palindromic numbers
C, 82 bytes



p(n,a,b,c,r){c=0;for(b=1;++b&lt;n-2;c+=r==n)for(a=n,r=0;a&gt;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&gt;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).