Permutapalindromic numbers
05AB1E, 15 14 13 bytes
Saved a byte thanks to Emigna! Code:
µNœvyJÂïÊP}_½
Explanation:
µ # c = 0, when c is equal to the input, print N.
N # Push N, the iteration variable.
œ # Push all permutations of N.
vyJ } # For each permutation...
 # Bifurcate, which is short for duplicate and reverse.
ï # Convert the seconds one to int, removing leading zeros.
Q # Check if they are not equal.
P # Product of the stack.
_ # Logical not.
½ # Pop a value, if 1 then increase c by 1.
Uses the CP-1252 encoding. Try it online!.
Brachylog, 19 bytes
~l<:1at.
.=pPrPl~l?
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Takes about 17 seconds for N = 270
.
Explanation
Main predicate:
~l Create a list whose length is Input. < The list is strictly increasing. :1a Apply predicate 1 to each element of the list. t. Output is the last element of the list.
Predicate 1:
.= Input = Output = an integer pPrP A permutation P of the Output is its own reverse l~l? The length of P is equal to the length of the Input
Pyth, 14
e.ff&_ITshT.p`
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Expansion:
e.ff&_ITshT.p`ZQ # Auto-fill variables
.f Q # Find the first input number of numbers that give truthy on ...
.p`Z # Take all the permutations of the current number
f& # Keep those that give a truthy value for both:
_IT # Invariance on reversing (is a palindrome)
shT # The integer value of the first digit (doesn't start with zero)
# A list with any values in it it truthy, so if any permutation matches
# these conditions, the number was a permutapalindrome
e # Take only the last number