Given a number, find the next palindrome on which this next palindrome is larger than the given number.

nextSmallestButLargerPalindrome.js

//We can check that the Recursive Version of Palindrome is way faster...
function palindromeRecursive(n) {
var nArr = n.split('');
  if(nArr.length <= 1) {
     return true;
  }
  else {
     if(nArr[0] === nArr[nArr.length-1]){
       return palindromeRecursive(n.slice(1, nArr.length-1));
     } else {
     	 return false;
     }
  }
}

function isPalindrome(word) {
  var wordArray = word.split('');
  var left = 0;
  var right = wordArray.length-1; 
  while (left < right) {
    if(wordArray[left] !== wordArray[right]) {
      return false;
    }
    left++;
    right--;
  }
  return true;
}

function nextPalLargerThanNumber (numberPal, callback) {
  console.time('how long?');
  var isPal = isPalindrome(numberPal.toString());
  while(!isPal) {
    numberPal++;
    isPal = isPalindrome(numberPal.toString());
  }
  console.timeEnd('how long?');
  console.log('next Palindrome:',numberPal);
  callback(numberPal);
}

function nextPalLargerThanNumberRecursive (numberPal) {
  console.time('how long RECURSIVE?');
  var isPal = palindromeRecursive(numberPal.toString());
  while(!isPal) {
    numberPal++;
    isPal = palindromeRecursive(numberPal.toString());
  }
  console.timeEnd('how long RECURSIVE?');
  console.log('next Palindrome:',numberPal);
  
}
var numberPal = 4545485848550; //with this one, should return 4554
nextPalLargerThanNumber(numberPal,nextPalLargerThanNumberRecursive);