Such numbers have considerable psycho-mathematical significance
 
 

Fascinating Palindromes
 

There are many fascinating palindromes involving consecutive digit sequences.
For example in base 10 we can construct palindromes with consecutive digits starting
with 1.

For example 121 is such a palindrome. It is divisible by 11 with the result = 11 which of course is also a palindrome.

(Note that in this example the highest digit 2 is even.)

With the requirement that the highest digit is even the next such palindrome is 1234321. Again it is divisible by 11 with the result = 112211.

Continuing on in this fashion the next such palindrome is 12345654321. Again it is divisible by 11 with the result = 1122332211.

This pattern will repeat itself with the final single even digit in base 10 as highest digit.

Thus 123456787654321 divided by 11 = 11223344332211.

There are several interesting features here.

Firstly - provided that the highest digit in the consecutive number sequence is even -these palindromes are always divisible by 11.

Secondly the resulting number is itself palindromic.

Thirdly there is a remarkable symmetry to these resulting palindromes. They are made up of sequences of consecutive digits - starting with I - with each digit occurring twice.

Also the highest digit in the resulting palindrome is always half of the highest digit in
the original.

This number behaviour can be extended to all number bases > 2.

In addition these resulting palindromes are in themselves again always divisible by 11 yielding fresh additional palindromes.

For example 112211 is divisible by 11 yielding 10201.

1122332211 is divisible by 11 giving 102030201.

These numbers contain the same digits in the same consecutive sequence as their source palindromes with each digit occurring only once in sequence and being always preceded by a '0'.
 
 

Another fascinating aspect to these original palindromes is that they all have an integral square root which is itself palindromic and consisting of a series of I's.

Thus the square root of 121 is 11, of 12321 is 111, of 1234321 is 1111 etc.

If the original palindrome has k digits, the square root will have (k + 1)/2 digits each being '1'.

Again this result can be extended to all number bases.
 

Yet again when we multiply these original palindromes by 11, we obtain once more highly symmetrical fresh palindromes.
 

121 * 11 = 1331; 112321 * 11 = 135531; 1234321 * 11 = 13577531, and 123454321 * 11 = 1357997531.

Here the resulting palindromes consist of an orderly ascending - descending sequence of the odd digits. Each odd digit corresponds in its ordinal ranking to the digits in the original palindrome, with the highest odd digit (which is n - 1) occurring twice.

This result extends to all number bases with the highest permitted odd digit = n - 1. Thus in hexadecimal, 123456787654321 * 11 = 13579ACEECA97531.
 

A variation on the above is to repeat the middle digit of the original ascending -descending sequence twice so that all digits in the palindrome occur twice. Thus - in base 10 - starting with 1221 we again find it is divisible by 11 giving 111.

The next palindrome in the sequence - this time with the digit being odd or even - is 123321.

Again it is divisible by 11 with the result 11211.

The next palindrome is 12344321. Again it is divisible by 11 with the result 1122211. The next palindrome is 1234554321. It is divisible by 11 with the result 112232211.

Once more there is a remarkable pattern unfolding.

Such palindromes are always divisible by 11. The resulting number in turn is always palindromic with a predictable structure.

For numbers with highest digit - in consecutive sequence - of original palindrome odd, the highest digit of resulting palindrome will be next even number divided by 2. All digits will be in consecutive sequence, and with exception of highest digit (which occurs once) repeat twice.

When the highest digit of original palindrome is even, the highest digit of resulting palindrome will be half this number. Again all digits will be in consecutive sequence and occur twice except for highest which occurs three times.

Again these results can be extended to all base numbers > 2.

However in these cases the resulting palindromes are not themselves divisible by 11.

Many other fascinating examples exist.

For example working with only two digits 0 and 1, if we square 101 we obtain 10201.

Next taking 10101 and squaring we obtain 102030201.

This symmetrical pattern keeps repeating with 1010101² = 1020304030201. Here we have (in the resulting palindrome,) an ascending - descending sequence of consecutive natural digits, each preceded by 0. The highest digit in the resulting palindrome will be equal to the number of l's in the original palindrome.

A similar pattern will result when we increase the frequency of 0's in the original number.

Thus 1001² = 1002001; 10001² = 100020001.

The number of 0's preceding the natural digits in resulting palindromes will equal the number of 0's in original source palindromes.
 
 

Frequency of Palindrome Distribution
 

There are simple formulae which can be given for the frequency of palindrome distribution in any base for numbers with any given level of digits.

If a number is in a given base system n and with k digits then the frequency of digits is given by the formulae;

(n - 1)^k/2 where k is even; (n - 1)^(k+1)/2 where k is odd.
 

Thus when the base is 10 and we are dealing with one digit numbers we will obtain 9 palindromes (i.e. all the natural digits from 1 - 9.)

With two digit numbers again there are 9 palindromes (i.e. 11, 22,……, 99.)

With three digit numbers there are 9² (i.e. 81) palindromes.

With four digit numbers there will again be 81 palindromes.

Finally - to illustrate - for both 9 and 10 digit numbers (in base 10,) there will be 9⁵ (i.e. 59049) palindromes.

The total number of palindomes of 10 digits or less is therefore 2(9 + 9² + 9³ + 9⁴ + 9⁵) which is 132858.

Of course we can extend these results to any base.

For example in base 2, with four digits we would expect just 2 palindromes. (These are in fact 1001 and 1111, which are 9 and 15 in base 10.)
 

 We can also derive formulae for the frequency of reverse numbers in any number base. Of course every number has a reverse, but some of these are not unique (i.e. palindromes.) Also in many cases the reverse of a number of k digits will be a number with less than k digits.

So what we are interested in determining is for any number of k digits the frequency of unique reverse numbers also of k digits.

Thus in terms of one digit numbers in any base there are no unique reverse numbers.

In the case of two digit numbers, we have to exclude the number of palindromes and all reverses that are in fact one digit numbers (e.g. the reverse of 80 is 08 which is a one digit number.)

In the case of base 10 two digit numbers we have 90 - 9 - 9 unique two digit reverse

numbers = 72 = 12 * 6.

In the case of three digit numbers - also in base 10 - we have 900 - 81 - 99 unique three digit reverses = 720 = 12 * 60.

In the case of four digit numbers in this base we have 9000 - 81 - 999 unique four digit reverses = 7920 = 12 * 660.

In the case of five digit numbers we nave 90000 - 729 - 9999 unique five digit reverses = 79272 = 12 * 6606

In general for a number in any base n with k digits the number of unique reverses of k digits is given by
 

(n - 1) *10^k-1 - [(n - 1)^k/2 + (10^k-1 - 1)] where k is even, and

(n - 1) * 10^k-1 -[(n - 1)^(k+1)/2 + (10^k-1 - 1)] where k is odd

There are general formulae which give the frequency of palindromes for all numbers

in any base n less than or equal to k digits.

If k is even, the formula = 2(n - l)[(n - 1)^k/2 - 1]
                                                          n - 2
 

If k is odd, formula = 2(n - l)[(n - 1)^(k+1)/2 - 1] - (n - 1)^(k+1)/2
                                                       n - 2
 
 

Thus the number of palindromes in base 10 with 4 digits or less

When n =2, formula = k[(k/2) + l]/2, where k is even,