Fascinating Palindrome Connection (1)

I wish to correct a recent assertion regarding the frequency of self generating numbers in different number bases.

What I mean in this context is a number that when multiplied by a positive integer (> 1) = its reverse.

So for example in base 8, 25 (original starting number) * 2 = 52 (i.e. its reverse number).

Now I claimed that where such a 2-digit number exists (in an appropriate base), that one example will then occur for every higher digit number (in the same number base).

However I have since discovered, through an interesting number connection, that this is not in fact strictly true.

Once again when we start with a positive integer n (> 1), then the reciprocal of n will result in a unique repeating 2-digit decimal sequence in the corresponding number bases

2n – 1, 3n – 1, 4n – 1, …    

So again for example when n = 3, then 1/n results in this unique 2-digit sequence in base 5, 8, 11, …

And then when we take these 2 digits as the original starting number and subtract it from its reverse, the same starting number will result.

Alternatively, when we multiply the starting number by 2, we obtain its reverse in these number bases.

Thus in base 5, 13 * 2 = 31; in base 8, 25 * 2 = 52; in base 11, 37 * 3 = 73, and so on.

However there is another revealing perspective with respect to these numbers.

Thus when we divide each of these numbers by the number that is one less than the number base in question i.e. 2n – 2, 3n – 2, 4n – 2, … respectively we obtain 2, 3, 4, …

Now once again we are excluding the 2-digit number in base 2 i.e. 01, as the first digit as a 0 is strictly redundant.

So 2 * 4 = 13 (in base 5); 3 * 7 = 25 (in base 8); 4 * 10 = 37 (in base 11) and so on.

Now a corresponding higher digit numbers (where starting number * 2 = its reverse number) can now be created with respect to the original starting number provided that its palindrome nature is preserved and where the inclusion of one or more zeros is allowed (between first and last digits).

So when the palindrome has 2 digits, then only one possibility arises.

Thus in base 5, 22 * 4 = 143 and 143 (starting number) * 2 = 341 (its reverse number).

In like manner, in base 8,  33 * 7 = 275 and 275 (starting number) * 2 = 572 (its reverse number).

And in base 11, 44 * 10 = 3A7 and 3A7 (starting number) * 2 = 7A3 (its reverse number).

Thus there is indeed only one example of a self-generating number (of this type) with respect to all 3-digit numbers in bases 5, 8, 11, …

However when the palindrome now has 3 digits, two possibilities exist

So with respect for example to base 8, we can have 333 or 303.

Thus in base 8, we have as our possible starting numbers 333 * 7 = 2775 and 2775 (starting number) * 2 = 5772 (its reverse number).

However, we also have 303 * 7 = 2525 and 2525 (starting number) * 2 = 5252 (its reverse number).

Thus for 4-digit numbers in bases 5, 8, 11, …, we have 2 examples of such self-generating numbers (i.e. where the starting number * 2 = its reverse number).

Now when the initial palindrome has 4 digits, again illustrating with respect to base 8, we have just two possibilities i.e. 3333 or 3003.

And in base 8, 3333 * 7 = 27775 and 27775 (starting number) * 2 = 57772 (its reverse).

However, we also have 3003 * 7 = 25025 and 25025 (starting number) * 2 = 52052 (its reverse).

So for 4 and 5-digit numbers, we have two examples in each case of such self-generating numbers.

Then when we go to 6 and 7-digit numbers, we have three examples in each case, with 8 and 9-digit numbers 4 examples, with 10 and 11-digit numbers five examples and so on.

Thus when k is an even integer we have k/2 examples of such numbers in the relevant number bases.

And when k is an odd integer, we have (k – 1)/2 examples.

And the cumulative number of such numbers up to and including k digits is k/2(k/2 + 1), where k is odd and k/2 * k/2 where k is even.

Therefore, for example the total number of such self generating numbers up to and including 6 digits (i.e. where the reverse is twice the original starting number), in the appropriate number bases  5, 8, 11, ..., = 3 * 3 = 9.