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.

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.

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