When n =
4, a unique repeating 2-digit sequence applies to 1/n in bases 2n –
1, 3n – 1, 4n – 1, …, so

that the original starting number (using these 2 digits) * 3 = its reverse number.

that the original starting number (using these 2 digits) * 3 = its reverse number.

And just as the k

^{th}term in the previous case “Fascinating Palindrome Connection 1”, where n = 3, is given by 3n^{2}– 2n, the k^{th}term (where n = 4) is given by 4n^{2}– 2n, resulting in the series,
2, 12, 30, 56, 90, …

So the relevant number bases, where this form of self-generating
behaviour applies is in the number bases 7, 11, 15, … (Once again we omit
the number base where n –
1 = 3, applying to the 1

^{st}term of the series i.e. 2, as this results in a redundant 1^{st}digit of 0 in the original starting number).
However the 2

^{nd}term, applying to base 7 is fully valid.
And 12 in base 7 = 15.

Therefore, 15 * 3 = 51 (in base 7).

However this can equally be expressed as 2 * (2n – 2) = 2 *
6 = 12 (in base 10), i.e. 15 in base 7.

So the subsequent 2-digit starting terms are given by 3 * 10
(= 28 in base 11), 4 * 14 (= 3B in base 15), 5 * 18 (= 4E in base 19) and so
on.

Thus 28 * 3 = 82 (in base 11)

3B * 3 = B3 (in base 15)

4E * 3 = E4 (in base 19)

…

Then one
3-digit term in each case is obtained from 22 * 6 in base 7, 33 * 10 (in base
11), 44 * 14 (in base 15), 55 * 18 (in base 19) and so on.

And as we
saw in “Fascinating
Palindrome Connection 1” one can then use the formulas k/2(k/2 + 1),
where k is odd and k/2 * k/2 where k is even, to calculate the total
number of self-generating numbers (of a particular type) up to (and including) k
digits.

So for example the total collection of self-generating numbers (where
the original starting number * 3 = its reverse) -up to an including 4 digits - in
each relevant number base, is

4/2 * 4/2 = 4.

And in base 7, these are 15, 165, 1515 and 1665 respectively.

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