If we start
with our customary base 10 system, through perhaps somewhat trivial, all of the
single-digit numbers from 1 - 9 (inclusive) can be viewed as palindromes.

The number
9 for example in clearly the same whether digit(s) are read from left to right
(or alternatively right to left).

Then with
respect to 2-digit numbers from 11 – 99 (inclusive) again we have 9 examples
(where the digits from 1 - 9 repeat).

Then with
respect to 3-digit palindromes, the 1

^{st}and last digits must be the same (as one of the 9 digits from 1 - 9). Then the middle digits can be any one of the 10 digits (from 0 to 9 inclusive).
Thus with
respect to 3-digit numbers, we have an additional 90 examples.

Then it is
just the same with respect to 4 digit numbers where the two middle digits must
be the same leaving again 10 options from 00 to 99.

Then with
5-digit numbers we will have 900 additional examples and another 900 with 6
digit numbers.

So we have 9 + 9 + 90 + 90 + 900 + 900 + …

= 18 + 180
+ 1800 + …

= 2 * 9(1 +
10 + 10

^{2}+ …)
Thus if we
let x = base number (which in this case = 10), then we have

2 (x –
1)(1 + x + x

^{2}+ …) = – 2(1 – x)( 1 + x + x^{2}+ …)
Therfore
from 1-digit to n-digit numbers (where n is even), the total no. of palindromes
is

2(x

^{n/2}– 1).
So where n = 6, we have

2(x

^{3 }– 1) which when x = 10, gives 2 * 999 = 1998 (i.e. 9 + 9 + 90 + 90 + 900 + 900).
When n is odd we get,

2{x

^{(n + 1)/2}} – {x – 1}x^{(n }^{–}^{ 1)/2}
So with n =
5 we get 2(x

^{3 }– 1) – {x – 1}x^{2}.
So again with x = 10 (as number base) we obtain

1998 – 900 = 1098 (i.e. 9 + 9 + 90 + 90 + 900).

Again this expresses the frequency of all palindromes for numbers up to 5 digits i.e. from 1 - 99999 (inclusive).

The 2nd part of the formula {x – 1}x

Then {x – 1}x

Thus when n = 6, the no. of palindromes (in base 10) = 9 * 10

^{}Again this expresses the frequency of all palindromes for numbers up to 5 digits i.e. from 1 - 99999 (inclusive).

The 2nd part of the formula {x – 1}x

^{(n }^{–}^{ 1)/2 }expresses the narrower notion of the number of n digit palindromes (where n is odd).Then {x – 1}x

^{(n }^{–}^{ 2)/2}^{ }expresses the corresponding notion of the number of n digit palindromes (where n is even).Thus when n = 6, the no. of palindromes (in base 10) = 9 * 10

^{2 }= 900.
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