Let q(ltr) represent a numerical value of the position in English alphabet of the letter ltr (ltr being one of the letters in ABC}.
So q(A)=1 , q(B)=2 , …q(K)=11 , …q(Z)=26.

Let's take 2k letters, 2 of each of the first k letters of the English ABC.
Now create a string in which between each of the identical letters say each of ltr there are q(ltr) other letters. The title string and its reversal - qualify as such concatenations for k=3.

Not for every k such solutions exist (try e.g. for k=5...).

It is impossible to create such a string for k=26 (either trust me or explain why), but adding one extra symbol, say #,- which will be counted as a letter with q(#)=0 - it can be done (try for k=5 as well)

Solve for k=26 adding "#" to 2*"ABC string".

13 14 15 16 17 18 19 20 21 22 23 24 25 26 13 11 14 01 15 01 16 07 17 09 18 05 19 11 20 07 21 05 22 09 23 08 24 10 25 12 26 03 06 00 08 03 04 02 10 06 02 04 12

 M  N  O  P  Q  R  S  T  U  V  W  X  Y  Z  M  K  N  A  O  A  P  G  Q  I  R  E  S  K  T  G  U  E  V  I  W  H  X  J  Y  L  Z  C  F  #  H  C  D  B  J  F  B  D  L

Note the beginning string 13 through 26  leaves a lot of spaces near the middle for most of the remaining odd numbers, the smaller evens are toward the end with a little help from the 3 and the 0 which is at the 44th position.  See my other post for why this is an even-numbered position.

Posted by Jer on 2017-10-09 13:25:55