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 Like CABACB (Posted on 2017-10-08)
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".

 No Solution Yet Submitted by Ady TZIDON No Rating

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 A solution for k=26 Comment 2 of 2 |
`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

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