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".
I'm not going to bother with converting numbers to letters. Its easier to keep track of numbers.
The title then is the solution for k=3, 312132
for k=4, the solution is 41312432
For other values of k, there are only solutions for k=0 or 3 (mod 4)
Suppose k=1 mod 4, say 4n+1
There are then 8n+2 positions to fill by 2n+1 pairs of odd numbers and 2n pairs of even numbers.
The positions themselves can be numbered and then classified as even or odd  there being 4n+1 of each type.
An even pair of numbers will fill an even and an odd type, whereas an odd pair will fill either two odds or two evens.
Placing all of the even pairs will fill 2n of each type leaving 2n+1 odd positions and 2n+1 even positions. Since these are both odd numbers, the odd numbers cannot fill them in pairs.
Suppose k=2 mod 4, say 4n+2
There are 8n+4 positions to fill by 2n+1 pairs of odd numbers and 2n+1 pairs of even numbers.
The positions are 4n+2 of each type.
Placing all the even pairs leaves 2n+1 of each type. Again, the odd numbers cannot fill them.
Suppose k=3 mod 4, say 4n1
There are 8n2 positions to be filled by 2n pairs of odd numbers and 2n1 pairs of even numbers.
The positions are 4n1 of each type.
Placing all the even pairs leaves 2n of each type. This time the odd numbers can fill them (or at least I haven't proven they cannot be filled.) Here's a k=7 solution: 45671415362732.
k=0 mod 4 proof is similar to above.
k=26 is 2 mod 4 so impossible by the above reasoning. Adding a single 0 gives an extra position and it is of the oddtype. Placing the 0 in an even position makes two more odds than evens. The above proof of impossibility no longer holds. I haven't tried 26 plus 0, but here are k=1,2,5,6 rendered possible:
101
20121
52402354131
3456314152602
(There are other solutions for the latter two.)
This doesn't prove k=26 plus a 0 is possible, but there are likely many solutions.

Posted by Jer
on 20171009 11:34:59 