A counting sequence is a sequence whose terms are also sequences. To be specific, its terms are sequences of natural numbers. [And if you are familiar with the look and say sequence, its very similar to it.]
Here is an example. $$ T_1: (5,1,2)$$ $$ T_2: (1,1,1,2,1,5)$$ $$ T_3: (4,1,1,2,1,5)$$ and so on... where, $$T_n=n^{th} term$$ NOTICE that the terms are created by counting the natural numbers inside it and ordering them as you list them in the next term.
Here is my problem: I want to make a proof of the following statement.
If the 1st term of the sequence is $$(m,m+1)$$, where $m\ge7$, then its $(m+1)^{th}$ term is $(m,1,1,2,1,3,...,1,m+1)$.
An example for this statement is: $ T_1: (7,8)$
$ T_2: (1,7,1,8)$
$ T_3: (2,1,1,7,1,8)$
$ T_4: (3,1,1,2,1,7,1,8)$
$ T_5: (4,1,1,2,1,3,1,7,1,8)$
$ T_6: (5,1,1,2,1,3,1,4,1,7,1,8)$
$ T_7: (6,1,1,2,1,3,1,4,1,5,1,7,1,8)$
$ T_8: (7,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8)$ which shows that this example satisfies that statement.
Any help is much appreciated. I have tried using induction on it but I'm having trouble with the induction hypothesis part.
If $T_1:(m,m+1)$, clearly, $T_2: (1,m,1,m+1)$, and $T_3: (2,1,1,m,1,m+1)$. So, $T_3$ follows the pattern $T_{n}:(\underbrace{n-1,1}, \underbrace{1,2},\underbrace{1,3},\ldots,\underbrace{1,n-2},\underbrace{1,m},\underbrace{1,m+1})$.
Let us assume that the pattern holds for $n\leq m$, so that $n-1$ and $m$ are distinct. Here, the number 1 is repeated $n-2+2=n$; the numbers $2,3,\ldots,n-1$ are repeated once; the numbers $m,m+1$ are repeated once. Thus, \begin{equation} T_{n+1}:(n,1,1,2,1,3\ldots,1,n-1,1,m,1,m+1). \end{equation} Hence, by induction, we prove that $T_{n}:(n-1,1,1,2,1,3,\ldots,1,n-2,1,m,1,m+1)$, for $n\leq m+1$. Finally, substituting $n=m+1$, we get the deisred result.