Abacaba D Abacaba

• Let $f(n)$ denote the sum of all the numbers in sequence $n$ .
• Let $\mu(n)$ be the arithmetic mean of sequence $n$ .
• Be aware that $f(n)=2f(n-1)+n$ $\begin{array}{l} f(0)&=&0&=&0\\ f(1)&=&1&=&1\\ f(2)&=&1+2+1&=&4\\ f(3)&=&1+2+1+3+1+2+1&=&11\\ f(4)&=&\cdots&=&26\\ f(5)&=&\cdots&=&57\\ f(6)&=&\cdots&=&120 \end{array}$
• To find a pattern, the first thing I'd do is the difference method. (This name is invented by me. I am referring to the section "Second Differences" on this page .)
• Let $f'(n)$ denote $f(n+1)-f(n)$ and $f"(n)$ denote $f'(n+1)-f'(n)$ . $\begin{array}{l} f'(0)&=&1-0&=&1\\ f'(1)&=&4-1&=&3\\ f'(2)&=&11-4&=&7\\ f'(3)&=&26-11&=&15\\ f'(4)&=&57-26&=&31\\ f'(5)&=&120-57&=&63\\ \end{array}$ $\begin{array}{l} f"(0)&=&3-1&=&2\\ f"(1)&=&7-3&=&4\\ f"(2)&=&15-7&=&8\\ f"(3)&=&31-15&=&16\\ f"(4)&=&63-31&=&32\\ \end{array}$ Now we are starting to see a pattern. $f"(n)=2^{n+1}$ $\begin{array}{rcl} f'(n)&=&1+\sum_{k=0}^{n-1}2^{k+1}\\ &=&1+\sum_{k=1}^n2^k\\ &=&1+\sum_{k=0}^n2^k-1\\ &=&2^{n+1}-1 \end{array}$ $\begin{array}{rcl} f(n)&=&\sum_{k=0}^{n-1}(2^{k+1}-1)\\ &=&\sum_{k=1}^{n}(2^k-1)\\ &=&\sum_{k=1}^{n}2^k-n\\ &=&\sum_{k=0}^{n}2^k-1-n\\ &=&2^{n+1}-2-n\\ \end{array}$ $\begin{array}{rcl} \mu(n)&=&\frac{2^{n+1}-2-n}{2^n-1}\\ &=&2-\frac n{2^n-1} \end{array}$ Since $\lim_{n\rightarrow\infty}\frac n{2^n-1}=\frac1{(\ln2)2^n}=0$ , the Cesaro Summation is 2, or $\frac21$ .

Since we are taking the arithmetic mean, we can "distribute" values in order to "flatten" the numbers as such:

$[1, 2, 1, 3, 1, 2, 1] \Rightarrow [1, 2, \textbf{2}, \textbf{2}, 1, 2, 1]$ , this resultant sequence having the same arithmetic mean. Here I subtracted $1$ from the center $3$ and added $1$ to the $2$ to its left, keeping the arithmetic mean the same.

$[1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1] \Rightarrow [1, 2, 2, 2, 1, 2, \textbf{2}, \textbf{2}, \textbf{2}, 2, 2, 2, 1, 2, 1]$ .

In the next iteration, the sequence will be repeated on both sides so that the center number will be $5$ , and so there will be $3$ $1$ 's that will be turned into $2$ 's. We can generalize this for any $n$ . Call the amount of $1$ 's at the $n$ th iteration (The first iteration being the sequence $[1]$ ) $A_n$ and the amount of $2$ 's $B_n$ .

$A_1 = 1, A_n = 2A_{n-1} - n + 2$ for $n > 1$ . In closed form: $A_n = n$

Because the total digits at $n$ is $2^n + 1$ , $B_n = 2^n + 1 - A_n = 2^n - n + 1$ .

Since $B_n$ grows exponentially while $A_n$ grows linearly, the arithmetic mean approaches $2 = \frac{2}{1}$ , so the answer is $2 + 1 = 3$ .

There exists a much better method than this one that a friend of a friend of mine found, but since he found it and not I, I will not post it.