The $n^{th}$ digits of $e+\pi$ and a periodic sequence

Question

Let $n$ be a positive integer greater than zero. I denote the $n^{th}$ digits of $e$ and $\pi$ by $e_n$ and $\pi_n$ respectively. Let $d(e_n+\pi_n)$ count the number of divisors of $e_n+\pi_n$ and set $$\alpha(n)=gcd\big(n,2^{d(e_n+\pi_n)}\big)$$

I want to prove and believe it to be true that $\alpha(n)$ is periodic with period $[1,2,1,4,1,2,1,8,1,2,1,4,1,2,1,16]$

I have verified the first 50 terms of $\alpha(n)$. A counter example would be nice. This is pure curiosity.

Answer 1

Your sequence is simply the largest power of $2$ dividing into $n$ up to $16$. It will fail when $e_n+\pi_n$ is prime and $n$ has a divisor of $8$ or $16$. It will also fail when $e_n+\pi_n=12$, which has $6$ factors, and $n$ is divisible by $32$. It will also fail when $e_n= \pi_n=0$ and $n$ is even. You just haven't looked far enough.

If you start counting with $e_1=7$, the digit behind the decimal point, your claim fails at $n=8$. We have $e_8=2, \pi_8=5,d(7)=2, \gcd=4$, not $8$ per your post. Please check your calculations. With the clarification that $n=1$ is the digit before the decimal point, it still fails at $n=32$. We have $e_{32}=6, \pi_{32}=5,d(11)=2, \gcd=4$, not $8$ per your post.

Answer 2

Not an answer but the reason why I believed it is true is based on the observation that of $n$ is odd then $\alpha(n)=1$. In fact we appear to have the following

• $\alpha(n)=1$ if $n$ is odd
• $\alpha(n)=2$ if $n=4k+2$
• $\alpha(n)=4$ if $n=8k+4$
• $\alpha(n)=8$ if $n=16k+8$
• $\alpha(n)=16$ if $n=16k$

This appears to be true for the first $50$ terms. That lead me to believe there was some structure here.

Answer 3

@rossmilman

You are correct, my claim is false. It holds only for the first 31 terms - where I had a hand written error. My bad ! I ran a code that calculated $\alpha(n)$ for the first $500\text{ }000$ digits of $e$ and $\pi$ and surely $\alpha(n)$ is never periodic ! I got the terms for $e$ and $\pi$ from NASA.