Infinite Arithmetic Progression
For a positive integer $n$, let $S_n=\{m\in\mathbb N: n|\tau(m)\}$. Find all $n$ so that $S_n$ has an infinite arithmetic progression as a sub-sequence.
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Fm Jakaria
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Re: Infinite Arithmetic Progression
I think all n works. Obviously n = 1 works. Suppose n > 1.
Now fix any prime p and set $s_m = p^{n-1}(p^{n-1}m + 1)$. Obviously $S = (s_m, m \in \mathbb{N})$ is an arithmetic progression with initial term $s_1$ and difference $s_2-s_1$. As $\tau$ is a multiplicative function, it follows that $S$ satisfies the desired conditions, being a subsequence of $S_n$.
Now fix any prime p and set $s_m = p^{n-1}(p^{n-1}m + 1)$. Obviously $S = (s_m, m \in \mathbb{N})$ is an arithmetic progression with initial term $s_1$ and difference $s_2-s_1$. As $\tau$ is a multiplicative function, it follows that $S$ satisfies the desired conditions, being a subsequence of $S_n$.
You cannot say if I fail to recite-
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.