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Michael Hardy
Michael Hardy
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Let $(a_{n}a_n)_{n\in N}=(1,2,3,4,6,8,9,12,\cdots)$ list the set$\{2^n3^m\mid m,n\in N\}$. Find α$α$ such that $(a_n)\alpha\pmod1$ is not equidistributed

Let $$(a_{n})_{n \in \mathbb{N}} = (1,2,3,4,6,8,9,12,16,18,\cdots)$$$$(a_n)_{n \in \mathbb{N}} = (1,2,3,4,6,8,9,12,16,18,\cdots)$$ be a sequence that is a listing of the set $$\{2^n3^m \mid m,n \in \mathbb{N}\}$$ We need to find an irrational number $\alpha$ such that $(a_n)\alpha\pmod1$ is not equidistributed.

I have a rough idea about it. I think that it may be related to normal number base 2$2$ and normal number base 3,$3,$ because we know that let $x \in [0,1]$, then $x$ is normal in base $2$ if and only if $2^n x$ mod $1$$2^n x \bmod 1$ is equidistributed. And same arguments work for numbers normal in base $3$. But I'm not sure how to proceed from here. Could anyone help me?

Let $(a_{n})_{n\in N}=(1,2,3,4,6,8,9,12,\cdots)$ list the set$\{2^n3^m\mid m,n\in N\}$. Find α such that $(a_n)\alpha\pmod1$ is not equidistributed

Let $$(a_{n})_{n \in \mathbb{N}} = (1,2,3,4,6,8,9,12,16,18,\cdots)$$ be a sequence that is a listing of the set $$\{2^n3^m \mid m,n \in \mathbb{N}\}$$ We need to find an irrational number $\alpha$ such that $(a_n)\alpha\pmod1$ is not equidistributed.

I have a rough idea about it. I think that it may be related to normal number base 2 and normal number base 3, because we know that let $x \in [0,1]$, then $x$ is normal in base $2$ if and only if $2^n x$ mod $1$ is equidistributed. And same arguments work for numbers normal in base $3$. But I'm not sure how to proceed from here. Could anyone help me?

Let $(a_n)_{n\in N}=(1,2,3,4,6,8,9,12,\cdots)$ list the set$\{2^n3^m\mid m,n\in N\}$. Find $α$ such that $(a_n)\alpha\pmod1$ is not equidistributed

Let $$(a_n)_{n \in \mathbb{N}} = (1,2,3,4,6,8,9,12,16,18,\cdots)$$ be a sequence that is a listing of the set $$\{2^n3^m \mid m,n \in \mathbb{N}\}$$ We need to find an irrational number $\alpha$ such that $(a_n)\alpha\pmod1$ is not equidistributed.

I have a rough idea about it. I think that it may be related to normal number base $2$ and normal number base $3,$ because we know that let $x \in [0,1]$, then $x$ is normal in base $2$ if and only if $2^n x \bmod 1$ is equidistributed. And same arguments work for numbers normal in base $3$. But I'm not sure how to proceed from here. Could anyone help me?

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Miranda
Miranda
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Miranda
Miranda
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Let $(a_{n})_{n\in N}=(1,2,3,4,6,8,9,12,\cdots)$ list the set$\{2^n3^m\mid m,n\in N\}$. Find α such that $(a_n)\alpha\pmod1$ is not equidistributed

Let $$(a_{n})_{n \in \mathbb{N}} = (1,2,3,4,6,8,9,12,16,18,\cdots)$$ be a sequence that is a listing of the set $$\{2^n3^m \mid m,n \in \mathbb{N}\}$$ We need to find an irrational number $\alpha$ such that $(a_n)\alpha\pmod1$ is not equidistributed.

I have a rough idea about it. I think that it may be related to normal number base 2 and normal number base 3, because we know that let $x \in [0,1]$, then $x$ is normal in base $2$ if and only if $2^n x$ mod $1$ is equidistributed. And same arguments work for numbers normal in base $3$. But I'm not sure how to proceed from here. Could anyone help me?