Unique partial sum of a finite prime number power series

Question

Are there any quick and simple ways to prove or disprove that

$$ \sum_{\substack{k\in G_1\\ G_1 \subset\mathbb{N}_n}} p^k \neq \sum_{\substack{j\in G_2 \\ G_2 \subset(\mathbb{N}_n\setminus G_1)}} p^j $$

where $p$ is a prime number and $\mathbb{N}_n = \{1, 2, 3,\ldots, n\}$.

score 2 · Accepted Answer · answered Sep 19 '15 at 01:18

2

These are not power series but finite sums.

Here I'm assuming $G_1$ and $G_2$ are nonempty disjoint subsets of $\{1, \dots, n\}$.

Let $k_0$ be the minimal element in $G_1 \cup G_2$. Say $k_0 \in G_2$. Then $p^{k_0+1}$ divides the sum over $G_1$ but not the one over $G_2$, so the sums are different.

answered Sep 19 '15 at 01:18

Bruno Joyal

54,711

Do you mean $p^{k_0 + d}$ where $k_1 = k_0 + d$ is the smallest element of $G_1$ ? – Gong-Yi Liao Sep 19 '15 at 01:24
@Gong-YiLiao That's also true, $p^{k_0+1}$ is enough to tell the sums apart. – Bruno Joyal Sep 19 '15 at 01:25
I see, since $\sum_{j\in G_2} p^{j-k_0} + 1$ will not be equal to $p^d[(\sum_{k\in G_1} p^{k - k_0 - d}) + 1]$ – Gong-Yi Liao Sep 19 '15 at 01:34

score 0 · Answer 2 · answered Sep 19 '15 at 09:52

Assuming $ G_1$ and $G_2$ are not both empty, let $m= \max G_1 \cup G_2$. If $m=1$ then one of the sums is at least $p$,which is at least $2$ ,and we follow the convention that the empty sum is $0$. If $ m>1$ then one of the sums is at least$$ A= p^m$$ and the other sum is at most $$B=\sum_{j=1}^{j=m-1} p^j =(p^m-1)/(p-1).$$ Now $$ A>B$$ $$\text {iff } (p-1)p^m>p^m-1$$ $$\text{ iff } p^{m+1}-2 p^m>-1$$ $$\text{iff } p^2-2 p+1>1-p^{-(m-1)}$$ $$\text{ iff } (p-1)^2 > 1-p^{-(m-1)}$$ which holds because $p-1 \ge 1$ and $m-1>0$.Observe that this is all valid if $p$ is any real number not less than $2$.

Unique partial sum of a finite prime number power series

2 Answers2