2

Let $A_i$ be the set of sums of consecutive numbers from $i$ to $i+n$ for each $n$, where the sum is greater than $2i+1$: $$A_i = \left\{ \sum_{k=i}^{i+n} k \mid n \in \mathbb{N}, \sum_{k=i}^{i+n} k \gt 2i+1\right\}$$ and $$A = \bigcup_{i=1}^\infty A_i$$

Then let $B$ be the set of all powers of two, greater than 2:

$$B=\{2^n\mid n\in\mathbb{N}, n \gt 1\}$$ and $$C = A \cup B$$ Now can we prove that $$\mathbb{P} = \mathbb{N} \ \backslash \ C$$ (where $\mathbb{P}$ is the set of all primes)?

Marijn
  • 1,027
  • 1
    Are you sure for $A$ ? Because it doesn't really mean something. Same for $B$. You can right $B$ like $$B={2^n\mid n\in\mathbb N}$$ or $$B={m\mid \exists n\in\mathbb N : m=2^n}$$ but not like you wrote it. – Surb Sep 28 '15 at 12:38
  • Your definition of $A_i$ is nonsense. It is very unclear what these sets are. – Crostul Sep 28 '15 at 12:43
  • @Surb Sorry, I'm still learning math.. Basically what I would like $A_i$ to be is the set of sums of consecutive numbers from $i$ to $i+n$ for each $n$ where the sum is greater than $2i + 1$. Does that make sense? And how would I write that down? I will edit $B$. Thanks – Marijn Sep 28 '15 at 12:52
  • Ok, I see. The answer is no, since $2\in B\cap\mathbb P$. – Surb Sep 28 '15 at 13:05
  • @Surb Ah yes, of course... I'll edit $B$ again to make $n > 1$. – Marijn Sep 28 '15 at 13:10
  • As long as n>1, the condition you stated for set $A_i$ is met. – porridgemathematics Sep 28 '15 at 13:49
  • $A-B$ in number theory often means ${a-b: a \in A, b \in B}$. – Aravind Sep 28 '15 at 13:58
  • @moorish 2+3+4=9 – Marijn Sep 28 '15 at 14:27
  • You are right, I will edit that bit out. Thanks. – porridgemathematics Sep 28 '15 at 14:33
  • Your question is the same thing as asking ' Is the set of all prime numbers equivalent to the set of all numbers that are not 2, not a power of 2, and cannot be written as the sum of three or more consecutive natural numbers. You can show that a number is prime if and only if it cannot be written as the sum of three consecutive integers. This should complete the proof. – porridgemathematics Sep 28 '15 at 14:38
  • 1
    Related but not identical. Somewhat similar reasoning though. – Jyrki Lahtonen Sep 28 '15 at 14:43
  • @moorish 4 cannot be written as the sum of three consecutive integers, same goes for 8.. – Marijn Sep 28 '15 at 14:48
  • You are right, I wasn't 100% clear, I should have said if and only if it cannot be written as the sum of three consecutive integers AND cannot be written as a power of 2. – porridgemathematics Sep 28 '15 at 14:55

1 Answers1

2

If $N$ is composite and not a power of 2, then $N=2^km$ where $m>1$ is odd. So we can write $N=\frac{(n+1)(n+2i)}{2}$, with $n+1=min\{2^{k+1},m\}$ and $n+2i=max\{2^{k+1},m\}$. It is easy to prove that primes cannot be written in this way, so you get: $\mathbb{P}=\mathbb{N} \setminus C$.

Aravind
  • 6,150