3

Looking at primes mod 3 starting at 5 we get a sequence 2 1 2 1 2 1 2 2 ... Counting the number of 1's minus the number of 2's we get a second sequence -1 0 -1 0 -1 0 -1 -2 .. . Then from the first sequence we derive a third sequence 1 2 1 2 1 2 2 .. by dropping the first term. We then add the second sequence to the third mod 3 and obtain the sequence

0 2 0 2 0 2 1 2 0 2 0 2 1 0 1 2 1 2 0 2 1 2 1 ..

My question is why is 0 never next to another 0 or 1 next to another 1 or 2 next to another 2. I have checked this for primes less than 200 thousand.

Stephen Wynn
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  • Any time you check something is true for thousands or millions of numbers there's always the possibility that a counterexample exists just beyond the range you checked. But in such a case, I would ask if maybe there's some very basic reason why it could be true for all numbers. – Robert Soupe May 15 '18 at 16:23

1 Answers1

3

Let $p_1,p_2,\dots$ the prime numbers starting at $5$. Then the final sequence is just

$$\quad p_1+p_2,\quad p_1+p_2+p_3,\quad p_1+p_2+p_3+p_4,\dots\pmod{3}$$

This is because the second sequence is $p_1,p_1+p_2,p_1+p_2+p_3,\dots\pmod3$ (note that $2\equiv-1\pmod3$), and the third is $p_2,p_3,p_4,\dots\pmod{3}$. Then two consecutive terms in the sequence cannot be equal, because in such a case

$$p_1+\cdots+p_k\equiv p_1+\cdots+p_k+p_{k+1}\pmod3 \implies p_{k+1}\equiv0\pmod{3}$$

but $3$ does not divide $p_{k+1}$ since it is prime.

user246336
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