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Let $n$ be a positive integer greater than zero. I write

$$a_n = \begin{cases} 1 , &\text{ if } n=0 \\ 1 , &\text{ if } n=1 \\ n(n-1), & \text{ if $2n-1$ is prime} \\ 3-n, & \text{ otherwise} \end{cases}$$

The sequence goes like this $$1,1,2,6,12,-2,30,42,-5,72,90,-8,132,-10,-11,\ldots$$ I would like to prove the following two claims.

claim 1 : If $a_n>0$ and ${a_n \above 1.5 pt 3} \notin \mathbb{Q}$ then $\sqrt{4a_n+1}$ is prime.

The table below illustrates what I am seeing:

\begin{array}{| l | l | l | l } \hline n & a_n & {a_n \above 1.5 pt 3} & \sqrt{4a_n+1}\\ \hline 0 & 1 & .333333.. & 2.2360679.. \\ 1 & 1 & .333333.. & 3 \\ 2 & 2 & .666666.. & 3 \\ 3 & 6 & 2 & 5 \\ 4 & 12 & 4 & 7 \\ 6 & 30 & 10 & 11 \\ 7 & 42 & 14 & 13 \\ 9 & 72 & 24 & 17 & \\ 10 & 90 & 30 & 19 \\ 12 & 132 & 44 & 23 \\ 15 & 210 & 70 & 29 \\ 16 & 240 & 80 & 31 \\ 19 & 342 & 114 & 37 \\ 21 & 420 & 140 & 41 \\ 22 & 462 & 154 & 43 \\ \hline \end{array}

claim 2: If $a_n<0$ then $2a_n+5$ is composite

  • 2
    How can $\frac{a_n}{3}$ not be rational. $a_n$ is an integer – 2017-01-05
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    I am trying to say that $a_n/3$ is not a fraction? – 2017-01-05
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    If $a_n$ is an integer, then $a_n/3$ is 'not a fraction' if and only if $a_n$ is a multiple of $3$, ie, iff $a_n \in 3\mathbb{Z}$. – 2017-01-05
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    I think what you mean is: $${a_n \above 1.5 pt 3} \notin \mathbb{Z}$$ – 2017-01-05
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    If $a_0 = a_1 = 1$, why are the respective entries in the fourth column for $\sqrt{4a_n + 1}$ different? – 2017-01-05
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    The second claim is bound to be false, since it would mean that one of $2n-1$ or $2n+5$ is prime for all $n>3$, which would mean that for all odd number $x$, either $x$ or $x+6$ is a prime. – 2017-01-05
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    It seems to be true for the first several hundred terms, though ? – 2017-01-05
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    for claim $1$, all of the $a_n$ in your table are divisible by $3$ Indeed, $2n-1$ n odd prime implies $a_n=n(n-1)$ is divisible by $3$....so what does your condition mean? – 2017-01-05
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    @lulu The table is to support the claim. If $n=5$ then $a_n=-2$ and $-2$ is not divisible by $3$. – 2017-01-05
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    But Claim $1$ seems tautological....as $\sqrt {4a_n+1}=2n-1$ which you already assumed was prime. – 2017-01-05
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    @lulu Are you suggesting I am going in a circle with respect to claim 1. So it is true by "defeinition" ? – 2017-01-05
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    If $a_n>0$ and $n>3$ then that means $2n-1$ is prime (since $a_n>0$) and and $\sqrt{4a_n+1}=\sqrt{4n^2-4n+1}=2n-1$, which you already know is prime - the same prime. – 2017-01-05
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    @AnthonyHernandez Yes. Claim $1$ has no content at all. – 2017-01-05
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    OK that settles claim 1. – 2017-01-05
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    And Claim $2$ is false (as the counterexample $n=13$ shows). – 2017-01-05
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    Ok, post edit $n=13$ is no longer a counterexample, but $n=17$ is: we have $2\times 17-1=33$ and, as $33$ is composite, $a_{17}=3-n=-14$, yes? But then $2\times a_{17}+5=-23$ which is surely not composite. – 2017-01-05
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    Right so this question is circuitous? It is just going in circles? The claims are true Implicitly ? – 2017-01-05
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    @quasi, that should be *is* a member of, not *is not* a member of. If the intended predicate is "is not a fraction." – 2017-01-06
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    @Wildcard -- yes, thanks, but I can't edit my comment to fix it. – 2017-01-06

4 Answers 4

3

$n=13$ is a counter example for claim 2.

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    Not sure why this was downvoted. Seems like a perfectly good counterexample to me (+1). – 2017-01-05
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    Please see the update to the the second claim due to a typo. Sorry about that. – 2017-01-05
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    While it may be difficult to keep pace with the "update... due to a typo", a fuller statement of the conclusion and reasoning behind it would be a significant improvement as Answer, benefiting future Readers even if the Question changes further. – 2017-01-05
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    Given that if $a_n<0$ then $a_n=3-n$ and $2a_n+5=11-2n$, It is hard to see why this is composite. However, $5-2a_n=2n-1$ is known to be composite, which means $5+2|a_n|$ is composite. – 2017-01-05
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    @ThomasAndrews Good catch. I expect this is what the OP intended. Tautological, again. – 2017-01-05
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Ignore $n = 0, 1$ since they're kind of irrelevant. Then $a_n \leq 0$ for all $n$ unless $2n - 1$ is prime, by the definition. In that case $a_n = n(n - 1)$. So,

$$4a_n + 1 = 4(n^2 - n) + 1 = (2n-1)^2$$

So, then $\sqrt{4a_n + 1} = 2n - 1$, which is prime.

0

Just to clarify I have that both claims are trivially true and are explicit in the definition of $a_n$.

For claim 1 we have from @callus:

proof of claim 1: $a_n \leq 0$ for all $n$ unless $2n - 1$ is prime, by the definition. In that case $a_n = n(n - 1)$. So, $4a_n + > 1 = 4(n^2 - n) + 1 = (2n-1)^2$. So $\sqrt{4a_n + 1} = 2n - 1$, which is prime.

For claim 2 we have from @Thomas Andrews

proof of claim 2: Given that if $a_n<0$ then $a_n=3−n$ and $2a_n+5=11−2n$, It is hard to see why this is composite. However, $5−2a_n=2n−1$ is known to be composite, which means $5+2|a_n|$ is composite.

-1

Claim 1 is vaccuously true, since given any $n$, $a_{n}\in\mathbb{Z}$.

It also looks like n=14 is a counterexample to claim 2.

Proof: $3-14=-11$ and $2(-11)+5=-17$ has positive divisors $1$ and $17$ only.