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Set $a$ and $b$ so that for each $k$, with $ a < k \leq a + b $,

there exist $ x $ and $ y $ with $ x \geq 2 $ and $ y \geq 1 $ such that $ k = 6xy + x -y $

Is it possible to proof that, set $a$, there is a value of $b$ for which this cannot happen?

For example by proving that $ b> c $

with $c$ maximum number of $k$ possible values that can be obtained for $ a < k \leq a + b $

Edit

I checked for $a <10000$ that $b < \sqrt{a}$

in fact for $$a<k \leq a+\sqrt{a}$$ the number of possible values for $k=6xy+x-y$ are lower than $c$ and $c<\sqrt{a}$

$c = \sum \limits_{\substack{x=2 \\ 6x-1 \text { prime}}}^{\left\lfloor{\frac{a+\lfloor\sqrt{a}\rfloor+1}{7}}\right\rfloor} {\left\lfloor \frac{\displaystyle a+\lfloor\sqrt{a}\rfloor-x}{ \displaystyle 6x-1}\right\rfloor } - \sum \limits_{\substack{x=2 \\ 6x-1 \text { prime}}}^{\left\lfloor{\frac{a+1}{7}}\right\rfloor} { \left\lfloor \frac{\displaystyle a-x}{ \displaystyle 6x-1}\right\rfloor}+ \sum \limits_{\substack{z=1 \\ 6z+1 \text { prime}}}^{\left\lfloor{\frac{a+\lfloor\sqrt{a}\rfloor-6}{35}}\right\rfloor} { \left\lfloor \frac{\displaystyle a+\lfloor\sqrt{a}\rfloor-(5z+1)}{ \displaystyle 6(5z+1)-1}\right\rfloor } -\sum \limits_{\substack{z=1 \\ 6z+1 \text { prime}}}^{\left\lfloor{\frac{a-6}{35}}\right\rfloor} {\left\lfloor \frac{\displaystyle a-(5z+1)}{ \displaystyle 6(5z+1)-1}\right\rfloor }$

Is it possible to find a value of the upper limit of $c$ to verify it is always less than $\sqrt{a}$?

user140242
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    You have $$ (6x-1)(6y+1) = 6k-1 $$ and your conditions give both factors larger than $6.$ This cannot occur when $6k-1$ is prime – Will Jagy Aug 28 '20 at 18:02
  • @WillJagy It is my intention to prove that there must be a prime number with $k>a$ – user140242 Aug 28 '20 at 18:51

1 Answers1

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$k = 6xy + x - y = (6x - 1)y + x$
imagining this number as if $x$ was fixed, and $y$ runs through the positive integers. With this logic we see that incrementing/decrementing $y$ by 1, the next value of $k$ is $6x - 1$ away.
Now if $1 < k - a < 6x - 1$, then it means $a$ is strictly smaller than $k$ and decrementing $y$ by 1 makes $k < a$, so that can not happen, the actual value of $k$ is the smallest to have $k > a$. On the other hand, $k-a$ always exists $\forall x, y$, as $x \geq 2 \Rightarrow 6x - 1 \geq 11$, $\forall x, y$, so we can have such values of $a$ $\forall x, y$.
For such values of $a$ there exist $b < k - a > 1 \Rightarrow a + b < k$, $\forall x, y$.

Dávid Laczkó
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