If there are two sequences, $t_k=ak+p$ and $t_l=bl+q$, how to find the equation for the sequence formed by intersection of them?

For eg, given two sequences $1,4,7,10,13,16,19,22,25... (3k+1)$ and $2,7,12,17,22,27... (5l+2)$, how to arrive at the common sequence $7,22,37... (15n+7?)$ ?

I think that first those two equations should be equated:



But from here, I don't know how to go forward.

  • 1
    $\begingroup$ Are the sequences integer sequences? If yes, it is essentially a Diophantine equation. One way to proceed is using modular arithmetic - you get that $3k\equiv 1\ (\textrm{mod}\ 5)$ which gives you $k\equiv 2\ (\textrm{mod}\ 5)$ or $k=5n+2$ which then gives you $3k+1=15n+7$. You can use then the initial terms of the sequences to find the initial term of the common sequence. $\endgroup$ Jun 13 at 15:19
  • $\begingroup$ If those sequences have no common elements, then the intersection is empty. Otherwise, let $t_0$ be common to both: that means the sequences are of the form $ak+t_0$ and $bl+t_0$, for $k,l\in\mathbb Z$, so the intersection will be $cm+t_0$ where $t_0$ is the same value as before, $c=\operatorname{lcm}(a,b)$ (the least common multiplier) and $m\in\mathbb Z$. $\endgroup$
    – user700480
    Jun 13 at 15:34
  • $\begingroup$ @StinkingBishop how to find $t_0$? Apart from just trial and error. Doesn't that again require to solve $3k=5l+1$ for integer solutions? $\endgroup$ Jun 13 at 16:29


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