4
$\begingroup$

Determine the missing number in the sequences below
$7, 9, 13, 17, ?, 29,... \\ 6, 8, ?, 16, 24, 28,...$

$8, 11, 29, 51, 125, ?, 293,... \\ 9, 11, 30, 51, 126, ?, 294,...$

I apologize for the simultaneous quantity, but I believe that the pairs have a similar pattern of beginning, so that resolving one will solve the others. Am I right?
These terms are important results to proceed with the solution of some integrals present in (Almost) Impossible Integrals, Sums, and Series (Problem Books in Mathematics) (English Edition)

$\endgroup$
3
  • $\begingroup$ To be clear, you want us to find the next terms in the sequences, not the terms represented by question marks? $\endgroup$
    – bobble
    Nov 15, 2020 at 16:15
  • 1
    $\begingroup$ @bobble The first sentence says "determine the missing number", therefore I'd assume you have to find the number represented by the question mark. $\endgroup$
    – Sleafar
    Nov 15, 2020 at 16:24
  • 1
    $\begingroup$ When I posted my first comment it said to find the "next terms", and it has been edited since to clarify the problem. $\endgroup$
    – bobble
    Nov 15, 2020 at 16:26

2 Answers 2

3
$\begingroup$

The answer to the last two sequences should be

171 and 171

Because

Hidden are the primes (in correct order) squared +4 or squared +2, for the first sequence. And the only thing that is different with the second sequence is +5 instead of +4. So in both cases we get the same answer.

So we get

$2^2 + 4 = 8, 3^2 + 2 = 11, 5^2 + 4 = 29, ...., 13^2 + 2 = 171,....$ and $2^2 + 5 = 9, 3^2 + 2 = 11, 5^2 + 5 = 30, ...., 13^2 + 2 = 171,.... $

$\endgroup$
1
  • 2
    $\begingroup$ Great find... and a funny coincidence too considering your name! $\endgroup$
    – PDT
    Nov 15, 2020 at 18:01
4
$\begingroup$

Partial Answer

First two sequences

If $p_n$ is the sequence of prime numbers, then the $n$th term of the first sequence is $2p_n+3$ and the second sequence is $2p_n+2$ so the complete sequences are $$7, 9, 13, 17, 25, 29,...$$ $$6, 8, 12, 16, 24, 28,...$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.