11
$\begingroup$

I found the problem here: http://www.artofproblemsolving.com/community/c7h1314296_2x15_is_divisible_by_x

$\large a_1 = 3$

$\large a_{n+1} = 2^{a_n -1} +5$

Prove:

$\large a_n|a_{n+1}$

Initially I tried inducting but found no way to make the statement carry on to the next recurrence.

This problem seems fairly interesting and I wonder if it can be done using elementary means.

The first four values of $a_n$ are as follows:

$3, 9, 261,$

$1852673427797059126777135760139006525652319754650249024631321344126610074238981$

Mathematica verifies the divisibility of $a_4$ by $a_3$, so the statement does appear to hold water.

$\endgroup$
  • 2
    $\begingroup$ Mathematica also verifies (using PowerMod) that $a_4\mid a_5$. $\endgroup$ – Greg Martin Oct 6 '16 at 6:22
  • 1
    $\begingroup$ Fiddling around with this function. $a_n$ is always odd because its is $2^n + 5 $. It will also be divisible by 3 because $2^n mod 3$ is 1 when n is even. $\endgroup$ – Q the Platypus Oct 7 '16 at 1:01
  • 1
    $\begingroup$ See mathoverflow.net/questions/251717/… for extensive discussion. $\endgroup$ – Max Alekseyev Oct 9 '16 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.