HELP! Sequence Problem!

i suppose you could try plugging in 1, 2, 3, 4 and seeing if there is a pattern

EDIT: nevermind i read want you wanted

$5^{a_{n+1}-a_n} - 1 = \dfrac{1}{n+\frac{2}{3}}, n\geq 1$

$k > 1, k\in \mathbb Z^+$ and $a_k$ is an integer.

Let us do something to the recursive definition of the sequence.

$\dfrac{5^{a_{n+1}}-5^{a_n}}{5^{a_n}} = \dfrac{1}{n+\frac{2}{3}}\\ 5^{a_{n+1}} = \left(\dfrac{3}{3n+2}+1\right) \cdot 5^{a_n}\\ 5^{a_{n+1}} = \left(\dfrac{3n+5}{3n+2}\right) \cdot 5^{a_n}\\ a_{n+1} = a_n + \log_5\left(\dfrac{3(n+1)+2}{3n+2}\right)$

You see somehow it is telescoping :P

Now list some values of a_n for small n.

$a_1 = 1\\ a_2 = 1 + \log_5\left(\dfrac{8}{5}\right) = \log_5(8)\\ a_3 = 1 + \log_5\left(\dfrac{8}{5}\right) + \log_5\left(\dfrac{11}{8}\right)=1+\log_5\left(\dfrac{11}{5}\right)=\log_5(11)\\$

You observe that: $a_n = \log_5\left(3n+2\right)$. 

Now we got the explicit formula for a_n. We are halfway done :) (Nice :D)

If a_k is an integer, then ${3k+2}$ must be a power of 5.

Now the question becomes a number theory question: What is the least value of k such that k > 1 and 3k + 2 is a power of 5?

Let $5^x = 3k+2$.

$5^x \equiv 2 \pmod 3$

$2^x \equiv 2 \pmod 3$

This congruence holds true when x is an odd integer.

So the least value of k can be obtained by the equation $5^3 = 3k+2$. (Why not 5¹?)

$3k + 2 = 125\\ k = \dfrac{123}{3} = 41$

Voilà!