# Digits of the square root of primes

Here is something interesting I discovered recently.

Let $$S_n$$ be the sum from the first decimal of $$\sqrt2+\sqrt{p_n}$$ to the $$p_n$$th decimal, where $$p_n$$ is the $$n$$th prime number. Then I conjecture that $$S_{n+1}>S_n$$ for all $$n>1$$.

Although I have no way of proving this and no feasible way to disprove this, I have a few comments that may be useful.

• The first few values of $$S_n$$ for $$n=2,3,4,5,6$$ are $$10,21,41,48,64$$. Clearly these are increasing.

• If we pick $$p_{n+1}$$ such that it is greater than a square number by a small amount, then it will have more zeroes, especially when $$n$$ is large. Therefore using this method there could be a higher chance that $$S_{n+1}\le S_n$$ and thus disproving the conjecture. I have done so for $$p_6=13$$ and $$p_7=17$$, and for $$p_{12}=37$$ and $$p_{11}=31$$, but to no avail since $$85=S_7>S_6=64$$ and $$149=S_{12}>S_{11}=123$$.

• I have a strong feeling that the conjecture is true, since every time we are adding more digits, and the chance of zeroes occurring that overcome the increasing sum is subjectively low.

Any advances on this are welcome!

If you do happen to find a value of $$n$$ that disprove the conjecture, please give a few more so that a pattern can hopefully be observed, if any.

• Your thought to have $p_{n+1}$ just greater than a square is a good one. It would also be good to have $p_n=p_{n+1}-2$, which means the square is a multiple of $36$ so you don't get so many digits added. This requires that the square be a multiple of $900$ because otherwise the square ends in $4$ or $6$ and one of the neighboring numbers is a multiple of $5$. I am sure there are people with code sitting around that can try millions or billions of cases easily. Dec 30, 2018 at 18:16
• @RossMillikan counterexample found Dec 30, 2018 at 18:25
• @Quintec: you should write it up as an answer. You just need to copy in the decimals and sums. Dec 30, 2018 at 18:36
• @RossMillikan Done. Dec 30, 2018 at 18:55

$$\begin{array}{r|r} p_n & S_n \\ \hline 2 & 10\\ 3 & 11 \\ 5 & 21 \\ 7 & 41 \\ 11 & 48 \\ 13 & 63 \\ 17 & 93 \\ \textbf{19} & \textbf{72} \\ \end{array}$$
As you can see, $$72 \not\gt 93$$.
Here is the source code with output up to the counterexample, and here is some more data that goes up to $$p_n < 100$$. The data includes $$p_n$$, $$\sqrt2+\sqrt{p_n}$$, and $$S_n$$. It also notes when another counterexample is found - as you can see, there are many.