# Sum of first n terms is prime.

$$a_n$$ is a non-decreasing sequence of positive integers. If an positive integer $$k$$ appears in $$a_n$$ exactly $$k$$ times and $$S_n$$ is the sum of the first n terms, find all $$N$$ such that $$S_N$$ is prime.

The sequence is $$1,2,2,3,3,3,4,4,4,4,5,5,5,5,5, \cdots$$ If $$n$$ is a triangular number, then the sum is just the sum of squares. So $$S_n=$$ $$\frac{n(n+1)(2n+1)}{6}-kn$$ for some non negative integer $$k. $$S_n=\frac{n(2n^2+3n+1-6k)}{6}$$ I have somewhat closed form, but what could I do to find the $$n$$ such that it's prime? Thanks!

Let $$n=m(m+1)/2 + j$$ with $$m$$ a positive integer and $$0\leq j \leq m$$.

Then $$S_n=\frac{(2m+1)(m+1)m}{6} + (m+1)j$$.

We write this as $$\frac{(2m^2+m+6j)(m+1)}{6}$$

Suppose $$m+1$$ is not a divisor of $$6$$, then the left side must be a divisor of $$6$$. This is clearly only possible if $$m=1$$, in which case $$j=0$$ gives $$S_n=1$$ and $$j=1$$ gives $$S_n=3$$.

In the other cases $$m+1$$ must be a divisor of $$6$$.

If $$m=2$$ then $$j=0$$ gives $$S_n= 5$$, $$j=1$$ gives $$8$$ and $$j=2$$ gives $$11$$.

If $$m=5$$ then then $$j=0$$ gives $$S_n = 55$$, $$j=1$$ gives,$$j=3$$ gives $$67$$ $$j=4$$ gives $$73$$, and $$j=5$$ gives $$79$$.

Hence the values of $$n$$ that you want are $$2, 3,5,16,17,18,19$$

That closed form gives you a factorization of the sum, unless the six on the bottom can eat at least one of the two terms. This can only happen for small $$n$$, so you should be able to get an upper bound and then check the remaining cases directly.

• So Case 1: one term is 1 and other is multiple of 6 Case 2: The other term is 1 and other is multiple of 6 Case 3: One term is 2 and other is Multiple of 3? How would you bound it (I can only see casework so far)? – Baker5680 Nov 20 '19 at 23:13