Problem 1269

Suppose that $S_{N}=0$ for some $N \geq 0$. Then $a_{N+1}=1$ because $N+1 \geq S_{N}$. The following table lists the values of $a_{k}$ and $S_{k}$ for a few $k \geq N$: $k$ & $a_{k}$ & $S_{k}$ $\text{}$$N$ & & 0 \$N+1$\text{1}$N+1$\$N+2$ & 1 & $2 N+3$ \$N+3$\text{-1}$N$\$N+4$ & 1 & $2 N+4$ \$N+5$\text{-1}$N-1$\$N+6$ & 1 & $2 N+5$ \$N+7$\text{-1}$N-2. We see inductively that, for every i $\geq$ 1$,$S_{N+2 i}=2 N+2+i$and$S_{N+1+2 i}=N+1-i$thus$S_{3 N+3}=0$is the next$k$for which$S_{k}=0$. The values of$k$for which$S_{k}=0 satisfy the recurrence relation p_{n+1}=3 p_{n}+3, and we compute that the first terms of the sequence are 0,3,12,39,120,363,1092$; hence 1092 is our answer.