Denote $a_n$ our expression. $a_n=0+1-2-3+5+6-7-8+10+11-\cdots$ Case 1: $a_{5k+1}=(0+1+5+6+10+11+15+16+\cdots+5k+5k+1)-(2+3+7+8+12+13+\cdots+5k-3+5k-2)=(1+10+1+20+\cdots+1+10k+1)-(5+10+5+20+\cdots+5+10(k-1)+5)=(k+1+10(\frac{k(k+1)}{2}))-(5k+10(\frac{k(k-1)}{2}))=-4k+1+10k=6k+1$ Case 2: $a_{5k+2}=a_{5k+1}-(5k+2)=6k+1-5k-2=k-1$ Case 3: $a_{5k+3}=a_{5k+2}-(5k+3)=k-1-5k-3=-4k-4$ Case 4: $a_{5k+4}=a_{5k+3}=-4k-4$ Case 5: $a_{5k+5}=a_{5k+4}+(5k+5)=-4k-4+5k+5=k+1$ So $13=6k+1$ $\iff$ $k=2$ $\iff$ $N=5k+1=12$ $13=k-1$ $\iff$ $k=14$ $\iff$ $N=5k+2=72$ $13=k+1$ $\iff$ $k=12$ $\iff$ $N=5k+2=62$