$*$my$\_$first$\_$i $=($n $/$ p$)*$ core$\_$id

$*$my$\_$last$\_$i $=*$my$\_$first$\_$i $+[($n $/$ p$)-1]$

If each call to compute$\_$next$\_$val requires different amounts of work $($For example, if k $=1$ and the first call $($i $=0)$ requires $2$ milliseconds, the second call $($i $=1)$ would require $4$ milliseconds, the third call $($i $=2)6$ milliseconds, and so on$),$ you would adjust the formulas for my$\_$first$\_$i and my$\_$last$\_$i accordingly