Select the best solution:

Write a function that will accept an empty vector of ints and an int and will fill the vector with the ulam sequence $\backslash (\{\}^\{1\}\backslash )$ beginning with the int. You may assume that the int is greater than $2.$

$\backslash (\{\}^\{1\}\backslash )$ The ulam sequence begins with an integer. If it is even, divide by $2.$ If it is odd, multiply by $3$ and add $1.$ Then apply the appropriate rule to the resulting integer. Continue in this way until the number produced is $1.$ This sequence will always, eventually produce a value of $1.$

$```$

void ulam$($vector &V$,$ int start$)$

$\{$

V$.$push$\_$back$($start$)$;

for $($int i $=$ start; i $>1$; i $+=($i$\%2))$

$\{$

if$($start $\%2==0)$

start $/=2$;

else

start $=$ start $*3+1$;

V$.$push$\_$back$($i$)$;

$\}$

$\}$

$``````$

vector &V;

int start;

cout $$ "Enter starting number" $$ endl;

cin $>>$ start;

V$.$push$\_$back$($start$)$;

while$($start $>1)$

$\{$

if$($start $\%2==0)$

start $/=2$;

else

start $=$ start $*3+1$;

V$.$push$\_$back$($start$)$;

$\}$

$```$

$```$

void ulam$($vector &V$,$ int start$)$

$\{$

V$.$push$\_$back$($start$)$;

while$($start $>1)$

$\{$

if$($start $\%2==0)$

start $/=2$;

else

start $=$ start $*3+1$;

V$.$push$\_$back$($start$)$;

$\}$

$\}$

$``````$

void ulam$($vector &V$,$ int start$)$

$\{$

V$.$push$\_$back$($start$)$;

while$($start $>1)$

$\{$

if$($start $\%2==1)$

start $/=2$;

else

start $=$ start $*3+1$;

V$.$push$\_$back$($start$)$;

$\}$

$\}$

$```$