Its just adding and subtracting

Let ${ S }_{ a }=0$

${ S }_{ a+1 }=a+1$

${ S }_{ a+2 }=2a+3$

${ S }_{ a+3 }=a$

${ S }_{ a+4 }=2a+4$

${ S }_{ a+5 }=a-1$

${ S }_{ a+6 }=2a+5$

${ S }_{ a+7 }=a-2$ and so on..

Notice ${ S }_{ a+(2m+1) }=a+1-m$ and that the minimum value for ${ S }_{ b }$ between $n+1\le b\le 2m+1$ between one $0$ and next $0$ is at $b=2m+1$

For $m=a+1$

${ S }_{ n+(2m+1) }={ S }_{ 3n+3 }=0$

Then

${ C }_{ k+1 }=3{ C }_{ k }+3$ and ${ C }_{ 1 }=3$

Solving this recurrence relation,we get

${ C }_{ k }=\frac { 3^{ k+1 }-3 }{ 2 }$

For $k=7$ , ${ C }_{ 7 }=3279$

Assume that $S_n=0$ . Then $S_{n+1}=n+1$ , $S_{n+2} = 2n+3$ and $S_{n+3} = n$ . This pattern will continue, with $S_{n+2k} = 2n+k+2$ and $S_{n+(2k+1)} = n - k + 1$ , until we reach $S_{n+(2m+1)} = 0$ , where $m=n+1$ . Thus, we can define $\langle C_k\rangle$ recursively by

• $C_1 = 3$
• $C_k = C_{k-1} + 2(C_{k-1}+1)+1 = 3C_{k-1} + 3$

Solving this recurrence relation yields $C_k = \frac{3^{k+1}-3}{2}$ and $C_7 = \frac{3^8-3}{2} = \boxed{3270}$ .