I'm looking for formula for series
$$0+0+0+0+0+0+0+0+0+0+0+0+0+107+0+0+0+0+0+0+0+0+0+0+0+0+0+107+...$$
I knew that e.g. series
$$0+0+1+0+0+1+0+0+1+...$$
can be described by formula
$$\sum_{n=1}^{\infty}(1+2n^2)\:mod\:3$$
I think the solution is taking series with one and multiply it by 107, but how to increase zero terms?
In general, for the sequence $\{a_n\}_{n=1}^\infty=\{\underbrace{0,\dots,0}_{k\text{ zeroes}},m,\dots\}$. I.e; $$n\not\equiv 0 \mod (k+1) \implies a_n=0,$$ $$n\equiv 0 \mod (k+1) \implies a_n=m,$$ you can represent $a_n$ as $$a_n=m\Big\lfloor\frac{n \mod (k+1)}{k}\Big\rfloor$$ where $\lfloor x\rfloor$ is the floor function. Thus, for your example where $(k,m)=(13,107)$, you could write the sum (despite it definitely not converging) as $$\sum_{n=1}^\infty 107\Big\lfloor\frac{n \mod (14)}{13}\Big\rfloor$$ $$= 0+0+0+0+0+0+0+0+0+0+0+0+0+107+0+0+0+0+0+0+0+0+0+0+0+0+0+107+...$$