What is the pattern for this sequence? 
I know that it increments by 1 until the (10n + 1)th term, where it increments by the term #. I don't know how to represent this entire pattern as an equation or summation of some sort.
 A: One way to approach the problem is to begin by looking at the subsequence of terms with indices of the form $10n+1$:
$$\begin{array}{rccc}
10n+1:&1&11&21&31&41\\
s_{10n+1}:&1&21&51&91&141\\
s_{10n+1}-(10n+1):&0&10&30&60&100
\end{array}$$
If you divide that last row of numbers by $10$, it might be familiar: $0,1,3,6,10,\ldots$ is the sequence of sums of consecutive integers.
$$\begin{array}{rccc}
10n+1:&1&11&21&31&41\\
s_{10n+1}:&1&21&51&91&141\\
s_{10n+1}-(10n+1):&0&10&30&60&100\\
\frac1{10}\big(s_{10n+1}-(10n+1)\big)&\sum_{k=0}^0k&\sum_{k=0}^1k&\sum_{k=0}^2k&\sum_{k=0}^3k&\sum_{k=0}^4k
\end{array}$$
It begins to look very much as if
$$s_{10n+1}-(10n+1)=10\sum_{k=0}^nk\;.\tag{1}$$
The gap between $s_m$ and $m$ remains constant as $m$ runs over the interval from $10n+1$ through $10n+10=10(n+1)$, so if $(1)$ is indeed correct, we must have
$$s_{10n+\ell}=10n+\ell+10\sum_{k=0}^nk\quad\text{for}\quad\ell=1,2,\ldots,10\;.\tag{2}$$
You can get rid of the summation by applying the familiar formula for the sum of consecutive integers (or more generally for the sum of an arithmetic progression). You can also get rid of the auxiliary quantity $\ell$ by noting that
$$m=10\left\lfloor\frac{m-1}{10}\right\rfloor+\left(m-10\left\lfloor\frac{m-1}{10}\right\rfloor\right)\;,$$
where
$$1\le m-10\left\lfloor\frac{m-1}{10}\right\rfloor\le 10\;,$$
so that if we set
$$n=\left\lfloor\frac{m-1}{10}\right\rfloor$$
and
$$\ell=m-10\left\lfloor\frac{m-1}{10}\right\rfloor\;,$$
we can rewrite $(2)$:
$$s_m=m+10\sum_{k=0}^{\left\lfloor\frac{m-1}{10}\right\rfloor}k\;.$$
I leave it to you to get rid of the summation and to convince yourself that $(1)$ really is correct in general.
A: Judging from what you have posted, it looks like you have
$$f(10)=10,\ f(20)=30,\ f(30)=60,\ f(40)=100$$
and so on.  A possible formula for this would be
$$f(10q)=10(1+2+\cdots+q)=5q(q+1)\ .$$
The numbers reading backwards from $10$ to $1$, from $20$ to $11$ and so on appear to just decrease by $1$ each time.  So these can be included by writing
$$f(10q-r)=5q(q+1)-r$$
for $q=1,2,3,\ldots$ and $r=0,1,2,\ldots,9$.
