#### Hide answer & extension

1, ~~3~~, 7, ~~12~~, 19, ...

1, 7, 19, ...

1=1; 1+7=8; 1+7+19=27; ...

1, 8, 27, ...

The final sequence is the cube numbers. To show why, let \(n\) be an integer and follow through the process.

Cross out every third number:

1, 2, ~~3~~, 4, 5, ~~6~~, ..., ~~3n~~, \(3n+1\), \(3n+2\), ...

1, 2, 4, 5, ..., \(3n+1\), \(3n+2\), ...

Find the cumulative sums:

$$1=1$$
$$1+2=1+2=3$$
$$1+2+4=1+2+3+4-3=7$$
$$1+2+4+5=1+2+3+4+5-3=12$$
$$...$$
$$1+2+4+5+...+(3n+1)=\sum_{i=1}^{3n+1}-\sum_{i=1}^{n}3i$$
$$=\frac{1}{2}(3n+1)(3n+2)-\frac{3}{2}n(n+1)$$
$$=3n^2+3n+1$$
$$1+2+4+5+...+(3n+2)=3n^2+3n+1+(3n+2)$$
$$=3n^2+6n+3$$
$$...$$

1, 3, 7, 12, ..., \(3n^2+3n+1\), \(3n^2+6n+3\), ...

Cross out every second number, starting with the second:

1, ~~3~~, 7, ~~12~~, ..., \(3n^2+3n+1\), ~~3n2+6n+3~~, ...

1, 7, ..., \(3n^2+3n+1\), ...

Find the cumulative sums. The \(m\)^{th} sum is:

$$\sum_{n=0}^{m}3n^2+3n+1$$
$$=3\sum_{n=0}^{m}n^2+3\sum_{n=0}^{m}n+\sum_{n=0}^{m}1$$
$$=\frac{3}{6}m(m+1)(2m+1)+\frac{3}{2}m(m+1)+m+1$$
$$=\frac{1}{2}(m+1)(m(2m+1)+3m+2)$$
$$=\frac{1}{2}(m+1)(2m^2+m+3m+2)$$
$$=\frac{1}{2}(m+1)(2m^2+4m+2)$$
$$=(m+1)(m^2+2m+1)$$
$$=(m+1)(m+1)^2$$
$$=(m+1)^3$$

Hence the numbers obtained are the cube numbers.

#### Extension

What happens if you cross out every third number starting at the second? Or every fifth number starting at the fifth? Or every \(n\)^{th} number starting at the \(m\)^{th}?