000

### 121

121 is a square number in every base larger than 3.

In base $n$, $$121_n = 1 + 2n + n^2$$ $$=(1+n)^2.$$

### Polya Strikes Out

Source: mscroggs.co.uk
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.

000

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