I wrote this sequence on a napkin a few weeks ago, found I don’t even remember where — it only said “it skips over perfect squares”:
So what happens? Look at its breakdown:
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1 | 2 | 2.5 | 2 |
2 | 3.1 | 3.6 | 3 |
3 | 4.7 | 5.3 | 5 |
4 | 6 | 6.5 | 6 |
5 | 7.2 | 7.9 | 7 |
6 | 8.4 | 8.9 | 8 |
7 | 9.6 | 10.1 | 10 |
8 | 10.8 | 11.3 | 11 |
This is such a fascinating sequence! How is it possible? How can we prove that it will skip over all perfect squares? This relies on a few facts that I discovered just by investigating this sequence:
- The amount of whole numbers in-between two perfect squares is always even. This is easily proven by
. Because we want it exclusive, i.e. not counting the number we land on, we still need to take away 1, which gives
.
- Of this even amount, the integer part of
is constant. Moreover, half of the outputs will be
and half
. That’s because half of the inputs are smaller than the inputs midpoint, and half are greater. And because
is monotonically increasing, order is preserved.
- The first output to exceed
in its decimal part gets bumped to the next integer by the
of the formula.