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This number sequence has numbers rounded to the nearest tenth.

1, 2, 4.5, ?, 26, ?

Find the pattern and the answers to the question marks rounded to the nearest tenth.

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Based on OP's hint in the comments I have a new answer

Looking at a binomial transform of the factorial numbers $$ a_n = \displaystyle \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} k!$$ for the numbers $n=2,3,4,\ldots$, we get the sequence 2, 4, 9, 21, 52, 134,...
Taking half of each of those numbers we get the sequence 1, 2, 4.5, 10.5, 26, 67,...
which seems to be the sequence we are looking for.
In summary, the $n$th term of the sequence is given by $$ a_n = \displaystyle \sum_{k=0}^{\lfloor (n+1)/2 \rfloor} \binom{n+1}{2k} \frac{k!}{2}$$

Previous Answer
Here's a recursive linear relationship (with rounding)

$a_{n+1} = 2.48 a_n - 0.5$ (rounded to the nearest tenth)

Worked out

$(2.48 \times 1) - 0.5 = 1.98 \rightarrow 2.0$
$(2.48 \times 2) - 0.5 = 4.46 \rightarrow 4.5$
$(2.48 \times 4.5) - 0.5 = 10.66 \rightarrow 10.7$
$(2.48 \times 10.7) - 0.5 = 26.036 \rightarrow 26.0$
$(2.48 \times 26) - 0.5 = 63.98 \rightarrow 64.0$

So the full sequence is

1, 2, 4.5, 10.7, 26, 64

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  • $\begingroup$ No, the numbers were close to correct, but theres a simpler direct function. Hint: rot13(Vg unf fbzrguvat gb qb jvgu snpgbevnyf.) $\endgroup$ – new QOpenGLWidget Dec 19 '20 at 0:53
  • $\begingroup$ Don't think too hard about binomials. There's a very simple function you missed, involving only powers and factorials. $\endgroup$ – new QOpenGLWidget Dec 19 '20 at 14:48
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Okay, it's been about 10 days since I asked this question, and @hexomino is close, but is not correct, so I'm just going to reveal it.

$$a_{n} =\frac{n^n}{n!}$$

And, plugging in, we get

1, 2, 4.5, 10.7, 26, 64.8

(Though I don't know how @hexomino managed to get such a close answer)

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