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gikf 1af6e7aa5a fix(curriculum): clean-up Project Euler 321-340 (#42988)
* fix: clean-up Project Euler 321-340

* fix: typo

* fix: corrections from review

Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com>

* fix: corrections from review

Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>

Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com>
Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
2021-07-29 11:59:06 -07:00

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id, title, challengeType, forumTopicId, dashedName
id title challengeType forumTopicId dashedName
5900f4b71000cf542c50ffc9 Problem 330: Euler's Number 5 301988 problem-330-eulers-number

--description--

An infinite sequence of real numbers a(n) is defined for all integers n as follows:

$$ a(n) = \begin{cases} 1 & n < 0 \\ \displaystyle \sum_{i = 1}^{\infty} \frac{a(n - 1)}{i!} & n \ge 0 \end{cases}



For example,

$$\begin{align}
  & a(0) = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots = e  1 \\\\
  & a(1) = \frac{e  1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots = 2e  3 \\\\
  & a(2) = \frac{2e  3}{1!} + \frac{e  1}{2!} + \frac{1}{3!} + \ldots = \frac{7}{2} e  6
\end{align}$$

with $e = 2.7182818\ldots$ being Euler's constant.

It can be shown that $a(n)$ is of the form $\displaystyle\frac{A(n)e + B(n)}{n!}$ for integers $A(n)$ and $B(n)$.

For example $\displaystyle a(10) = \frac{328161643e  652694486}{10!}$.

Find $A({10}^9)$ + $B({10}^9)$ and give your answer $\bmod 77\\,777\\,777$.

# --hints--

`eulersNumber()` should return `15955822`.

```js
assert.strictEqual(eulersNumber(), 15955822);
```

# --seed--

## --seed-contents--

```js
function eulersNumber() {

  return true;
}

eulersNumber();
```

# --solutions--

```js
// solution required
```