1af6e7aa5a
* 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>
62 lines
1.3 KiB
Markdown
62 lines
1.3 KiB
Markdown
---
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id: 5900f4b71000cf542c50ffc9
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title: 'Problem 330: Euler''s Number'
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challengeType: 5
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forumTopicId: 301988
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dashedName: problem-330-eulers-number
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---
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# --description--
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An infinite sequence of real numbers $a(n)$ is defined for all integers $n$ as follows:
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$$ a(n) =
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\begin{cases}
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1 & n < 0 \\\\
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\displaystyle \sum_{i = 1}^{\infty} \frac{a(n - 1)}{i!} & n \ge 0
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\end{cases}
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$$
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For example,
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$$\begin{align}
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& a(0) = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots = e − 1 \\\\
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& a(1) = \frac{e − 1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots = 2e − 3 \\\\
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& a(2) = \frac{2e − 3}{1!} + \frac{e − 1}{2!} + \frac{1}{3!} + \ldots = \frac{7}{2} e − 6
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\end{align}$$
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with $e = 2.7182818\ldots$ being Euler's constant.
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It can be shown that $a(n)$ is of the form $\displaystyle\frac{A(n)e + B(n)}{n!}$ for integers $A(n)$ and $B(n)$.
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For example $\displaystyle a(10) = \frac{328161643e − 652694486}{10!}$.
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Find $A({10}^9)$ + $B({10}^9)$ and give your answer $\bmod 77\\,777\\,777$.
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# --hints--
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`eulersNumber()` should return `15955822`.
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```js
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assert.strictEqual(eulersNumber(), 15955822);
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```
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# --seed--
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## --seed-contents--
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```js
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function eulersNumber() {
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return true;
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}
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eulersNumber();
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```
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# --solutions--
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```js
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// solution required
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```
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