2018-10-10 18:03:03 -04:00
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---
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id: 5900f4ed1000cf542c50fffe
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2020-12-16 00:37:30 -07:00
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title: 问题384:Rudin-Shapiro序列
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2018-10-10 18:03:03 -04:00
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challengeType: 5
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videoUrl: ''
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2021-01-13 03:31:00 +01:00
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dashedName: problem-384-rudin-shapiro-sequence
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2018-10-10 18:03:03 -04:00
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---
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2020-12-16 00:37:30 -07:00
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# --description--
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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将序列a(n)定义为n的二进制展开(可能重叠)中相邻的1对的数量。例如:a(5)= a(1012)= 0,a(6)= a(1102)= 1,a(7)= a(1112)= 2
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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定义序列b(n)=( - 1)a(n)。该序列称为Rudin-Shapiro序列。还要考虑b(n)的总和序列:。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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这些序列的前几个值是:n 0 1 2 3 4 5 6 7 a(n)0 0 0 1 0 0 1 2 b(n)1 1 1 -1 1 1 -1 1 s(n)1 2 3 2 3 4 3 4
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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序列s(n)具有显着特性,即所有元素都是正的,并且每个正整数k恰好出现k次。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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定义g(t,c),其中1≤c≤t,作为s(n)中的索引,其中t在s(n)中出现第c次。例如:g(3,3)= 6,g(4,2)= 7,g(54321,12345)= 1220847710。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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设F(n)为由下式定义的斐波那契数:F(0)= F(1)= 1且F(n)= F(n-1)+ F(n-2),n> 1。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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定义GF(t)= g(F(t),F(t-1))。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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找到ΣGF(t)为2≤t≤45。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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# --hints--
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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`euler384()`应返回3354706415856333000。
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2018-10-10 18:03:03 -04:00
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```js
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2020-12-16 00:37:30 -07:00
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assert.strictEqual(euler384(), 3354706415856333000);
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2018-10-10 18:03:03 -04:00
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```
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2020-08-13 17:24:35 +02:00
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2021-01-13 03:31:00 +01:00
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# --seed--
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## --seed-contents--
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```js
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function euler384() {
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return true;
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}
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euler384();
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```
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2020-12-16 00:37:30 -07:00
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# --solutions--
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2021-01-13 03:31:00 +01:00
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```js
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// solution required
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```
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