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							|  |  |  | id: 5900f53c1000cf542c51004e | 
					
						
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										 |  |  | title: '問題 463: 風変わりな漸化式' | 
					
						
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										 |  |  | challengeType: 5 | 
					
						
							|  |  |  | forumTopicId: 302138 | 
					
						
							|  |  |  | dashedName: problem-463-a-weird-recurrence-relation | 
					
						
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							|  |  |  | # --description--
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										 |  |  | すべての正の整数に対して関数 $f$ が次のように定義されます。 | 
					
						
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							|  |  |  | $$\begin{align} & f(1) = 1 \\\\ & f(3) = 3 \\\\ & f(2n) = f(n) \\\\ & f(4n + 1) = 2f(2n + 1) - f(n) \\\\ & f(4n + 3) = 3f(2n + 1) - 2f(n) \end{align}$$ | 
					
						
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										 |  |  | 関数 $S(n)$ は $\sum_{i=1}^{n} f(i)$ と定義されます。 | 
					
						
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										 |  |  | $S(8) = 22$, $S(100) = 3604$ です。 | 
					
						
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										 |  |  | $S(3^{37})$ を求めなさい。 回答は、下位 9 桁とすること。 | 
					
						
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							|  |  |  | # --hints--
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										 |  |  | `weirdRecurrenceRelation()` は `808981553` を返す必要があります。 | 
					
						
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							|  |  |  | ```js | 
					
						
							|  |  |  | assert.strictEqual(weirdRecurrenceRelation(), 808981553); | 
					
						
							|  |  |  | ``` | 
					
						
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							|  |  |  | # --seed--
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							|  |  |  | ## --seed-contents--
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							|  |  |  | ```js | 
					
						
							|  |  |  | function weirdRecurrenceRelation() { | 
					
						
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							|  |  |  |   return true; | 
					
						
							|  |  |  | } | 
					
						
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							|  |  |  | weirdRecurrenceRelation(); | 
					
						
							|  |  |  | ``` | 
					
						
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							|  |  |  | # --solutions--
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							|  |  |  | ```js | 
					
						
							|  |  |  | // solution required | 
					
						
							|  |  |  | ``` |