"피보나치 수열의 짝수항"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) |
Pythagoras0 (토론 | 기여) |
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| 1번째 줄: | 1번째 줄: | ||
==개요== | ==개요== | ||
| − | * 피보나치 수열 | + | * 피보나치 수열 <math>\{F_{n}\}_{n\geq 0}</math>의 짝수항으로 주어지는 수열, 즉 <math>\{F_{2n}\}_{n\geq 0}</math> |
| − | + | :<math> | |
1, 1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181,\cdots | 1, 1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181,\cdots | ||
| − | + | </math> | |
| − | * 점화식 | + | * 점화식 <math>a_{n+2}=3a_{n+1}-a_{n}</math> |
* 다음을 만족한다 | * 다음을 만족한다 | ||
| − | + | :<math> | |
a_{n+2}=\frac{1+a_{n+1}^2}{a_{n}} | a_{n+2}=\frac{1+a_{n+1}^2}{a_{n}} | ||
| − | + | </math> | |
* 불변량 | * 불변량 | ||
| − | + | :<math> | |
\frac{a_{n-1}^2+a_{n}^2+1}{a_{n-1}a_{n}}=3 | \frac{a_{n-1}^2+a_{n}^2+1}{a_{n-1}a_{n}}=3 | ||
| − | + | </math> | |
==일반적인 초기조건== | ==일반적인 초기조건== | ||
* 점화식 | * 점화식 | ||
| − | + | :<math> | |
a_{n+2}=\frac{1+a_{n+1}^2}{a_{n}}, a_{0}=\alpha, a_{1}=\beta | a_{n+2}=\frac{1+a_{n+1}^2}{a_{n}}, a_{0}=\alpha, a_{1}=\beta | ||
| − | + | </math> | |
* 다음의 선형점화식을 만족한다 | * 다음의 선형점화식을 만족한다 | ||
| − | + | :<math> | |
a_{n+2}=\frac{1+\alpha ^2+\beta ^2}{\alpha \beta }a_{n+1}-a_{n} | a_{n+2}=\frac{1+\alpha ^2+\beta ^2}{\alpha \beta }a_{n+1}-a_{n} | ||
| − | + | </math> | |
* 처음 몇 개의 항은 다음과 같다 | * 처음 몇 개의 항은 다음과 같다 | ||
| − | + | :<math> | |
\alpha ,\beta ,\frac{1+\beta ^2}{\alpha },\frac{\alpha ^2+\left(1+\beta ^2\right)^2}{\alpha ^2 \beta },\frac{\alpha ^4+2 \alpha ^2 \left(1+\beta ^2\right)+\left(1+\beta ^2\right)^3}{\alpha ^3 \beta ^2},\frac{\alpha ^6+3 \alpha ^2 \left(1+\beta ^2\right)^2+\left(1+\beta ^2\right)^4+\alpha ^4 \left(3+2 \beta ^2\right)}{\alpha ^4 \beta ^3} | \alpha ,\beta ,\frac{1+\beta ^2}{\alpha },\frac{\alpha ^2+\left(1+\beta ^2\right)^2}{\alpha ^2 \beta },\frac{\alpha ^4+2 \alpha ^2 \left(1+\beta ^2\right)+\left(1+\beta ^2\right)^3}{\alpha ^3 \beta ^2},\frac{\alpha ^6+3 \alpha ^2 \left(1+\beta ^2\right)^2+\left(1+\beta ^2\right)^4+\alpha ^4 \left(3+2 \beta ^2\right)}{\alpha ^4 \beta ^3} | ||
| − | + | </math> | |
* 불변량 | * 불변량 | ||
| − | + | :<math> | |
\frac{a_{n-1}^2+a_{n}^2+1}{a_{n-1}a_{n}}=\frac{\alpha ^2+\beta ^2+1}{\alpha \beta } | \frac{a_{n-1}^2+a_{n}^2+1}{a_{n-1}a_{n}}=\frac{\alpha ^2+\beta ^2+1}{\alpha \beta } | ||
| − | + | </math> | |
| 40번째 줄: | 40번째 줄: | ||
==관련논문== | ==관련논문== | ||
| − | * Alperin, Roger C. 2011. “Integer Sequences Generated by | + | * Alperin, Roger C. 2011. “Integer Sequences Generated by <math>x_{n+1}=\frac {x^2_n+A}{x_{n-1}}</math>.” The Fibonacci Quarterly. The Official Journal of the Fibonacci Association 49 (4): 362–365. http://www.math.sjsu.edu/~alperin/IntegerA-Sequences.pdf |
* Caldero, Philippe, and Andrei Zelevinsky. 2006. “Laurent Expansions in Cluster Algebras via Quiver Representations.” Moscow Mathematical Journal 6 (3): 411–429, 587. | * Caldero, Philippe, and Andrei Zelevinsky. 2006. “Laurent Expansions in Cluster Algebras via Quiver Representations.” Moscow Mathematical Journal 6 (3): 411–429, 587. | ||
[[분류:수열]] | [[분류:수열]] | ||
2020년 11월 13일 (금) 10:39 기준 최신판
개요
- 피보나치 수열 \(\{F_{n}\}_{n\geq 0}\)의 짝수항으로 주어지는 수열, 즉 \(\{F_{2n}\}_{n\geq 0}\)
\[ 1, 1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181,\cdots \]
- 점화식 \(a_{n+2}=3a_{n+1}-a_{n}\)
- 다음을 만족한다
\[ a_{n+2}=\frac{1+a_{n+1}^2}{a_{n}} \]
- 불변량
\[ \frac{a_{n-1}^2+a_{n}^2+1}{a_{n-1}a_{n}}=3 \]
일반적인 초기조건
- 점화식
\[ a_{n+2}=\frac{1+a_{n+1}^2}{a_{n}}, a_{0}=\alpha, a_{1}=\beta \]
- 다음의 선형점화식을 만족한다
\[ a_{n+2}=\frac{1+\alpha ^2+\beta ^2}{\alpha \beta }a_{n+1}-a_{n} \]
- 처음 몇 개의 항은 다음과 같다
\[ \alpha ,\beta ,\frac{1+\beta ^2}{\alpha },\frac{\alpha ^2+\left(1+\beta ^2\right)^2}{\alpha ^2 \beta },\frac{\alpha ^4+2 \alpha ^2 \left(1+\beta ^2\right)+\left(1+\beta ^2\right)^3}{\alpha ^3 \beta ^2},\frac{\alpha ^6+3 \alpha ^2 \left(1+\beta ^2\right)^2+\left(1+\beta ^2\right)^4+\alpha ^4 \left(3+2 \beta ^2\right)}{\alpha ^4 \beta ^3} \]
- 불변량
\[ \frac{a_{n-1}^2+a_{n}^2+1}{a_{n-1}a_{n}}=\frac{\alpha ^2+\beta ^2+1}{\alpha \beta } \]
매스매티카 파일 및 계산 리소스
- https://docs.google.com/file/d/0B8XXo8Tve1cxUl9YZnh0V0pOLTQ/edit?usp=drivesdk
- http://oeis.org/A001519
관련논문
- Alperin, Roger C. 2011. “Integer Sequences Generated by \(x_{n+1}=\frac {x^2_n+A}{x_{n-1}}\).” The Fibonacci Quarterly. The Official Journal of the Fibonacci Association 49 (4): 362–365. http://www.math.sjsu.edu/~alperin/IntegerA-Sequences.pdf
- Caldero, Philippe, and Andrei Zelevinsky. 2006. “Laurent Expansions in Cluster Algebras via Quiver Representations.” Moscow Mathematical Journal 6 (3): 411–429, 587.