Xn+1==(sqrt(5*Xn*Xn+4*(-1)^n)+Xn)/2

rt

Re: Xn+1==(sqrt(5*Xn*Xn+4*(-1)^n)+Xn)/2

first for bound=1e5,using matirx power to compute x0,x1 x[bound],x[bound+1] ,x[2*bound],x[2*bound+1]..... upper bound is 2*p using map to record (x[i*bound],x[i*bound+1])

I use this formula and discrete log algo to find candidate of xn+1

then continue to compute xn-1 xn-2 .... until we find (xi,xi+1) in the map then it is valid

otherwise it is invalid

complex : sqrt(p*log(p))

*Edited by author 02.12.2017 15:17*

Re: Xn+1==(sqrt(5*Xn*Xn+4*(-1)^n)+Xn)/2

Posted by

rausen 6 Mar 2018 19:13

Could you do me a favor to explain the meaning of this problem?

My teammates and me are working on this but we can hardly get the point.

Thx for helping anyway.

p.s. Chinese would be better XD

Re: Xn+1==(sqrt(5*Xn*Xn+4*(-1)^n)+Xn)/2

给定 斐波那契 的某一项 %p的值x,求它的下一项 %p的所有可能值

Re: Xn+1==(sqrt(5*Xn*Xn+4*(-1)^n)+Xn)/2

Posted by

LJFan 17 Aug 2018 21:26

请问斐波那契的一阶递推式是怎样推导的？