mathe

由此我们可以得到
$b(n)="round"({r^{-1}(r-1)}/{(t+1)r-2t}r^n)$
而抛硬币n次出现连续t次正面的概率为
$p(n)=1-{"round"({r-1}/{(t+1)r-2t}r^{n+1})}/{2^n}; r>1&&r^{t+1}-2r^t+1=0$

mathe

Now we could summarize how the formula of t-step Fibonacci sequence is got.
In 9#, we got the characteristic polynomial of
the Linear recurrence equation of the t-step Fibonacci sequence: ${x^{t+1}-2x^t+1}/{x-1}$.
According to link solutions of equation $x^m=2-x^{-n}$, we could use Rouché's Theorem to prove that there's
only one root of polynomial $x^{t+1}-2x^t+1$ whose norm is greater than 1 and t-1 roots whose norms are less than 1. And obviously, the root with greatest norm is a real root greater than 1.
Let's assuming that the real root greater than 1 is r, and besides from $1/{z_0}=1$ and $1/{z_1}=r$, there're another t-1 roots of polynomial $x^{t+1}-2x^t+1$. Let's assuming they're $1/{z_2},1/{z_3},...,1/{z_t}$. And it is obvious there're no multiple roots.
So that we could write the n'th item of the t-step Fibonacii sequence in the form: $u_1z_1^{-n}+u_2z_2^{-n}+...+u_{t}z_t^{-n}$, and we got that $u_s={z_s-1}/{2t-(t+1)z_s^{-1}}$
Now let's assuming $e_n = F_n^{(t)} - u_1 z_1^{-n} = u_2z_2^{-n}+...+u_{t}z_t^{-n}$, so we know that the characteristic polynomial of $e(n)$ is ${x^{t+1}-2x^t+1}/{(x-1)(x-r)}$.
In 30#, a probability modeling is used to prove that $|e(n)|<1/2$.
Let's assuming a man is randomly walking in an axis. He starts from a positive integer location in the axis and each time he random walks towards the positive direction by 1 or towards the negative direction by t with equivalent probability.
And he will stop when he reached a point whose coordinate is no more than 1. We will give a score $e(x)$ to him when he finally stops at coordinate x.
We know the probablity that he will finally stop is 1 when t is no less than 1. It is interesting that we could find that the expected score is $e_n$ when he starts from point whose coordinate is n.
In 29# it is proved that the absolute values of all those final scores ($e(-t+2)$ to $e(1)$) are less than $1/2$, so we proved that $|e_n|<1/2$ for all $n>=-t+2$
So we know now that $F_n^{(t)}$ could be represented by $"round"(u_1 z_1^{-n})$, or $F_n^{(t)}="round"({r-1}/{(t+1)r-2t}r^{n-1})$ for any $n>=-t+2$, where r is t-bonacci constant.

zhangjianquan

高手,我想问一个问题,如果此个题100次趋于无穷大,请求出多大次数的连续正面的概率为最大?

mathe

你这个是个不同的问题,很简单.连续出现t次正面的概率为$1/{2^t}$,所以出现一次的概率最大,为1/2

shangwen_ren

whoops... i just did something silly... fortunately the posted comment can be edited...