Is there a pattern to the golden ratio number figures?

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The golden ratio or phi is 1.6180339887498948482045... I am wondering if there is a pattern in the numbers so given a certain set of figures, you are able to figure out the rest of the figures aaccurately? Essentially is there a pattern that someone could follow and say you were given 1.6180339887 and from the pattern formulated you can figure that the next numbers are 498948...?

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You can calculate all of the digits of $\phi$ without having any of the previous digits by just evaluating $\frac{1+\sqrt{5}}{2}$. However, if you are trying to find a pattern in the digits, this is an unsolved problem. If the digits of a number are evenly distributed and every digit is just as likely as any other digit, then the number is called normal. Proving that a number is normal is extremely difficult unless the number was defined to have uniformly distributed digits. You can read more here.

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Given a finite sequence of numbers, it is impossible to answer the question what is the next number of the sequence. Such questions usually rely on there being a fairly simple recognizable pattern, but there is always an implicit assumption that the pattern will continue. However, there is no guarantee that this is true.

Suppose that a sequence starts: $1,4,7$. If asked for the next number of the sequence, most people would respond $10$. However, if you pick any positive integer $n$, there is a degree 3 polynomial $P$ such that $P(1)=1, P(2)=4, P(3)=7$ and $P(4)=n$. So your question is ill conceived.

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The Golden ratio is well-known for being written with repeating patterns, such as : $$\phi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \ddots}}}$$ or $$\phi = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}\,.$$ It can be written as a series: $$\phi=\frac{13}{8}+\sum_{n=0}^{\infty}\frac{(-1)^{(n+1)}(2n+1)!}{(n+2)!n!4^{(2n+3)}}\,.$$ It can be derived from the Fibonacci sequence, where several patterns exist. Yet, I am not unware of a pattern in its digits in base $10$.