- #1
MevsEinstein
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- TL;DR Summary
- I found a super cool property of n-bonacci numbers but it isn't really worth publishing. Ho do I make it known?
Hi PF!
Everyone knows that: $${\varphi }^2 - \varphi - 1 = 0$$ But guess what? $${\varphi}^3-2{\varphi}^2+1=0$$ Generalizing this for all n-bonacci numbers: $$x^{n+1}+1 = 2x^n$$ where ##x## is the n-bonacci number and ##n## is the degree of the polynomial that the n-bonacci number is a root of, i.e: ##x^n-x^{n-1}-...-x-1=0##. To prove this property, what we do is to use the fact that all the terms of lower degree than ##n## are in a geometric series with ratio ##x## and with a beginning term ##-1##: $$x^n-\frac{1-x^n}{1-x}=0$$$$x^n=\frac{1-x^n}{1-x}$$$$x^n-x^{n+1}=1-x^n$$$$[{2x^n=x^{n+1}+1}]$$ Now this is a very small property that isn't worth writing a paper on it (I think), but it's cool. Writing in Physics Forums is pretty good already, but how do I spread it quicker?
Everyone knows that: $${\varphi }^2 - \varphi - 1 = 0$$ But guess what? $${\varphi}^3-2{\varphi}^2+1=0$$ Generalizing this for all n-bonacci numbers: $$x^{n+1}+1 = 2x^n$$ where ##x## is the n-bonacci number and ##n## is the degree of the polynomial that the n-bonacci number is a root of, i.e: ##x^n-x^{n-1}-...-x-1=0##. To prove this property, what we do is to use the fact that all the terms of lower degree than ##n## are in a geometric series with ratio ##x## and with a beginning term ##-1##: $$x^n-\frac{1-x^n}{1-x}=0$$$$x^n=\frac{1-x^n}{1-x}$$$$x^n-x^{n+1}=1-x^n$$$$[{2x^n=x^{n+1}+1}]$$ Now this is a very small property that isn't worth writing a paper on it (I think), but it's cool. Writing in Physics Forums is pretty good already, but how do I spread it quicker?
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