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mathematicalgemstones.com Valuation

US$50,751,386

Last updated: 2023-05-14 11:16:22

mathematicalgemstones.com has Semrush global rank of 208,552. mathematicalgemstones.com has an estimated worth of US$ 50,751,386, based on its estimated Ads revenue. mathematicalgemstones.com receives approximately 5,855,930 unique visitors each day. Its web server is located in United States, with IP address 172.67.169.208. According to SiteAdvisor, mathematicalgemstones.com is safe to visit.

Traffic & Worth Estimates

Purchase/Sale Value	US$50,751,386
Daily Ads Revenue	US$46,848
Monthly Ads Revenue	US$1,405,423
Yearly Ads Revenue	US$16,865,076
Daily Unique Visitors	390,396
Note: All traffic and earnings values are estimates.

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Mathematical Gemstones On the quest for mathematical beauty and truth Search Main menu Skip to primary content Skip to secondary content Home All posts Contribute! About this blog Maria Gillespie’s Home Page Post navigation ← Older posts Sorting sets of binary strings into threes Posted on August 7, 2022 by Maria Gillespie Reply Time for a just-for-fun combinatorial problem! Thanks to my brother Kenneth Monks for suggesting it. Consider the numbers of the form $2^{2^n}-1$ for positive integers $n$. The first few, for $n=1,2,3,4,\ldots$, are: $$3, 15, 255, 65535, \ldots$$ One thing that all of these numbers have in common is that they are divisible by $3$. This is not hard to prove by induction; the first entry is divisible by $3$, and if $2^{2^n}-1$ is divisible by $3$, then $2^{2^{n+1}}-1=(2^{2^n})^2-1=(2^{2^n}-1)(2^{2^n}+1)$ is also divisible by $3$. But is there a combinatorial proof? In particular, take the most natural combinatorial interpretation of $2^n$, as the number of

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Domain Name: MATHEMATICALGEMSTONES.COM
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mathematicalgemstones.com valuation and analysis