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I created a table to find primes What I did was understand that if you divide every number in half we get a unique number. e.g. 5 is 5/2, and 3 is 3/2, so by including more columns we create a full range of unique identities in a spectrum by dividing by 3 then 4 and so on.

Denominator and numerator values lead to primes after very few calculations.

How I made this table is quite simple. For the case of 1 I created columns like this: 1 , 1 + 1/2, 1 + 1/2 + 1/3, 1 + 1/2 + 1/3 + 1/4 and 1 + 1/2 + 1/3 +1/4 + 1/5 ....

and for 5 it was the same thing I created similarly identical columns: 5, 5 + 5/2, 5 + 5/2 + 5/3, 5 + 5/2 + 5/3 + 5/4, 5 + 5/1 + 5/2 + 5/3 + 5/4 + 5/5 ...

SPECTRAL NUMBER LINE

In the first column is easy to see every number is over 1, then in the following, we have a change between 1s and 2s which does have a binary feel to it. Then the sequence gets more mysterious. Yes this is simply a harmonic sequence but I added more values for my numerator which bridges the gap and helps us locate primes.

Larry Revell
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  • The $n$th column is just $n$ times the first column – Zoe Allen May 04 '23 at 15:41
  • Thats the rule for primes. Then tell me the rule for the composites. – Larry Revell May 04 '23 at 15:45
  • Until "In the first column is easy to see every number is over 1" your post is understandable, although trivial. After that sentence it goes beyond any meanings. I don't see the point in creating such table: it's just the harmonic series multiplied by 2,3,4,5 and so on. – Crostul May 04 '23 at 16:01
  • The same table shows the relationship, between all numbers. – Larry Revell May 04 '23 at 16:15

2 Answers2

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If you want to analyze the numerator and denominator of the series $$H_n=1+\cdots +\frac{1}{n},$$ you can do so by noticing that $$H_{n+1}-H_{n}=\frac{1}{n+1}$$ and put $$H_n=\frac{N_n}{n!}$$ yielding $$\frac{N_{n+1}}{(n+1)!}-\frac{N_n}{n!}=\frac{1}{n+1}$$ or $$N_{n+1}=n!+(n+1)N_n.$$ You can find a closed form for this one in terms of sums. Here's a list of values: $$\begin{pmatrix}n & N_n\\ 1&1\\ 2& 3\\ 3& 11\\ 4&50\\ 5& 274\\ 6& 1764\\ 7&13068\\ \text{&c.}&\text{&c.}\end{pmatrix}$$

Alexander Conrad
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The entries in the $n$'th row are exactly $n$ times the corresponding entries in the first row. Since you're displaying these as fully reduced fractions, common divisors of $n$ and the denominators in the first row cancel, and what's left of the denominator typically has fewer prime factors and is bound to be prime sometimes. If you feel like there's more to it, maybe you can elaborate a bit on what exactly you mean by "numerator and denominator lead to primes".

Achim Krause
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  • Prime pattern breaks on the number it is, if prime. So if you follow these directions: start at 3/1 and right to 9/2, down to 15/2, right to 55/6, down to 77/6, right to 175/12, down to 275/12, right to 1507/60, down to 1781/60, right to 637/20, down to 833/20, right to 6171/140, down to 6897/140. Where down is a search for a denominator identical, and right is the specific places to move along with success. – Larry Revell May 04 '23 at 16:07
  • By definition of primes : Numerator/denominator cannot simplify in your staff for rows indexed with a prime, and they simplify if index is not a prime. – Lourrran May 04 '23 at 16:22