My Math Forum Instant Prime Identifier Table: How does it do it?

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 March 18th, 2018, 11:07 AM #1 Newbie   Joined: Mar 2018 From: Belize Posts: 7 Thanks: 0 Instant Prime Identifier Table: How does it do it? Using our base 10 numerical system adjacent to a base 4, I find that all primes pair with another and identify each other further up the table without the need for a formula (example 17=41). Not one prime is missed. What is going on? Even semi-primes are sniffed out by primes, such as 35 and 91. this is way too weird for me to figure out, but seems like it is beneficial to investigate more thoroughly. First few rows of my table below: 0=0 1=1 2=2 3=3 4=10 5=11 6=12 7=13 8=20 9=21 10=22 11=23 12=30 13=31 14=32 15=33 16=40 17=41 18=42 19=43 20=50 21=51 22=52 23=53 24=60 25=61 26=62 27=63 28=70 29=71 30=72 31=73 32=80 33=81 34=82 35=83 36=90 37=91 38=92 39=93 40=100 41=101 42=102 43=103 44=110 45=111 46=112 47=113 I have run this table up to beyond 4000 with a 100% success rate at identifying primes. It finds every one for you once you have the first few marked. A second table using only 4 rows consecutively for the base 4 system (being more appropriate) also gives similar extraordinary pairings of primes and semis, though limited by fewer integers. It can be used to cross-reference with the main table above. Example below: 0=0 1=1 2=2 3=3 4=10 5=11 6=12 7=13 8=20 9=21 10=22 11=23 12=30 13=31 14=32 15=33 16=100 17=101 18=102 19=103 20=110 21=111 22=112 23=113 24=120 25=121 26=122 27=123 28=130 29=131 30=132 31=133 32=200 33=201 34=202 35=203 36=210 37=211 38=212 39=213 40=220 41=221 42=222 43=223 44=230 45=231 46=232 47=233 Could a formula be created to express what is going on here? No other base syncing seems to produce similar results. Our base 10 decimal system appears to be the one with the prime anomaly, simply because we have 10 fingers. LoL Am I going mad?
 March 18th, 2018, 05:36 PM #2 Newbie   Joined: Mar 2018 From: UK Posts: 3 Thanks: 0 First table doesn't work. 37(prime) -> 91 = 7*13. Second table doesn't work. 31(prime) -> 133 = 7*19.
 March 18th, 2018, 05:38 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 383 Thanks: 207 Math Focus: Dynamical systems, analytic function theory, numerics You are exhausting every positive integer so of course every prime occurs. The problem is that when a prime appears, instead of seeing a counterexample, you see that your prime has now selected a semi-prime instead. It's just confirmation bias combined with the fact that primes and semi-primes are quite common among small integers. Last edited by skipjack; March 19th, 2018 at 07:57 AM.
 March 19th, 2018, 06:51 PM #4 Newbie   Joined: Mar 2018 From: Belize Posts: 7 Thanks: 0 Occultations: As I explained, both tables also identify semiprimes (composites whose divisors are a pair of primes). You rightly point out 91 and 133 - as I did above. So both tables DO work.
 March 19th, 2018, 07:19 PM #5 Newbie   Joined: Mar 2018 From: Belize Posts: 7 Thanks: 0 SDK: thanks for the reply. Every prime occurs and more peculiar, each pairs with another. No other numerical base will line up with base 10 and hit every single prime number, plus the occasional super semis (two between 0 and 100), every single prime however is paired. No prime number is hanging out with a composite between 0 and 4000 except for specific non-equal divisory semiprimes. Just from knowing that 5 is a prime number, I now see on my table that 11, 23, 53, 131, etc. are also primes. But the pair 27 and 63 are not. Nor are 39 and 93. This is the bit that excited me. How does it do it? Agreed, the small integers have many primes and semiprimes. Semiprimes galore, pairing up a lot, that is not my interest; like, 49 which has a single prime derivative. My table is not interested either. So, I will extend my table beyond 6 digits and get back to you on this. Again, thanks for your input...
March 21st, 2018, 05:45 AM   #6
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Quote:
 Using our base 10 numerical system adjacent to a base 4, I find that all primes pair with another and identify each other further up the table without the need for a formula (example 17=41).
you have used a formula because a formula is necessary in order to write the base 4 representation of a base 10 number, the digits of the base 4 representation of the base 10 numbers are the coefficients of the 4-adic expansion of that number.

Take a combinatorial approach to see that an infinite set of base n representations of numbers,if taken to be natural numbers are then subsets of any infinite set of base m representations of numbers if m>n.

So yes, of course we will see patterns for corresponding numbers in two different number base representations, we calculate one from the other. I have written such a formula, and I'm sure there are other ways of number base conversion that others use, like I said, simply extracting the coefficients of a numbers p-adic expansion and putting them along side one another gives the base p representation of the number.

But as far as any special relationship between specifically base 10 and base 4 (other than what will be a direct consequence of what I have said) no you are actually going insane. But if you want to read about a subject that is as close as possible as the kind of relationship in the sense of it being orientated around a specific natural number, I would recommend Quadratic Reciprocity, then later when you want you delusions of grandeur concerning 2 squared to be shattered, move onto Generalized Reciprocity Laws.

But anyway I see that your 2nd table is correct in the left hand side of the = is the base 10 and the right hand side is the base 4 for a given number, but the first table is not base 10 adjacent to base 4 representations of numbers, and you have not provided any intelligent explanation as to how your table was constructed, so I just cant help you with your project until you do. But if you still don't believe me I can post about any number of arrays of bois that have my eyebrows twitching at night, (granted, there has to be some sort of foundation for the heuristic belief I am adopting as I study those patterns) and if you are prepared to multiply your workload as per needed to assist me, then yes, we can be friends, welcome brother.

Last edited by Adam Ledger; March 21st, 2018 at 06:01 AM. Reason: extra anger felt

 March 21st, 2018, 12:47 PM #7 Senior Member   Joined: Sep 2016 From: USA Posts: 383 Thanks: 207 Math Focus: Dynamical systems, analytic function theory, numerics I was intrigued by this example which serves as a great example of the "law of small numbers" which is a tongue in cheek statement that there are "too many" small numbers to fill all the roles required of them so many coincidences crop up. Here is a proof that your construction is a coincidence, though it is a neat example anyway and I may steal it to give as an assignment the next time i teach computational math. Suppose $p$ is a prime in your enumeration which is in the 2nd column. Note that these pairs all have the form $(4k + 1,10k + 1)$ for some $k \in \mathbb{N}$ so there are two cases. If $n$ is the left member in its pairing, then the right member is $10\frac{p-1}{4}$ which is also prime. On the other hand, if $n$ is the right member, then its partner is $4\frac{p-1}{10}$. A similar analysis shows that if $p = 3 \mod 4$, then its partner lies in $[4 \frac{p-1}{10}, 10 \frac{p-1}{4}]$. We will count (asymptotically) all of the instances where $p$ is paired with another prime and show that there are too many primes in this interval for every prime to have a prime partner. Let $\pi(x)$ be the prime counting function and note that the prime number theorem gives the asymptotic estimate $\pi(x) \to \frac{x}{\log x}.$ Fix $k,M \in \mathbb{N}$, and let $I = [M,kM]$ and $J = [M/4,10kM]$. By the prime number theorem, the number of primes in $I$ is given (asymptotically) by $\pi(kM) - \pi(M) \approx \frac{M \log(\frac{M^{k-1}}{k})}{\log M \log(kM)} := c_1$ and the number of primes in $I \setminus J$ is $\pi(10kM) - \pi(kM) + \pi(M) - \pi(M/4) \approx \frac{\frac{M}{4} \log\left(\frac{\frac{M}{4}^{40k-1}}{40k}\right)}{\log(kM) \log(\frac{M}{4})} - c_1 := c_2$ We will estimate $\frac{c_2}{c_1}$ and we notice that if your scheme truly pairs primes together, then primes in $I$ are in bijection with primes from $I\setminus J$ so we would have to have $\frac{c_2}{c_1} \to 1$ as $k,M \to \infty$. To prove this isn't true, we have the following simple estimates from calculus. Estimate 1: For any constant $c \in \mathbb{R}$, we have $\lim\limits_{M \to \infty} \frac{\log cM}{\log M} = 1.$ This is easily proved using L'Hospital's rule. Estimate 2: For any constants $a,b,\kappa_1,\kappa_2$, we have $\lim\limits_{M \to \infty} \frac{\log (\frac{M^a}{\kappa_1})}{\log(\frac{M^b}{\kappa_2}) } = \frac{a}{b}.$ This follows from the decomposition $\log(\frac{M^a}{\kappa_1}) = a \log(M) - \log(\kappa_1)$ which is a polynomial in $\log(M)$ and likewise for the denominator. Now, we want to estimate the limit $\lim\limits_{M \to \infty} \frac{c_2}{c_1} = \lim\limits_{M \to \infty} \frac{\frac{M}{4} \log\left(\frac{\frac{M}{4}^{40k-1}}{40k}\right) \log M \log(kM)} {\log(kM) \log(\frac{M}{4}) M \log(\frac{M^{k-1}}{k})} - 1.$ Applying estimate 1, we can take $M$ large enough so that $\frac{\log M \log (kM)}{\log (kM) \log(\frac{M}{4})} \approx 1$ so we have $\lim\limits_{M \to \infty} \frac{c_2}{c_1} = \lim\limits_{M \to \infty} \frac{\log \left(\frac{\frac{M}{4}^{40k -1}}{40k}\right)}{4 \log \left(\frac{M^{k-1}}{k}\right)} - 1.$ Applying estimate 2, we can take $M$ large enough so that $\frac{c_2}{c_1} \approx \frac{(40k-1) \log(\frac{M}{4})}{4(k-1) \log M} - 1$ where another application of estimate 1 gives us the limit $\lim\limits_{M \to \infty} \frac{c_2}{c_1} = \frac{40k-1}{4(k-1)} - 1 > 9.$ Hence, if $M$ is taken large enough, we have the lower bound $\frac{c_2}{c_1} \geq 9$ which holds for all $k$. Note that this proves there are too many primes in $I \setminus J$ for the primes in $I$ to cover them all i.e. lots of primes can't possibly be paired up with other primes. In fact, if you work a bit harder, you can justify exchanging the limit $k \to \infty$ with $M \to \infty$ and show that the density of primes which are paired to other primes in your scheme goes to zero i.e. almost every prime appearing in your list will be paired with a non-prime. Thanks from Adam Ledger and AlanRGR
 March 21st, 2018, 12:53 PM #8 Banned Camp   Joined: Apr 2016 From: Australia Posts: 244 Thanks: 29 Math Focus: horses,cash me outside how bow dah, trash doves that's going to take me a while to digest oh dear I don't know If approve of this calculus it feels heretical some how but I will have to make a note for next week to sit down and take a look at this one I don't have a proof of coincidences in my bag of tricks yet
 April 8th, 2018, 12:56 PM #9 Newbie   Joined: Mar 2018 From: Belize Posts: 7 Thanks: 0 Thanks to you, SDK and others for your input. I have done extensive research in the last 10 days, including reference to Fermat's Little Theorem, producing my own modular arithmetic formula for picking out prime numbers into the millions and have now completed a whole paper on the subject (though still a work in progress). I am very grateful to all who have responded and I have tried to interact with all comments here on My Math Forum regarding my post, in the paper. My paper is too large to present here at the forum so wonder if it is acceptable to place a link to my website and the page on Prime Identifiers. If so, I paste it below: Prime Number Identifier incredibly Pairs and Links even semiprimes I will continue to work on this subject and continue to check back with you all. Of course, any other math topic that stumps me will certainly come your way. Alan
 April 8th, 2018, 01:08 PM #10 Newbie   Joined: Mar 2018 From: Belize Posts: 7 Thanks: 0 OK, brilliant and thanks a million, SDK. Actually I just wrote a full paper on the subject and refer specifically to primes in the millions. Remarkably my new modular arithmetic notations pick out primes and link them down time and again to integers of less than one hundred. Please see my last response to Adam Ledger above for a link to my paper (if possible). What a great reply to my conundrum. Thank you so much.

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