February 2020 - mathfun@mailman.xmission.com

[math-fun] about p(n) and pi(n)
by Simon Plouffe 23 Feb '20

23 Feb '20

Hello everybody, I have been working on prime numbers recently and I have computed a formula for p(n) and pi(n) using Lambert W function. I used the formula from Dusart (2010) which states that pi(n) = n/(log(n)-1) when n > 5393. By inverting this formula, one can find a formula for p(n) : the n'th prime number. p(n) = -n*W(-1,-e/n). It is quite precise, when n=10^24, the approx is 0.999 from the real value, much better than p(n) = n*log(n) (the classical formula). >From that, I worked on a more precise expression by analyzing the remainder which <looks> like a logarithmic curve. Since the remainder is more or less logarithmic, I put a second term in the first formula which is n/W(0,n) or n/W(n) for short. It gives then p(n) = -n*(W(-1,-exp(1)/n)) - n/W(0,n) Again, by analyzing the remainder, one can find that we still have a logarithmic type of curve (this is like matrioshkas), the russian puppets. Since it looks like a log curve again then I pose p(n) = -n*(W(-1,-exp(1)/n)) - X*n/W(0,n) and solve for each n, here typically X = 0.8 to 1. I use newton bisection method to find X. Once X is found, I calculate the LLS : logarithmic least square curve for X. By using a table for p(n) up to 10^17 , I could calculate (for example ) that p(n) = -n*(W(-1,-exp(1)/n)) - c(n)*n/W(0,n) where c(n) = 0.00074741174603301665420395275429537*ln(n)+0.88596350453664534160747106131754 Once found we can now calculate a few values to see that : it is more precise than the Cipolla-Salvy formula for the n'th prime. Now, since I have a formula for p(n) : if I take the functional inverse I can produce a formula for pi(n) : In short we have : p(n) = -n*(W(-1,-exp(1)/n)) - c(n)*n/W(0,n) pi(n) = {-n*(W(-1,-exp(1)/n)) - c(n)*n/W(0,n)} ^ (-1). The calculation of pi(n) is easy and fast. It is also : more precise than Li(n) , and even <almost> precise as Riemann-R or Gram series. Using a table of p(n) and pi(n) up to 744 billion from 100000 (10^5 ) to 744 billion. the formula for pi(n) has a better average than Riemann R or Gram series. For the scale 10 billion (10^10) up to 10^16 : almost, not as Riemann-R or Gram, but definitively more precise than Li(n). Here are the details of this : http://plouffe.fr/NEW/Une%20formule%20pour%20les%20nombres%20premiers%20II.… The article is in french, I am working now on the english version. Best regards, Simon Plouffe

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[math-fun] Occurrences
by Éric Angelini 23 Feb '20

23 Feb '20

If I'm not wrong, there are more occurrences of <occurence> than occurrences of <occurrence> in the OEIS. https://www.merriam-webster.com/dictionary/occurrence É. Catapulté de mon aPhone à+ É. Catapulté de mon aPhone

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[math-fun] sum of 2 triangular numbers is a triangular number
by Neil Sloane 22 Feb '20

22 Feb '20

Let T(i) = i(i+1)/2. Given n, let k be smallest number such that T(n) + T(k) = T(m) for some m. The k and m values are in A082183 and A082184. It must be classical that k and m always exist. - can someone supply a reference or a proof? The graph of the k values is quite irregular. Is there an upper bound?

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[math-fun] Imbalance
by Éric Angelini 21 Feb '20

21 Feb '20

Hello Math-Fun, In this country, digits go by closed boxes containing 10 distinct digits each. To write the number 2019 you can open a single box and extract the digits 2, 0, 1 and 9. We call "imbalance" the quantity of unused digit in one (or more) opened box(es). So, writing 2019 leaves an imbalance of 6 (the 6 unused digits in the box). Now, if we want to write 2020 we must open two boxes (...) The question: "Is S infinite?" is explained there, on my personal blog page: http://cinquantesignes.blogspot.com/2020/02/integer-imbalance.html Best, É.

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[math-fun] little memories
by françois mendzina essomba2 21 Feb '20

21 Feb '20

Solve without using Bezout let's take this example 879*U[0]-13723*V[0]=1 Step1 13723 : 879 = 15,6120 879(15)-13723(1)=-532 879(15,6*10)-13723(1*10)=-106 879(15,62*100)-13723(1*100)=698 Step 2 698 :106=6,5 We take as integers 6 and 7 a) 879(156*6)-13723(10*6)=-636 879(1562)-13723(100)=698 ------------------------------------------ 879(2498)-13723(160)=62 b) 879(156*7)-13723(10*7)=-742 879(1562)-13723(100)=698 ------------------------------------------ 879(2654)-13723(170)=-44 a+b) 879(2498)-13723(160)=62 879(2654)-13723(170)=-44 ---------------------------------------- 879(5152)-13723(330)=18 Step 3 44 :18=2,4 We use integers 2 and 3 c) 879(5152*2)-13723(330*2)=36 879(2654)-13723(170)=-44 ------------------------------------- 879(12958)-13723(830)=-8 d) 879(5152*3)-13723(330*3)=54 879(2654)-13723(170)=-44 ------------------------------------- 879(18110)-13723(1160)=10 c+d) 879(12958)-13723(830)=-8 879(18110)-13723(1160)=10 -------------------------------------- 879(31068)-13723(1990)=2 After simplification by 2 879(15534)-13723(995)=1

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Re: [math-fun] Varun, is your Euler Circle still doing combinatorics?
by Bill Gosper 20 Feb '20

20 Feb '20

On Wed, Feb 19, 2020 at 9:07 PM Simon Rubinstein-Salzedo < complexzeta(a)gmail.com> wrote: > We aren't doing things like that in Varun's class. But Varun, you have my > material from the analytic number theory class from last spring. This > problem is similar to some that are discussed in chapters 8 and especially > 9 of that book. > > Simon > Thanks, Simon! Varun, from Chapter 9 can you get (Awesome) Julian's result? He says "It's a number theory question. The answer is roughly 1.78*log(D)/log(N), for D small compared to N (using Mertens' theorem). This is a poor approximation for your case, though, which has D close to √N; in that case, 1 – ((log log√N) – log log D) should be a decent approximation (again using Mertens' theorem)." This fits my experience! [1.78 ~ e^𝜸.] —Bill > > On Wed, 19 Feb 2020 at 19:25, Bill Gosper <billgosper(a)gmail.com> wrote: > >> On Wed, Feb 19, 2020 at 2:25 PM Varun Rajkumar <pi.fermet(a)gmail.com> >> wrote: >> >>> Bill, my eulercircle class is doing combinatorics. But your email was >>> not asking that question. >>> >> >> Varun, you may be right. My question was >> " What is the probability that N is prime if it's coprime to D! ?" >> >> What subject is this if not Combinatorics? Number Theory? >> Combinatoric Number Theory?? >> —Bill >> On Thu, Feb 13, 2020 at 9:30 PM Rudy Rucker <rudy(a)rudyrucker.com> wrote: >> >>> From the peanut gallery. >>> >>> When I can't sleep I factor all the numbers less than a hundred, usually >>> working down from 99. And going off on tangents. And if I went to sleep >>> in the 70s the day before I restart there. If I feel extra ambitious I >>> jump up into numbers less than 200, but blessed sleep soon comes. In the >>> same vein, I had a bout of computing larger and larger Fibonacci series >>> numbers this fall. >>> >>> Factoring 5 and 6 digit numbers in my head? Not yet... >>> On 2/13/2020 7:43 PM, Bill Gosper wrote: >>> >>> Here's a question I often ask myself but am too lazy to attack. (And >>> I'm working on a 3D print.) >>> I like to factor 5 or 6 digit numbers in my head. There's a bag of >>> tricks for finding small factors, >>> but eventually you wind up slogging through division by consecutive >>> primes. When I've reached, >>> say, ¾ √N, I often get bored|tired and then cheat by trying PrimeQ, and >>> then when it says True, I'm >>> glad I cheated. >>> It seems to me that this happens improbably often. What is the >>> probability that N is prime if it's >>> coprime to D! ? —Bill >>> >>> -- >>> Rudy Rucker | rudy(a)rudyrucker.com | >>> >> On Thu, 13 Feb 2020 at 22:48, Bill Gosper <billgosper(a)gmail.com> wrote: >>> >>>> http://www.purplestreak.com/g4g13/mentalfactoring.pdf >>>> >>>> >>>>>

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Re: [math-fun] Varun, is your Euler Circle still doing combinatorics?
by Bill Gosper 20 Feb '20

20 Feb '20

On Wed, Feb 19, 2020 at 2:25 PM Varun Rajkumar <pi.fermet(a)gmail.com> wrote: > Bill, my eulercircle class is doing combinatorics. But your email was not > asking that question. > Varun, you may be right. My question was " What is the probability that N is prime if it's coprime to D! ?" What subject is this if not Combinatorics? Number Theory? Combinatoric Number Theory?? —Bill On Thu, Feb 13, 2020 at 9:30 PM Rudy Rucker <rudy(a)rudyrucker.com> wrote: > From the peanut gallery. > > When I can't sleep I factor all the numbers less than a hundred, usually > working down from 99. And going off on tangents. And if I went to sleep > in the 70s the day before I restart there. If I feel extra ambitious I > jump up into numbers less than 200, but blessed sleep soon comes. In the > same vein, I had a bout of computing larger and larger Fibonacci series > numbers this fall. > > Factoring 5 and 6 digit numbers in my head? Not yet... > On 2/13/2020 7:43 PM, Bill Gosper wrote: > > Here's a question I often ask myself but am too lazy to attack. (And I'm > working on a 3D print.) > I like to factor 5 or 6 digit numbers in my head. There's a bag of tricks > for finding small factors, > but eventually you wind up slogging through division by consecutive > primes. When I've reached, > say, ¾ √N, I often get bored|tired and then cheat by trying PrimeQ, and > then when it says True, I'm > glad I cheated. > It seems to me that this happens improbably often. What is the > probability that N is prime if it's > coprime to D! ? —Bill > > -- > Rudy Rucker | rudy(a)rudyrucker.com | > On Thu, 13 Feb 2020 at 22:48, Bill Gosper <billgosper(a)gmail.com> wrote: > >> http://www.purplestreak.com/g4g13/mentalfactoring.pdf >> >> >>>

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[math-fun] Fixed Point Question
by Tom Karzes 20 Feb '20

20 Feb '20

Does a function which is a scaled proper rotation plus an offset in R^n always have a fixed point? More precisely, let R be a proper rotation matrix in R^n. Reflections are excluded. So R is an orthogonal matrix with determinant 1. Let s be a scale factor, 0 < s < 1. Let A be the matrix s*R, combining the scaling and the rotation. Let b be a vector in R^n. Then: f(v) = A*v + b It's clear that f can have at most one fixed point, since f changes the distances between points by a factor of s. Since the distance between two fixed points must remain constant, there can be at most one. I want to know if f always has a fixed point, or if there are examples where it does not. I.e., can I always find a fixed point vf that satisfies: vf = A*vf + b So we need vf to satisfy: (I - A)*vf = b where I is the identity matrix. If the determinant of (I - A) is zero, then there should be either no solution or multiple solutions. But we already know that there can be at most one solution, so I believe the question is equivalent to asking whether: det(I - A) = 0 If so, then I believe there is no fixed point. Otherwise we have: vf = inv(I - A)*b Now consider the following. If b is the zero vector, then vf exists and is also the zero vector, for any A that satisfies the problem definition. This implies that det(I - A) cannot be zero for any A that satisfies the problem definition. This is independent of the value of b. So (I - A) always has an inverse, and f always has exactly one fixed point. Note that this argument works if R has determinant -1 as well, i.e. if R includes a reflection, so R can be any orthogonal matrix. The argument also works for s > 1. Does this reasoning sound correct? Or did I miss something? Thanks, Tom

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[math-fun] Palindromes multiplied
by Éric Angelini 19 Feb '20

19 Feb '20

Hello Math-Fun, is this seq finite (and correct)? S = 1,2,3,11,4,22,101,5,111,6,1001,7,88,77,8,1111,9,... (my knowledges about repunits and palindromes are close to the palindrome 0 themselves) Definition: Lexicographically earliest seq of distinct palindromes such that the product of two successive palindromes is a palindrome. Best, É.

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[math-fun] Fibonacci
by Brent Meeker 18 Feb '20

18 Feb '20

My son just told me that his new Fibonacci joke is as bad as the previous two Fibonacci jokes put together. Brent

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