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[math-fun] Arkansas Events in March.
by Brad Klee 23 Feb '20

23 Feb '20

Flyer: https://0x0.st/iNl4.pdf . Thanks to the Arkansas Arthropod museum, we are also starting to get better entomological data, see for example: https://0x0.st/iNIj.jpg (the genus Phanaeus is an incredibly interesting case study for evolution, having travelled all across the Americas, into 50+ species). --Brad

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[math-fun] Progress in k-imperfect numbers
by Allan Wechsler 23 Feb '20

23 Feb '20

I'm including Michel Marcus because (a) I don't know if he's on this mailing list, and (b) he has helped me a lot in this research. The search for k-perfect numbers is well-known. We consider the function sigma(n), the sum of the divisors of n. Because sigma(n) >= n (with equality only when n = 1), it makes sense to look for numbers n such that sigma(n) = kn for some k; such a number is called k-perfect. (The "classical" perfect numbers are 2-perfect in this terminology.) The k-perfect numbers, for all indices k, are presented at OEIS sequence A007691. The following variation is due to Greg Martin, who presented it at the Western Number Theory Conference at Asilomar in 1999; it is "Western Number Theory problem 99:08". (See Myerson's compendium of WNT problems at https://www.math.colostate.edu/~achter/wntc/problems/problems2001.pdf .) Douglas Iannucci made some progress on the problem in his 2006 paper -- see http://math.colgate.edu/~integers/g41/g41.pdf . I'm using Iannucci's notation and nomenclature. Suppose d is a divisor of n. Consider the number of prime factors of the codivisor n/d, counting multiplicity. Call this the index of d in n. For example, the index of 2 in 12 is 2, because 12 = 2*(2*3). The index of 10 in 1000 is 4, because 1000 = 10*2*2*5*5. Now compute a _weighted_ sum of the divisors of n, where the weight of a divisor is 1 if the divisor has even index, and -1 if its index is odd. Call this weighted sum rho(n). For example, rho(12) = 12 - 6 - 4 + 3 + 2 - 1 = 6. Note that rho(n) <= n (with equality only when n = 1), so it makes sense to look for numbers n such that k*rho(n) = n for some k; Iannucci calls such a number k-imperfect. (I personally would have preferred "k-contraperfect", but that ship has sailed.) Because 12 = 2*rho(12), 12 is 2-imperfect. The function rho(n) is multiplicative, with rho(p^e) = p^e - p^(e-1) + ... +/- 1; the sign of the trailing unit depends on the parity of e. Like k-perfect numbers, there are lots of known examples of k-imperfect numbers. Many of these are listed at https://oeis.org/A127724 . Because they get big so quickly, we soon lose track of whether they are consecutive; we are pretty sure we know the smallest 50, but after that, the gaps have not been searched exhaustively. About 1800 examples are known; part of my recent research has added around 200 new ones. While k-perfect numbers are known for all 1 <= k <= 11, for k-imperfect numbers, we only had examples of k = 1, 2, 3, or 4 ... until earlier today, when I found one with k = 5. Martin and Iannucci had only found values of k up to 3, but many examples with k = 4 were subsequently found by Corneth, Johnson, Lelechenko, and Marcus (Michel Marcus found more than a thousand). We didn't know for sure if there were any 5-imperfect numbers until this afternoon. The smallest known 4-imperfect number is 993803899780063855042560 (24 digits). My discovery today is so far the only known 5-imperfect number; it has 208 digits, and the exact value is 1947793410288108579327587698415272737289992039373107522449638016140636142596276017072442826236838486130285072853813458640948347868163450516845165654669897619318450994951647517899051394662400000000000000000000. Michel Marcus and I have both checked it (he with a package he wrote in Pari, I with a complete cludge in Emacs Lisp). I would welcome additional confirmation. I have just emerged from a "dry valley" in my search space, and expect to find at least a few more 5-imperfect numbers in the next few days, but I thought I shouldn't let the occasion of the discovery of the first example go unmentioned.

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[math-fun] a formula for primes
by Simon Plouffe 23 Feb '20

23 Feb '20

Hello, Here is the english version of the article about primes : http://plouffe.fr/NEW/a%20formula%20for%20primes.pdf Best regards Simon Plouffe

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[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 22 Feb '20

22 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 20 Feb '20

20 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 19 Feb '20

19 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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