playmath

a boy’s math playground; math experiment

Cody Luo(cody@ustc.edu)

https://github.com/a-boy/playmath
https://nbviewer.jupyter.org/github/a-boy/playmath/tree/master/

Records:

• :star: 2019-09-02, I proved Goldbach’s Conjecture! stage12-try to prov Goldbach Conjecture.ipynb
  Note: in Sagemath env to run the codes

  Goldbach Conjecture Inequality 1: gold(n) < prime_pi(n)+sigma(n,0)
  gold(n): the min non-negative integer g makes that both n-g and n+g are primes
  prime_pi(n): the count of primes in 1..n
  sigma(n,0): the count of n.divisors()

  gold(n) < prime_pi(n), while n>344

  gold(n) < prime_pi(n)*4395/3449751 ≈ prime_pi(n)*0.0013, while n>57989356

  Goldbach Conjecture Inequality 2: gold(n) < prime_pi(prime_pi(n)+n)

• :star: 2019-10-28, I solved 3n+1 Problem! http://a-boy.tk/playmath/stage26-3n%2B1conjecture/proving-3n%2B1-conjecture.html
  

# Syracuse function g(n)
def g(n):
    while n%2==0 : n/=2
    n=3*n+1
    while n%2==0 : n/=2
    return n

Collatz-Syracuse Decent Theorem: For any odd positive integer n=2*k+1, it exists s1,s1<=g(n)<=(3*n+1)/2 to make nest(g,s1,n)==1;
Except n=27 or 31, it will get a less number m, m<n before n times iterately calling g(x).

(n,f(n),g(n),(3*n+1)/2,s(n))  
(27, 82, 41, 41, (37, 41))  
(31, 94, 47, 47, (35, 39))  

Collatz-Odd-Tree1.png

Collatz正奇数回归树生成规则(Collatz-Odd-Tree Generation Rule):

1. x在完全的Collatz-Odd-Tree中是叶节点 iff (x%3==0)
2. 每个节点的长子由 v(x)=(2*x-1)/3 or (4*x-1)/3 产生
3. 其余每个小兄弟由 h(x)=4*x+1 迭代陆续产生, 因为 3(4x+1)+1 = 12x+4 = 4(3x+1) 。

证明3n+1猜想成立也就只需证明Collatz-Odd-Tree中逆向生成了所有的正奇数。 显然，从x0=1出发，通过 h(x)=4*x+1 和 v(x)=(2*x-1)/3 or (4*x-1)/3 反复迭代，会生成所有形如4k+1和4k-1的数，即所有正奇数。Collatz猜想证明完毕□

• :star: 2018-03-06, I proved Twin Primes Conjecture! stage9-Prime gap subsegment if repeats twice then infinitely times.nb

证明孪生质数猜想，并提出更普遍的规律：任意质数阶差子段如果出现了两遍就会继续出现无数遍，例如{2},{4},{2*n},{6,6},{2,4,2},......

• :star: 2010-02-04, I discovered Prime-Gap-Inequality: The i-th prime gap p[i+1]-p[i]<=i In other words, range(n,n+primepi(n)) contains one or more primes. So, the i-th prime p[i]<=1+2+...+ i-1 + p[1] = i*(i-1)/2 +2

• :star: I discovered Bread Curve and Bread Model by chance in 2011:

def r(theta):= nest(sin,theta,1000)
polar_plot(r(theta),(theta,0,2*PI))

• before2011/果中的泪滴.png
  

• :star: 2022-02-07, I improved Oppermann’s Conjecture!

NextPrime-Of-Square-Inequality: next_prime(n^2) - n^2 <= 1+euler_phi(n)

idea:(mailto:a_boy@live.com)

• Goldbach-Triangle: every item is the average of top and right primes.

3
4	5
5	6	7
7	8	9	11
8	9	10	12	13
10	11	12	14	15	17
11	12	13	15	16	18	19
13	14	15	17	18	20	21	23
16	17	18	20	21	23	24	26	29
17	18	19	21	22	24	25	27	30	31
20	21	22	24	25	27	28	30	33	34	37

• n>=3, let p=nextprime(n!)-n! , then p is always prime or 1, because p is less than ` nextprime(n)^2 , very often p<n^2 ` .

• denote S(k,v):=RamseyNumber(k+1,v+1)-1 . Conjecture: for any integer v>=2, S(2,v)%5 in {0,2,3}, here S(2,v)=RamseyNumber(3,v+1)-1 . That is to say, RamseyNumber(3,v+1) can NOT be the form 5k or 5k+2

m	n	R(m,n)	Reference
3	3	6	Greenwood and Gleason 1955
3	4	9	Greenwood and Gleason 1955
3	5	14	Greenwood and Gleason 1955
3	6	18	Graver and Yackel 1968
3	7	23	Kalbfleisch 1966
3	8	28	McKay and Min 1992
3	9	36	Grinstead and Roberts 1982
3	10	[40, 43]	Exoo 1989c, Radziszowski and Kreher 1988
3	11	[46, 51]	Radziszowski and Kreher 1988
3	12	[52, 59]	Exoo 1993, Radziszowski and Kreher 1988, Exoo 1998, Lesser 2001
3	13	[59, 69]	Piwakowski 1996, Radziszowski and Kreher 1988

5	5	[43, 49]	Exoo 1989b, McKay and Radziszowski 1995
6	6	[102, 165]	Kalbfleisch 1965, Mackey 1994
7	7	[205, 540]	Hill and Irving 1982, Giraud 1973

Guess: for any ineteger n>=1, RamseyNumber(n+1,n+1)-1 = S(n,n) contains only the factors of Fermat Numbers ` F[m]=2^2^m+1 `, {1,2,3,5,17,257,641,65537,…} S(2,2)=5, S(3,3)=17, I guess S(4,4)=45

• try to prove Twin Prime Conjecture ```
  1. Method 1: If a prime gap subsequence repeats twice, then it will occur infinitely times. such as {2}, {4},{2*k},{2,4,2},{6,6}…
  2. Method 2: by using modern database function groupby on prime gap sequence, …
     Observing prime gap frequency distribution for primes up to some big integer N0 .
     Peaks occur at multiples of 6. And the ratio of {2}s to {6}s will be great than a const(0.5 ?). 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4, 6, 2, 10, 6, 6, 6, 2, 6, 4, 2, … (sequence A001223 in the OEIS). ```

• define the integer sequence x[n+1]:=x[n]^2+1, if take x[0]>1, then x[5] is very often composite.

• 定义:二密分解 n=q1*q2 , q1取小于或等于√n的最大因数, q2取大于或等于√n的最小因数。 是否值得尝试，使用二密分解或p-密分解的一些性质证明费马大定理? 抛开Wiles的复杂理论和过程?

• OPPC(Odd Primes Position Constant) 奇质位常数，2*k-1如果是质数则二进制小数点后第k位为1，否则为0

const OPPC = .0111 0110 1101 0011
               1357 9

• 可能将圆周率Pi的二进制值分段，形成一个有规律的数组吗？如果可以的话，计算Pi的值就存在与之对应的极速算法。

  1   2   4    135     60753
  1|1.0|0100|10000111|1110110101010001|00010000101101000110000100011010011000100110001100110001010001011100000001101110000011100110100010010100100000010010011100000100010001010011001111100110001110100000000100000101110111110101001100011101100010011100110110010001001
  

  Question

• Ultra-Primes-Conjecture ``` There exists infinitely many primes formed of 2^(F[n!+1])-1 . here 2^p-1 is Mersenne number, F[n+2]=F[n+1]+F[n], 0,1,1,2,3,5,8,… is Fibonacci number, n! is factorial.

https://math.stackexchange.com/questions/3503947

2^F[2!+1]-1 = 3, 2^F[3!+1]-1= 2^13-1= 8191 are two known Ultra-Primes! I guess after 3, 8191 there are more prime heros! ```

• I think there exists Inertia Law and Polar Method for integer sequences. 整数数列的惯性定理和极性定理

math tools

• Mathematica : .nb is Mathematica notebook
• Sagemath : .ipynb is Jupyter notebook, most of *.ipynb files in a-boy/playmath are used Sagemath, to view by Jupyter nbviewer, but to run codes only in Sagemath env; .sagews is Sagemath worksheet
• Maple
• GeoGebra
• Octave
• MathType

赞助支持

感谢你的赞助和支持，希望我们每天都能做得更好！！！