# a-boy / playmath

a boy's math playground; math experiment | QQ群 499866721 |

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# playmath

a boy's math playground; math experiment

## Records:

``````# 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=2k+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). To be brief, it exists `s0,s0<n` to make `m= nest(g,s0,n)<n` except n=27 or n=31;
(n,f(n),g(n),(3
n+1)/2,s(n))
(27, 82, 41, 41, (37, 41))
(31, 94, 47, 47, (35, 39))

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

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

• 2010-02-04, I discovered Prime-Gap-Inequality: The i-th prime gap `p[i+1]-p[i]<=i`

• 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))

``````  ## idea:(mailto:a_boy@live.com)

• Prime-Gap-Inequality: The i-th prime gap `p[i+1]-p[i]<=i` . In other words, `range(n,n+primepi(n)+1)` contains one or more primes. So, the i-th prime `p[i]<=1+2+...+ i-1 + p = i*(i-1)/2 +2`
• `range(n^2, n*(n+1))` contains at least one primes. This is because: as to the array `{n*n, n*n+1, n*n+2, ... ,next_prime(n*n)-1 }`, we can dispatch distinct real factors for every item, these real factors are in `{2,3,4,...,n}` , Pigeonhole principle shows `next_prime(n*n)< n*(n+1)` . Samely, `range(n*(n+1), (n+1)*(n+1))` contains at least one prime. so, this is one sentence proof for
``````"still unsolved Legendre's conjecture asks whether for every n > 1, there is a prime p, such that n^2 < p < (n + 1)^2 "

https://en.wikipedia.org/wiki/Legendre%27s_conjecture
``````

In fact,it holds true that

NextPrime-Square-Inequality; `next_prime_delta(n^2) <= 1+euler_phi(n)`

• Try to prove Goldbach's Conjecture
``````    Except n=344,for any integer n>=2, there exists g=gold(n) , 0=<g<=primepi(n)+1, satisfies that both n-g and n+g are primes.

?? because we can dispatch distinct prime factors for {n+1, n+2, ... n+g}.replace( isprime(n+x) => n-x) ?? why n=344 is an exception?
``````
• 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` . Conjecture: for any n>=3, `n! - prevprime(...(prevprime(n!)))` is always prime or 1, the count of nest `prevprime` can be from 1 to floor(sqrt(n)) times.

• 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 . 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,...} 现在地球居民都知道 S(2,2)=5, S(3,3)=17, 我猜 S(4,4)=45, S(5,5)=

• 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}...
1. 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>1, then x is always composite, never be prime. I guess that x[k] can always be written as another form of two square sum, k>=5in.

• 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 = 0.0111 0110 1101 0011
1357 9
``````

## Question

• sum(1/(p[2k-1]*p[2k]), k, 1, oo)==16/75 ?
``````1/(2*3)+1/(5*7)+1/(11*13)+1/(17*19)+1/(23*29)+... == sum(1/(p[2k-1]*p[2k]), k, 1, oo)==16/75 ?
sum(1/k^2, k, 1, oo)==1/6*pi^2  /*Basel problem solved by Euler*/
``````

https://math.stackexchange.com/questions/3503866

• 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 is the only 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. But I can not express them clear! 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
• SketchUp

#### 赞助支持  