Product with polynomial roots

by fungarwai, Mar 18, 2024, 11:35 AM

Vandermonde polynomial type

$\displaystyle \prod_{1\le i<j\le m} (x_j-x_i)^2 =(-1)^{\binom{m}{2}}\prod_{j=1}^m f'(x_j), ~f(x)=\prod_{i=1}^m (x-x_i)$

Proof

Example
polynomial

The application of Synthetic division in Polynomial theory

by fungarwai, Feb 9, 2023, 4:40 AM

Power sum of polynomial roots

Synthetic division can be applied to find the power sums of roots of a polynomial $f(x)$
by dividing $f’(x)$ with $f(x)$

$f(x)=x^2+2x-3=0$ has two real roots $x_1=-3,~x_2=1$
$f’(x)=2x+2$

$\begin{array}{c|cccccccccccc} ~ & 2 & 2 \\ -2 & ~ & -4 & 4 & -20 & 52 & -164 & 484 & -1460 & 4372 & -13124 \\ 3 & ~ & ~ & 6 & -6 & 30 & -78 & 246 & -726 & 2190 & -6558 & 19686 \\ \hline ~ & 2 & -2 & 10 & -26 & 82 & -242 & 730 & -2186 & 6562 & -19682 & 19686 \end{array}$

The coefficients of the quotient show that
$\displaystyle \sum_{k=1}^2 x_k^0=2,~\sum_{k=1}^2 x_k^1=-2,~\sum_{k=1}^2 x_k^2=10,~\sum_{k=1}^2 x_k^3=-26,\dots$

This result is caused by the formula $\dfrac{f'(x)}{f(x)}=\sum_{k=0}^\infty \dfrac{1}{x^{k+1}}\left(\sum_{j=1}^n x_j^k\right)$

Reference: The application of Synthetic division in Polynomial theory (Chinese) Section 1.1.5

Proof

Example

Python program to generate Synthetic division in LaTeX code

  1. dividend=[27,-18,0];#input by user, 27x^2-18x for example
  2. divisor=[9,-9,0,1];#input by user, 9x^3-9x^2+1 for example
  3.  
  4. def roundint(x):
  5.  if x == int(x):
  6.   return str(int(x))
  7.  else:
  8.   for k in range(1,100):
  9.    T=0
  10.    if x*k == int(x*k):
  11.     T=1;break;
  12.   if T==1:
  13.    return "\dfrac{"+str(int(x*k))+"}{"+str(k)+"}"
  14.   else:
  15.    return str(round(x,3))
  16. qlen=11;e=0;
  17. dlen=len(divisor)-1;
  18. ddlen=len(dividend);
  19. dd="~ "
  20. for k in range(0,ddlen):
  21.  dividend[k]=dividend[k]/divisor[0]
  22.  dd=dd+"& "+roundint(dividend[k])+" "
  23. for k in range(0,qlen-ddlen+1):
  24.  dividend.append(0)
  25. dd=dd+"\\\\\n"
  26. for k in range(1,dlen+1):
  27.  divisor[k]=-divisor[k]/divisor[0]
  28. q=[]
  29. q.append(dividend[0])
  30. qq="~ & "+roundint(q[0])+" "
  31. s="$\\begin{array}{c|"
  32. s2=[]
  33. for k in range(1,dlen+1):
  34.  s2.append(roundint(divisor[k])+" ")
  35. for k in range(0,dlen):
  36.  for i in range(0,k+1):
  37.   s2[k]=s2[k]+"& ~ "
  38. for i in range(0,qlen-dlen):
  39.  t=dividend[i+1]
  40.  for k in range(0,dlen):
  41.   tt=q[i]*divisor[k+1]
  42.   s2[k]=s2[k]+"& "+roundint(tt)+" "
  43.   if i-k >= 0:
  44.    t=t+q[i-k]*divisor[k+1]
  45.  q.append(t)
  46.  if t==0:
  47.   e=e+1
  48.  else:
  49.   e=0
  50.  if e>=dlen and i>=ddlen-1:
  51.   break
  52.  qq=qq+"& "+roundint(t)+" "
  53. for i in range(qlen-dlen,qlen-1):
  54.  if e>=dlen and i>=ddlen-1:
  55.   break
  56.  t=dividend[i+1]
  57.  for k in range(0,dlen):
  58.   if i-k < qlen-dlen:
  59.    t=t+q[i-k]*divisor[k+1]
  60.  q.append(t)
  61.  qq=qq+"& "+roundint(t)+" "
  62. for k in range(0,qlen+1):
  63.  s=s+"c"
  64. s=s+"}\n"+dd
  65. for k in range(0,dlen):
  66.  s=s+s2[k]+"\\\\\n"
  67. s=s+"\\hline\n"+qq+"\\end{array}$"
  68. print(s)




pywindow code

			

			
This post has been edited 1 time. Last edited by fungarwai, Feb 9, 2023, 4:57 AM

				

				
number theory
polynomial
Summation

				

			

				

					
				

				

				

			

		


	

		
GCD of 2^n-1 and 2^m+1

		
by fungarwai, Feb 23, 2022, 1:46 PM 

		


			
$(x^n-1,x^m-1)=x^{(n,m)}-1$

Proof



$(n,p^k m)=\begin{cases}
(n,m) & \nu_p(n)\le \nu_p(m)\\
p^k(n,m) & \nu_p(n)\ge \nu_p(m)+k
\end{cases}$

Proof



If $(m_1,m_2)=p^k$,

$\quad(n,m_1 m_2)=\begin{cases}
p^{-\nu_p(n)}(n,m_1)(n,m_2) & \nu_p(n)\le\nu_p(m_1),~\nu_p(n)\le\nu_p(m_2)\\
(n,m_1)(n,m_2) & \nu_p(n)\ge \nu_p(m_1)+\nu_p(m_2)
\end{cases}$

Proof



$(2^n-1,2^m+1)=\begin{cases}
1 & \nu_2(n)\le \nu_2(m)\\
2^{(n,m)}+1 & \nu_2(n)\ge \nu_2(m)+1
\end{cases}$

Proof

Example



$(3^n-1,3^m+1)=\begin{cases}
2 & \nu_2(n)\le \nu_2(m)\\
3^{(n,m)}+1 & \nu_2(n)\ge \nu_2(m)+1\end{cases}$

Proof



$(x^n-1,x^m+1)=\begin{cases}(x-1,2) & \nu_2(n)\le \nu_2(m)\\
x^{(n,m)}+1 & \nu_2(n)\ge \nu_2(m)+1\end{cases}$

Proof

			

			
This post has been edited 3 times. Last edited by fungarwai, Feb 25, 2022, 12:32 PM

				

				
number theory

				

			

				

					
				

				

				

			

		


	

		
modular arithmetic on 1^1+2^2+3^3+...+n^n

		
by fungarwai, Dec 13, 2021, 1:45 PM 

		


			
To find the modular residue  $1^1+2^2+3^3+\dots+n^n \pmod{p^s}$,



the main idea is rewriting $1^1+2^2+3^3+\dots+n^n$ as

$1^1+2^2+\dots+(p-1)^{p-1}+p^p$

$+(p+1)^1+(p+2)^2+\dots+(2p-1)^{2p-1}+(2p)^{2p}+\dots$

$=\sum_{r=1}^{p-1}\sum_{k=1}^N (pk-p+r)^{pk-p+r}+\sum_{k=1}^N (pk)^{pk}+\sum_{k=N+1}^n k^k$



While $s$ is enough small, $\sum_{k=1}^N (pk)^{pk}\pmod{p^s}$ would be simply zero.



$\sum_{k=N+1}^n k^k$ has to be calculated manually. Make it as short as it can.



Here $\sum_{r=1}^{p-1}\sum_{k=1}^N (pk-p+r)^{pk-p+r}\pmod{p^s}$ is the main problem.



$p^{s-1}\mid N\implies \sum_{k=1}^N (pk-p+1)^{pk-p+1}
\equiv (1-\frac{p}{2})N\equiv 
\begin{cases} 0 \pmod{p^s} & p=2\\
N\pmod{p^s} & p\neq 2\end{cases}$



Proof



$\varphi(p^s)=p^{s-1}(p-1)\mid N \Rightarrow \sum_{k=1}^N (pk-p+r)^{pk-p+r}
\equiv 0 \pmod{p^s} (2\le r\le p-1)$



Proof



Example



Python program for general cases



input values for n, m to get $\sum_{k=1}^n k^k\pmod{m}$




  1. n = int(input("n="))
  2. m = int(input("m="))
  3. sum=0;
  4. for k in range(1,n+1):
  5.  p=1
  6.  for t in range(1,k+1):
  7.   p=(p*k) % m
  8.  sum=(sum+p) % m
  9. print(sum)




input values for N, p, s to get $\sum_{k=1}^N (pk-p+r)^{pk-p+r}\pmod{p^s}$ for $1\le r\le p-1$




  1. N = int(input("N="))
  2. p = int(input("p="))
  3. s = int(input("s="))
  4. sum=0;m=p**s;
  5. for r in range(1,p):
  6.  for k in range(1,N+1):
  7.   P=1
  8.   for t in range(1,p*(k-1)+r+1):
  9.    P=(P*(p*(k-1)+r)) % m
  10.   sum=(sum+P) % m
  11.  print(sum, "when r=",r)

			

			
This post has been edited 2 times. Last edited by fungarwai, Dec 14, 2021, 12:15 PM

				

				
Finite difference
number theory
Summation

				

			

				

					
				

				

				

			

		


	

		
Coincidence on Linear Regression Analysis with president election data

		
by fungarwai, Feb 16, 2021, 12:58 AM 

		


			
The Linear Regression model with president election data indicates that a candidate has stolen other candidates averagely 120 votes on each ward in Milwaukee

Here are data sources from Milwaukee

11-3-20 General and Presidential Election - Unofficial Results

My Regression Model is

Candidate's vote = Candidate's voting rate × Total votes

which is no constant term expected



Form data at Nov. 4, 2020

Biden's vote = 5.3191 + 0.5821 × Total votes

R^2=0.9087

Trump's vote = -6.6928 + 0.4002 × Total votes

R^2=0.8255

No constant term is expected.

This constant term is possibly caused by Biden stealing other candidate's vote.

However, the constant term lies on the convincible interval involving 0.



Without constant term, my Linear Regression model at Nov. 4, 2020 becomes:

Biden's vote = 0.5877 × Total votes

R^2=0.9085

Trump's vote = 0.3931 × Total votes

R^2=0.8262

which is Candidate's vote = Candidate's voting rate × Total votes



Guess what?



Form data at Nov. 5, 2020

Biden's vote = 121.6115 + 0.5629 × Total votes

R^2=0.8128

Trump's vote = -120.573 + 0.4171 × Total votes

R^2=0.7051

The constant term lies on the convincible interval far from 0.

My Linear Regression model indicates that Biden has stolen other candidates averagely 120 votes on each ward.



What a Coincidence!



That's all.

Here is my video concerning this Linear Regression Analysis if you're interested.



Possible leakage of this Linear Regression model

Suppose there are two different groups of votes (mail-in ballots and in-person ballots) with different voting rate

$\begin{cases}y_A=\hat{\beta_A} x_A,~\hat{\alpha_A}=\bar{y_A}-\bar{x_A}\hat{\beta_A}=0\\
y_B=\hat{\beta_B} x_B,~\hat{\alpha_B}=\bar{y_B}-\bar{x_B}\hat{\beta_B}=0\end{cases}$

$\hat{\alpha}=\frac{p_A p_B \bar{x_A}\bar{x_B}(\hat{\beta_B}-\hat{\beta_A})}
{p_A D(x_A)+p_B D(x_B)}
(\bar{x_A}-\bar{x_B}+\frac{D(x_A)}{\bar{x_A}}-\frac{D(x_B)}{\bar{x_B}})$

$\hat{\alpha}=y-x\hat{\beta}=0$ holds when $E(x_A)=E(x_B)~\text{and}~D(x_A)=D(x_B)$

The nature of mail-in ballots and in-person ballots possibly fluctrates the constant term.



Proof

			
Attachments:

			
This post has been edited 3 times. Last edited by fungarwai, Feb 18, 2021, 12:40 AM

				

				
statistics

				

			

				

					
				

				

				

			

		


	

		
Subsequent factor of binomial coefficients

		
by fungarwai, Sep 20, 2020, 11:31 AM 

		


			
$n!\mid\binom{mn}{m,m,\dots,m}=\frac{(mn)!}{(m!)^n}$



Proof



$(m,n+1)=1\Rightarrow (n+1-m)\mid\binom{n}{m}$



Proof



Example



$(m+1,n+1)=1\Rightarrow (m+1)\mid\binom{n}{m}$



Proof



$(a,b)=1\Rightarrow n \mid \binom{n}{a}\binom{n}{b}$, specially $n \mid \binom{n}{m}\binom{n}{m+1}$



Proof



Example

			

			
This post has been edited 7 times. Last edited by fungarwai, Mar 20, 2021, 12:52 PM

				

				
combinatorics
number theory

				

			

				

					
				

				

				

			

		


	

		
The product of fraction with arithmetic sequence

		
by fungarwai, Sep 20, 2020, 2:38 AM 

		


			
An inequality concerning limit of this product



$\prod_{k=0}^n \frac{ak+b}{ak+c}<\left(\frac{b}{an+a+b}\right)^{\frac{c-b}{a}}$



Proof



Example



An identity of this product



$\prod_{n=1}^\infty \frac{(an-b)(an+b)}{(an-c)(an+c)}
=\frac{c\sin (b\pi/a)}{b\sin (c\pi/a)}$



Proof



Example



Summation of this product



$\sum_{n=0}^N\prod_{k=0}^n\frac{ak+b}{ak+c}
=\frac{b}{c-a-b}-\frac{a(N+1)+b}{c-a-b}\prod_{k=0}^N\frac{ak+b}{ak+c}$



Proof



Example

			

			
This post has been edited 7 times. Last edited by fungarwai, May 19, 2021, 6:07 AM

				

				
Inequality
Product
Summation

				

			

				

					
				

				

				

			

		


	

		
Summation with polynomial roots

		
by fungarwai, May 26, 2020, 12:52 PM 

		


			
Reciprocal type



$\sum_{i=1}^m \frac{1}{x-x_i}=\frac{f'(x)}{f(x)},~f(x)=\prod_{i=1}^m (x-x_i)$



Credit to the idea posted by ccmmjj in mathchina forum



Proof



$\sum_{i=1}^m \frac{-1}{(x-x_i)^2}=\frac{f''(x)f(x)-[f'(x)]^2}{[f(x)]^2}$



$\sum_{i=1}^m \frac{2}{(x-x_i)^3}=\frac{f'''(x)[f(x)]^2-3f''(x)f'(x)f(x)+2[f'(x)]^3}{[f(x)]^3}$



Identity of a conditional symmetric polynomial



Let $\displaystyle S_{n,m}=\sum_{r_1+r_2+\cdots+r_m=n} x_1^{r_1}x_2^{r_2}\cdots x_m^{r_m},~\sigma_{k,m} =\sum_{1\le j_1 < j_2 < \cdots < j_k \le m} x_{j_1} \dots x_{j_k}$

$\displaystyle \boxed{S_{j,m}=\sum_{k=1}^{m-1} (-1)^{k-1} \sigma_{k,m} S_{j-k,m}+(-1)^{m-1}\sigma_{m,m} S_{j-m,m}}$

$\displaystyle \text{or}~ \boxed{S_{j,m}=\sum_{k=1}^{j-1} (-1)^{k-1} \sigma_{k,m} S_{j-k,m}+(-1)^{j-1}\sigma_{j,m}}$



Proof



Example



$\displaystyle S_{j,K}=\sum_{k=1}^{M-1} (-1)^{k-1}
\sigma_{k,M} S_{j-k,K}+(-1)^{M-1}\sigma_{M,M}S_{j-M,K},~\forall K\le M\le m$



Proof



Example



Lagrange polynomial type



$S_{n,m}=\sum_{i=1}^m x_i^n \prod_{j=1\atop i\neq j}^m\frac{1}{x_i-x_j}
=\begin{cases}0 & 0\le n\le m-2\\1 & n=m-1\end{cases}$



Proof



$S_{n,m}=\sum_{i=1}^m x_i^n \prod_{j=1\atop i\neq j}^m\frac{1}{x_i-x_j}
=\begin{cases}0 & 0\le n\le m-2\\1 & n=m-1\\
\displaystyle
\boxed{\sum_{r_1+r_2+\cdots+r_m=n-m+1} x_1^{r_1}x_2^{r_2}\cdots x_m^{r_m}}
 & n\ge m
\end{cases}$



$S_{m,m}=\sum a$

$S_{m+1,m}=\sum a^2+\sum ab$

$S_{m+2,m}=\sum a^3+\sum a^2 b+\sum abc$

$S_{m+3,m}=\sum a^4+\sum a^3 b+\sum a^2 b^2+\sum a^2 bc+\sum abcd$



Best Proof Credit to the idea with four variables posted by natmath at #2



Proof Credit to the idea with three variables posted by gasgous at #10



Other Proof



$S_{n,m}=\sum_{i=1}^m x_i^n \prod_{j=1\atop i\neq j}^m\frac{1}{x_i-x_j}=\begin{cases}0 & 0\le n\le m-2\\1 & n=m-1\\
\displaystyle
\boxed{\sum_{r_1 + 2r_2 + \cdots + mr_m = s \atop r_1\ge 0, \ldots, r_m\ge 0} (-1)^n \frac{(r_1 + r_2 + \cdots + r_m)!}{r_1!r_2! \cdots r_m!} \prod_{i=1}^m (-\sigma_{i,m})^{r_i}} & n=s+m-1\ge m
\end{cases}$



where $\sigma_{k,m} =\sigma_k (x_1 , \ldots , x_m )=\sum_{1\le  j_1 < j_2 < \cdots < j_k \le m} x_{j_1} \dots x_{j_k}$



$S_{m,m}=\sigma_{1,m}$

$S_{m+1,m}=\sigma_{1,m}^2-\sigma_{2,m}$

$S_{m+2,m}=\sigma_{1,m}^3-2\sigma_{1,m}\sigma_{2,m}+\sigma_{3,m}$

$S_{m+3,m}=\sigma_{1,m}^4-3\sigma_{1,m}^2\sigma_{2,m}+\sigma_{2,m}^2-2\sigma_{1,m}\sigma_{3,m}+\sigma_{4,m}$



Proof



Example

			

			
This post has been edited 23 times. Last edited by fungarwai, May 19, 2023, 12:07 PM

				

				
polynomial
Summation

				

			

				

					
				

				

				

			

		


	

		
Chromatic polynomial of a 3×n grid

		
by fungarwai, Apr 26, 2020, 1:26 AM 

		


			
Denote Chromatic polynomial of a m×n grid as $P(G_{m,n},t)$



Chromatic polynomial of a 2×n grid has been found and named as ladder graph



$P(G_{2,n},t)=t(t-1)(t^2-3t+3)^{n-1}$



My recurrence equation found for Chromatic polynomial of a 3×n grid is as follow



$P(G_{3,n},t)=(t^3-5t^2+11t-10)P(G_{3,n-1},t)-(t^4-7t^3+19t^2-24t+11)P(G_{3,n-2},t)$

$~=(t-2)(t^2-3t+5)P(G_{3,n-1},t)-(t-1)(t^3-6t^2+13t-11)P(G_{3,n-2},t)$



$P(G_{3,1},t)=t(t-1)^2,~P(G_{3,2},t)=t(t-1)(t^2-3t+3)^2$

$P(G_{3,3},t)=(t^3-5t^2+11t-10)t(t-1)(t^2-3t+3)^2
-(t^4-7t^3+19t^2-24t+11)t(t-1)^2$

$~=t(t-1)(t^7-11t^6+55t^5-161t^4+298t^3-350t^2+244t-79)$



[asy]
unitsize(5mm);
int i, j, k;
k=0;
for(i=0; i<k+7; i=i+1){
if(i!=2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
draw((0,j)--(k+6,j));
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$=$", (7,1));
k=8;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+6,2));
for(i=k; i<k+7; i=i+1){
if(i!=10){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (15,1));
k=16;
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
draw((k,j)--(k+5,j));
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
dot((k+6,0.5));dot((k+6,1.5));
draw((k+5,0)--(k+6,0.5));
draw((k+5,1)--(k+6,0.5));
draw((k+5,1)--(k+6,1.5));
draw((k+5,2)--(k+6,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
label("$~$", (0,1));label("$=$", (15,1));
k=16;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+6,2));
draw((k+6,1)--(k+6,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+6,0));dot((k+6,1));dot((k+6,2));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (23,1));
k=24;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+6,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
draw((k+6,0)--(k+6,2));dot((k+6,0));dot((k+6,2));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (31,1));
k=32;
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
draw((k,j)--(k+5,j));
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
dot((k+6,0.5));dot((k+6,1.5));
draw((k+5,0)--(k+6,0.5));
draw((k+5,1)--(k+6,0.5));
draw((k+5,1)--(k+6,1.5));
draw((k+5,2)--(k+6,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
label("$~$", (0,1));label("$=[(t-1)^3-(t-2)^2]$", (17,1));
k=22;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (28,1));
k=29;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+6,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
draw((k+6,0)--(k+6,2));dot((k+6,0));dot((k+6,2));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
label("$~$", (0,1));label("$=(t^3-4t^2+7t-5)$", (17,1));
k=22;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (28,1));
k=29;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+6,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
draw((k+6,0)--(k+6,2));dot((k+6,0));dot((k+6,2));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy][asy]
unitsize(5mm);
int i, j, k;
k=0;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+6,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
draw((k+6,0)--(k+6,2));dot((k+6,0));dot((k+6,2));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$=$", (7,1));
k=8;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+6,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+6,0));dot((k+6,2));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (15,1));
k=16;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
draw((k+5,2)--(k+6,0));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+6,0));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy][asy]
unitsize(5mm);
int i, j, k;
label("$~$", (-7,1));label("$=$", (7,1));
k=8;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+6,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+6,0));dot((k+6,2));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (15,1));
k=16;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+6,0));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$+$", (23,1));
k=24;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
draw((k+5,0)..(k+6,1)..(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
label("$~$", (-14,1));label("$=[(t-1)^2-(t-1)]$", (3,1));
k=8;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$+$", (15,1));
k=16;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
draw((k+5,0)..(k+6,1)..(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
label("$~$", (-11,1));label("$=(t-1)(t-2)$", (3,1));
k=7;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$+$", (14,1));
k=15;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
draw((k+5,0)..(k+6,1)..(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
k=0;
for(i=0; i<k+7; i=i+1){
if(i!=2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
draw((0,j)--(k+6,j));
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$=[(t-1)^3-(t-2)^2]$", (11,1));
k=16;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (22,1));
k=23;
draw((k,0)--(k+6,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+6,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
draw((k+6,0)--(k+6,2));dot((k+6,0));dot((k+6,2));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
label("$~$", (-10,1));label("$=[(t-1)^3-(t-2)^2-(t-1)^2+(t-1)]$", (11,1));
k=20;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (26,1));
k=27;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
draw((k+5,0)..(k+6,1)..(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
label("$~$", (0,1));label("$=(t^3-5t^2+10t-7)$", (11,1));
k=16;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (22,1));
k=23;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
draw((k+5,0)..(k+6,1)..(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
k=0;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
draw((k+5,0)..(k+6,1)..(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$=$", (7,1));
k=8;
draw((k,0)--(k+5,0));
draw((k,1)--(k+5,1));
draw((k,2)--(k+5,2));
for(i=k; i<k+6; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (15,1));
k=16;
draw((k,0)--(k+4,0));
draw((k,1)--(k+6,1));
draw((k,2)--(k+4,2));
draw((k+4,2)--(k+6,1));
draw((k+4,0)--(k+6,1));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+5,1));dot((k+6,1));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
k=0;
draw((k,0)--(k+4,0));draw((k+4,0)--(k+6,1));
draw((k,1)--(k+6,1));
draw((k,2)--(k+4,2));draw((k+4,2)--(k+6,1));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+5,1));dot((k+6,1));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$=$", (7,1));
k=8;
draw((k,0)--(k+4,0));draw((k+4,0)--(k+6,1));
draw((k,1)--(k+4,1));draw((k+5,1)--(k+6,1));
draw((k,2)--(k+4,2));draw((k+4,2)--(k+6,1));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+5,1));dot((k+6,1));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-$", (15,1));
k=16;
draw((k,0)--(k+4,0));draw((k+4,0)--(k+6,1));
draw((k,1)--(k+6,1));
draw((k,2)--(k+4,2));draw((k+4,2)--(k+6,1));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+6,1));
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);int i, j, k;
label("$=$", (7,1));
k=8;draw((k,0)--(k+4,0));draw((k+4,0)--(k+6,1));
draw((k,1)--(k+4,1));draw((k+5,1)--(k+6,1));
draw((k,2)--(k+4,2));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+5,1));dot((k+6,1));
label("...", (k+2,0.5));label("...", (k+2,1.5));
label("$-$", (15,1));
k=16;draw((k,0)--(k+4,0));
draw((k,1)--(k+4,1));draw((k+4,2)--(k+5,1));
draw((k,2)--(k+4,2));
draw((k+4,0)..(k+6,1)..(k+4,2));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+5,1));
label("...", (k+2,0.5));label("...", (k+2,1.5));
label("$-$", (23,1));
k=24;draw((k,0)--(k+4,0));
draw((k,1)--(k+4,1));draw((k,2)--(k+4,2));
draw((k+4,2)--(k+6,1));draw((k+4,0)--(k+6,1));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+6,1));
label("...", (k+2,0.5));label("...", (k+2,1.5));
label("$+$", (31,1));
k=32;draw((k,0)--(k+4,0));draw((k,1)--(k+4,1));draw((k,2)--(k+4,2));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);int i, j, k;
label("$=$", (7,1));
k=8;draw((k,0)--(k+4,0));draw((k+4,0)--(k+6,1));
draw((k,1)--(k+4,1));draw((k+5,1)--(k+6,1));
draw((k,2)--(k+4,2));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
dot((k+5,1));dot((k+6,1));
label("...", (k+2,0.5));label("...", (k+2,1.5));
label("$-$", (15,1));
k=16;draw((k,0)--(k+4,0));
draw((k,1)--(k+4,1));draw((k+4,2)--(k+5,1));
draw((k,2)--(k+4,2));
draw((k+4,0)..(k+6,1)..(k+4,2));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}dot((k+5,1));
label("...", (k+2,0.5));label("...", (k+2,1.5));label("$-$", (23,1));
k=24;draw((k,0)--(k+4,0));
draw((k,1)--(k+4,1));draw((k,2)--(k+4,2));
draw((k+4,0)--(k+6,1));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){if(i!=k+2){dot((i,j));}}}
dot((k+6,1));label("...", (k+2,0.5));label("...", (k+2,1.5));label("$+$", (31,1));
k=32;draw((k,0)--(k+4,0));draw((k,1)--(k+4,1));draw((k,2)--(k+4,2));
draw((k+4,0)..(k+5,1)..(k+4,2));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));label("...", (k+2,1.5));
label("$+$", (39,1));
k=40;draw((k,0)--(k+4,0));draw((k,1)--(k+4,1));draw((k,2)--(k+4,2));
for(i=k; i<k+5; i=i+1){if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));label("...", (k+2,1.5));
[/asy]





[asy]
unitsize(5mm);
int i, j, k;
label("$~$", (-14,1));label("$=[(t-1)^2-(t-1)+1]$", (3,1));
k=8;
draw((k,0)--(k+4,0));
draw((k,1)--(k+4,1));
draw((k,2)--(k+4,2));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-(t-2)$", (14,1));
k=16;
draw((k,0)--(k+4,0));
draw((k,1)--(k+4,1));
draw((k,2)--(k+4,2));
draw((k+4,0)..(k+5,1)..(k+4,2));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

[asy]
unitsize(5mm);
int i, j, k;
label("$~$", (-10,1));label("$=(t^2-3t+3)$", (3,1));
k=6;
draw((k,0)--(k+4,0));
draw((k,1)--(k+4,1));
draw((k,2)--(k+4,2));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
label("$-(t-2)$", (12,1));
k=14;
draw((k,0)--(k+4,0));
draw((k,1)--(k+4,1));
draw((k,2)--(k+4,2));
draw((k+4,0)..(k+5,1)..(k+4,2));
for(i=k; i<k+5; i=i+1){
if(i!=k+2){draw((i,0)--(i,2));}
for(j=0; j<3; j=j+1){
if(i!=k+2){dot((i,j));}}}
label("...", (k+2,0.5));
label("...", (k+2,1.5));
[/asy]

$a_n=(t^3-5t^2+10t-7)a_{n-1}-b_{n-1}$

$b_n=a_n-(t^2-3t+3)a_{n-1}+(t-2)b_{n-1}$



$b_{n-1}=(t^3-5t^2+10t-7)a_{n-1}-a_n$

$b_n=(t^3-5t^2+10t-7)a_n-a_{n+1}$

$(t^3-5t^2+10t-7)a_n-a_{n+1}=a_n-(t^2-3t+3)a_{n-1}+(t-2)[(t^3-5t^2+10t-7)a_{n-1}-a_n]$

$a_{n+1}=(t^3-5t^2+11t-10)a_n-(t^4-7t^3+19t^2-24t+11)a_{n-1}$



[asy]
unitsize(5mm);int i, j, k;
label("$a_1:$", (0,1));
k=2;
for(i=k; i<k+1; i=i+1){
draw((i,0)--(i,2));
for(j=0; j<3; j=j+1){dot((i,j));}}
label("$a_2:$", (5,1));
k=7;
for(i=k; i<k+2; i=i+1){
draw((i,0)--(i,2));
for(j=0; j<3; j=j+1){
draw((k,j)--(k+1,j));
dot((i,j));}}
label("$a_3:$", (11,1));
k=13;
for(i=k; i<k+3; i=i+1){
draw((i,0)--(i,2));
for(j=0; j<3; j=j+1){
draw((k,j)--(k+2,j));
dot((i,j));}}
[/asy]



$a_1=t(t-1)^2,~a_2=t(t-1)(t^2-3t+3)^2$

$a_3=(t^3-5t^2+11t-10)t(t-1)(t^2-3t+3)^2-(t^4-7t^3+19t^2-24t+11)t(t-1)^2$

$~=t(t-1)(t^7+11t^6+55t^5-161t^4+298t^3-350t^2+244t-79)$

			

			
This post has been edited 17 times. Last edited by fungarwai, Aug 27, 2021, 12:50 PM

				

				
graph theory
combinatorics
polynomial

				

			

				

					
				

				

				

			

		


	

		
The module distribution for Pascal's triangle row

		
by fungarwai, Sep 13, 2019, 12:14 PM 

		


			
Let $n=\sum_{k=0}^\infty n_k p^k, N_r=\{k|n_k=r\}$



For p=2,



There are $2^{|N_1|}$ different ways to choose $m$ for $\binom{n}{m}\equiv 1\pmod{2}$



For p=3,



There are $2^{|N_1|}\cdot \frac{3^{|N_2|}-1}{2}$ different ways to choose $m$ for $\binom{n}{m}\equiv 2\pmod{3}$ and

$2^{|N_1|}\cdot \frac{3^{|N_2|}+1}{2}$ different ways to choose $m$ for $\binom{n}{m}\equiv 1\pmod{3}$



Proof



For p=5,



The number of $\binom{n}{m}\equiv 1\pmod{5}$ is $
\frac{1}{4}2^{|N_1|}5^{|N_4|}(3^{|N_2|}4^{|N_3|}+0^{|N_3|})
+\frac{1}{2}2^{|N_1|}5^{\frac{|N_2|}{2}}2^{\frac{3|N_3|}{2}}
\cos\left(|N_2|\tan^{-1}\frac{1}{2}-|N_3|\frac{\pi}{4}\right)$

The number of $\binom{n}{m}\equiv 2\pmod{5}$ is $
\frac{1}{4}2^{|N_1|}5^{|N_4|}(3^{|N_2|}4^{|N_3|}-0^{|N_3|})
+\frac{1}{2}2^{|N_1|}5^{\frac{|N_2|}{2}}2^{\frac{3|N_3|}{2}}
\sin\left(|N_2|\tan^{-1}\frac{1}{2}-|N_3|\frac{\pi}{4}\right)$

The number of $\binom{n}{m}\equiv 3\pmod{5}$ is $
\frac{1}{4}2^{|N_1|}5^{|N_4|}(3^{|N_2|}4^{|N_3|}-0^{|N_3|})
-\frac{1}{2}2^{|N_1|}5^{\frac{|N_2|}{2}}2^{\frac{3|N_3|}{2}}
\sin\left(|N_2|\tan^{-1}\frac{1}{2}-|N_3|\frac{\pi}{4}\right)$

The number of $\binom{n}{m}\equiv 4\pmod{5}$ is $
\frac{1}{4}2^{|N_1|}5^{|N_4|}(3^{|N_2|}4^{|N_3|}+0^{|N_3|})
-\frac{1}{2}2^{|N_1|}5^{\frac{|N_2|}{2}}2^{\frac{3|N_3|}{2}}
\cos\left(|N_2|\tan^{-1}\frac{1}{2}-|N_3|\frac{\pi}{4}\right)$



Proof



Example

			

			
This post has been edited 1 time. Last edited by fungarwai, Sep 13, 2019, 12:15 PM

				

				
number theory
combinatorics

				

			

				

					
				

				

				

			

		


	

		
Notable algebra methods with proofs and examples

		
avatar

		
			fungarwai
		
	

		
Shouts

	

		

			
			
Submit
			

		


	

	

		
  • 
	Nice blog!
	
    by
	Inconsistent, Mar 18, 2024, 2:41 PM
  • 
	hey, nice blog! really enjoyed the content  here and thank you for this contribution to aops. Sure to subscribe! :)
	
    by
	thedodecagon, Jan 22, 2022, 1:33 AM
  • 
	thanks for this
	
    by
	jasperE3, Dec 3, 2021, 10:01 PM
  • 
	I am working as accountant and studying as ACCA student now.
    
I graduated applied mathematics at bachelor degree in Jinan University but I still have no idea to find a specific job with this..
	
    by
	fungarwai, Aug 28, 2021, 4:54 AM
  • 
	Awesome algebra blog :)
	
    by
	Euler1728, Mar 22, 2021, 5:37 AM
  • 
	I wonder if accountants need that kind of math tho
	
    by
	Hamroldt, Jan 14, 2021, 10:55 AM
  • 
	Nice!!!!
	
    by
	Delta0001, Dec 12, 2020, 10:20 AM
  • 
	this is very interesting i really appericate it :)
	
    by
	vsamc, Oct 29, 2020, 4:42 PM
  • 
	this is god level
	
    by
	Hamroldt, Sep 4, 2020, 7:48 AM
  • 
	Super Blog! You are Pr0! :)
	
    by
	Functional_equation, Aug 23, 2020, 7:43 AM
  • 
	Great blog!
	
    by
	freeman66, May 31, 2020, 5:40 AM
  • 
	cool thx! :D
	
    by
	erincutin, May 18, 2020, 4:55 PM
  • 
	How so op???
	
    by
	Williamgolly, Apr 30, 2020, 2:42 PM
  • 
	Beautiful
	
    by
	Al3jandro0000, Apr 25, 2020, 3:11 AM
  • 
	Nice method :)
	
    by
	Feridimo, Jan 23, 2020, 5:05 PM
  • 
	This is nice!
	
    by
	mufree, May 26, 2019, 6:40 AM
  • 
	Wow! So much Algebra.
	
    by
	AnArtist, Mar 15, 2019, 1:19 PM
  • 
	:omighty: :omighty:
	
    by
	AlastorMoody, Feb 9, 2019, 5:17 PM
  • 
	31415926535897932384626433832795
	
    by
	lkarhat, Dec 25, 2018, 11:53 PM
  • 
	rip 0 shouts and 0 comments until now
	
    by
	harry1234, Nov 17, 2018, 8:56 PM

		20 shouts
	

	

		
Tags

	

	

	

		
number theory
Summation
polynomial
combinatorics
Finite difference
Product
graph theory
Inequality
matrix
statistics
trigonometric function

	

	

		
About Owner

	

	

	

		
    
			
  • 
				Posts:
				863
			
    • 
			
  • 
				Joined:
				Mar 11, 2017
			
    • 
		

	

	

		
Blog Stats

	

	

	

		
    
			
  • 
				Blog created:
				Sep 15, 2018
			
    • 
			
  • 
				Total entries:
				18
			
    • 
			
  • 
				Total visits:
				6389
			
    • 
			
  • 
				Total comments:
				8
			
    • 

		

	

	

		

		

		
Search Blog

	

	

	

		

			

				
				


			

		


	

	

	


	





	








	
a