# Research: Looking for a sequence that produce variation's of Pascal's triangle

Prologue

I am an undergraduate so if my terminology or approach seem inappropriate/confusing please explain in the comments.

I created a notation where

$$F(0 \rightarrow n,x) = [\hspace{1mm}F(0 ,x),\hspace{2mm}F(1, x),\hspace{2mm}F(2, x), \ldots,\hspace{2mm} F(n-2,x),\hspace{2mm}F(n-1,x),\hspace{2mm}F( n,x)\hspace{1mm}]$$

Last semester in Calculus 1 I created a function noted as $\mu(e,f)$ where $f$,$e$ are just positive integers including 0 and if c is any numerical value then $\mu(c,c)=0$ and $\frac{d \mu(c,c)}{dx}=0$. The interesting part was $$\frac{d \mu(e,f)}{dx}=\mu(e+1,f)+\mu(e,f+1)$$ I then figured out by playing around with the function that if $f>e + c$ for any real c then: $$\frac{d^\lambda \mu(e,f)}{(dx)^\lambda}=\sum_{i=0}^{\lambda}\binom{\lambda}{i}\mu(e+i,f+\lambda-i)$$

I figured this out by putting the Function into Pascal's Triangle, where each row represents the derivative of the row before it. I will call this triangle Pascal's Triangle for Functions (PTF). For clarity I made some stuff that makes the patterns easier to see. In the equation $\frac{d \mu(e,f)}{dx}=\mu(e+1,f)+\mu(e,f+1)$, The $\mu(e+1,f)$ will be presented as $\swarrow$ and $\mu(e,f+1)$ will be presented as $\searrow$. Let $G(j,k) = \mu(e+j,f+k)$ therefore $\frac{d G(j,k)}{dx}=G(j+1,k)+G(j,k+1)$.

$$G(0,0)\\ \swarrow\searrow \\ G(1,0)+G(0,1)\\ \swarrow\searrow\hspace{12mm}\swarrow\searrow\\ G(2,0)+2G(1,1)+G(0,2)\\ \swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow \\ G(3,0)+3G(2,1)+3G(1,2)+G(0,3)\\ \swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow \\ G(4,0)+4G(3,1)+6G(2,2)+4G(1,3)+G(0,4)\\ \swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow \\ G(5,0)+5G(4,1)+10G(3,2)+10G(2,3)+5G(1,4)+G(0,5)\\ \ldots$$

Anyways knowing $\mu(c,c)=0$ and $e$,$f$ are finite then let $\alpha = e-f$ so $\mu(f+\alpha,f)$. I as the writer am going to ignorantly for the sake of clarity assume if a function ever becomes 0 that it no longer exists and therefore loses it's significant. Therefore 0 will be replaced with white space for these illustrations.

If $\alpha=1$ and $e =0$

$$\hspace{5mm}\mu(0,1)\\ \hspace{5mm}\swarrow\searrow\\ [\hspace{1mm}0=\mu(1,1)\hspace{1mm}]+\mu(0,2)\\ \hspace{26mm}\swarrow\searrow\\ \hspace{25mm}\mu(1,2)+\mu(0,3)\\ \hspace{28mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\\ \hspace{15mm}[\hspace{1mm}0=\mu(2,2)\hspace{1mm}]+2\mu(1,3)+\mu(0,4)\\ \hspace{50mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\\ \hspace{51mm}2\mu(2,3)+3\mu(1,4)+\mu(0,5)\\ \hspace{58mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{10mm}\swarrow\searrow\\ \hspace{45mm}2[\hspace{1mm}0=\mu(3,3)\hspace{1mm}]+2\mu(2,4)+3\mu(1,5)+\mu(0,6)\\ \hspace{78mm}\swarrow\searrow\hspace{12mm}\swarrow\searrow\hspace{10mm}\swarrow\searrow\\ \hspace{75mm}2\mu(3,4)+5\mu(2,5)+4\mu(1,6)+\mu(0,7)\\ \ldots$$

if $\alpha=2,e=0$: $$\hspace{5mm}\mu(0,2)\\ \hspace{5mm}\swarrow\hspace{5mm}\searrow\\ \hspace{5mm}\mu(1,2)+\mu(0,3)\\ \hspace{8mm}\swarrow\hspace{5mm}\searrow\hspace{2mm}\swarrow\hspace{5mm}\searrow\\ \hspace{0mm}[\hspace{1mm}0=\mu(2,2)\hspace{1mm}]+2\mu(1,3)+\mu(0,4)\\ \hspace{28mm}\swarrow\hspace{5mm}\searrow\hspace{2mm}\swarrow\hspace{5mm}\searrow\\ \hspace{30mm}2\mu(2,3)+3\mu(2,4)+\mu(0,5)\\ \hspace{30mm}\swarrow\hspace{5mm}\searrow\hspace{3mm}\swarrow\hspace{5mm}\searrow\hspace{3mm}\swarrow\hspace{5mm}\searrow\\ \hspace{15mm}2[\hspace{1mm}0=\mu(3,3)\hspace{1mm}]+5\mu(2,4)+4\mu(1,5)+\mu(0,6)\\ \hspace{50mm}\swarrow\hspace{5mm}\searrow\hspace{3mm}\swarrow\hspace{5mm}\searrow\hspace{3mm}\swarrow\hspace{5mm}\searrow\\ \hspace{50mm}5\mu(3,4)+9\mu(2,5)+5\mu(1,6)+\mu(0,7)\\ \hspace{55mm}\swarrow\hspace{5mm}\searrow\hspace{6mm}\swarrow\hspace{5mm}\searrow\hspace{6mm}\swarrow\hspace{5mm}\searrow\hspace{6mm}\swarrow\hspace{5mm}\searrow\\ \hspace{40mm}5[\hspace{1mm}0=\mu(4,4)\hspace{1mm}]+14\mu(3,5)+14\mu(2,6)+6\mu(1,7)+\mu(0,8)\\ \ldots$$

I created a function noted as $\Upsilon_\alpha(\beta,R)$, where $\beta=0,1,2...$ represents the row number, $\alpha=1,2...$ represents variation, $R$ represents the column of a row. using the illastrations above I created some examples but I noted $\Upsilon_\alpha(\beta,R)$as $\binom{\beta}{R}_\alpha$ and I did not include the arrows.

If $\alpha=1$ and $e =0$

$$\hspace{5mm}\binom{0}{0}_1\mu(0,1)\\ [\hspace{1mm}0=\mu(1,1)\hspace{1mm}]+\binom{0}{0}_1\mu(0,2)\\ \hspace{25mm}\binom{1}{0}_1\mu(1,2)+\binom{1}{1}_1\mu(0,3)\\ \hspace{15mm}[\hspace{1mm}0=\mu(2,2)\hspace{1mm}]+\binom{1}{0}_1\mu(1,3)+\binom{1}{1}_1\mu(0,4)\\ \hspace{51mm}\binom{2}{0}_1\mu(2,3)+\binom{2}{1}_1\mu(1,4)+\binom{2}{2}_1\mu(0,5)\\ \hspace{45mm}2[\hspace{1mm}0=\mu(3,3)\hspace{1mm}]+\binom{2}{0}_1\mu(2,4)+\binom{2}{1}_1\mu(1,5)+\binom{2}{2}_1\mu(0,6)\\ \hspace{65mm}\binom{3}{0}_1\mu(3,4)+\binom{3}{1}_1\mu(2,5)+\binom{3}{2}_1\mu(1,6)+\binom{3}{3}_1\mu(0,7)\\ \ldots$$

if $\alpha=2,e=0$: $$\hspace{5mm}\binom{0}{0}_2\mu(0,2)\\ \hspace{5mm}\binom{1}{0}_2\mu(1,2)+\binom{1}{1}_2\mu(0,3)\\ \hspace{0mm}[\hspace{1mm}0=\mu(2,2)\hspace{1mm}]+\binom{1}{0}_2\mu(1,3)+\binom{1}{1}_2\mu(0,4)\\ \hspace{30mm}\binom{2}{0}_2\mu(2,3)+\binom{2}{1}_2\mu(2,4)+\binom{2}{2}_2\mu(0,5)\\ \hspace{15mm}[\hspace{1mm}0=\mu(3,3)\hspace{1mm}]+\binom{2}{0}_2\mu(2,4)+\binom{2}{1}_2\mu(1,5)+\binom{2}{2}_2\mu(0,6)\\ \hspace{50mm}\binom{3}{0}_2\mu(3,4)+\binom{3}{1}_2\mu(2,5)+\binom{3}{2}_2\mu(1,6)+\binom{3}{3}_2\mu(0,7)\\ \hspace{30mm}[\hspace{1mm}0=\mu(4,4)\hspace{1mm}]+\binom{3}{0}_2\mu(3,5)+\binom{3}{1}_2\mu(2,6)+\binom{3}{2}_2\mu(1,7)+\binom{3}{3}_2\mu(0,8)\\ \ldots$$

if $\alpha=c,e=0$: $$\binom{0}{0}_c\mu(0,c)\\ \binom{1}{0}_c\mu(1,c)+\binom{1}{1}_c\mu(0,c+1)\\ \ldots\\ \binom{c-1}{0}_c\mu(c-1,c)+\binom{c-1}{1}_c\mu(c-2,c+1)+\ldots+\binom{c-1}{c-2}_c\mu(1,2c-2)+\binom{c-1}{c-1}_c\mu(0,2c-1)\\ [\mu(c,c)=0]+\binom{c-1}{0}_c\mu(c-1,c+1)+\ldots+\binom{c-1}{c-2}_c\mu(1,2c-1)+\binom{c-1}{c-1}_c\mu(0,2c)\\$$

I then realized that if I allow.

$$R_{max}=\left\{ \begin{array}{@{}ll@{}} \frac{\beta+\alpha+mod_2(\beta+\alpha)}{2}, & \text{if }\beta\geq\alpha \\ \beta, & \text{if }\beta<\alpha \end{array}\right.$$

AND

$$\Gamma=\left\{ \begin{array}{@{}ll@{}} \frac{\beta-\alpha+mod_2(\beta-\alpha)}{2}, & \text{if }\beta\geq\alpha \\ 0, & \text{if }\beta<\alpha \end{array}\right.$$

Then the sum can be summarized by:

$$\frac{d^\beta \mu(e,e+\alpha)}{(dx)^\beta}=\sum_{i=0}^{R_{max}}\Upsilon_\alpha(\beta,R_{max}-i)\mu(e+i+\Gamma,e+\alpha+\beta-i)$$

And since I've always called $\binom{i}{R}$ the Pascal's Triangle notation I will call $\Upsilon_\alpha(\beta,R)$ the Variant of Pascal's Triangle notation (VPT)

another notable behavior is When $\beta<\alpha$ then $R_{max}=\beta, \Gamma=0$, remembering $f = e+\alpha$ then $$\frac{d^\beta \mu(e,e+\alpha)}{(dx)^\beta}=\frac{d^\beta \mu(e,f)}{(dx)^\beta}$$ $$\sum_{i=0}^{\beta}\Upsilon_\alpha(i,\beta-i)\mu(e+i,f+\beta-i)=\sum_{i=0}^{\beta}\binom{\beta}{\beta-i}\mu(e+i,f+\beta-i)$$ therefore $\Upsilon_\alpha(\beta,R_{max}-i)=\binom{\beta}{\beta-i}$ for $\beta<\alpha$

BUT If $\beta\geq\alpha$ then all of the Pascal's rules apply for all $R$'s in $\beta$ EXCEPT when $\beta=\alpha+1$ a the horizontal column of blank space called the "You Shall Not Pass Line", YSNPL, prevents any new functions from becoming created at the beyond or on the line. This is because of the property $\mu(c,c)=0$ and $\frac{d \mu(c,c)}{dx}=0$ prevent any function from going beyond or on that line of 0's.

examples of $\Upsilon_\alpha(\beta,R)$

In order to understand what I just wrote I created 3 examples classified as functions called Pascal's Triangle Numerical Version (PTN$(\alpha)$), Variant of Pascal's Triangle Numerical Version (VPTN$(\alpha)$), and Variant of Pascal's Triangle $\Upsilon_\alpha$ form (VPTY$(\alpha)$).

Both PTN$(\alpha)$ and VPTN$(\alpha)$ will produce 6 rows of their given sequence.

PTN$(\alpha)$ will produce Pascal's Triangle with $\alpha$ number circles around the left most 1 on a given row, starting with the top row. The circles represent that part of Pascal's Triangle will not change after it encounter the given VPT.

VPTN$(\alpha)$ will show the VPT in numerical form.

VPTY$(\alpha)$ will produce the VPT in $\Upsilon_\alpha$ form

$\hspace{10mm}\text{PTN(0):}\hspace{54mm}\text{VPTN(0):}\hspace{30mm}\text{VPTY(0):}\\ \hspace{12mm}(1)\hspace{70mm}1\hspace{40mm}\Upsilon_0(0 \rightarrow 0,0)\\ \hspace{12mm}1,1\hspace{72mm}1\hspace{38mm}\Upsilon_0(0 \rightarrow 0,1)\\ \hspace{10mm}1,2,1\hspace{68mm}1,1\hspace{36mm}\Upsilon_0(0 \rightarrow 1,2)\\ \hspace{8mm}1,3,3,1\hspace{68mm}2,1\hspace{34mm}\Upsilon_0(0 \rightarrow 1,3)\\ \hspace{6mm}1,4,6,4,1\hspace{64mm}2,3,1\hspace{32mm}\Upsilon_0(0 \rightarrow 2,4)\\ \hspace{2mm}1,5,10,10,5,1\hspace{62mm}5,4,1\hspace{30mm}\Upsilon_0(0 \rightarrow 2,5)$

$\hspace{10mm}\text{PTN(1):}\hspace{55mm}\text{VPTN(1):}\hspace{30mm}\text{VPTY(1):}\\ \hspace{12mm}(1)\hspace{70mm} 1\hspace{40mm}\Upsilon_1(0 \rightarrow 0,0)\\ \hspace{10mm}(1),1\hspace{66mm}1,1\hspace{38mm}\Upsilon_1(0 \rightarrow 1,1)\\ \hspace{10mm}1,2,1\hspace{68mm}2,1\hspace{35mm}\Upsilon_1(0 \rightarrow 1,2)\\ \hspace{8mm}1,3,3,1\hspace{64mm}2,3,1\hspace{33mm}\Upsilon_1(0 \rightarrow 2,3)\\ \hspace{6mm}1,4,6,4,1\hspace{64mm}5,4,1\hspace{31mm}\Upsilon_1(0 \rightarrow 2,4)\\ \hspace{2mm}1,5,10,10,5,1\hspace{58mm}5,9,5,1\hspace{28mm}\Upsilon_1(0 \rightarrow 3,5)\\$

$\hspace{10mm}\text{PTN(2):}\hspace{55mm}\text{VPTN(2):}\hspace{30mm}\text{VPTY(2):}\\ \hspace{12mm}(1)\hspace{70mm}1\hspace{40mm}\Upsilon_2(0 \rightarrow 0,0)\\ \hspace{10mm}(1),1\hspace{66mm}1,1\hspace{38mm}\Upsilon_2(0 \rightarrow 1,1)\\ \hspace{8mm}(1),2,1\hspace{62mm}1,2,1\hspace{36mm}\Upsilon_2(0 \rightarrow 2,2)\\ \hspace{8mm}1,3,3,1\hspace{64mm}3,3,1\hspace{33mm}\Upsilon_2(0 \rightarrow 2,3)\\ \hspace{6mm}1,4,6,4,1\hspace{60mm}3,6,4,1\hspace{31mm}\Upsilon_2(0 \rightarrow 3,4)\\ \hspace{2mm}1,5,10,10,5,1\hspace{58mm}9,10,5,1\hspace{26mm}\Upsilon_2(0 \rightarrow 3,5)\\$

$\hspace{10mm}\text{PTN(3):}\hspace{55mm}\text{VPTN(3):}\hspace{30mm}\text{VPTY(3):}\\ \hspace{12mm}(1)\hspace{70mm}1\hspace{40mm}\Upsilon_3(0 \rightarrow 0,0)\\ \hspace{10mm}(1),1\hspace{66mm}1,1\hspace{38mm}\Upsilon_3(0 \rightarrow 1,1)\\ \hspace{8mm}(1),2,1\hspace{62mm}1,2,1\hspace{36mm}\Upsilon_3(0 \rightarrow 2,2)\\ \hspace{6mm}(1),3,3,1\hspace{58mm}1,3,3,1\hspace{34mm}\Upsilon_3(0 \rightarrow 3,3)\\ \hspace{6mm}1,4,6,4,1\hspace{60mm}4,6,4,1\hspace{31mm}\Upsilon_3(0 \rightarrow 3,4)\\ \hspace{2mm}1,5,10,10,5,1\hspace{52mm}4,10,10,5,1\hspace{26mm}\Upsilon_3(0 \rightarrow 3,5)\\$

Remember $\alpha$ represents a variation of Pascal's Triangle, $\beta$ represents a row in $\alpha$ and $R$ represents a column in $\beta$. One should have noticed for each $\alpha$ there is a invisible line where no new number exists on or to the left of that line, YSNPL. One should have noticed for every column in $\beta$ the top right column and the top left column sum together in the row underneath, $\beta+1$, to create a new column in between both of the original columns.

While researching this phenomena I produced some results for $\alpha=1,2,\infty$

$$\Upsilon_2(\beta+1,R)=\Upsilon_1(\beta,R) \text{ can be found in }\left\{ \begin{array}{@{}ll@{}} \frac{R+1}{\beta+1}\binom{\beta+1}{(\frac{\beta-R}{2})}, & \text{if }R<\beta \text{ and } mod_2(\beta-R)=0\\ , & \text{otherwise} \end{array}\right.$$

let $\eta=2R-mod_2(\beta+\alpha-1)$

if maximuim $R$ = $R_{max}$

$$\Upsilon_1(R,\beta)=\Upsilon_2(R,\beta+1)=\frac{\eta+1}{\beta+1}\binom{\beta+1}{\frac{\beta-\eta}{2}}$$

I found this by creating my own variant from the formula given in A008313

$\Upsilon_3$ in A026009 EQ UNKNOWN

$\Upsilon_4$ in A026022 EQ UNKNOWN

Since YSNPL only exists if $\beta=\alpha+1$ and $\beta\neq\infty$ if $\beta$ is finite. Therefore the following statement must be true.$$\Upsilon_\infty(R,\beta) = \binom{R}{\beta}$$ My Question: Is this a known pattern? Are there other known $\alpha$'s? This code can find any triangle with a given $\alpha,\beta$

def tri(alpha,beta) : if alpha > 0 : D = [0]*((beta)//2+alpha); D[0] = 1 b = False; h = 1 for i in range(beta) : print([D[z] for z in range(0,h,1)]) if i < alpha-1 : for R in range(h-1, -1, -1) : D[R+1] += D[R] h += 1 else : if b : for R in range(h, -1, -1) : D[R] += D[R-1] h += 1 else : for R in range(0, h-1, 1) : D[R] += D[R+1] b = not b tri(1,10)

• Where you wrote $\alpha{\leq}\beta{\leq}i$ I changed it to $\alpha\leq\beta\leq i$ simply by changing the code from \alpha{\leq}\beta{\leq}i to \alpha\leq\beta\leq i. Why do you make that extra effort to put in those {curly braces} whose only effect is to prevent proper spacing? $\qquad$ – Michael Hardy Jun 14 '16 at 3:42
• @MichaelHardy I thought I always had to use the brackets to separate known words from characters and numbers which I believed included known words. But thank's you for the assistance. – Jon Jun 14 '16 at 3:47
• To write $\int_0^{12}$ you write \int_0^{12}, and if you omit the braces you see $\int_0^12$. To write $\frac 1 {12}$ you can write \frac 1 {12} and if you omit the braces you see $\frac 1 12$. For $\int_0^\infty$ you can write \int_0^\infty but for $\int_0^{+\infty}$ you write \int_0^{+\infty} and if you omit the braces you see $\int_0^+\infty$. Those are some examples of what braces do. $\qquad$ – Michael Hardy Jun 14 '16 at 4:25
• Binary relation symbols and binary operators have some space before and after them, as in $5+3=8$, coded as 5+3=8, and if you write 5{+}3{=}8 then there's nothing before or after the "+" or the "=" so that space does not appear and you see $5{+}3{=}8$ instead of $5+3=8$. $\qquad$ – Michael Hardy Jun 14 '16 at 4:27
• I don't see where you actually define $\Upsilon_\alpha(\beta,\gamma)$. – joriki Jun 14 '16 at 4:33

$$\Upsilon_\alpha(\beta,R)=\left\{ \begin{array}{@{}ll@{}} \binom{\beta}{R}, & \text{if }\beta\leq\alpha \\ \binom{\beta}{R}, & \text{if }\beta\leq R+\alpha \\ \binom{\beta}{R}-\binom{\beta-\alpha}{R}, & \text{otherwise} \end{array}\right.$$