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Clark's Triangle


A Number Triangle created by setting the Vertex equal to 0, filling one diagonal with 1s, the other diagonal with multiples of an Integer $f$, and filling in the remaining entries by summing the elements on either side from one row above. Call the first column $n=0$ and the last column $m=n$ so that

$\displaystyle c(m,0)$ $\textstyle =$ $\displaystyle fm$ (1)
$\displaystyle c(m,m)$ $\textstyle =$ $\displaystyle 1,$ (2)

then use the Recurrence Relation
\end{displaymath} (3)

to compute the rest of the entries. For $n=1$, we have
\end{displaymath} (4)

\end{displaymath} (5)

For arbitrary $m$, the value can be computed by Summing this Recurrence,
c(m,1)=f\left({\,\sum_{k=1}^{m-1} k}\right)+1={\textstyle{1\over 2}}fm(m-1)+1.
\end{displaymath} (6)

Now, for $n=2$ we have
\end{displaymath} (7)

c(m,2)-c(m-1,2)=c(m-1,1)={\textstyle{1\over 2}}f(m-1)m+1,
\end{displaymath} (8)

so Summing the Recurrence gives

$\displaystyle c(m,2)$ $\textstyle =$ $\displaystyle \sum_{k=1}^m [{\textstyle{1\over 2}}fk(k-1)+1]=\sum_{k=1}^m({\textstyle{1\over 2}}fk^2-{\textstyle{1\over 2}}fk+1)$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}f[{\textstyle{1\over 6}}m(m+1)(2m+1)]-{\textstyle{1\over 2}}f[{\textstyle{1\over 2}}m(m+1)]+m$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 6}}(m-1)(fm^2-2fm+6).$ (9)

Similarly, for $n=3$ we have

c(m,3)-c(m-1,3)=c(m-1,2)={\textstyle{1\over 6}} fm^3-fm^2+({\textstyle{11\over 6}}f+1)m-(f+2).
\end{displaymath} (10)

Taking the Sum,
c(m,3)=\sum_{k=2}^m {\textstyle{1\over 6}} fk^3-fk^2+({\textstyle{11\over 6}}f+1)k-(f+2).
\end{displaymath} (11)

Evaluating the Sum gives
c(m,3)={\textstyle{1\over 24}}(m-1)(m-2)(fm^2-3fm+12).
\end{displaymath} (12)

So far, this has just been relatively boring Algebra. But the amazing part is that if $f=6$ is chosen as the Integer, then $c(m,2)$ and $c(m,3)$ simplify to

$\displaystyle c(m,2)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 6}}(m-1)(6m^2-12m+6)$  
  $\textstyle =$ $\displaystyle (m-1)^3$ (13)
$\displaystyle c(m,3)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}(m-1)^2(m-2)^2,$ (14)

which are consecutive Cubes $(m-1)^3$ and nonconsecutive Squares $n^2=[(m-1)(m-2)/2]^2$.

See also Bell Triangle, Catalan's Triangle, Euler's Triangle, Leibniz Harmonic Triangle, Number Triangle, Pascal's Triangle, Seidel-Entringer-Arnold Triangle, Sum


Clark, J. E. ``Clark's Triangle.'' Math. Student 26, No. 2, p. 4, Nov. 1978.

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© 1996-9 Eric W. Weisstein