## Lacklustre Bio

No longer a final year Math student at CUHK.

## Thousand Separators

The program I attempted to use to finish Pandigital Numbers Divisible by All Integers up to 18, but alas it is off by a factor of $19$.

Since I am not putting $>600 \times 3$ lines of verification in the page, here is the program whose skeleton can be extracted for similar programs in the future.

#include <stdio.h>

int main()
{
long int i,j;
int count = 0;
for (i=6; i<43; i++)
{
j=232792560;
j*=i;
if (i%5!=0)
{
count++;
printf("{{eqn | l = %ld N\n      | r = ", i);
printf("%ld \\, ", j/1000000000);
if ((j/1000000)%1000<100)
printf("0");
if ((j/1000000)%1000<10)
printf("0");
printf("%ld \\, ", (j/1000000)%1000);
if ((j/1000)%1000<100)
printf("0");
if ((j/1000)%1000<10)
printf("0");
printf("%ld \\, ", (j/1000)%1000);
if (j%1000<100)
printf("0");
if (j%1000<10)
printf("0");
printf("%ld\n}}\n", j%1000);
}
}
printf("%d",count);
}


Its predecessor has been used to find the largest and smallest pandigital squares.

## Primes Expressible as $x^2+ny^2$

I'll put this here for (my) future reference:

#include <stdio.h>
int main(){int n,x,y;for(n=1;n<11;n++)for(x=1;x<33;x++)for(y=1;y<33;y++)if(x*x+y*y*n==1009)printf("1009=%d^2+%dx%d^2\n",x,n,y);}


also

1129=20^2+1x27^2
1129=27^2+1x20^2
1129=29^2+2x12^2
1129=19^2+3x16^2
1129=27^2+4x10^2
1129=2^2+5x15^2
1129=23^2+6x10^2
1129=11^2+7x12^2
1129=29^2+8x6^2
1129=20^2+9x9^2
1129=33^2+10x2^2

## Sum of Product of $k$ Consecutive Integers

 $\ds \sum_{i \mathop = m} ^ n \prod_{k \mathop = 0} ^ l (i + k)$ $=$ $\ds \sum_{i \mathop = m} ^ n i (i + 1) \cdots (i + l)$ $= [ m (m + 1) \cdots (m + l) ] + [ (m + 1) (m + 2) \cdots (m + l + 1) ] + \cdots + [ n (n + 1) \cdots (n + l) ]$ $\ds$ $=$ $\ds \sum_{i \mathop = m} ^ n \dfrac {(i + l) !} {(i - 1) !}$ Definiton of Factorials $\ds$ $=$ $\ds \sum_{i \mathop = m} ^ n \dfrac {(i + l) !} {(i - 1) ! (l + 1) !} \times (l + 1) !$ $\ds$ $=$ $\ds \sum_{i \mathop = m} ^ n \binom {l + i} {l + 1} (l + 1) !$ Definiton of Binomial Coefficients $\ds$ $=$ $\ds (l + 1) ! \paren {\sum_{i \mathop = 1} ^ n \binom {l + i} {l + 1} - \sum_{i = 1} ^ {m - 1} \binom {l + i} {l + 1} }$ $\ds$ $=$ $\ds (l + 1) ! \paren {\binom {l + n + 1} {l + 2} - \binom {l + m} {l + 2} }$ Rising Sum of Binomial Coefficients $\ds$ $=$ $\ds (l + 1) ! \paren {\dfrac {(l + n + 1) !} {(l + 2) ! (n - 1) !} - \dfrac {(l + m) !} {(l + 2) ! (m - 2) !} }$ Definiton of Binomial Coefficients $\ds$ $=$ $\ds \dfrac 1 {l + 2} \paren {\dfrac {(l + n + 1) !} {(n - 1) !} - \dfrac {(l + m) !} {(m - 2) !} }$ Definiton of Factorials $\ds$ $=$ $\ds \dfrac 1 {l + 2} \paren {n (n + 1) \cdots (n + l + 1) - (m - 1) (m) (m + 1) \cdots (m + l) }$ Definiton of Factorials

$\blacksquare$

In particular, when $m = 1$, the theorem takes a nice form:

 $\ds \sum_{i \mathop = 1}^n \prod_{k \mathop = 0} ^ l (i + k)$ $=$ $\ds \sum_{i \mathop = 1}^n i (i + 1) \cdots (i + l)$ $\ds = \frac {n (n + 1) \cdots (n + l + 1)} {l + 2}$

For $l = 0$, $\ds \sum_{i \mathop = 1} ^ n i = \dfrac {n (n + 1)} 2$ which is Closed Form for Triangular Numbers

For $l = 1$, $\ds \sum_{i \mathop = 1} ^ n i (i + 1) = \dfrac {n (n + 1) (n + 2)} 3$ which is Sum of Sequence of Products of Consecutive Integers

Credit: The relationship between binomial coefficients and products of consecutive integers was noticed by my friend Oscar, whose observation allowed for a intuitive view to the validity of the equation.

My original proof for the cases $m = 0$ was an induction on $n$ with a fixed $l$.

For $n = 1$, $RHS = \dfrac {1 (2) \cdots (1 + l + 1)} {l + 2} = (l + 1) ! = LHS$

For the induction step, assume $\ds \sum_{i \mathop = 1} ^ m i (i + 1) \cdots (i + l) = \dfrac {m (m + 1) \cdots (m + l + 1)} {l + 2}$.

Then

 $\ds \sum_{i \mathop = 1} ^ {m + 1} i (i + 1) \cdots (i + l)$ $=$ $\ds \dfrac {m (m + 1) \cdots (m + l + 1)} {l + 2} + (m + 1) (m + 2) \cdots (m + l + 1)$ $\ds$ $=$ $\ds \dfrac {(m + 1) (m + 2) \cdots (m + l + 1)} {l + 2} (m + (l + 2))$

The general theorem results from subtraction.

$\blacksquare$

### Solution to Problem $1290$, mmmaa

See 1988: Problem 1290 (Math. Mag. Vol. 61, no. 1: pp. 46 – 59)  www.jstor.org/stable/2690332

For positive integers $n$ and $r$, find a closed form expression for $\ds \sum_{r \mathop = 1}^k r \binom {n + r - 1} r$.
 $\ds \sum_{r \mathop = 1}^k r \binom {n + r - 1} r$ $=$ $\ds \sum_{r \mathop = 1}^k r \binom {n + r - 1} {n - 1}$ Symmetry Rule for Binomial Coefficients $\ds$ $=$ $\ds \sum_{r \mathop = 1}^k r \frac {\paren {n + r - 1} \paren {n + r - 2} \dots \paren {1 + r} } {\paren {n - 1}!}$ Definition of Binomial Coefficient $\ds$ $=$ $\ds \frac 1 {\paren {n - 1}!} \sum_{r \mathop = 1}^k r \paren {r + 1} \dots \paren {r + n - 1}$ $\ds$ $=$ $\ds \frac 1 {\paren {n - 1}!} \times \frac {k \paren {k + 1} \cdots \paren {k + n} } {n + 1}$ taking $i = r, l = n - 1, n = k$ in the theorem above in 'nice form' $\ds$ $=$ $\ds \frac n {\paren {n + 1}!} \times \frac {\paren {k + n}!} {\paren {k - 1}!}$ $\ds$ $=$ $\ds n \binom {n + k} {n + 1}$ Definition of Binomial Coefficient

One may, of course, avoid the use of this theorem by proceeding from the third step as follows:

 $\ds \frac 1 {\paren {n - 1}!} \sum_{r \mathop = 1}^k r \paren {r + 1} \dots \paren {r + n - 1}$ $=$ $\ds \frac n {n!} \sum_{r \mathop = 1}^k \frac {\paren {r + n - 1}!} {\paren {r - 1}!}$ $\ds$ $=$ $\ds n \sum_{r \mathop = 1}^k \binom {n + r - 1} n$ Definition of Binomial Coefficient $\ds$ $=$ $\ds n \sum_{r \mathop = 0}^{k - 1} \binom {n + r} n$ Translation of Index Variable of Summation $\ds$ $=$ $\ds n \binom {n + k} {n + 1}$ Rising Sum of Binomial Coefficients

In their notation $\ds \sequence {n \atop r}:= \binom {n + r - 1} r$, we will have:

$\ds \sum_{r \mathop = 1}^k r \sequence {n \atop r} = n \sequence {k \atop {n + 1} }$

$\blacksquare$

I wish to generalize this result to arithmetic progressions at some point.

### Potential Corollary and its Generalisation

We have:

 $\ds 1 + 2$ $=$ $\ds 3$ $\ds 4 + 5 + 6$ $=$ $\ds 7 + 8$ $\ds 9 + 10 + 11 + 12$ $=$ $\ds 13 + 14 + 15$ $\ds$ $:$ $\ds$

Can't find related result - a visual proof was given by RBN in PWW.

 $\ds T_1 + T_2 + T_3$ $=$ $\ds T_4$ $\ds T_5 + T_6 + T_7 + T_8$ $=$ $\ds T_9 + T_{10}$ $\ds T_{11} + T_{12} + T_{13} + T_{14} + T_{15}$ $=$ $\ds T_{16} + T_{17} + T_{18}$ $\ds$ $:$ $\ds$ Sum of Adjacent Sequences of Triangular Numbers