## Lacklustre Bio

Currently a final year Math student at CUHK.

## 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

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

$\blacksquare$

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

 $\displaystyle \sum_{i = 1} ^ n \prod_{k = 0} ^ l (i + k)$ $=$ $\displaystyle \sum_{i = 1} ^ n i (i + 1) \cdots (i + l)$ $\displaystyle = \dfrac {n (n + 1) \cdots (n + l + 1)} {l + 2}$

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

For $l = 1$, $\displaystyle \sum_{i = 1} ^ n i (i + 1) = \dfrac {n (n + 1) (n + 2)} 3$ 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 $\displaystyle \sum_{i = 1} ^ m i (i + 1) \cdots (i + l) = \dfrac {m (m + 1) \cdots (m + l + 1)} {l + 2}$.

Then

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

The general theorem results from subtraction.

$\blacksquare$

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

### Potential Corollary and its Generalisation

We have:

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

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

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

Then do we have:

 $\displaystyle H_1 + H_2 + H_3 + H_4$ $=$ $\displaystyle H_5$ $\displaystyle H_6 + H_7 + H_8 + H_9 + H_{10}$ $=$ $\displaystyle H_{11} + H_{12}$ $\displaystyle H_{13} + H_{14} + H_{15} + H_{16} + H_{17} + H_{18}$ $=$ $\displaystyle H_{19} + H_{20} + H_{21}$ $\displaystyle$ $:$ $\displaystyle$

or

 $\displaystyle P_1 + P_2 + P_3 + P_4 + P_5$ $=$ $\displaystyle P_6$ $\displaystyle P_7 + P_8 + P_9 + P_{10} + P_{11} + P_{12}$ $=$ $\displaystyle P_{13} + P_{14}$ $\displaystyle$ $:$ $\displaystyle$

For tetrahedral numbers and pentatope numbers, etc.?

This is possibly equivalent to:

 $\displaystyle \map S {l, 1} + \dots + \map S {l, l + 2}$ $=$ $\displaystyle \map S {l, l + 3}$ $\displaystyle \map S {l, l + 4} + \dots + \map S {l, 2 l + 6}$ $=$ $\displaystyle \map S {l, 2 l + 7} + \map S {l, 2 l + 8}$ $\displaystyle \map S {l, 2 l + 9} + \dots + \map S {l, 3 l + 12}$ $=$ $\displaystyle \map S {l, 3 l + 13} + \map S {l, 3 l + 14} + \map S {l, 3 l + 15}$ $\displaystyle$ $:$ $\displaystyle$

Where $\map S {l, m} = \displaystyle \sum_{i \mathop = 1}^m i (i + 1) \cdots (i + l)$.

For which, if true, its proof will be a wonderful cacophony of sums and algebra.