User:RandomUndergrad

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Lacklustre Bio

Currently a final year Math student at CUHK.

Random Proofs (Poorly Formatted)

23...3^2 = 54...428...89

Cyclic Permutations Preserve Divisibility

Number of Partitions with no Multiple of n equals Number of Partitions where Parts appear less than n times

Generalisation of Nicomachus's Theorem

Volume of Unit Hypersphere

Square as Difference between Square and Square of Reversal

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.