User:RandomUndergrad

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Lacklustre Bio

No longer 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

Potential Solution to Dudeney Problem 109

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