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On Claims Derived from Shifted Legendre Polynomial Coefficient Formulas

January 7th, 2018 No comments

A few combinatorics arguments... perhaps you can think of some proofs? Hint: I did not approach this via induction proofs:

General Multiplicative Notions

Corollary

 \binom{n}{k-1} \binom{n+k-1}{k-1} = k n \left( \frac{1}{n-k} \right) \left(\frac{1}{n-k+1} \right) \prod_{m \neq n, k} \left| \frac{n}{m-k} -1 \right|

for n \neq k and n \neq k - 1

Corollary

 \binom{2n - 1}{n-1} = \binom{2n - 1}{n} = \prod_{m \neq n} \frac{2n-m}{n-m}

Corollary

 \binom{2n}{n} = \left(n + 1 \right) \prod_{m \neq n} \frac{2n - m + 1}{n + m + 1} = \prod_{m} \frac{2n - m + 1}{n - m + 1} = \prod_m \frac{2n - m + 1}{m}

Corollary: Binomial Multiplicative Formula

 \binom{n}{k} = \prod_{m = 1}^k \frac{n-m+1}{k-m+1} = \prod_{m = 1}^k \frac{n-m+1}{m}

Corollary

 2 \prod_{m \neq n} \frac{2n - m}{n - m} = \left( n + 1 \right) \prod_{m \neq n} \frac{2n - m + 1}{n - m + 1}

Catalan Numbers

Corollary

 \frac{1}{n+1} \binom{2n}{n} = \prod_{m \neq n} \frac{2n - m + 1}{n - m + 1}

Corollary

 \prod_{m = 2}^n \frac{n+m}{m} = \prod_{m \neq n} \frac{2n - m + 1}{n - m + 1} = \frac{2}{n+1} \prod_{m \neq n} \frac{2n - m}{n - m}