Proof of Bertrand's postulate: Difference between revisions - Wikipedia

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since <math>\tbinom{2n}{n}</math> is the largest term in the sum in the right-hand side, and the sum has <math>2n</math> terms (including the initial <math>2</math> outside the summation).

For a fixed prime <math>p</math>, define <math>R:=R(p,n,p)</math> to be the [[p-adic order]] of <math>\tbinom{2n}{n}</math>, that is, the largest natural number <math>r</math> such that <math>p^r</math> divides <math>\tbinom{2n}{n}</math>.

===Lemma 2===