LashaBukhnikashvili's blog

By LashaBukhnikashvili, 6 years ago, In English

Help me to prove: C(2*n,n) | lcm(1,2,3,...,2*n) for all natural n

 
 
 
 
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6 years ago, # |
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Sorry I was wrong :(

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6 years ago, # |
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Let p be a prime number, and assume that k is max number such that p^k <= 2n. Let v_p(n) be the max number t, such that p^t | n. We must prove:

v_p(C(2n, n)) <= v_p(lcm(1, 2, ... , 2*n))

It's easy to see that v_p(lcm(1, 2, ..., 2*n)) = k, and (http://en.wikipedia.org/wiki/Legendre%27s_formula):

v_p(C(2n, n)) = ([2n/p]-2*[n/p]) + ([2n/p^2]-2[n/p^2]) + ... + ([2n/p^k]-2[n/p^k])

But we can easly prove that [2x/y] — 2[x/y] <= 1 for any positive integer x,y, and that's all :)