Help me to prove: C(2*n,n) | lcm(1,2,3,...,2*n) for all natural n
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Help me to prove: C(2*n,n) | lcm(1,2,3,...,2*n) for all natural n
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Sorry I was wrong :(
dude your wrong :| lcm of m numbers isnt their multipication divided by their gcd
o_O
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 :)
Thanks I got it ;)
That's the man I expected to answer that question :D