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Please help me to find solution or prove that it does not exist.

$$ C^{2^k}_{2^n} < 2^{2^k (n - k)}, 1<k<n, k \in \mathbb{N}, n \in \mathbb{N} $$

I tried to find solution numerically, but when $n$ > 20, the numbers grows very fast, so it seems like it has not solution (because I didn't find it when $n$ < 20). I am looking for analytical methods to prove this.

  • I take it that symbol on the left is $2^k$-factorial choose $2^n$-factorial? But if $n>k$, then this symbol evaluates to zero. – Gerry Myerson Feb 02 '18 at 07:57
  • Wow. Never saw the notation used that way. – Gerry Myerson Feb 02 '18 at 11:15
  • @gerry-myerson https://en.wikipedia.org/wiki/Binomial_coefficient#History_and_notation –  Feb 02 '18 at 11:21
  • @gerry-myerson You was right, it definitely my mistake, I fixed equation, so the left hand side must be $\frac{2^n!}{2^k!(2^n-2^k)!}$ –  Feb 02 '18 at 12:48

1 Answers1

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I've found an answer.

Firstly we can introduce $N = 2^n$, $K = 2^k$. So the inequality will be rewritten as

$C_N^K < 2^{K(\log_2(N)-\log_2(K))}$

or

$\log_2(C_N^K) < K \log_2(N/K)$.

Secondly, according to Best upper and lower bound for a binomial coeficient we can use inequality

$C_N^K \ge (N / K)^K$.

It shows that there are no solutions.