For each positive integer $ n$, define $ f(n)$ as the exponent of the $ 2$ in the decomposition in prime factors of the number $ n!$. Prove that the equation $ n-f(n)=a$ has infinitely many solutions for any positive integer $ a$.
Using the well-known Legendre's formula for $ f(n)$ it is easy to see that $ n-f(n)=2n-f(2n)$ thus if there is a solution for $ n-f(n)=a$ than there are infinitly many. To find one solution apply induction on a. it's obvious that $ n_0=1$ is a solution when $ a=1$ . Suppose we have $ b-f(b)=a-1$ for some positive integer a. Let $ k$ be the minimum number such that $ b<2^{k}$. Then $ n=2^{k}+b$ is a solution for $ n-f(n)=a$ and we are done.
Claim:: $ n=2^a-1$ works.
Proof:Using Legrendre`s formula $f(n)=[n/2]+[n/4]+...$ we obtain that $ f(n)=2^a-1-a$ .
Let the set ${2^a-1=2^k-1 , k\ge a} $. Easily this satisfies the conditions.