MathDB

8

Part of 1983 Austrian-Polish Competition

Problems(1)

c_{n}=(4n-6)/n c_{n-1} is integer, prod (2^{n+1-i}-1)^{2^i} divides(2^{n+1}-1)!

Source: Austrian Polish 1983 APMC

4/30/2020
(a) Prove that (2n+1−1)!(2^{n+1}-1)! is divisible by ∏i=0n(2n+1−i−1)2i \prod_{i=0}^n (2^{n+1-i}-1)^{2^i }, for every natural number n (b) Define the sequence (cnc_n) by c1=1c_1=1 and cn=4n−6ncn−1c_{n}=\frac{4n-6}{n}c_{n-1} for n≥2n\ge 2. Show that each cnc_n is an integer.
number theoryrecurrence relationSequencedividesfactorialdivisibleProduct