MathLabs

Problem 4

Find all pairs (k,n)(k,n) of positive integers such that k!=(2n−1)(2n−2)(2n−4)⋯(2n−2n−1).k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}).
Step 2 of 6: Compute the 2-adic valuation of the product
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Ln=20+1+⋯+(n−1)∏i=0n−1(2n−i−1),v2(Ln)=n(n−1)2L_n=2^{0+1+\cdots+(n-1)}\prod_{i=0}^{n-1}(2^{n-i}-1),\qquad v_2(L_n)=\frac{n(n-1)}2
Detailed analysis

Factor 2i2^i from the ii-th factor: 2n−2i=2i(2n−i−1)2^n-2^i=2^i(2^{n-i}-1). Every remaining factor is odd, so v2(Ln)=0+1+⋯+(n−1)=n(n−1)/2v_2(L_n)=0+1+\cdots+(n-1)=n(n-1)/2.