A359945 Largest k < n such that n! / k! = m! = A000142(m) for some m.
0, 1, 1, 1, 1, 5, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 23, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1
Keywords
Examples
For n = 1, the largest k < n is k = 0 and indeed, 1! / 0! = 1! is a factorial number, so a(1) = 0. Similarly, for all n in A000142, i.e., n = m!, the largest k < n is k = n-1 and n! / (n-1)! = n = m!, so a(n = m!) = n-1. For n = 10, 10! / 9! = 10 and 10! / 8! = 90 aren't factorial numbers, but 10! / 7! = 10*9*8 = 2*3*4*5*6, so a(10) = 7.
Links
- Paul Erdős, Problems and results on number theoretic properties of consecutive integers and related questions, Proc. 5th Manitoba Conf. Numerical Math., Congress. Num. 16 (1975), 25-44.
- Laurent Habsieger, Explicit bounds for the diophantine equation A!B! = C!, arXiv:1903.08370, March 2019.
Programs
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PARI
a(n)={my(m=1, f=n!); while(n-->m, while(m!*n!
Formula
a(n) > n/2 unless a(n) = 1 or n = 1.
a(n) = n-1 iff n is in A000142 = factorial numbers.
Comments