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%I #18 Jan 31 2023 08:54:19
%S 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,
%T 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,
%U 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
%N Largest k < n such that n! / k! = m! = A000142(m) for some m.
%C For n = 0 there is no k < 0 for which k! would be defined, therefore the sequence starts at offset n = 1.
%C Surányi conjectured (cf. Erdős and Habsieger references and A003135) that a(10) = 7 corresponding to 10! = 7! * 6! is the only nontrivial solution, i.e., other than a(n) = n-1 for n = m! and a(n) = 1 otherwise.
%H Paul Erdős, <a href="http://www.renyi.hu/~p_erdos/1976-39.pdf">Problems and results on number theoretic properties of consecutive integers and related questions</a>, Proc. 5th Manitoba Conf. Numerical Math., Congress. Num. 16 (1975), 25-44.
%H Laurent Habsieger, <a href="https://arxiv.org/abs/1903.08370">Explicit bounds for the diophantine equation A!B! = C!</a>, arXiv:1903.08370, March 2019.
%F a(n) > n/2 unless a(n) = 1 or n = 1.
%F a(n) = n-1 iff n is in A000142 = factorial numbers.
%e For n = 1, the largest k < n is k = 0 and indeed, 1! / 0! = 1! is a factorial number, so a(1) = 0.
%e 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.
%e 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.
%o (PARI) a(n)={my(m=1, f=n!); while(n-->m, while(m!*n!<f,m++); n!*m!==f && return(n)); f>1;}
%Y Cf. A000142 (factorial numbers).
%Y Cf. A003135 (n! is a nontrivial product of factorials).
%K nonn
%O 1,6
%A _M. F. Hasler_, Jan 19 2023
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