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All solutions to phi(x)=n! are listed in the n-th row of A165773 (when written as table with row lengths A055506). Thus this sequence gives the last element in these rows, and therefore A165774(n) = A165773(sum(A055506(k),k=1..n)).
All terms in this sequence are even, since if x is an odd solution to phi(x)=n!, then 2x is a larger solution because phi(2x)=phi(2)*phi(x)=phi(x).
Most terms (and any term divisible by 4) are divisible by 3, since if x=2^k*y is a solution with k>1 and gcd(y,2*3)=1, then x*3/2 = 2^(k-1)*3*y is a larger solution because phi(2^(k-1)*3)=2^(k-2)*(3-1)=2^(k-1)=phi(2^k).
For the same reason, most terms are divisible by 5, since if x=2^k*y is a solution with k>2 and gcd(y,2*5)=1, then x*5/4 is a larger solution.
Also, any term of the form x=2^k*3^m*y with k,m>1 must be divisible by 7 (else x*7/6 would be a larger solution), and so on.
It seems that all solutions to phi(x)=n! are in the interval [n!,(n+1)! ]. Clearly, A055487(n) is by definition larger than n! for all n>1. Experimentally, A165774(n) = c(n)*(n+1)! with a coefficient c(n)~2^(-n/10) (c(1)=c(2)=1, c(10)~0.5)
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