OFFSET
1,1
COMMENTS
All solutions to phi(x)=n! belong to the interval [n!,(n+1)!] and 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, i.e., a(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.
Experimentally, a(n) = c(n)*(n+1)! with a coefficient c(n) ~ 2^(-n/10) (e.g., c(1) = c(2) = 1, c(10) ~ 0.5)
LINKS
Max A. Alekseyev, Computing the Inverses, their Power Sums, and Extrema for Euler's Totient and Other Multiplicative Functions. Journal of Integer Sequences, Vol. 19 (2016), Article 16.5.2
EXAMPLE
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Oct 04 2009
EXTENSIONS
Edited and terms a(12)-a(28) added by Max Alekseyev, Jan 26 2012, Jul 09 2014
STATUS
approved