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 A055487 Least m such that phi(m) = n!. 8
 1, 3, 7, 35, 143, 779, 5183, 40723, 364087, 3632617, 39916801, 479045521, 6227180929, 87178882081, 1307676655073, 20922799053799, 355687465815361, 6402373865831809, 121645101106397521, 2432902011297772771, 51090942186005065121, 1124000727844660550281, 25852016739206547966721, 620448401734814833377121, 15511210043338862873694721, 403291461126645799820077057, 10888869450418352160768000001, 304888344611714964835479763201 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Erdős believed (see Guy reference) that phi(x) = n! is solvable. Factorial primes of the form p = A002981(m)! + 1 = k! + 1 give the smallest solutions for some m (like m = 1,2,3,11) as follows: phi(p) = p-1 = A002981(m)!. According to Tattersall, in 1950 H. Gupta showed that phi(x) = n! is always solvable. - Joseph L. Pe, Oct 01 2002 A123476(n) is a solution to the equation phi(x)=n!. - T. D. Noe, Sep 27 2006 From M. F. Hasler, Oct 04 2009: (Start) Conjecture: Unless n!+1 is prime (i.e., n in A002981), a(n)=pq where p is the least prime > sqrt(n!) such that (p-1) | n! and q=n!/(p-1)+1 is prime. Probably "least prime > sqrt(n!)" can also be replaced by "largest prime <= ceiling(sqrt(n!))". The case "= ceiling(...)" occurs for n=5, sqrt(120) = 10.95..., p=11, q=13. a(n) is the first element in row n of the table A165773, which lists all solutions to phi(x)=n!. Thus a(n) = A165773((Sum_{k

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Last modified February 1 08:20 EST 2023. Contains 359992 sequences. (Running on oeis4.)