1,2

Erdos believed (see Guy reference) that Phi[x] = n! is solvable.

Factorial primes of p = A002981[m]!+1 = k!+1 form give 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 (joseph_l_pe(AT)hotmail.com), Oct 01 2002

A123476(n) is a solution to the equation phi(x)=n! - T. D. Noe (noe(AT)sspectra.com), Sep 27 2006

Contribution from M. F. Hasler (MHasler(AT)univ-ag.fr), 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 <= ceil(sqrt(n!))". The case "= ceil(...)" occurs for n=5, sqrt(120)=10.95..., p=11, q=13.

A055487(n) is the first element in row n of the table A165773, which lists all solutions to phi(x)=n!. Thus A055487(n)=A165773(sum(A055506(k),k<n)+1). The last element of each row (i.e. the largest solution to phi(x)=n!) is given in A165774. (End)

R. K. Guy, (1981): Unsolved problems In Number Theory, Springer - page 53.

Tattersall, J. "Elementary Number Theory in Nine Chapters", Cambridge University Press, 2001, p. 162.

P. Erdos and J. Lambek, Problem 4221, Amer. Math. Monthly, 55 (1948), 103.

a(n) = Min{m : Phi[m] = n!} = Min{m : A000010(m) = A000142(n)}

(PARI) A055487(n)={ my( f=n!, p=sqrtint(f)); isprime(f+1) && return(f+(n>1)); until( isprime(f/p+1), while( f%p=nextprime(p+2)-1, )); (p+1)*(f/p+1) } /* based on the conjecture */ [From M. F. Hasler (MHasler(AT)univ-ag.fr), Oct 04 2009]

Cf. A055486-A055489, A055506, A000010, A000142.

Sequence in context: A047907 A145874 A147681 this_sequence A121130 A006099 A053530

Adjacent sequences: A055484 A055485 A055486 this_sequence A055488 A055489 A055490

nonn

Labos E. (labos(AT)ana.sote.hu), Jun 28 2000

More terms from djr(AT)nk.ca, Nov 05 2001