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A055487 Least m such that phi(m) = n!. 8

%I #46 Nov 27 2019 05:32:19

%S 1,3,7,35,143,779,5183,40723,364087,3632617,39916801,479045521,

%T 6227180929,87178882081,1307676655073,20922799053799,355687465815361,

%U 6402373865831809,121645101106397521,2432902011297772771,51090942186005065121,1124000727844660550281,25852016739206547966721,620448401734814833377121,15511210043338862873694721,403291461126645799820077057,10888869450418352160768000001,304888344611714964835479763201

%N Least m such that phi(m) = n!.

%C Erdős believed (see Guy reference) that phi(x) = n! is solvable.

%C 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)!.

%C According to Tattersall, in 1950 H. Gupta showed that phi(x) = n! is always solvable. - _Joseph L. Pe_, Oct 01 2002

%C A123476(n) is a solution to the equation phi(x)=n!. - _T. D. Noe_, Sep 27 2006

%C From _M. F. Hasler_, Oct 04 2009: (Start)

%C 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.

%C 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.

%C 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<n} A055506(k)) + 1). The last element of each row (i.e., the largest solution to phi(x)=n!) is given in A165774. (End)

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

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

%H Max A. Alekseyev, <a href="https://www.emis.de/journals/JIS/VOL19/Alekseyev/alek5.html">Computing the Inverses, their Power Sums, and Extrema for Euler's Totient and Other Multiplicative Functions</a>. Journal of Integer Sequences, Vol. 19 (2016), Article 16.5.2.

%H P. Erdős and J. Lambek, <a href="https://www.jstor.org/stable/2305755">Problem 4221</a>, Amer. Math. Monthly, 55 (1948), 103.

%F a(n) = Min{m : phi(m) = n!} = Min{m : A000010(m) = A000142(n)}.

%t Array[Block[{k = 1}, While[EulerPhi[k] != #, k++]; k] &[#!] &, 10] (* _Michael De Vlieger_, Jul 12 2018 *)

%Y Cf. A055486, A055488, A055489, A055506, A000010, A000142.

%Y Cf. A123476, A165773, A165774.

%K nonn

%O 1,2

%A _Labos Elemer_, Jun 28 2000

%E More terms from _Don Reble_, Nov 05 2001

%E a(21)-a(28) from _Max Alekseyev_, Jul 09 2014

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Last modified May 2 04:48 EDT 2024. Contains 372178 sequences. (Running on oeis4.)