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A055487
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Least m such that phi(m) = n!.
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8
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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
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OFFSET
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1,2
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COMMENTS
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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
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<n} A055506(k)) + 1). The last element of each row (i.e., the largest solution to phi(x)=n!) is given in A165774. (End)
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REFERENCES
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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.
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LINKS
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P. Erdős and J. Lambek, Problem 4221, Amer. Math. Monthly, 55 (1948), 103.
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FORMULA
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MATHEMATICA
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Array[Block[{k = 1}, While[EulerPhi[k] != #, k++]; k] &[#!] &, 10] (* Michael De Vlieger, Jul 12 2018 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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