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A319008
Let k = A000396(n) be the n-th perfect number, a(n) is the least number m such that k*d*m + 1 is prime for all of the proper divisors d of k so their product is a Carmichael number.
1
1, 2136, 13494274080, 216818853118725
OFFSET
1,2
COMMENTS
Chernick proved that (6m + 1)*(12m + 1)*(18m + 1) is a Carmichael number, if all the 3 factors are primes (A033502, A046025).
Lieuwens generalized it to Product_{i} (k*d(i)*m + 1), for k a perfect number.
a(1) corresponds to 6. It was found by Jack Chernick in 1939.
a(2) corresponds to 28. It was found by Dubner in 1996. Lieuwens evaluated that the least corresponding Carmichael number > 10^27.
a(3) corresponds to 496. It was found by Jim Fougeron in 2002 (Dubner found a larger value: 474382033125).
a(4) corresponds to 8128. It was found by Phil Carmody in 2002.
The corresponding Carmichael numbers are 1729, 599966117492747584686619009, 1.631... * 10^126, 4.559... * 10^260, ...
REFERENCES
Harold Davenport, The Higher Arithmetic, Cambridge University Press, 7th ed., 1999, exercise 8.4.
Harvey Dubner, Carmichael numbers and Egyptian fractions, Mathematica japonicae, Vol. 43, No. 2 (1996), pp. 411-419.
LINKS
Jack Chernick, On Fermat's simple theorem, Bulletin of the American Mathematical Society, Vol. 45, No. 4 (1939), pp. 269-274.
Claude Goutier, De l'utilité d'une vieille curiosité grecque, Crux Mathematicorum, Vol. 46, No. 8 (2020), pp. 397-403.
G. L. Honaker, Jr. and Chris Caldwell, Prime Curios!  59996...19009 (27-digits).
Erik Lieuwens, Fermat pseudo primes, Doctoral Thesis, Delft University of Technology, 1971, pp. 29-30.
Carlos Rivera, Puzzle 171. Perfect & Carmichael numbers, The Prime Puzzles & Problems Connection.
EXAMPLE
28 = 1 + 2 + 4 + 7 + 14 is the second perfect number. 2136 is the least number m such that 28*1*333 + 1 = 59809, 28*2*2136 + 1 = 119617, 28*4*2136 + 1 = 239233, 28*7*2136 + 1 = 418657 and 28*14*2136 + 1 = 837313 are all primes, therefore 59809*119617*239233*418657*837313 = 599966117492747584686619009 is a Carmichael number.
MATHEMATICA
ms = {2, 3, 5, 7, 13}; ns = Length[ms]; M[p_] := 2^(p - 1)*(2^p - 1); L[m_] := Module[{}, d = Most[Divisors[m]]*m; aQ[n_] := AllTrue[d*n + 1, PrimeQ]; n=1; While[!aQ[n], n++]; n]; s={}; Do[m = M[ms[[k]]]; b = L[m]; AppendTo[s, b], {k, 1, ns}]; s
CROSSREFS
KEYWORD
nonn,bref,more
AUTHOR
Amiram Eldar, Sep 07 2018
STATUS
approved