In 1939, Jack Chernick proved that, for n ≥ 3 and m ≥ 1:

   U(n, m) = (6m + 1) * (12m + 1) * Product_{i=1..n-2} (2^i * 9m + 1)

is a Carmichael number if all the factors are primes and, for n > 4, m is a multiple of 2^(n-4).



Example
   U(3, m) = (6m + 1) * (12m + 1) * (18m + 1)
   U(4, m) = U(3, m) * (2^2 * 9m + 1)
   U(5, m) = U(4, m) * (2^3 * 9m + 1)
   ...
   U(n, m) = U(n-1, m) * (2^(n-2) * 9m + 1)

 The smallest Chernick's Carmichael number with 3 prime factors, is: U(3, 1) = 1729.
 The smallest Chernick's Carmichael number with 4 prime factors, is: U(4, 1) = 63973.
 The smallest Chernick's Carmichael number with 5 prime factors, is: U(5, 380) = 26641259752490421121.


For n = 5, the smallest number m that satisfy Chernick's conditions, is m = 380, therefore U(5, 380) is the smallest Chernick's Carmichael number with 5 prime factors.

U(5, 380) is a Chernick's Carmichael number because m = 380 is a multiple of 2^(n-4), where n = 5, and the factors { (6*380 + 1), (12*380 + 1), (18*380 + 1), (36*380 + 1), (72*380 + 1) } are all prime numbers.



Task
For n ≥ 3, let a(n) be the smallest Chernick's Carmichael number with n prime factors.

 Compute a(n) for n = 3..9.
 Optional: find a(10).

Note: it's perfectly acceptable to show the terms in factorized form:

 a(3) = 7 * 13 * 19
 a(4) = 7 * 13 * 19 * 37
 a(5) = 2281 * 4561 * 6841 * 13681 * 27361
 ...



See also
 Jack Chernick, On Fermat's simple theorem (PDF)
 OEIS A318646: The least Chernick's "universal form" Carmichael number with n prime factors


 Related tasks
 Carmichael 3 strong pseudoprimes


ClearAll[PrimeFactorCounts, U]
PrimeFactorCounts[n_Integer] := Total[FactorInteger[n][[All, 2]]]
U[n_, m_] := (6 m + 1) (12 m + 1) Product[2^i 9 m + 1, {i, 1, n - 2}]
FindFirstChernickCarmichaelNumber[n_Integer?Positive] := 
 Module[{step, i, m, formula, value},
  step = Ceiling[2^(n - 4)];
  If[n > 5, step *= 5];
  i = step;
  formula = U[n, m];
  PrintTemporary[Dynamic[i]];
  While[True,
   value = formula /. m -> i;
   If[PrimeFactorCounts[value] == n,
    Break[];
    ];
   i += step
   ];
  {i, value}
  ]
FindFirstChernickCarmichaelNumber[3]
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