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


package main
 
import (
    "fmt"
    "math/big"
)
 
var (
    zero = new(big.Int)
    prod = new(big.Int)
    fact = new(big.Int)
)
 
func ccFactors(n, m uint64) (*big.Int, bool) {
    prod.SetUint64(6*m + 1)
    if !prod.ProbablyPrime(0) {
        return zero, false
    }
    fact.SetUint64(12*m + 1)
    if !fact.ProbablyPrime(0) { // 100% accurate up to 2 ^ 64
        return zero, false
    }
    prod.Mul(prod, fact)
    for i := uint64(1); i <= n-2; i++ {
        fact.SetUint64((1<<i)*9*m + 1)
        if !fact.ProbablyPrime(0) {
            return zero, false
        }
        prod.Mul(prod, fact)
    }
    return prod, true
}
 
func ccNumbers(start, end uint64) {
    for n := start; n <= end; n++ {
        m := uint64(1)
        if n > 4 {
            m = 1 << (n - 4)
        }
        for {
            num, ok := ccFactors(n, m)
            if ok {
                fmt.Printf("a(%d) = %d\n", n, num)
                break
            }
            if n <= 4 {
                m++
            } else {
                m += 1 << (n - 4)
            }
        }
    }
}
 
func main() {
    ccNumbers(3, 9)
}