OFFSET
1,2
COMMENTS
Given a number n and a prime number p such that n*p is a Lucas-Carmichael number, then (p+1)|(n-1), so the number of prime solution p given n is bounded by the number of divisors of (n-1).
The number of solutions is 0, 1, 2, 4, 5, 6, 7, 8, 10, 13, 15, 32.
EXAMPLE
21 has one prime number, 19, such that 21*19 = 399 is a Lucas-Carmichael number. 55 has 2 prime numbers, 17 and 53, such that 55*17 = 935 and 55*53 = 2915 are Lucas-Carmichael numbers.
MATHEMATICA
lucasCarmichaelQ[n_]:=!PrimeQ[n] && Union[Transpose[FactorInteger[n]][[2]]] == {1} && Union[Mod[n + 1, Transpose[FactorInteger[n]][[1]]+1]]=={0};
numSol[n_]:=Module[{m = 0}, ds = Divisors[n-1]; Do[p = ds[[k]]-1; If[!PrimeQ[p], Continue[]]; If[! lucasCarmichaelQ[p*n], Continue[]]; m++, {k, 1, Length[ds]}]; m]; numSolmax = -1; seq = {}; nums = {};
Do[m = numSol[n]; If[m > numSolmax, AppendTo[seq, n]; AppendTo[nums, m]; Print[{n, m}]; numSolmax = m], {n, 1, 100000}]; seq
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Amiram Eldar, Sep 15 2017
STATUS
approved