OFFSET
1,2
COMMENTS
Product_{i} (k*d(i)*m + 1) is a Carmichael number.
Chernick proved that (6m + 1)*(12m + 1)*(18m + 1) is a Carmichael number, if all the 3 factors are prime (A033502, A046025).
Lieuwens generalized it to Product_{i} (k*d(i)*m + 1), for k a perfect number (A000396), e.g., A067199 for k = 28.
Rotkiewicz generalized it to any number k with a subset of its proper divisors that sums to k.
The corresponding generated Carmichael numbers are 1729, 393575432565765601, 599966117492747584686619009, 17167430884969, 11744090279809908081796578516491199598397832961, 3680409480386689, 617027029751094776871101828064081267143041, 7622722964881, 700705956080852569, 90694625332467786841, 24182595473200959889, 553229304821570521, 3915654940974324169, 9215447790472998049, 4890416189580986381506017143209122707839833885365268481, 1746281192537521, ...
Supersequence of A319008.
LINKS
Jack Chernick, On Fermat's simple theorem, Bulletin of the American Mathematical Society, Vol. 45, No. 4 (1939), pp. 269-274.
Bill Daly, Perfect numbers and Carmichael numbers - a hidden relation, transcription of a network conversation at David Eppstein's Egyptian Fractions web site.
Erik Lieuwens, Fermat pseudo primes, Doctoral Thesis, Delft University of Technology, 1971, pp. 29-30.
Andrzej Rotkiewicz, On some problems of W. Sierpinski, Acta Arithmetica, Vol. 21 (1972), pp. 251-259.
EXAMPLE
20 = 1 + 4 + 5 + 10 is the sum of a single subset of the proper divisors of 20. 333 is the least number such that 20*1*333 + 1 = 6661, 20*4*333 + 1 = 26641, 20*5*333 + 1 = 33301, and 20*10*333 + 1 = 66601 are all primes, therefore 6661*26641*33301*66601 = 393575432565765601 is a Carmichael number.
MATHEMATICA
okQ[n_] := Module[{d = Most[Divisors[n]]}, SeriesCoefficient[Series[Product[1 + x^i, {i, d}], {x, 0, n}], n] == 1]; s = Select[Range[300], okQ]; divSubset[n_] := Module[{d = Most[Divisors[n]]}, divSets = Subsets[d]; ns = Length[divSets];
Do[divs = divSets[[k]]; If[Total[divs] == n, Break[]], {k, 1, ns}]; divs]; leastMultiplier[n_] := Module[{divs = divSubset[n]}, m = 1;
While[! AllTrue[n*m*divs + 1, PrimeQ], m++]; m]; seq = {}; Do[s1 = s[[k]]; m = leastMultiplier[s1]; AppendTo[seq, m], {k, 1, Length[s]}]; seq (* after Harvey P. Dale at A064771 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, Sep 07 2018
STATUS
approved