A318646 The least Chernick's "universal form" Carmichael number with n prime factors.

1729, 63973, 26641259752490421121, 1457836374916028334162241, 24541683183872873851606952966798288052977151461406721, 53487697914261966820654105730041031613370337776541835775672321, 58571442634534443082821160508299574798027946748324125518533225605795841

Offset

3,1

Comments

Chernick proved that U(k, m) = (6m + 1)*(12m + 1)*Product_{i = 1..k-2} (9*(2^i)m + 1), for k >= 3 and m >= 1 is a Carmichael number, if all the factors are primes and, for k >= 4, 2^(k-4) divides m. He called U(k, m) "universal forms". This sequence gives a(k) = U(k, m) with the least value of m. The least values of m for k = 3, 4, ... are 1, 1, 380, 380, 780320, 950560, 950560, 3208386195840, 31023586121600, ...

Links

Amiram Eldar, Table of n, a(n) for n = 3..11

Jack Chernick, On Fermat's simple theorem, Bulletin of the American Mathematical Society, Vol. 45, No. 4 (1939), pp. 269-274.

Daniel Suteu, C++ program

Samuel S. Wagstaff, Jr., Large Carmichael numbers, Mathematical Journal of Okayama University, Vol. 22, (1980), pp. 33-41.

Example

For k=3, m = 1, a(3) = U(3, 1) = (6*1 + 1)*(12*1 + 1)*(18*1 + 1) = 1729.
For k=4, m = 1, a(4) = U(4, 1) = (6*1 + 1)*(12*1 + 1)*(18*1 + 1)*(36*1 + 1) = 63973.
For k=5, m = 380, a(5) = U(5, 1) = (6*380 + 1)*(12*380 + 1)*(18*380 + 1)*(36*380 + 1)*(72*380 + 1) =  26641259752490421121.

Mathematica

fc[k_] := If[k < 4, 1, 2^(k - 4)]; a={}; Do[v = Join[{6, 12}, 2^Range[k-2]*9];
w = fc[k]; x = v*w; m = 1; While[! AllTrue[x*m + 1, PrimeQ], m++]; c=Times @@ (x*m + 1); AppendTo[a, c], {k, 3, 9}]; a

Crossrefs

Cf. A002997, A033502 (3 prime factors), A206024 (4 prime factors), A206349 (5 prime factors), A126797.

Sequence in context: A272798 A212920 A317126 * A182087 A327787 A352970

Adjacent sequences: A318643 A318644 A318645 * A318647 A318648 A318649

Keyword

nonn

Author

Amiram Eldar, Aug 31 2018

Status

approved