A382809 a(n) = (6*n + 1)*(12*n + 1)*(18*n + 1).

1, 1729, 12025, 38665, 89425, 172081, 294409, 464185, 689185, 977185, 1335961, 1773289, 2296945, 2914705, 3634345, 4463641, 5410369, 6482305, 7687225, 9032905, 10527121, 12177649, 13992265, 15978745, 18144865, 20498401, 23047129, 25798825, 28761265, 31942225, 35349481

Offset: 0

Author

Stefano Spezia, Apr 05 2025

Keywords

• nonn
• easy

Comments

a(n) is a Carmichael number if all the three factors (6*n + 1), (12*n + 1), and (18*n + 1) are prime (see Chernick and Ribenboim).

References

• Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 101.
• James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 146.

Links

• Jack Chernick, On Fermat's simple theorem, Bulletin of the American Mathematical Society, Vol. 45, No. 4 (1939), pp. 269-274.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A002997, A033502, A046025.

Cf. A002476, A002997, A016921, A017533, A068228, A161705.

Programs

Mathematica

LinearRecurrence[{4,-6,4,-1},{1,1729,12025,38665},31]

Formula

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3.

G.f.: (1 + 1725*x + 5115*x^2 + 935*x^3)/(1 - x)^4.

E.g.f.: exp(x)*(1 + 1728*x + 4284*x^2 + 1296*x^3).

a(n) = A016921(n) * A017533(n) * A161705(n).

a(n) == 1 (mod 72).