cp's OEIS Frontend

Sierpiński primes of the form n^n + 1 are {2,5,257,...} = A121270. The prime p divides a((p-1)/2) for p = {5,7,13,23,29,31,37,47,53,61,71,...} = A003628 Primes congruent to {5, 7} mod 8. p^2 divides a((p-1)/2) for prime p = {29,37,3373,...}. - Alexander Adamchuk, Sep 11 2006

n divides a(n-1) for even n, or 2n divides a(2n-1). a(2n-1)/(2n) = A124899(n) = {1, 7, 521, 102943, 38742049, 23775972551, 21633936185161, 27368368148803711, 45957792327018709121, ...}. 2^n divides a(2^n-1). A014566[2^n - 1] / 2^n = A081216[2^n - 1] = A122000[n] = {1, 7, 102943, 27368368148803711, 533411691585101123706582594658103586126397951, ...}. p+1 divides a(p) for prime p. a(p)/(p+1) = A056852[n] = {7, 521, 102943, 23775972551, 21633936185161, ...}. p^2 divides a((p-1)/2) for prime p = {29, 37, 3373} = A121999(n). - Alexander Adamchuk, Nov 12 2006

a(n) is prime for n = {3, 5, 17, 157} = A056826(n) Primes p such that (p^p + 1)/(p + 1) is a prime. Prime a(n) are {7, 521, 45957792327018709121, ...}. Bisection of a(n) is Sierpinski quotient a(2n-1) = A124899(n) = ((2n-1)^(2n-1) + 1)/(2n) = A014566(2n-1)/(2n). - Alexander Adamchuk, Nov 12 2006

This is related to the dimension of the primitive middle cohomology of Dwork hypersurfaces x1**n+x2**n+...+xn**n=n*psi*x1*x2*...*xn. [F. Chapoton, Dec 11 2009]

Conjecture: for n > 2, a(n) is a Fermat pseudoprime to base n.

If p is an odd prime, then a(p) is a Cipolla pseudoprime to base p.

Is a(m) a Fermat pseudoprime to base m for every composite m?

Amiram Eldar confirmed this up to m = 3800.

From Jianing Song, Aug 28 2022: (Start)

a(n) = Product_{d|(2n),d>2} Phi(d,n), where Phi(n,x) is the d-th cyclotomic polynomial. Note that Phi(n,x) > 1 for x >= 2 unless (n,x) = (1,2): suppose that n >= 3 and x >= 2, then Phi(n,x) = Product_{1<=j<=n,gcd(j,n)=1} (x - exp(2*j*Pi*i/n)) = Product_{1<=j<=n/2,gcd(j,n)=1} (x^2 - 2*cos(2*j*Pi/n)*x + 1) = Product_{1<=j<=n/2,gcd(j,n)=1} ((x - cos(2*j*Pi/n))^2 + (sin(2*j*Pi/n))^2) > 1 since x - cos(2*j*Pi/n) > 1. This shows that a(n) is composite for n > 2.

For n > 2, a(n) is a Fermat pseudoprime to base n, since n^(2*n) == 1 (mod a(n)) and 2*n divides a(n)-1 = n^2*(n^(2*n-2)-1)/(n^2-1): if n is even, then 2*n | n^2; if n is odd, then n | n^2 and 2 | n^2+1 = (n^4-1)/(n^2-1) | (n^(2*n-2)-1)/(n^2-1). (End)