cp's OEIS Frontend

GCD(b,n)=1 and b^(n+1) == 1 (mod n).

The sequence lists the squarefree composite numbers n such that every prime divisor p of n satisfies (p-1)|(n+1) (similar to Korselt's criterion).

The sequence can be considered as an extension of k-Knödel numbers to k negative, in this case equal to -1.

Numbers n > 3 such that b^(n+2) == b (mod n) for every integer b. Also, numbers n > 3 such that A002322(n) divides n+1. Are there infinitely many such numbers? It seems that such numbers n > 35 have at least three prime factors. - Thomas Ordowski, Jun 25 2017

Proof that 15 and 35 are the only numbers with only two prime factors (and so all others have >= 3): Since n is squarefree and composite, it has at least two prime factors, p and q. If these are the only two, n = p*q. Then the criterion is that (p-1)|(n+1) -> (p-1)|pq+1, and q-1|pq+1. Write pq+1 = j*(p-1) = k*(q-1). Rearranging, p*(j-q)=j+1 and q*(k-p)=k+1. Since j = (pq+1)/(p-1), for large n, j~=q, and k~=p. But we see that p divides j+1~=q, and q divides k+1~=p. For large n this is only possible if p and q are roughly equal, so j-q=k-p=1. Otherwise, we have j+1 >= 2*p and k+1 >= 2*q, and which puts upper bounds on p and q. Enumerating these gives (p,q)=(3,5) and (p,q)=(5,7) as the only solutions. - Alex Meiburg, Oct 03 2024

Quasi-Carmichael numbers are squarefree composites n with the property that for every prime factor p of n, p+b divides n+b positively with b being any integer besides 0.

If b is negative, then it is always larger than 0 minus the square root of the corresponding Quasi-Carmichael number. But if b is positive, how large can it be in relation to its corresponding Quasi-Carmichael number? Conjecture: It is always smaller than the square root of the corresponding Quasi-Carmichael number.

Are 1885 and 1886 the only two consecutive integers such that both numbers are Quasi-Carmichael numbers?

From Robert G. Wilson v, Dec 05 2015: (Start)

The conjecture that b < sqrt(n) is false. Look at n = 87061 = 13*37*181, 87365 = 5*101*173, and 96473 = 13*41*181. Their b values are 299, 331, and 351, while the corresponding sqrt(n) values are 295, 295, and 310, respectively.

For b to result in (n+b)/(p+b) > 0 with n = P_1*p_2*...*p_i and P_1 < p_2 < ... < p_i, -p_1 < b < |(n-p_i^2)/p_i|. (n+b)/(p+b) >= b+1. Solve for b.

Less than 0.5% are even (A262252). Of course they are == 2 (mod 4).

Least k-almost prime quasi-Carmichael number with k>1: 35, 165, 6545, 179998, 7509579, ..., .

(End)

a(n)*(2a(n)+1)*(3a(n)+2) are Lucas-Carmichael numbers for n > 1. Analogous to A174734 as A006972 (Lucas-Carmichael numbers) is analogous to A002997 (Carmichael numbers). - Amiram Eldar, Aug 11 2017