cp's OEIS Frontend

Fermat pseudoprimes to base 2 are also called Poulet numbers.

Most of the terms shown can be written in one of the following two ways:

(1) p*(p*(n + 1) - n)*(p*(m + 1) - m);

(2) p*(p*n - (n + 1))*(p*m - (m + 1)),

where p is the smallest of the three prime factors and n, m natural numbers.

Exempli gratia for Poulet numbers from the first category:

10585 = 5*29*73 = 5*(5*7 - 6)*(5*18 - 17);

13741 = 7*13*151 = 7*(7*2 - 1)*(7*25 - 24);

13981 = 11*31*41 = 11*(11*3 - 2)*(11*4 - 3);

29341 = 13*37*61 = 13*(13*3 - 2)*(13*5 - 4);

137149 = 23*67*89 = 23*(23*3 - 2)*(23*4 - 3).

Exempli gratia for Poulet numbers from the second category:

6601 = 7*23*41 = 7*(7*4 - 5)*(7*7 - 8).

Note: from the numbers from the sequence above, just the numbers 30889, 88561 and 91001 can't be written in one of the two ways.

What these three numbers have in common: they all have a prime divisor q of the form 30*k + 23 (i.e. 23, 53, 83) and can be written as q*((r + 1)*q - r), where r is a natural number.

Conjecture: Any Poulet number P with three or more prime divisors has at least one prime divisor q for that can be written as P = q*((r + 1)*q - r), where r is a natural number.

Note: it can be proved that a Carmichael number can be written this way for any of its prime divisors - see the sequence A213812.

Note: there are also many Poulet numbers with two prime divisors that can be written this way, but here are few exceptions: 7957, 23377, 42799, 49981, 60787.

The conjecture fails for a(80) = 617093 = 43 * 113 * 127. - Charles R Greathouse IV, Dec 07 2014

First differs from A074380 at n=56. - Amiram Eldar, Jun 28 2019

Fermat pseudoprimes are listed in A001567.

The correspondent p for the numbers from the sequence above: 645, 1729, 2821, 3277, 4369, 7957, 10261, 10585, 12801, 13741, 13981, 16705, 18705, 25761, 31621, 49981, 65281, 75361, 83665, 88357, 93961, 101101.

Note that for 22 of the first 80 Poulet numbers, we obtained through this formula another Poulet number!

The formula could be generalized this way: Poulet numbers that can be written as (n + 1)*p^2 - n*p, where n is natural, n > 0, and p is another Poulet number.

For n = 1, that formula becomes the formula set out for the sequence above.

For n = 2, that formula becomes 3*p^2 - 2*p, from which the Poulet numbers 348161 (for p = 341) and 1246785 (for p = 645) were obtained.

For n = 3, that formula becomes 4*p^2 - 3*p, from which the Poulet number 119273701 (for p = 5461) was obtained.

For n = 4, that formula becomes 5*p^2 - 4*p, from which the Poulet numbers 2077545 (for p = 645) and 9613297 (for p = 1387) were obtained.

Conjecture: there are infinitely many Poulet numbers that can be written as (n + 1)*p^2 - n*p, where n is natural, n > 0, and p is another Poulet number.

Finally, considering, e.g., that for the Poulet number 645, Poulet numbers were obtained for n = 1, 2, 4 (i.e., 831405, 1246785, 2077545), yet another conjecture: For any Poulet number p, there are infinitely many Poulet numbers that can be written as (n + 1)*p^2 - n*p, where n is natural, n > 0.

A very interesting observation due to Peter Bala: about half of the terms from the sequence have the form p*(4*p - 3), where p is prime. For this form of Fermat pseudoprimes see the sequences A213812 and A215343.

After a(22) = 95423329, no more terms through 10^8.

The corresponding (p,n): (341,3), (645,2), (645,3), (341,11), (645,5), (561,8), (1729,2), (1387,5), (341,120), (561,44), (1905,5), (645,47), (3277,2), (2701,3), (2047,9), (4369,2), (341,384), (2821,6), (2047,13), (2465,12), (3277,7), (4369,5).

Conjecture: For any Fermat pseudoprime p to base 2 there are infinitely many Fermat pseudoprimes to base 2 equal to p^2*n - p*n + p, where n is a positive integer.

See the sequence A215343: the generalized formula from there is p^2*n - p*n + p^2, which suggests an extrapolated formula for obtaining some Fermat pseudoprime to base 2 from another: p^2*n - p*n + p^k.

Conjecture: For any Fermat pseudoprime p to base 2 and any positive integer k, there are infinitely many Fermat pseudoprimes to base 2 equal to p^2*n - p*n + p^k, where n is a positive integer.