A350421 Numbers p^2*q, p > q odd primes such that q does not divide p-1, and q does not divide p+1.

245, 845, 847, 1445, 1859, 2023, 2527, 2645, 3179, 3703, 3757, 3971, 4693, 6137, 6727, 6845, 6877, 8993, 9245, 9251, 9583, 10051, 10571, 10933, 11045, 12493, 14045, 14297, 15059, 15463, 15979, 16337, 17797, 18259, 18491, 19343, 19663, 21853, 22103, 22445, 23273

Offset: 1

Author

Bernard Schott, Dec 30 2021

Keywords

• nonn

Comments

As odd prime q does not divide p-1 and does not divide also p+1, then q >= 5, so p >= 7.

For these terms m, there are precisely 2 groups of order m, so this is a subsequence of A054395.

The 2 groups are abelian; they are C_{p^2*q} and (C_p X C_p) X C_q, where C means cyclic groups of the stated order and the symbol X means direct product.

Examples

			245 = 7^2 * 5, 5 and 7 are odd primes, 5 does not divide 7-1 = 10 and does not divide 7+1 = 8, hence 245 is a term.
		

References

• Pascal Ortiz, Exercices d'Algèbre, Collection CAPES / Agrégation, Ellipses, problème 1.35, pp. 70-74, 2004.

Crossrefs

Equals A350422 \ A350332.

Subsequence of A051532, A054395, A054753, A060687 and A350322.

Other subsequences of A054753 linked with order of groups: A079704, A143928, A349495, A350115, A350245.

Programs

Magma

f:=Factorisation; [n:n in [3..24000 ]|#PrimeDivisors(n) eq 2 and  f(n)[1][1] lt f(n)[2][1] and f(n)[1][2] eq 1 and f(n)[2][2] eq 2  and (f(n)[2][1]-1) mod f(n)[1][1] ne 0 and (f(n)[2][1]+1) mod f(n)[1][1] ne 0]; // Marius A. Burtea, Dec 30 2021

Mathematica

q[n_] := Module[{f = FactorInteger[n], p, e}, p = f[[;; , 1]]; e = f[[;; , 2]]; e == {1, 2} && ! Or @@ Divisible[p[[2]] + {-1, 1}, p[[1]]]]; Select[Range[1, 24000, 2], q] (* Amiram Eldar, Dec 30 2021 *)

PARI

isok(m) = my(f=factor(m)); if (f[, 2] == [1, 2]~, my(p=f[2, 1], q=f[1, 1]); ((p-1) % q) && ((p+1) % q)); \\ Michel Marcus, Dec 30 2021

Python

from sympy import integer_nthroot, primerange
def aupto(limit):
    aset, maxp = set(), integer_nthroot(limit**2, 3)[0]
    for p in primerange(3, maxp+1):
        pp = p*p
        for q in primerange(1, min(p, limit//pp+1)):
            if (p-1)%q != 0 and (p+1)%q != 0:
                aset.add(pp*q)
    return sorted(aset)
print(aupto(24000)) # Michael S. Branicky, Dec 30 2021

Extensions

More terms from Marius A. Burtea and Hugo Pfoertner, Dec 30 2021