OFFSET
1,1
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.
REFERENCES
Pascal Ortiz, Exercices d'Algèbre, Collection CAPES / Agrégation, Ellipses, problème 1.35, pp. 70-74, 2004.
EXAMPLE
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.
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 *)
PROG
(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
(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
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Dec 30 2021
EXTENSIONS
More terms from Marius A. Burtea and Hugo Pfoertner, Dec 30 2021
STATUS
approved