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A350638
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Numbers of the form p^2*q, with odd primes p > q, such that q divides p-1.
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3
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147, 507, 605, 1083, 2883, 4107, 4805, 5547, 5819, 5887, 8405, 11163, 12943, 13467, 15987, 18605, 18723, 25205, 28227, 31827, 35287, 35643, 36517, 48387, 49379, 50807, 51005, 57963, 68403, 73947, 79707, 81133, 85805, 87131, 89383, 98283, 100949, 111747, 112903
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OFFSET
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1,1
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COMMENTS
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For these terms m there are precisely (q+9)/2 groups of order m.
Only two of these groups are abelian: C_{p^2*q} and (C_p X C_p) X C_q. The (q+5)/2 groups that are nonabelian are C_{p^2} : C_q and the (q+3)/2 semidirect products of the form (C_p X C_p) : C_q that are not isomorphic, where C means cyclic groups of the stated order, the symbols X and : mean direct and semidirect products respectively.
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REFERENCES
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Pascal Ortiz, Exercices d'Algèbre, Collection CAPES / Agrégation, Ellipses, problème 1.35, pp. 70-74, 2004.
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LINKS
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EXAMPLE
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147 = 7^2 * 3, 3 and 7 are odd primes, 3 divides 7-1 = 6, hence 147 is a term.
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MATHEMATICA
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q[n_] := Module[{f = FactorInteger[n], p, e}, p = f[[;; , 1]]; e = f[[;; , 2]]; e == {1, 2} && Divisible[p[[2]] - 1, p[[1]]]]; Select[Range[1, 120000, 2], q] (* Amiram Eldar, Jan 11 2022 *)
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PROG
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(Python)
from sympy import integer_nthroot, primerange
def aupto(limit):
aset, maxp = set(), integer_nthroot(limit**2, 3)[0]
for p in primerange(5, maxp+1):
pp = p*p
for q in primerange(3, min(p, limit//pp+1)):
if (p-1)%q == 0:
aset.add(pp*q)
return sorted(aset)
(PARI) isok(m) = if (m%2, my(f=factor(m)); if (f[, 2] == [1, 2]~, my(p=f[1, 1], q=f[2, 1]); ((q-1) % p) == 0)); \\ Michel Marcus, Jan 11 2022
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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