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A352287 Numbers k such that, for every prime p dividing k, k has a nontrivial divisor which is congruent to 1 (mod p).

%I A352287 #27 May 05 2022 07:14:53
%S A352287 1,12,24,30,36,48,56,60,72,80,90,96,105,108,112,120,132,144,150,160,
%T A352287 168,180,192,210,216,224,240,252,264,270,280,288,300,306,315,320,324,
%U A352287 336,351,360,380,384,392,396,400,420,432,448,450,480,495,504,520,525,528,540,546,552,560,576,600
%N A352287 Numbers k such that, for every prime p dividing k, k has a nontrivial divisor which is congruent to 1 (mod p).
%C A352287 When considering whether an integer k is the order of a finite simple group, the first thing one checks is whether the number of p-Sylow subgroups is forced to be 1 for some p dividing k. This occurs if the only divisor of k which is 1 (mod p) is 1 itself. This sequence consists of the numbers that survive this test.
%H A352287 Peter Luschny, <a href="/A352287/b352287.txt">Table of n, a(n) for n = 1..10000</a>
%H A352287 Mariano Suárez-Álvarez, <a href="https://mathoverflow.net/questions/403728/">On the density of the orders excluded by the Sylow theorems for simple groups</a>, MathOverflow, 2021.
%H A352287 <a href="/index/Gre#groups">Index entries for sequences related to groups</a>.
%e A352287 105 is in the sequence, since it is divisible by 7 which is 1 (mod 3), 21 which is 1 (mod 5), and 15 which is 1 (mod 7).
%t A352287 divq[n_, p_] := AnyTrue[Rest @ Divisors[n], Mod[#, p] == 1 &]; q[1] = True; q[n_] := AllTrue[FactorInteger[n][[;; , 1]], divq[n, #] &]; Select[Range[600], q] (* _Amiram Eldar_, May 05 2022 *)
%o A352287 (PARI) isok(k) = {my(f=factor(k), d=divisors(f)); for (i=1, #f~, if (vecsum(apply(x->((x % f[i,1]) == 1), d)) == 1, return(0)); ); return(1);} \\ _Michel Marcus_, Mar 11 2022
%o A352287 (Sage)
%o A352287 print([ n for n in range(1, 601)
%o A352287         if set( prime_factors(n) )
%o A352287         == set( p for p in prime_factors(n)
%o A352287                 for d in divisors(n)
%o A352287                 if d > 1 and d < n
%o A352287                 if p.divides(d - 1)
%o A352287       ) ] )  # _Peter Luschny_, Mar 14 2022
%Y A352287 Cf. A001034, A056866, A060793, A338853.
%K A352287 nonn
%O A352287 1,2
%A A352287 _David Speyer_, Mar 10 2022