

A352287


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


1



1, 12, 24, 30, 36, 48, 56, 60, 72, 80, 90, 96, 105, 108, 112, 120, 132, 144, 150, 160, 168, 180, 192, 210, 216, 224, 240, 252, 264, 270, 280, 288, 300, 306, 315, 320, 324, 336, 351, 360, 380, 384, 392, 396, 400, 420, 432, 448, 450, 480, 495, 504, 520, 525, 528, 540, 546, 552, 560, 576, 600
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OFFSET

1,2


COMMENTS

When considering whether an integer k is the order of a finite simple group, the first thing one checks is whether the number of pSylow 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.


LINKS



EXAMPLE

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).


MATHEMATICA

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 *)


PROG

(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
(Sage)
print([ n for n in range(1, 601)
if set( prime_factors(n) )
== set( p for p in prime_factors(n)
for d in divisors(n)
if d > 1 and d < n
if p.divides(d  1)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



