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

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

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 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.

Links

Peter Luschny, Table of n, a(n) for n = 1..10000

Mariano Suárez-Álvarez, On the density of the orders excluded by the Sylow theorems for simple groups, MathOverflow, 2021.

Index entries for sequences related to groups.

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)
      ) ] )  # Peter Luschny, Mar 14 2022

Crossrefs

Cf. A001034, A056866, A060793, A338853.

Sequence in context: A364532 A350056 A334799 * A364462 A369182 A108938

Adjacent sequences: A352284 A352285 A352286 * A352288 A352289 A352290

Keyword

nonn

Author

David Speyer, Mar 10 2022

Status

approved