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
Keywords
Examples
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).
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.
Programs
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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 *)
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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
Comments