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A181941
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Numbers n such that some group of order n has a non-cyclic commutator group.
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0
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12, 18, 24, 32, 36, 48, 50, 54, 56, 60, 64, 72, 75, 80, 81, 84, 90, 96, 98, 100, 108, 112, 120, 126, 128, 132, 144, 147, 150, 156, 160, 162, 168, 180, 192, 196, 198, 200, 204, 216, 224, 225, 228, 234, 240, 242, 243, 250, 252, 256, 264, 270, 276, 280, 288, 294, 300, 306, 312, 320, 324, 336, 338, 342
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OFFSET
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1,1
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COMMENTS
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The complementary sequence 1, 2, 3, 4... is much denser and contains all n such that each group of order n has a cyclic commutator group.
Let the factorization of n into powers of squarefree mutually coprime numbers n_1, n_2, n_3, n_4, n_5,... be n = n_1 *n_2^2 *n_3^3 *n_4^4 * n_5^5*..., see A051903.
Then the complementary sequence contains n of the form n = n_1*n_2^2*n_3^3*n_4^4 under the constraints:
(i) n_4=1 or n_4=2
(ii) gcd(n, psi(n_2^2*n_3^3*n_4^4)) =1 where psi(k) = abs(A153038(k)) .
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LINKS
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EXAMPLE
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1) Does not contain 10 = 10*1*1*1 where n_4=1 and gcd(10,|A153038(1)|)=1.
Both groups of order 10 have cyclic commutator groups: D10 has C5 and C10 has E.
2) Contains 12 = 3*2^2 where n_4=1 and gcd(12,|A153038(4)|) >1.
The group A4 of order 12 has a commutator group C2 x C2 which is not cyclic.
3) Contains 18 = 2*3^2 where n_4=1 and gcd(18,|A153038(9)|) >1.
The group (C3 x C3) : C2 of order 18 has a commutator group C3 x C3 which is not cyclic. (Gap notation, SmallGroup(18,4), where the colon is the semidirect product)
4) Contains 24 = 3*1*2^3 where n_4=1 and gcd(24,|A153038(9)|) >1.
5) Contains 32 = 1*1*1*1*2^5 where n_5>1.
6) Contains 48 = 3*1*1*2^4 where n_4=2 and gcd(48,|A153038(16)|)>1.
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MAPLE
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nsq := proc(n)
local f, L ;
L := [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] ;
if n = 1 then
return L;
else
for f in ifactors(n)[2] do
p := op(1, f) ;
e := op(2, f) ;
i := e ;
L := subsop(i=op(i, L)*p^e, L) ;
end do:
return L ;
end if;
end proc:
Pazdn4 := proc(L)
if nops(L) <4 then
1;
else
sqrt(sqrt(op(4, L))) ;
end if;
end proc:
hihno1 := proc(L)
i := 0 ;
for j from 1 to nops(L) do
if op(j, L) > 1 then
i := j ;
end if;
end do:
i ;
end proc:
for n from 1 to 600 do
nf := nsq(n) ;
n4 := Pazdn4(nf) ;
psarg := op(2, nf)*op(3, nf)*op(4, nf) ;
if ( n4 =1 or n4 =2) and gcd(n, abs(A153038(psarg))) = 1 and hihno1(nf) < 5 then
;
else
printf("%d, ", n) ;
end if;
end do:
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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