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%I #64 Nov 02 2023 12:31:43
%S 60,120,168,180,240,300,336,360,420,480,504,540,600,660,672,720,780,
%T 840,900,960,1008,1020,1080,1092,1140,1176,1200,1260,1320,1344,1380,
%U 1440,1500,1512,1560,1620,1680,1740,1800,1848,1860,1920,1980,2016,2040
%N Orders of non-solvable groups, i.e., numbers that are not solvable numbers.
%C A number is solvable if every group of that order is solvable.
%C This comment is about the three sequences A001034, A060793, A056866: The Feit-Thompson theorem says that a finite group with odd order is solvable, hence all numbers in this sequence are even. - Ahmed Fares (ahmedfares(AT)my-deja.com), May 08 2001 [Corrected by _Isaac Saffold_, Aug 09 2021]
%C Insoluble group orders can be derived from A001034 (simple non-cyclic orders): k is an insoluble order iff k is a multiple of a simple non-cyclic order. - _Des MacHale_
%C All terms are divisible by 4 and either 3 or 5. - _Charles R Greathouse IV_, Sep 11 2012
%C Subsequence of A056868 and hence of A060652. - _Charles R Greathouse IV_, Apr 16 2015, updated Sep 11 2015
%C The primitive elements are A257146. Since the sum of the reciprocals of the terms of that sequence converges, this sequence has a natural density and so a(n) ~ k*n for some k (see, e.g., Erdős 1948). - _Charles R Greathouse IV_, Apr 17 2015
%C From _Jianing Song_, Apr 04 2022: (Start)
%C Burnside's p^a*q^b theorem says that a finite group whose order has at most 2 distinct prime factors is solvable, hence all terms have at least 3 distinct prime factors.
%C Terms not divisible by 12 are divisible by 320 and have at least 4 distinct prime factors (cf. A257391). (End)
%H Charles R Greathouse IV, <a href="/A056866/b056866.txt">Table of n, a(n) for n = 1..10000</a> (first 2240 terms from T. D. Noe)
%H R. Brauer, <a href="http://www.pnas.org/cgi/reprint/47/12/1891.pdf">Investigation on groups of even order, I</a>.
%H R. Brauer, <a href="http://www.pnas.org/cgi/reprint/55/2/254.pdf">Investigation on groups of even order, II</a>.
%H P. Erdős, <a href="http://www.renyi.hu/~p_erdos/1948-06.pdf">On the density of some sequences of integers</a>, Bull. Amer. Math. Soc. 54 (1948), pp. 685-692. See p. 685.
%H W. Feit and J. G. Thompson, <a href="http://www.pnas.org/content/48/6/968">A solvability criterion for finite groups and consequences</a>, Proc. N. A. S. 48 (6) (1962) 968.
%H J. Pakianathan and K. Shankar, <a href="http://www.math.ou.edu/%7Eshankar/papers/nil2.pdf">Nilpotent numbers</a>, Amer. Math. Monthly, 107, August-September 2000, 631-634.
%H Cindy Tsang, Qin Chao, <a href="https://arxiv.org/abs/1901.10636">On the solvability of regular subgroups in the holomorph of a finite solvable group</a>, arXiv:1901.10636 [math.GR], 2019.
%H <a href="/index/Gre#groups">Index entries for sequences related to groups</a>
%F A positive integer k is a non-solvable number if and only if it is a multiple of any of the following numbers: a) 2^p*(2^(2*p)-1), p any prime. b) 3^p*(3^(2*p)-1)/2, p odd prime. c) p*(p^2-1)/2, p prime greater than 3 such that p^2 + 1 == 0 (mod 5). d) 2^4*3^3*13. e) 2^(2*p)*(2^(2*p)+1)*(2^p-1), p odd prime.
%t ma[n_] := For[k = 1, True, k++, p = Prime[k]; m = 2^p*(2^(2*p) - 1); If[m > n, Return[False], If[Mod[n, m] == 0, Return[True]]]]; mb[n_] := For[k = 2, True, k++, p = Prime[k]; m = 3^p*((3^(2*p) - 1)/2); If[m > n, Return[False], If[Mod[n, m] == 0, Return[True]]]]; mc[n_] := For[k = 3, True, k++, p = Prime[k]; m = p*((p^2 - 1)/2); If[Mod[p^2 + 1, 5] == 0, If[m > n, Return[False], If[Mod[n, m] == 0, Return[True]]]]]; md[n_] := Mod[n, 2^4*3^3*13] == 0; me[n_] := For[k = 2, True, k++, p = Prime[k]; m = 2^(2*p)*(2^(2*p) + 1)*(2^p - 1); If[m > n, Return[False], If[Mod[n, m] == 0, Return[True]]]]; notSolvableQ[n_] := OddQ[n] || ma[n] || mb[n] || mc[n] || md[n] || me[n]; Select[ Range[3000], notSolvableQ] (* _Jean-François Alcover_, Jun 14 2012, from formula *)
%o (PARI) is(n)={
%o if(n%5616==0,return(1));
%o forprime(p=2,valuation(n,2),
%o if(n%(4^p-1)==0, return(1))
%o );
%o forprime(p=3,valuation(n,3),
%o if(n%(9^p\2)==0, return(1))
%o );
%o forprime(p=3,valuation(n,2)\2,
%o if(n%((4^p+1)*(2^p-1))==0, return(1))
%o );
%o my(f=factor(n)[,1]);
%o for(i=1,#f,
%o if(f[i]>3 && f[i]%5>1 && f[i]%5<4 && n%(f[i]^2\2)==0, return(1))
%o );
%o 0
%o }; \\ _Charles R Greathouse IV_, Sep 11 2012
%Y Subsequence of A000977 and A056868.
%Y Cf. A003277, A051532, A056867, A001034, A060652.
%K nonn,easy,nice
%O 1,1
%A _N. J. A. Sloane_, Sep 02 2000
%E More terms from _Des MacHale_, Feb 19 2001
%E Further terms from Francisco Salinas (franciscodesalinas(AT)hotmail.com), Dec 25 2001
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