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%I #18 Sep 08 2022 08:45:32
%S 1,2,4,8,12,19,26,37,48,63,78,98,117,142,166,196,225,261,295,337,377,
%T 425,471,526,578,640,699,768,834,911,984,1069,1150,1243,1332,1434,
%U 1531,1642,1748,1868,1983,2113,2237,2377,2511,2661,2805,2966,3120,3292,3457
%N Number of metacyclic groups of order 2^n.
%C For number of metacyclic groups of order p^n, prime p >= 3, see A136185.
%H Klaus Brockhaus, <a href="/A136184/b136184.txt">Table of n, a(n) for n=1..1000</a> [Values computed with MAGMA]
%H Steven Liedahl, <a href="http://dx.doi.org/10.1006/jabr.1996.0381">Enumeration of metacyclic p-groups</a>, J. Algebra 186 (1996), no. 2, 436-446.
%H MAGMA Computational Algebra System, V2.14-1, <a href="http://magma.maths.usyd.edu.au/magma/htmlhelp/text701.htm#6850">Metacyclic p-groups</a>
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-1,-2,-1,2,1,-1).
%F G.f.: -x*(x^10 + x^9 - x^8 + x^6 - x^3 - x - 1)/((x - 1)^4*(x + 1)^2*(x^2 + x + 1)).
%F For n > 3, a(n) = (n^3 + 48*n^2 - c*n + d)/72, where c = 168 or 177 for n even/odd, and d = 432, 416 or 424 for n = 0, 1 or 2 (mod 3), according to the Liedahl paper. Since this would yield (4,4,5) for n=1,2,3, one can simply add [n<4]*(n-4) to get a formula valid for all n. - _M. F. Hasler_, Jan 13 2015
%e a(3) = 4 since there are four metacyclic groups of order 2^3; they have invariants <3, 0, 0, 3, [ 8 ], >, <1, 2, 1, 1, [ 2, 4 ], >, <1, 1, 1, 2, [ 2 ], Dihedral> and <1, 1, 1, 2, [ 2 ], Quaternion> resp. (for details see MAGMA link).
%t LinearRecurrence[{1,2,-1,-2,-1,2,1,-1},{1,2,4,8,12,19,26,37,48,63,78},60] (* _Harvey P. Dale_, May 31 2019 *)
%o (Magma) [ NumberOfMetacyclicPGroups(2, n): n in [1..51] ];
%o (PARI) A136184(n)=if(n<4,2^(n-1),(((n+48)*n-[168, 177][1+n%2])*n+[432, 416, 424][1+n%3])/72) \\ _M. F. Hasler_, Jan 13 2015
%Y Cf. A136185.
%K nonn
%O 1,2
%A _Klaus Brockhaus_, Dec 19 2007
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