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A049332
Number of conjugacy classes in Clifford group CL(n).
3
2, 4, 5, 10, 17, 34, 65, 130, 257, 514, 1025, 2050, 4097, 8194, 16385, 32770, 65537, 131074, 262145, 524290, 1048577, 2097154, 4194305, 8388610, 16777217, 33554434, 67108865, 134217730, 268435457, 536870914, 1073741825, 2147483650
OFFSET
0,1
COMMENTS
Floretion Algebra Multiplication Program, FAMP Code: 4tesforseq[ (- .25'i - .25i' - .25'ii' + .25'jj' + .25'kk' + .25'jk' + .25'kj' - .25e)*( + .5'i + .5i' + .5'ii' + .5'jk' + .5'kj' + .5e ) ], 1vesforseq = (1,1,1,1,1,1,1). (Dement)
REFERENCES
B. Simon, Representations of Finite and Compact Groups, Amer. Math. Soc., 1996, p. 69.
FORMULA
a(n+2) - A101622(n+1) = 4. - Creighton Dement, Mar 07 2005
From Colin Barker, Apr 18 2012: (Start)
a(n) = (3/2 - (-1)^n/2 + 2^n).
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3).
G.f.: (2-5*x^2)/((1-x)*(1+x)*(1-2*x)). (End)
E.g.f.: cosh(x) + cosh(2*x) + 2*sinh(x) + sinh(2*x). - Stefano Spezia, May 27 2022
a(n) = 2*A000975(n+1) -5*A000975(n-1). - R. J. Mathar, Oct 12 2022
MAPLE
A049332 := proc(n) if n mod 2 = 0 then 2^n+1 else 2^n+2; fi; end;
MATHEMATICA
CoefficientList[Series[(2-5*x^2)/((1-x)*(1+x)*(1-2*x)), {x, 0, 40}], x] (* Vincenzo Librandi, Apr 27 2012 *)
Table[2^n + Mod[n, 2] + 1, {n, 0, 31}] (* Jean-François Alcover, Feb 11 2014 *)
LinearRecurrence[{2, 1, -2}, {2, 4, 5}, 40] (* Harvey P. Dale, Nov 29 2014 *)
PROG
(Magma) [(3/2-(-1)^n/2+2^n): n in [0..40]]; // Vincenzo Librandi, Apr 27 2012
(PARI) a(n) = 3/2 - (-1)^n/2 + 2^n \\ Charles R Greathouse IV, Feb 10 2017
CROSSREFS
Cf. A101622, A014551 (first differences)
Sequence in context: A138856 A333188 A018401 * A335206 A096570 A046430
KEYWORD
nonn,nice,easy
STATUS
approved