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Even moments of the trace of elements of the binary icosahedral group.
1

%I #18 May 20 2019 14:23:58

%S 1,1,2,5,14,42,133,442,1534,5525,20502,77826,300357,1172770,4616054,

%T 18267797,72556958,288881562,1152001669,4598779210,18370959022,

%U 73420590101,293516780262,1173633626610,4693399603269,18770627198002,75074730049958,300278555188757

%N Even moments of the trace of elements of the binary icosahedral group.

%C a(n) is the expected value of |tr(U)|^{2n} where U is drawn uniformly at random from the 120-element binary icosahedral group, viewed as a subgroup of SU(2) (or equivalently, the unit quaternion group). Note that |tr(U)| takes values in 0, phi^{-1}, 1, phi, 2 (with phi the golden ratio) with probabilities 1/4, 1/5, 1/3, 1/5, 1/60 respectively.

%C Is a reasonably good match with A000108 which corresponds to the case where U is drawn from all of SU(2) with the Haar distribution.

%H Colin Barker, <a href="/A308329/b308329.txt">Table of n, a(n) for n = 0..1000</a>

%H Terence Tao, <a href="https://terrytao.wordpress.com/2019/05/17/a-function-field-analogue-of-riemann-zeta-statistics/#comment-516485">A function field analogue of Riemann zeta statistics</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (8,-20,17,-4).

%F a(n) = 0^n/4 + (phi^{2n} + phi^{-2n})/5 + 1/3 + 4^n/60.

%F From _Colin Barker_, May 20 2019: (Start)

%F G.f.: (1 - 7*x + 14*x^2 - 8*x^3 + x^4) / ((1 - x)*(1 - 4*x)*(1 - 3*x + x^2)).

%F a(n) = 8*a(n-1) - 20*a(n-2) + 17*a(n-3) - 4*a(n-4) for n>4.

%F a(n) = (20 + 4^n + 3*2^(2-n)*((3-sqrt(5))^n + (3+sqrt(5))^n)) / 60 for n>0.

%F (End)

%e a(1) = (phi^2 + phi^{-2})/5 + 1/3 + 4/60 = 3/5 + 1/3 + 1/15 = 1.

%o (PARI) Vec((1 - 7*x + 14*x^2 - 8*x^3 + x^4) / ((1 - x)*(1 - 4*x)*(1 - 3*x + x^2)) + O(x^30)) \\ _Colin Barker_, May 20 2019

%Y Cf. A000108 (Catalan numbers), A001622 (phi).

%K easy,nonn

%O 0,3

%A _Terence Tao_, May 20 2019