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%I #33 Nov 13 2022 09:33:18
%S 1,3,12,36,144,432,1728,5184,20736,62208,248832,746496,2985984,
%T 8957952,35831808,107495424,429981696,1289945088,5159780352,
%U 15479341056,61917364224,185752092672,743008370688,2229025112064,8916100448256,26748301344768,106993205379072,320979616137216
%N Start with 1; multiply alternately by 3 and 4.
%C Satisfies Benford's law.
%H Colin Barker, <a href="/A282022/b282022.txt">Table of n, a(n) for n = 0..1000</a>
%H Arno Berger and Theodore P. Hill, <a href="https://doi.org/10.1007/s00283-010-9182-3">Benford's law strikes back: no simple explanation in sight for mathematical gem</a>, The Mathematical Intelligencer 33.1 (2011): 85-91. Also <a href="https://digitalcommons.calpoly.edu/rgp_rsr/72/">at CalPoly</a>.
%H <a href="/index/Be#Benford">Index entries for sequences related to Benford's law</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,12).
%F From _Ilya Gutkovskiy_, Feb 09 2017: (Start)
%F O.g.f.: (1 + 3*x)/(1 - 12*x^2).
%F E.g.f.: sqrt(3)*sinh(2*sqrt(3)*x)/2 + cosh(2*sqrt(3)*x).
%F (End)
%F From _Colin Barker_, Feb 09 2017: (Start)
%F a(n) = 2^n * 3^(n/2) for n even.
%F a(n) = 2^(n-1) * 3^((n+1)/2) for n odd.
%F a(n) = 12*a(n-2) for n>1.
%F (End)
%t LinearRecurrence[{0,12},{1,3},30] (* _Harvey P. Dale_, Jun 19 2021 *)
%o (PARI) Vec((1 + 3*x) / (1 - 12*x^2) + O(x^30)) \\ _Colin Barker_, Feb 09 2017
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Feb 08 2017