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%I #13 Jan 25 2020 09:19:12
%S 1,12,90,594,3726,22842,138510,835434,5025726,30193722,181280430,
%T 1088036874,6529284126,39178893402,235082926350,1410526255914,
%U 8463243628926,50779720053882,304679095164270,1828076895508554,10968468346620126
%N a(n) = b(n)*Integral_{x=0..1/(4^n)} (1 - sqrt(x)) dx, where b(n) = 3*24^n.
%C The integral is 1/4^n - 1/(3*2^(3n-1)). - _R. J. Mathar_, Nov 24 2008
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (9,-18)
%F a(n) = 3*6^n - 2*3^n. - _R. J. Mathar_, Nov 24 2008
%F From _Philippe Deléham_, Nov 26 2008: (Start)
%F a(n) = 9*a(n-1) - 18*a(n-2), n > 1; a(0)=1, a(1)=12.
%F G.f.: (1+3*x)/(1-9*x+18*x^2). (End)
%e For n=3, Integral_{x=0..1/(4^n)} (1 - sqrt(x)) dx = 594/41472, and b(3) = 3*24^3 = 41472, so a(3) = (594/41472)*41472 = 594.
%t Table[3*6^n-2*3^n,{n,0,20}] (* or *) LinearRecurrence[{9,-18},{1,12},30] (* _Harvey P. Dale_, Jan 23 2019 *)
%K nonn,easy
%O 0,2
%A Al Hakanson (hawkuu(AT)excite.com), May 18 2004
%E Changed offset to 0. Adapted definition to the fact that these are not reduced numerators of the integral. - _R. J. Mathar_, Nov 24 2008
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