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%I #13 Jan 21 2018 18:37:57
%S 1,5,11,47,101,425,911,3827,8201,34445,73811,310007,664301,2790065,
%T 5978711,25110587,53808401,225995285,484275611,2033957567,4358480501,
%U 18305618105,39226324511,164750562947,353036920601,1482755066525,3177332285411,13344795598727
%N a(n) = a(n-1) + 9*a(n-2) - 9*a(n-3), where a(0) = 1, a(1) = 5, a(2) = 11.
%C Conjecture: a(n) = least positive whose base-3 up-variation is n; see A297441.
%H Clark Kimberling, <a href="/A297445/b297445.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,9,-9)
%F a(n) = a(n-1) + 9*a(n-2) - 9*a(n-3) for n>2. - Corrected by _Colin Barker_, Jan 21 2018
%F G.f.: (1 + 4 x - 3 x^2)/(1 - x - 9 x^2 + 9 x^3).
%F a(n) = (-1 - (-3)^n + 2*3^(1+n)) / 4. - _Colin Barker_, Jan 21 2018
%p seq(-1/4-(-3)^n/4+3*3^n/2, n=0..40); # _Robert Israel_, Jan 21 2018
%t LinearRecurrence[{1, 9, -9}, {1, 5, 11}, 40]]
%o (PARI) Vec((1 + 4*x - 3*x^2) / ((1 - x)*(1 - 3*x)*(1 + 3*x)) + O(x^40)) \\ _Colin Barker_, Jan 21 2018
%Y Cf. A297442.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Jan 21 2018
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