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a(n) = 20*a(n-1) - 64*a(n-2) - 45 for n>1; a(0) = 399, a(1) = 5695.
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%I #11 Oct 21 2022 22:10:12

%S 399,5695,88319,1401855,22384639,357974015,5726863359,91626930175,

%T 1466019348479,23456263438335,375300030463999,6004799749226495,

%U 96076793034833919,1537228676746182655,24595658780694282239

%N a(n) = 20*a(n-1) - 64*a(n-2) - 45 for n>1; a(0) = 399, a(1) = 5695.

%C Related to Reverse and Add trajectory of 318 in base 4: A075153(6*n+3) = 15*a(n).

%C lim_{n -> infinity} a(n)/a(n-1) = 16.

%H G. C. Greubel, <a href="/A166915/b166915.txt">Table of n, a(n) for n = 0..500</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (21, -84, 64).

%F a(n) = (1024*16^n + 176*4^n - 3)/3.

%F G.f.: (399 - 2684*x + 2240*x^2)/((1-x)*(1-4*x)*(1-16*x)).

%F From _G. C. Greubel_, May 28 2016: (Start)

%F a(n) = 21*a(n-1) - 84*a(n-2) + 64*a(n-3).

%F E.g.f.: (1/3)*(1024*exp(16*x) + 176*exp(4*x) - 3*exp(x)). (End)

%t LinearRecurrence[{21, -84, 64}, {399, 5695, 88319}, 50] (* _G. C. Greubel_, May 28 2016 *)

%o (PARI) m=15; v=concat([399, 5695], vector(m-2)); for(n=3, m, v[n]=20*v[n-1]-64*v[n-2]-45); v

%Y Cf. A075153, A166912, A166913, A166914, A166916, A166917, A006105, A167031, A167032, A167033.

%K nonn,easy

%O 0,1

%A _Klaus Brockhaus_, Oct 27 2009