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a(n) = 20*a(n-1) - 64*a(n-2) - 15 for n > 1; a(0) = 357, a(1) = 5525.
10

%I #11 Oct 21 2022 22:10:17

%S 357,5525,87637,1399125,22373717,357930325,5726688597,91626231125,

%T 1466016552277,23456252253525,375299985724757,6004799570269525,

%U 96076792319006037,1537228673882871125,24595658769241036117

%N a(n) = 20*a(n-1) - 64*a(n-2) - 15 for n > 1; a(0) = 357, a(1) = 5525.

%C Related to Reverse and Add trajectory of 318 in base 4: A075153(6*n+4) = 30*a(n).

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

%H G. C. Greubel, <a href="/A166916/b166916.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 + 48*4^n - 1)/3.

%F G.f.: (357 - 1972*x + 1600*x^2)/((1-x)*(1-4*x)*(1-16*x)).

%F a(0)=357, a(1)=5525, a(2)=87637, a(n)=21*a(n-1)-84*a(n-2)+64*a(n-3). - _Harvey P. Dale_, Sep 24 2012

%F E.g.f.: (1/3)*(1024*exp(16*x) + 48*exp(4*x) - exp(x)). - _G. C. Greubel_, May 28 2016

%t LinearRecurrence[{21,-84,64},{357,5525,87637},20] (* _Harvey P. Dale_, Sep 24 2012 *)

%o (PARI) m=15; v=concat([357, 5525], vector(m-2)); for(n=3, m, v[n]=20*v[n-1]-64*v[n-2]-15); v

%Y Cf. A075153, A166912, A166913, A166914, A166915, A166917, A006105, A167031, A167032, A167033.

%K nonn,easy

%O 0,1

%A _Klaus Brockhaus_, Oct 27 2009