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a(n) = 20*a(n-1) - 64*a(n-2) + 2 for n > 1; a(0) = 1, a(1) = 21.
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%I #17 Sep 08 2022 08:45:48

%S 1,21,358,5818,93450,1496650,23952202,383258442,6132227914,

%T 98116017994,1569857773386,25117730316106,401883708825418,

%U 6430139436277578,102882231360724810,1646115703292731210,26337851258768236362

%N a(n) = 20*a(n-1) - 64*a(n-2) + 2 for n > 1; a(0) = 1, a(1) = 21.

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

%H Vincenzo Librandi, <a href="/A167032/b167032.txt">Table of n, a(n) for n = 0..200</a>

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

%F a(n) = (257*16^n - 85*4^n + 8)/180.

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

%F From _G. C. Greubel_, May 30 2016: (Start)

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

%F E.g.f.: (1/180)*(257*exp(16*x) - 85*exp(4*x) + 8*exp(x)). (End)

%p A167032:=n->(257*16^n - 85*4^n + 8)/180: seq(A167032(n), n=0..25); # _Wesley Ivan Hurt_, May 30 2016

%t LinearRecurrence[{21, -84, 64}, {1,21,358}, 50] (* _G. C. Greubel_, May 30 2016 *)

%t RecurrenceTable[{a[0]==1,a[1]==21,a[n]==20a[n-1]-64a[n-2]+2},a,{n,20}] (* _Harvey P. Dale_, Oct 27 2019 *)

%o (Magma) [ n le 2 select 20*n-19 else 20*Self(n-1)-64*Self(n-2)+2: n in [1..17] ];

%Y Cf. A006105, A166915, A166916, A167031, A167033.

%K nonn,easy

%O 0,2

%A _Klaus Brockhaus_, Oct 27 2009