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a(n) = (9^n - (-7)^n)/(9 - (-7)).
2

%I #27 Aug 29 2021 12:14:54

%S 1,2,67,260,4741,25862,350407,2330120,26735881,200269322,2084899147,

%T 16786765580,164922177421,1387410586382,13164918350287,

%U 113736703642640,1056863263353361,9279138856193042,85140663303647827,754867074547457300

%N a(n) = (9^n - (-7)^n)/(9 - (-7)).

%C Theon from Smyrna used a(n+1)=2a(n)+a(n-1), a(1)=1, a(2)=2, to determine sqrt(2).

%C F(n) = (r^n - s^n)/(r - s) where r is different from s will generate Fibonacci-type sequences.

%H Vincenzo Librandi, <a href="/A160958/b160958.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,63).

%F a(n) = 2a(n-1)+63a(n-2), a(1)=1 a(2)=2.

%F G.f.: x/((1-9x)(1+7x)). - _R. J. Mathar_, Jun 22 2009

%F a(n+1) = Sum_{k = 0..n} A238801(n,k)*8^k. - _Philippe Deléham_, Mar 07 2014

%p A160958 := proc(n) (9^n-(-7)^n)/16 ; end: seq(A160958(n),n=1..30) ; # _R. J. Mathar_, Jun 22 2009

%p a := proc (n) options operator, arrow: (1/16)*9^n-(1/16)*(-7)^n end proc: seq(a(n), n = 1 .. 20); # _Emeric Deutsch_, Jun 21 2009

%t Table[(9^n - (-7)^n)/(9 - (-7)), {n, 20}] (* _Wesley Ivan Hurt_, Mar 07 2014 *)

%t CoefficientList[Series[1/((1 - 9 x) (1 + 7 x)), {x, 0, 30}], x] (* _Vincenzo Librandi_, Mar 08 2014 *)

%t LinearRecurrence[{2,63},{1,2},20] (* _Harvey P. Dale_, Aug 29 2021 *)

%K nonn,easy

%O 1,2

%A _Sture Sjöstedt_, May 31 2009

%E Edited by _N. J. A. Sloane_, Jun 07 2009

%E Extended by _Emeric Deutsch_ and _R. J. Mathar_, Jun 22 2009