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A053455 a(n) = ((8^n) - (-6)^n)/14.

%I A053455 #48 Jul 02 2025 16:01:59
%S A053455 0,1,2,52,200,2896,15392,169792,1078400,10306816,72376832,639480832,
%T A053455 4753049600,40201179136,308548739072,2546754076672,19903847628800,
%U A053455 162051890937856,1279488468058112,10337467701133312,82090381869056000,660379213392510976,5261096756499709952,42220395755839946752
%N A053455 a(n) = ((8^n) - (-6)^n)/14.
%C A053455 Previous name was: A linear recursive sequence.
%D A053455 A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
%H A053455 Vincenzo Librandi, <a href="/A053455/b053455.txt">Table of n, a(n) for n = 0..200</a>
%H A053455 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,48).
%F A053455 a(n) = 2*a(n-1) + 48*a(n-2), n>=2; a(0)=0, a(1)=1.
%F A053455 a(n) = ((8^n)-(-6)^n)/14 = (2^(n-1))*((4^n) - (-3)^n)/7 = 2^(n-1)*A053404(n).
%F A053455 G.f.: x/((1+6*x)*(1-8*x)). - _Harvey P. Dale_, Nov 28 2011
%F A053455 a(n) = A080921(n). - _Philippe Deléham_, Mar 05 2014
%F A053455 a(n+1) = Sum_{k=0..n} A238801(n,k)*7^k. - _Philippe Deléham_, Mar 07 2014
%p A053455 A053455:=n->((8^n)-(-6)^n)/14; seq(A053455(n), n=0..30); # _Wesley Ivan Hurt_, Mar 07 2014
%t A053455 LinearRecurrence[{2,48},{1,2},30] (* _Harvey P. Dale_, Nov 28 2011 *)
%t A053455 CoefficientList[Series[x / (1 - 2 x - 48 x^2), {x, 0, 20}], x] (* _Vincenzo Librandi_, Mar 08 2014 *)
%o A053455 (PARI) a(n) = ((8^n)-(-6)^n)/14; \\ _Joerg Arndt_, Mar 08 2014
%Y A053455 Cf. A053404, A051958, A015441, A080921.
%K A053455 easy,nonn
%O A053455 0,3
%A A053455 _Barry E. Williams_, Jan 13 2000
%E A053455 More terms from _James Sellers_, Feb 02 2000
%E A053455 New name (from formula), _Joerg Arndt_, Mar 05 2014