A090040 - OEIS

%I #32 Sep 08 2022 08:45:12

%S 1,5,28,164,976,5840,35008,209984,1259776,7558400,45349888,272098304,

%T 1632587776,9795522560,58773127168,352638746624,2115832446976,

%U 12694994616320,76169967566848,457019805138944,2742118830309376

%N (3*6^n + 2^n)/4.

%C A090040 is the Q-residue of the triangle A175840, where Q is the triangular array (t(i,j)) given by t(i,j)=1; see A193649 for the definition of Q-residue. - _Clark Kimberling_, Aug 07 2011

%H Vincenzo Librandi, <a href="/A090040/b090040.txt">Table of n, a(n) for n = 0..300</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-12).

%F G.f.: (1-3*x)/((1-2*x)*(1-6*x)).

%F E.g.f.: (3*exp(6*x)+exp(2*x))/4 = exp(4*x)*(cosh(2*x)+sinh(2*x)/2).

%F a(n) = 8*a(n-1) -12*a(n-2), a(0)=1, a(1)=5.

%F a(n) = (3*6^n+2^n)/4.

%F a(n)=6*a(n-1)-2^(n-1). - _Paul Curtz_, Jan 09 2009

%F Fourth binomial transform of (1, 1, 4, 4, 16, 16, ...). a(n)=sum{k=1..floor(n/2), C(n, 2k)4^(n-k-1)}. - _Paul Barry_, Nov 22 2003

%F a(n) = A019590 (mod 4), proof via a(n)=8*a(n-1)-12*a(n-2). - _R. J. Mathar_, Feb 25 2009

%F a(n) = Sum_{k, 0<=k<=n} A117317(n,k)*3^k. - _Philippe Deléham_, Jan 28 2012

%t LinearRecurrence[{8,-12},{1,5},30] (* _Harvey P. Dale_, Nov 23 2014 *)

%o (Magma) [(3*6^n+2^n)/4: n in [0..30]]; // _Vincenzo Librandi_, Jun 10 2011

%o (PARI) a(n)=(3*6^n+2^n)/4 \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A081335.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Nov 20 2003