cp's OEIS Frontend

A108981 a(n) = 3a(n-1) + 4a(n-2), a(0) = 1, a(1) = 5.

%I A108981 #46 Apr 22 2025 12:17:35
%S A108981 1,5,19,77,307,1229,4915,19661,78643,314573,1258291,5033165,20132659,
%T A108981 80530637,322122547,1288490189,5153960755,20615843021,82463372083,
%U A108981 329853488333,1319413953331,5277655813325,21110623253299
%N A108981 a(n) = 3a(n-1) + 4a(n-2), a(0) = 1, a(1) = 5.
%C A108981 The Hankel transform of this sequence is [1,-6,0,0,0,0,0,0,0,0,...]. - _Philippe Deléham_, Apr 15 2008
%C A108981 Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n-1) = charpoly(A,2). - _Milan Janjic_, Jan 26 2010
%H A108981 Vincenzo Librandi, <a href="/A108981/b108981.txt">Table of n, a(n) for n = 0..200</a>
%H A108981 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,4).
%F A108981 Inverse binomial transform of A003948.
%F A108981 a(2n) = 4a(2n-1) - 1; a(2n+1) = 4a(2n) + 1.
%F A108981 O.g.f.: (1+2*x)/((1+x)(1-4*x)). - _R. J. Mathar_, Apr 02 2008
%F A108981 Sum_{k=0..n} a(k) = A037481(n+1). - _Philippe Deléham_, Apr 15 2008
%t A108981 LinearRecurrence[{3,4},{1,5},30] (* _Harvey P. Dale_, Feb 16 2014 *)
%o A108981 (PARI) Vec((1+2*x)/(1+x)/(1-4*x)+O(x^99)) \\ _Charles R Greathouse IV_, Jan 11 2012
%o A108981 (Magma) I:=[1,5]; [n le 2 select I[n] else 3*Self(n-1)+4*Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Feb 17 2014
%o A108981 (Python)
%o A108981 def A108981(n): return ((4<<(m:=n+1<<1))|2)//5-((1<<m)|2)//5>>1 # _Chai Wah Wu_, Apr 22 2025
%Y A108981 Cf. A003948, A037481.
%K A108981 nonn,easy
%O A108981 0,2
%A A108981 _Philippe Deléham_, Jul 23 2005
%E A108981 Corrected by _T. D. Noe_, Nov 07 2006
%E A108981 Edited by _N. J. A. Sloane_ at the suggestion of _R. J. Mathar_, Apr 14 2008