cp's OEIS Frontend

A049072 Expansion of 1/(1 - 3*x + 4*x^2).

%I A049072 #58 Sep 08 2022 08:44:58
%S A049072 1,3,5,3,-11,-45,-91,-93,85,627,1541,2115,181,-7917,-24475,-41757,
%T A049072 -27371,84915,364229,753027,802165,-605613,-5025499,-12654045,
%U A049072 -17860139,-2964237,62547845,199500483,348310069,246928275,-652455451,-2945079453,-6225416555
%N A049072 Expansion of 1/(1 - 3*x + 4*x^2).
%C A049072 From Sharon Sela (sharonsela(AT)hotmail.com), Jan 22 2002: (Start)
%C A049072 a(n) is the determinant of the following tridiagonal n X n matrix:
%C A049072 [3 2 0 0 .... ]
%C A049072 [2 3 2 0 .... ]
%C A049072 [0 2 3 2 0 .. ]
%C A049072 [. 0 2 3 2 .. ]
%C A049072 [. . . . .... ]
%C A049072 [. . . 2 3 2 0]
%C A049072 [. . . 0 2 3 2]
%C A049072 [. . . 0 0 2 3]
%C A049072 (End)
%C A049072 With offset 1 (a(0) = 0, a(1) = 1) this is a divisibility sequence. - _R. K. Guy_, May 19 2015
%C A049072 With offset 1 (a(0) = 0, a(1) = 1), then this is the Lucas sequence U_n(P, Q) = U_n(3, 4). V_n(P, Q) = V_n(3, 4) = A247563(n). Again with offset 1 (a(0) = 0, a(1) = 1), then (A247563(n)/2)^2 + 7(a(n)/2)^2 = 4^n. This is a specific case of the Lucas sequence identity (V_n/2)^2 - D*(U_n/2)^2 = Q^n where V_n = (a^n + b^n), U_n = (a^n - b^n)/(a - b), Q = (a*b) = 4 and D = (a - b)^2 = -7; a = (3 + sqrt(-7))/2 and b = (3 - sqrt(-7))/2. - _Raphie Frank_, Dec 04 2015
%H A049072 Reinhard Zumkeller, <a href="/A049072/b049072.txt">Table of n, a(n) for n = 0..1000</a>
%H A049072 B. R. Myers, <a href="http://dx.doi.org/10.1137/1017045">On Spanning Trees, Weighted Compositions, Fibonacci Numbers, and Resistor Networks</a>, SIAM Rev., 17 (1975), 465-474.
%H A049072 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,-4).
%F A049072 G.f.: 1/(1 - 3*x + 4*x^2).
%F A049072 a(n) = (-1)^n * Sum_{k=0..n} binomial(2n-k+1, k)*(-2)^k. - _Paul Barry_, Jan 17 2005
%F A049072 a(n) = 3*a(n-1) - 4*a(n-2); a(0)=1, a(1)=3. - _Sergei N. Gladkovskii_, Mar 14 2013
%F A049072 G.f.: 1/(1/Q(0)+2*x^3) where Q(k) = 1 + k*(2*x+1) + 8*x - 2*x*(k+1)*(k+5)/Q(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Mar 14 2013
%F A049072 a(n) = - a(-2-n) * 4^(n+1) for all n in Z. - _Michael Somos_, Jun 03 2015
%F A049072 a(n - 1) = (((3 + sqrt(-7))/2)^n - ((3 - sqrt(-7))/2)^n)/(((3 + sqrt(-7))/2) - ((3 - sqrt(-7))/2)). - _Raphie Frank_, Dec 04 2015
%e A049072 G.f.: 1 + 3*x + 5*x^2 + 3*x^3 - 11*x^4 - 45*x^5 - 91*x^6 - 93*x^7 + ...
%p A049072 A049072:=n->(-1)^n*add(binomial(2*n-k+1,k)*(-2)^k, k=0..n): seq(A049072(n), n=0..40); # _Wesley Ivan Hurt_, Dec 05 2015
%t A049072 Join[{a=1,b=3},Table[c=3*b-4*a;a=b;b=c,{n,100}]] (* _Vladimir Joseph Stephan Orlovsky_, Jan 17 2011 *)
%t A049072 a[ n_] := ChebyshevU[ n, 3/4] 2^n; (* _Michael Somos_, Jun 03 2015 *)
%t A049072 a[ n_] := Module[ {m = n + 1, s = 1}, If[ m < 0, {m, s} = -{m, 4^m}]; s SeriesCoefficient[ x / (1 - 3 x + 4 x^2), {x, 0, m}]]; (* _Michael Somos_, Jun 03 2015 *)
%o A049072 (PARI) {a(n) = 2^n * subst( -3*poltchebi(n+1) + 4*poltchebi(n), 'x, 3/4) * 4/7}; /* _Michael Somos_, Sep 15 2005 */
%o A049072 (PARI) {a(n) = if(n<0, 0, matdet(matrix(n, n, i, j, if(abs(i-j)<2, 3-abs(i-j)))))} /* _Michael Somos_, Sep 15 2005 */
%o A049072 (PARI) {a(n) = polchebyshev(n, 2, 3/4) * 2^n}; /* _Michael Somos_, Jun 03 2015 */
%o A049072 (Sage) [lucas_number1(n,3,4) for n in range(1, 34)] # _Zerinvary Lajos_, Apr 23 2009
%o A049072 (Haskell)
%o A049072 a049072 n = a049072_list !! n
%o A049072 a049072_list = 1 : 3 :
%o A049072     zipWith (-) (map (* 3) $ tail a049072_list) (map (* 4) a049072_list)
%o A049072 -- _Reinhard Zumkeller_, Oct 25 2013
%o A049072 (Magma) I:=[1,3]; [n le 2 select I[n] else 3*Self(n-1)-4*Self(n-2): n in [1..50]]; // _Vincenzo Librandi_, Jun 12 2015
%o A049072 (PARI) x='x+O('x^100); Vec(1/(1-3*x+4*x^2)) \\ _Altug Alkan_, Dec 04 2015
%Y A049072 Cf. A006516, A025170, A247563.
%K A049072 sign,easy
%O A049072 0,2
%A A049072 _N. J. A. Sloane_