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A071101 Expansion of (5 + 6*x + 3*x^2 - 2*x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4) in powers of x.

%I A071101 #42 Oct 03 2024 16:17:34
%S A071101 5,16,45,130,377,1088,3145,9090,26269,75920,219413,634114,1832625,
%T A071101 5296384,15306833,44237570,127848949,369490320,1067846845,3086134658,
%U A071101 8919094697,25776662080,74495936025,215297250946,622220603405,1798250918672,5197041610021
%N A071101 Expansion of (5 + 6*x + 3*x^2 - 2*x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4) in powers of x.
%C A071101 Number of tilings of the 2-mod-4 pillow of order n is a perfect square times a(n). [Propp, 1999, p. 272]
%D A071101 J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 12).
%H A071101 A.H.M. Smeets, <a href="/A071101/b071101.txt">Table of n, a(n) for n = 0..2169</a>
%H A071101 J. Propp, <a href="http://jamespropp.org/update.pdf">Updated article</a>
%H A071101 J. Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), <a href="http://www.msri.org/publications/books/Book38/contents.html">New Perspectives in Algebraic Combinatorics</a>
%H A071101 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,2,-1).
%F A071101 G.f.: (5 + 6*x + 3*x^2 - 2*x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4).
%F A071101 a(-n) = a(-6 + n). a(-1) = 2, a(-2) = 1, a(-3) = 0. a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4). - _Michael Somos_, Dec 15 2011
%F A071101 A112835(2*n + 3) = a(n).
%F A071101 Lim_{n -> inf} a(n)/a(n-1) = A318605. - _A.H.M. Smeets_, Sep 12 2018
%e A071101 G.f. = 5 + 16*x + 45*x^2 + 130*x^3 + 377*x^4 + 1088*x^5 + 3145*x^6 + 9090*x^7 + ...
%p A071101 seq(coeff(series((5+6*x+3*x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4),x,n+1), x, n), n = 0 .. 30); # _Muniru A Asiru_, Sep 12 2018
%t A071101 Table[Abs[Fibonacci[n+3, 1+I]]^2, {n,0,30}] (* _Vladimir Reshetnikov_, Oct 05 2016 *)
%t A071101 CoefficientList[Series[(5+6*x+3*x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4), {x, 0, 30}], x] (* _Stefano Spezia_, Sep 12 2018 *)
%t A071101 LinearRecurrence[{2,2,2,-1},{5,16,45,130},30] (* _Harvey P. Dale_, Oct 03 2024 *)
%o A071101 (PARI) {a(n) = my(m = abs(n+3)); polcoeff( (x - x^3) / (1 - 2*x - 2*x^2 - 2*x^3 + x^4) + x * O(x^m), m)};  /* _Michael Somos_, Dec 15 2011 */
%o A071101 (PARI) x='x+O('x^33); Vec((5+6*x+3*x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4)) \\ _Altug Alkan_, Sep 12 2018
%o A071101 (Python)
%o A071101 from math import log
%o A071101 a0,a1,a2,a3,n = 130,45,16,5,3
%o A071101 print(0,a3)
%o A071101 print(1,a2)
%o A071101 print(2,a1)
%o A071101 print(3,a0)
%o A071101 while log(a0)/log(10) < 1000:
%o A071101     a0,a1,a2,a3,n = 2*(a0+a1+a2)-a3,a0,a1,a2,n+1
%o A071101     print(n,a0) # _A.H.M. Smeets_, Sep 12 2018
%o A071101 (GAP) a:=[5,16,45,130];; for n in [5..30] do a[n]:=2*a[n-1]+2*a[n-2] +2*a[n-3]-a[n-4]; od; a; # _Muniru A Asiru_, Sep 12 2018
%o A071101 (Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (5+6*x+3*x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4) )); // _G. C. Greubel_, Jul 29 2019
%o A071101 (Sage) ((5+6*x+3*x^2-2*x^3)/(1-2*x-2*x^2-2*x^3+x^4)).series(x, 30).coefficients(x, sparse=False) # _G. C. Greubel_, Jul 29 2019
%Y A071101 Cf. A112835, A138573.
%K A071101 nonn,easy
%O A071101 0,1
%A A071101 _N. J. A. Sloane_, May 28 2002