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A121449 Expansion of (1 - 3*x + 2*x^2)/(1 - 4*x + 3*x^2 + x^3).

%I A121449 #41 Feb 03 2023 01:36:17
%S A121449 1,1,3,8,22,61,170,475,1329,3721,10422,29196,81797,229178,642125,
%T A121449 1799169,5041123,14124860,39576902,110891905,310712054,870595599,
%U A121449 2439354329,6834918465,19151015274,53659951372,150351841201,421276495414,1180390506681,3307380699281
%N A121449 Expansion of (1 - 3*x + 2*x^2)/(1 - 4*x + 3*x^2 + x^3).
%C A121449 From _Roman Witula_, Aug 07 2012: (Start)
%C A121449 In the cited Witula-Slota-Warzynski paper three so-called quasi-Fibonacci numbers A(n;d), B(n;d) and C(n;d), where n = 0,1,...,d \in C are discussed. These numbers are created by each of the following relations:
%C A121449 (1+d*c(j))^n = A(n;d) + B(n;d)*c(j) + C(n;d)*c(2*j), for every j=1,2,4, where c(j):=2*cos(2*Pi*j/7).
%C A121449 In fact all these "numbers" are integer polynomials of the argument d.
%C A121449 In the sequel for d=-1 we obtain A(n;-1)=a(n), B(n+1;-1)=-A085810(n).
%C A121449 Moreover, we have A(n;1)=A077998(n), B(n;1)=A006054(n+1), C(n;1)=A006054(n), and A(n;2)=A121442(n).
%C A121449 We note that the elements of the sequences A(n;-1), B(n;-1), and C(n;-1) satisfy the following system of recurrence equations:
%C A121449 A(0;-1)=1, B(0;-1)=C(0;-1)=0,
%C A121449 A(n+1;-1)=A(n;-1)-2*B(n;-1)+C(n;-1),
%C A121449 B(n+1;-1)=-A(n;-1)+B(n;-1), C(n+1;-1)=-B(n;-1)+2*C(n;-1).
%C A121449 It is proved that binomial transforms of the sequences: A(n;1), B(n;1), and C(n;1) are equal to the following sequences:
%C A121449 A(n;1)*(A(n;-1)-C(n;-1))-B(n;1)*(B(n;-1)+C(n;-1))+C(n;1)*B(n;-1), -A(n;1)*C(n;-1)+B(n;1)*(A(n;-1)-C(n;-1))-C(n;1)*(B(n;-1)-C(n;-1)), and
%C A121449 A(n;1)*(B(n;-1)-C(n;-1))-B(n;1)*B(n;-1)+C(n;1)*(A(n;-1)-B(n;-1)+C(n;-1)), respectively, whereas we have
%C A121449 A(n;-1) = Sum_{k=0..n} binomial(n,k)*(A(k;1)*A(n-k;1)-A(k;1)*B(n-k;1)-B(k;1)*C(n-k;1)-A(n-k;1)*C(k;1)+2*B(n-k;1)*C(k;1)-C(k;1)*C(n-k;1)),
%C A121449 B(n;-1) = Sum_{k=0..n} binomial(n,k)*(-A(k;1)*B(n-k;1)+A(k;1)*C(n-k;1)+B(k;1)*B(n-k;1)-A(n-k;1)*C(k;1)+B(n-k;1)*C(k;1)-C(k;1)*C(n-k;1)), and
%C A121449 C(n;-1) = Sum_{k=0..n} binomial(n,k)*(-A(k;1)*B(n-k;1)+A(n-k;1)*B(k;1)+B(k;1)*B(n-k;1)-B(k;1)*C(n-k;1)-A(n-k;1)*C(k;1)) (see identities (3.50-52) and (3.61-63) in the Witula-Slota-Warzynski paper).
%C A121449 (End)
%H A121449 G. C. Greubel, <a href="/A121449/b121449.txt">Table of n, a(n) for n = 0..1000</a>
%H A121449 Paul Barry, <a href="https://arxiv.org/abs/2104.01644">Centered polygon numbers, heptagons and nonagons, and the Robbins numbers</a>, arXiv:2104.01644 [math.CO], 2021.
%H A121449 Roman Witula, Damian Slota and Adam Warzynski, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Slota/slota57.html">Quasi-Fibonacci Numbers of the Seventh Order</a>, J. Integer Seq., 9 (2006), Article 06.4.3.
%H A121449 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-3,-1).
%F A121449 a(0)=a(1)=1, a(2)=3, a(n+1) = 4*a(n) - 3*a(n-1) - a(n-2) for n>=2.
%F A121449 7*a(n) = (2-c(4))*(1-c(1))^n + (2-c(1))*(1-c(2))^n + (2-c(2))*(1-c(4))^n = (s(2))^2*(1-c(1))^n + (s(4))^2*(1-c(2))^n + (s(1))^2*(1-c(4))^n, where c(j):=2*cos(2*Pi*j/7) and s(j):=2*sin(2*Pi*j/7) -- it is the special case, for d=-1, of the Binet's formula for the respective quasi-Fibonacci number A(n;d) discussed in the Witula-Slota-Warzynski paper. - _Roman Witula_, Aug 07 2012
%t A121449 CoefficientList[Series[(1 - 3*x + 2*x^2)/(1-4*x + 3*x^2 + x^3), {x, 0, 200}], x] (* _Stefan Steinerberger_, Sep 11 2006 *)
%t A121449 LinearRecurrence[{4,-3,-1},{1,1,3},50] (* _Roman Witula_, Aug 07 2012 *)
%o A121449 (Magma) I:=[1,1,3]; [n le 3 select I[n] else 4*Self(n-1)-3*Self(n-2)-Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Sep 18 2015
%o A121449 (PARI) x='x+O('x^30); Vec((1-3*x+2*x^2)/(1-4*x+3*x^2+x^3)) \\ _G. C. Greubel_, Apr 19 2018
%Y A121449 Cf. A085810, A077998, A006054.
%K A121449 nonn,easy
%O A121449 0,3
%A A121449 _Philippe Deléham_, Sep 06 2006
%E A121449 More terms from _Stefan Steinerberger_, Sep 11 2006
%E A121449 a(27)-a(29) from _Vincenzo Librandi_, Sep 18 2015