cp's OEIS Frontend

A267633 Expansion of (1 - 4t)/(1 - x + t x^2): a Fibonacci-type sequence of polynomials.

%I A267633 #48 Apr 01 2024 16:07:54
%S A267633 1,-4,1,-4,1,-5,4,1,-6,8,1,-7,13,-4,1,-8,19,-12,1,-9,26,-25,4,1,-10,
%T A267633 34,-44,16,1,-11,43,-70,41,-4,1,-12,53,-104,85,-20,1,-13,64,-147,155,
%U A267633 -61,4,1,-14,76,-200,259,-146,24
%N A267633 Expansion of (1 - 4t)/(1 - x + t x^2): a Fibonacci-type sequence of polynomials.
%H A267633 Tom Copeland, <a href="http://tcjpn.wordpress.com/2015/10/12/the-elliptic-lie-triad-kdv-and-ricattt-equations-infinigens-and-elliptic-genera/">Addendum to Elliptic Lie Triad</a>
%F A267633 O.g.f. G(x,t) = (1 - 4t)/(1 - x + t x^2) = a /  [t (x - (1+sqrt(a))/(2t))(x - (1-sqrt(a))/(2t))] with a = 1-4t.
%F A267633 Recursion P(n,t) = -t P(n-2,t) + P(n-1,t) with P(-1,t)=0 and P(0,t) = 1-4t.
%F A267633 Convolution of the Fibonacci polynomials of signed A011973 Fb(n,-t) with coefficients of (1-4t), e.g., (1-4t)Fb(5,-t) = (1-4t)(1-3t+t^2) = 1-7t+13t^2-4t^3, so, for n>=1 and k<=(n-1), T(n,k) = (-1)^k [-4*binomial(n-(k-1),k-1) - binomial(n-k,k)] with the convention that 1/(-m)! = 0 for m>=1, i.e., let binomial(n,k) = nint[n!/((k+c)!(n-k+c)!)] for c sufficiently small in magnitude.
%F A267633 Third column is A034856, and the fourth, A000297. Embedded in the coefficients of the highest order term of the polynomials is A008586 (cf. also A008574).
%F A267633 With P(0,t)=0, the o.g.f. is H(x,t) = (1-4t) x(1-tx)/[1-x(1-tx)] = (1-4t) Linv(Cinv(tx)/t), where Linv(x) = x/(1-x) with inverse L(x) = x/(1+x) and Cinv(x) = x (1-x) is the inverse of the o.g.f. of the shifted Catalan numbers A000108, C(x) = [1-sqrt(1-4x)]/2. Then Hinv(x,t) = C[t L(x/(1-4t))]/t = {1 - sqrt[1-4t(x/(1-4t))/[1+x/(1-4t)]]}/2t = {1-sqrt[1-4tx/(1-4t+x)]}/2t = 1/(1-4t) + (-1+t)/(1-4t)^2 x + (1-2t+2t^2)/(1-4t)^3 x^ + (-1+3t-6t^2+5t^3)/(1-4t)^4 + ..., where the numerators are the signed polynomials of A098474, reverse of A124644.
%F A267633 Row sums (t=1) are periodic -3,-3,0,3,3,0, repeat the six terms ... with o.g.f. -3 - 3x (1-x) / [1-x(1-x)]. Cf. A084103.
%F A267633 Unsigned row sums (t=-1) are shifted A022088 with o.g.f. 5 + 5x(1+x) / [x(1+x)].
%e A267633 Row polynomials:
%e A267633 P(0,t) = 1 - 4t
%e A267633 P(1,t) = 1 - 4t = [-t(0) + (1-4t)] = -t(0) + P(0,t)
%e A267633 P(2,t) = 1 - 5t + 4t^2 = [-t(1-4t) + (1-4t)] = -t P(0,t) + P(1,t)
%e A267633 P(3,t) = 1 - 6t + 8t^2 = [-t(1-4t) + (1-5t+4t^2)] = -t P(1,t) + P(2,t)
%e A267633 P(4,t) = 1 - 7t + 13t^2 - 4t^3 = [-t(1-5t+4t^2) + (1-6t+8t^2)]
%e A267633 P(5,t) = 1 - 8t + 19t^2 - 12t^3 = [-t(1-6t+8t^2) + (1-7t+13t^2)]
%e A267633 P(6,t) = 1 - 9t + 26t^2 - 25t^3 + 4t^4
%e A267633 P(7,t) = 1 - 10t + 34t^2 - 44t^3 + 16t^4
%e A267633 P(8,t) = 1 - 11t + 43t^2 - 70t^3 + 41t^4 - 4t^5
%e A267633 P(9,t) = 1 - 12t + 53t^2 - 104t^3 + 85t^4 - 20t^5
%e A267633 P(10,t) = 1 - 13t + 64t^2 - 147t^3 + 155t^4 - 61t^5 + 4t^6
%e A267633 P(11,t) = 1 - 14t + 76t^2 - 200t^3 + 259t^4 - 146t^5 + 24t^6
%e A267633 ...
%e A267633 Apparently: The odd rows for n>1 are reversed rows of A140882 (mod signs), and the even rows for n>0, the 9th, 15th, 21st, 27th, etc. rows of A228785 (mod signs). The diagonals are reverse rows of A202241.
%t A267633 p = (1 - 4 t) / (1 - x + t x^2) + O[x]^12 // CoefficientList[#, x] &;
%t A267633 CoefficientList[#, t] & /@ p // Flatten (* _Andrey Zabolotskiy_, Mar 07 2024 *)
%Y A267633 Cf.  A000108, A000297, A008574, A008586, A011973, A022088 A034856, A084103, A098474, A124644, A140882, A202241, A228785.
%K A267633 easy,sign,tabf
%O A267633 0,2
%A A267633 _Tom Copeland_, Jan 18 2016
%E A267633 Data corrected by _Andrey Zabolotskiy_, Mar 07 2024