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

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, 34, -44, 16, 1, -11, 43, -70, 41, -4, 1, -12, 53, -104, 85, -20, 1, -13, 64, -147, 155, -61, 4, 1, -14, 76, -200, 259, -146, 24

Offset: 0

Author

Tom Copeland, Jan 18 2016

Keywords

• easy
• sign
• tabf

Examples

			Row polynomials:
P(0,t) = 1 - 4t
P(1,t) = 1 - 4t = [-t(0) + (1-4t)] = -t(0) + P(0,t)
P(2,t) = 1 - 5t + 4t^2 = [-t(1-4t) + (1-4t)] = -t P(0,t) + P(1,t)
P(3,t) = 1 - 6t + 8t^2 = [-t(1-4t) + (1-5t+4t^2)] = -t P(1,t) + P(2,t)
P(4,t) = 1 - 7t + 13t^2 - 4t^3 = [-t(1-5t+4t^2) + (1-6t+8t^2)]
P(5,t) = 1 - 8t + 19t^2 - 12t^3 = [-t(1-6t+8t^2) + (1-7t+13t^2)]
P(6,t) = 1 - 9t + 26t^2 - 25t^3 + 4t^4
P(7,t) = 1 - 10t + 34t^2 - 44t^3 + 16t^4
P(8,t) = 1 - 11t + 43t^2 - 70t^3 + 41t^4 - 4t^5
P(9,t) = 1 - 12t + 53t^2 - 104t^3 + 85t^4 - 20t^5
P(10,t) = 1 - 13t + 64t^2 - 147t^3 + 155t^4 - 61t^5 + 4t^6
P(11,t) = 1 - 14t + 76t^2 - 200t^3 + 259t^4 - 146t^5 + 24t^6
...
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.
		

Links

• Tom Copeland, Addendum to Elliptic Lie Triad

Crossrefs

Cf. A000108, A000297, A008574, A008586, A011973, A022088 A034856, A084103, A098474, A124644, A140882, A202241, A228785.

Programs

Mathematica

p = (1 - 4 t) / (1 - x + t x^2) + O[x]^12 // CoefficientList[#, x] &;
CoefficientList[#, t] & /@ p // Flatten (* Andrey Zabolotskiy, Mar 07 2024 *)

Formula

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.

Recursion P(n,t) = -t P(n-2,t) + P(n-1,t) with P(-1,t)=0 and P(0,t) = 1-4t.

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.

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).

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.

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.

Unsigned row sums (t=-1) are shifted A022088 with o.g.f. 5 + 5x(1+x) / [x(1+x)].

Extensions

Data corrected by Andrey Zabolotskiy, Mar 07 2024