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%I #38 Jun 07 2026 07:26:56
%S 1,0,3,1,0,-3,0,2,0,3,1,0,-5,0,-3,0,1,0,8,0,3,1,0,-6,0,-11,0,-3,0,0,0,
%T 14,0,14,0,3,1,0,-6,0,-25,0,-17,0,-3,0,-1,0,20,0,39,0,20,0,3,1,0,-5,0,
%U -45,0,-56,0,-23,0,-3,0,-2,0,25,0,84,0,76,0,26,0,3
%N Triangle read by rows: the coefficients of the polynomials P(n, x) = -x * P(n-1, x) + P(n-2, x), where P(0, x) = 1 and P(1, x) = 3*x, ordered by ascending powers.
%C Define F(t, x) = (phi(x)^(-t) - (-phi(x))^t) / (2*phi(x) - x) and L(t, x) = (phi(x)^(-t) + (-phi(x))^t), where phi(x) = (x + sqrt(x^2 + 4))/2. On this page and in A396651, we examine LF(t, x) = (1/2) * L(t, x) + (7/2)*x*F(t, x). In expanded form this leads to a generalized Binet-type representation:
%C LF(t, x) = ((3*x + phi(x))*phi(x)^(-t) + (-phi(x))^t*(-4*x + phi(x)))/(2*phi(x) - x).
%C We dub this linear combination of the Lucas and the Fibonacci polynomials the Lucanacci polynomials, and LF(t, 1) the Lucanacci function. The coefficients of the polynomials LF(-n, x) are for n >= 0 on the present page and for LF(n, x) in A396651.
%C The Lucanacci function yields the sequences LF(n, 1) = A396201(n) and LF(-n, 1) = A000285(n) for n >= 0, which are the row sums of the corresponding triangles of coefficients.
%H Peter Luschny, <a href="/A396650/a396650.png">Illustration of the A396650 polynomials</a>.
%F T(n, k) are the coefficients of P(n, x) = U(n, -x/2) + 4*x*U(n - 1, -x/2), where U(n, x) are the Chebyshev U polynomials.
%F The connection with the Lucas and Fibonacci polynomials:
%F P(n, x) = (-1)^n*(Lucas(n, x) - 7*x*Fibonacci(n, x)) / 2.
%F The Binet representation of the polynomials:
%F P(n, x) = mu(x)*phi(-x)^n + (1-mu(x))*(-x-phi(-x))^n, where phi(x) = (x + sqrt(x^2+4))/2, and mu(x) = (1 + 7*x/sqrt(x^2+4))/2. Note that phi(1) = A001622, and mu(1) = A189961.
%F G.f.: (1 + 4*x*y) / (1 + x*y - y^2).
%F E.g.f.: exp(-x*y/2)*(cosh(h*y) + g*sinh(h*y)), where h = phi(x) - x/2, and g = 2*mu(x) - 1. Notice that the only difference between this and the EGF of A396651 is the minus sign in the leading exponential factor.
%F T(n, k) = 0 if n - k is odd, T(0, 0 ) = 1, T(n, n) = (-1)^(n + 1)*3 for n > 0, and otherwise T(n, k) = (-1)^n*(n - 7*k)*binomial((n + k)/2 - 1, k) / (n - k). (See A396649.)
%F Sum_{k=0..n} T(n, k) = A396201(n).
%e Table starts:
%e 0: 1;
%e 1: 0, 3;
%e 2: 1, 0, -3;
%e 3: 0, 2, 0, 3;
%e 4: 1, 0, -5, 0, -3;
%e 5: 0, 1, 0, 8, 0, 3;
%e 6: 1, 0, -6, 0, -11, 0, -3;
%e 7: 0, 0, 0, 14, 0, 14, 0, 3;
%e 8: 1, 0, -6, 0, -25, 0, -17, 0, -3;
%e 9: 0, -1, 0, 20, 0, 39, 0, 20, 0, 3;
%e 10: 1, 0, -5, 0, -45, 0, -56, 0, -23, 0, -3;
%e 11: 0, -2, 0, 25, 0, 84, 0, 76, 0, 26, 0, 3;
%p flat := ListTools:-FlattenOnce:
%p PolyTriangle := proc(len, start, pos) local c,i,a,b,row;
%p c := [[1], [0, start]];
%p for i from 3 to len do
%p a := flat([op([0]), c[i-1]]); b := flat([c[i-2], op([0, 0])]);
%p if pos then row := zip((x, y) -> y + x, a, b);
%p else row := zip((x, y) -> y - x, a, b) fi;
%p c := [op(c), row]; od; c end:
%p # Using the above template function:
%p A396650poly := n -> PolyTriangle(n, 3, false):
%p for row in A396650poly(12) do lprint(row) od;
%p # Alternative:
%p A396650poly := proc(n, x) option remember;
%p if n = 0 then 1 elif n = 1 then 3*x else -x*procname(n-1, x) + procname(n-2, x) fi end:
%p Trow := n -> local k; seq(coeff(A396650poly(n, x), x, k), k = 0..n):
%p seq(print(Trow(n)), n = 0..11);
%t P[0] = 1; P[1] = 3 x; P[n_] := P[n] = Expand[-x P[n - 1] + P[n - 2]];
%t Table[CoefficientList[P[n], x], {n, 0, 11}]
%t (* Alternative: Chebyshev polynomials.*)
%t P[n_, x_] := Expand[I^n * ChebyshevU[n, (I*x)/2] + 4*x * I^(n - 1) * ChebyshevU[n - 1, (I*x)/2]]
%t Table[CoefficientList[P[n, x], x], {n, 0, 11}]
%t (* Alternative: The Lucas/Fibonacci connection. *)
%t P[n_, x_] := (-1)^n * (LucasL[n, x] - 7 x Fibonacci[n, x]) / 2
%t Table[CoefficientList[P[n, x], x], {n, 0, 11}]
%o (Python)
%o def PolyTriangle(len: int, start: int, pos: bool) -> list[list[int]]:
%o c = [[1], [0, start]]
%o for i in range(2, len):
%o row1 = [0] + c[i-1] if pos else [0] + [-v for v in c[i-1]]
%o row2 = c[i-2] + [0, 0]
%o row = [t1 + t2 for t1, t2 in zip(row1, row2)]
%o c.append(row)
%o return c
%o A396650poly = lambda n: PolyTriangle(n, 3, False)
%o for row in A396650poly(12): print(row)
%Y Cf. A396201 (row sums), A396649, A396651, A168561, A114525, A001622, A189961.
%Y Diagonals, |T(n + m, n)|, n >= 0, modulo offset: A122553 (m=0), A016789 (m=2), A095794 (m=4), A005564 (m=6), A059302 (m=8).
%K sign,tabl,easy
%O 0,3
%A _Peter Luschny_, Jun 01 2026