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A322942

Jacobsthal triangle, coefficients of orthogonal polynomials J(n, x) where J(n, 0) are the Jacobsthal numbers (A001045 with a(0) = 1). Triangle read by rows, T(n, k) for 0 <= k <= n.

1, 1, 1, 1, 2, 1, 3, 5, 3, 1, 5, 12, 10, 4, 1, 11, 27, 28, 16, 5, 1, 21, 62, 75, 52, 23, 6, 1, 43, 137, 193, 159, 85, 31, 7, 1, 85, 304, 480, 456, 290, 128, 40, 8, 1, 171, 663, 1170, 1254, 916, 480, 182, 50, 9, 1, 341, 1442, 2793, 3336, 2750, 1652, 742, 248, 61, 10, 1

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OFFSET

0,5

COMMENTS

The name 'Jacobsthal triangle' used here is not standard.

LINKS

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

FORMULA

J(n, x) = (x+1)*J(n-1, x) + 2*J(n-2, x) for n >= 3.

T(n, k) = [x^k] J(n, x).

Equals the Riordan square (cf. A321620) generated by (2*x^2-1)/((x + 1)*(2*x - 1)).

Sum_{k=0..n} T(n, k) = A152035(n).

Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).

From G. C. Greubel, Sep 20 2023: (Start)

T(n, k) = [x^k]( [n=0] + (i*sqrt(2))^n*(ChebyshevU(n, (x+1)/(2*sqrt(2)*i)) + ChebyshevU(n-2, (x+1)/(2*sqrt(2)*i))) ).

G.f.: (1 - 2*t^2)/(1 - (x+1)*t - 2*t^2).

Sum_{k=0..floor(n/2)} T(n-k, k) = (2/3)*[n=0] + A006138(n-1).

Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = 2*[n=0] + Fibonacci(n-2). (End)

EXAMPLE

The first few polynomials are:

J(0, x) = 1;

J(1, x) = x + 1;

J(2, x) = x^2 + 2*x + 1;

J(3, x) = x^3 + 3*x^2 + 5*x + 3;

J(4, x) = x^4 + 4*x^3 + 10*x^2 + 12*x + 5;

J(5, x) = x^5 + 5*x^4 + 16*x^3 + 28*x^2 + 27*x + 11;

J(6, x) = x^6 + 6*x^5 + 23*x^4 + 52*x^3 + 75*x^2 + 62*x + 21;

The triangle starts:

[0] 1;

[1] 1, 1;

[2] 1, 2, 1;

[3] 3, 5, 3, 1;

[4] 5, 12, 10, 4, 1;

[5] 11, 27, 28, 16, 5, 1;

[6] 21, 62, 75, 52, 23, 6, 1;

[7] 43, 137, 193, 159, 85, 31, 7, 1;

[8] 85, 304, 480, 456, 290, 128, 40, 8, 1;

[9] 171, 663, 1170, 1254, 916, 480, 182, 50, 9, 1;

MAPLE

J := proc(n) option remember;

`if`(n < 3, [1, x+1, x^2 + 2*x + 1][n+1], (x+1)*J(n-1) + 2*J(n-2));

sort(expand(%)) end: seq(print(J(n)), n=0..11); # Computes the polynomials.

seq(seq(coeff(J(n), x, k), k=0..n), n=0..11);

MATHEMATICA

J[n_, x_]:= J[n, x]= If[n<3, (x+1)^n, (x+1)*J[n-1, x] + 2*J[n-2, x]];

T[n_, k_]:= Coefficient[J[n, x], x, k];

Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Jean-François Alcover, Jun 17 2019 *)

(* Alternative: *)

A322942[n_, k_]:= Coefficient[Series[Boole[n==0] + (I*Sqrt[2])^n*(ChebyshevU[n, (x+1)/(2*Sqrt[2]*I)] + ChebyshevU[n-2, (x+ 1)/(2*Sqrt[2]*I)]), {x, 0, 50}], x, k];

Table[A322942[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Sep 20 2023 *)

PROG

(SageMath) # use[riordan_square from A321620]

riordan_square((2*x^2 - 1)/((x + 1)*(2*x - 1)), 9)

(Magma)

R<x>:=PowerSeriesRing(Rationals(), 30);

g:= func< n, x | (&+[Binomial(n-k, k)*2^k*(x+1)^(n-2*k): k in [0..Floor(n/2)]]) >;

f:= func< n, x | n le 1 select (x+1)^n else g(n, x) - 2*g(n-2, x) >;

A322942:= func< n, k | Coefficient(R!( f(n, x) ), k) >;

[A322942(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 20 2023

CROSSREFS

Row sums are A152035, alternating row sums are A000007, values at x=1/2 are A323232, values at x=0 (first column) are A152046.