A322942 -id:A322942 - cp's OEIS frontend

A323232 a(n) = 2^n*J(n, 1/2) where J(n, x) are the Jacobsthal polynomials as defined in A322942.

1, 3, 9, 51, 225, 1083, 5049, 23811, 111825, 525963, 2472489, 11625171, 54655425, 256967643, 1208146329, 5680180131, 26705711025, 125558574123, 590321410569, 2775432824691, 13048869758625, 61350071873403, 288441173689209, 1356124096054851, 6375901677678225

Offset: 0

Peter Luschny, Jan 07 2019

Is it true that p prime and p not 2 or 5 implies that a(p) is squarefree?

			The first few prime factorizations of a(n):
   1| 3;
   2| 3^2;
   3| 3   * 17;
   4| 3^2 * 5^2;
   5| 3   * 19^2;
   6| 3^3 * 11 * 17;
   7| 3   * 7937;
   8| 3^2 * 5^2 * 7 * 71;
   9| 3   * 17 * 10313;
  10| 3^2 * 19^2 * 761;
  11| 3   * 3875057;
  12| 3^3 * 5^2 * 11 * 17 * 433;
  13| 3   * 85655881;
  14| 3^2 * 13 * 1301 * 7937;
  15| 3   * 17 * 19^2 * 308521;
  16| 3^2 * 5^2 * 7 * 71 * 79 * 3023;
  17| 3   * 67 * 624669523;
  18| 3^4 * 11 * 17 * 3779 * 10313;
  19| 3   * 419 * 2207981563;

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,8).

Cf. A015525, A322942.

[1] cat [n le 2 select 3^n else 3*Self(n-1) +8*Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 27 2021

a := proc(n) option remember:
    if n < 3 then return [1, 3, 9][n+1] fi;
    8*a(n-2) + 3*a(n-1) end:
seq(a(n), n=0..24);

Mathematica
```
LinearRecurrence[{3, 8}, {1, 3, 9}, 25]
```

def a():
    yield 1
    yield 3
    c = 3; b = 9
    while True:
        yield b
        a = (b << 2) + (c << 3) - b
        c = b
        b = a
A323232 = a()
[next(A323232) for _ in range(30)]

a(n) = 3*a(n-1) + 8*a(n-2) for n >= 3.
a(n) is an odd integer and 3 | a(n) if n > 0.
a(n) = Sum_{k=0..n} 2^(n - k)*A322942(n, k).
a(n) = [x^n] (8*x^2 - 1)/(8*x^2 + 3*x - 1).
Let s = sqrt(41), u = -1/(s+3) and v = 1/(s-3); then
a(n) = (3/s)*16^n*(v^n - u^n) for n >= 1.
a(n) = n! [x^n](1 + (6*exp(3*x/2)*sinh(s*x/2))/s).
a(n) = n! [x^n](1 + (3/s)*(exp((3 + s)*x/2) - exp((3 - s)*x/2))).
a(n)/a(n+1) -> 2/(sqrt(41) + 3) = (sqrt(41) - 3)/16 for n -> oo.

A330794 Inverse of the Jacobsthal triangle (A322942). Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, -1, 1, 1, -2, 1, -1, 1, -3, 1, 1, 4, 2, -4, 1, -1, -7, 10, 4, -5, 1, 1, -14, -25, 16, 7, -6, 1, -1, 65, -21, -55, 21, 11, -7, 1, 1, -24, 196, -8, -98, 24, 16, -8, 1, -1, -367, -204, 400, 42, -154, 24, 22, -9, 1, 1, 774, -963, -688, 666, 148, -222, 20, 29, -10, 1

Offset: 0

Peter Luschny, Jan 03 2020

The inverse matrix of the Riordan square (cf. A321620) generated by (1 - 2*x^2)/((1 + x)*(1 - 2*x)).

			Triangle starts:
[0]   1;
[1]  -1,    1;
[2]   1,   -2,    1;
[3]  -1,    1,   -3,    1;
[4]   1,    4,    2,   -4,    1;
[5]  -1,   -7,   10,    4,   -5,    1;
[6]   1,  -14,  -25,   16,    7,   -6,    1;
[7]  -1,   65,  -21,  -55,   21,   11,   -7,    1;
[8]   1,  -24,  196,   -8,  -98,   24,   16,   -8,    1;
[9]  -1, -367, -204,  400,   42, -154,   24,   22,   -9,    1;

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

Cf. A152947, A256893, A321620, A322942.

m=30;
A322942:= CoefficientList[CoefficientList[Series[(1-2*t^2)/(1-(x+1)*t-2*t^2), {x,0,m}, {t,0,m}], t], x];
M:= M= Table[If[k<=n, A322942[[n+1,k+1]], 0], {n,0,m}, {k,0,m}];
g:= g= Inverse[M];
A330794[n_, k_]:= g[[n+1,k+1]];
Table[A330794[n,k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 20 2023 *)

# uses[riordan_array from A256893]
Jacobsthal = (2*x^2 - 1)/((x + 1)*(2*x - 1))
riordan_array(Jacobsthal, Jacobsthal, 10).inverse()

From G. C. Greubel, Sep 15 2023: (Start)
T(n, 0) = (-1)^n.
T(n, n) = 1.
T(n, n-1) = -n.
T(n, n-2) = A152947(n-1). (End)

A321620 The Riordan square of the Riordan numbers, triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 3, 5, 5, 3, 1, 1, 6, 13, 10, 7, 4, 1, 1, 15, 29, 26, 16, 9, 5, 1, 1, 36, 73, 61, 42, 23, 11, 6, 1, 1, 91, 181, 157, 103, 61, 31, 13, 7, 1, 1, 232, 465, 398, 271, 156, 83, 40, 15, 8, 1, 1, 603, 1205, 1040, 702, 419, 221, 108, 50, 17, 9, 1, 1

Offset: 0

Peter Luschny, Nov 15 2018

If gf is a generating function of the sequence a then by the 'Riordan square of a' we understand the integer triangle given by the Riordan array (gf, gf). This mapping can be seen as an operator RS: Z[[x]] -> Mat[Z].
For instance A039599 is the Riordan square of the Catalan numbers, A172094 is the Riordan square of the little Schröder numbers and A063967 is the Riordan square of the Fibonacci numbers. The Riordan square of the simplest sequence of positive numbers (A000012) is the Pascal triangle.
The generating functions used are listed in the table below. They may differ slightly from those defined elsewhere in order to ensure that RS(a) is invertible (as a matrix). We understand this as a kind of normalization.
  ---------------------------------------------------------------------------
  Sequence a         | RS(a)   | gf(a)
  ---------------------------------------------------------------------------
  Catalan            | A039599 | (1 - sqrt(1 - 4*x))/(2*x).
  1, Riordan         | A321620 | 1 + 2*x/(1 + x + sqrt(1 - 2*x - 3*x^2)).
  Motzkin            | A321621 | (1 - x - sqrt(1 - 2*x - 3*x^2))/(2*x^2).
  Fine               | A321622 | 1 + (1 - sqrt(1 - 4*x))/(3 - sqrt(1 - 4*x)).
  large Schröder     | A321623 | (1 - x - sqrt(1 - 6*x + x^2))/(2*x).
  little Schröder    | A172094 | (1 + x - sqrt(1 - 6*x + x^2))/(4*x).
  Lucas              | A321624 | 1 + x*(1 + 2*x)/(1 - x - x^2).
  swinging factorial | A321625 | (1 + x/(1 - 4*x^2))/sqrt(1 - 4*x^2).
  ternary trees      ||A109956|| u = sqrt(x*3/4); sin(arcsin(3*u)/3)/u.
  central trinomial  | A116392 | 1/sqrt(1 - 2*x - 3*x^2)
  Bell               | A154380 | Sum_{k>=0} x^k/Product_{j=1..k}(1 - j*x).
  (2*n-1)!!          | A321627 | 1/(1-x/(1-2*x/(1-3*x/(1-4*x/(1-5*x/...
  powers of 2        | A038208 | 1/(1 - 2*x).
  the all 1's seq.   | A007318 | 1/(1 - x).
  Fibonacci          | A063967 | 1/(1 - x - x^2).
  tribonacci         | A187889 | 1/(1 - x - x^2 - x^3).
  tetranacci         | A353593 | 1/(1 - x - x^2 - x^3 - x^4).
  Jacobsthal         | A322942 | (2*x^2-1)/((x + 1)*(2*x - 1))

			The triangle starts:
   [ 0]   1
   [ 1]   1    1
   [ 2]   0    1    1
   [ 3]   1    1    1   1
   [ 4]   1    3    2   1   1
   [ 5]   3    5    5   3   1   1
   [ 6]   6   13   10   7   4   1   1
   [ 7]  15   29   26  16   9   5   1  1
   [ 8]  36   73   61  42  23  11   6  1  1
   [ 9]  91  181  157 103  61  31  13  7  1 1
   [10] 232  465  398 271 156  83  40 15  8 1 1

Peter Luschny, The Riordan Square Transform, 2018.

Row sums are A007971 and |A167022| shifted one place left.

Cf. A039599, A321621, A321622, A321624, A172094, A321623, A321624, A321625, A109956, A154380, A038208, A007318, A063967, A187889, A321629, A322942, A236376.

RiordanSquare := proc(d, n, exp:=false) local td, M, k, m, u, j;
    series(d, x, n+1); td := [seq(coeff(%, x, j), j = 0..n)];
    M := Matrix(n); for k from 1 to n do M[k, 1] := td[k] od;
    for k from 1 to n-1 do for m from k to n-1 do
       M[m+1, k+1] := add(M[j, k]*td[m-j+2], j = k..m) od od;
    if exp then u := 1;
       for k from 1 to n-1 do u := u * k;
          for m from 1 to k do j := `if`(m = 1, u, j/(m-1));
             M[k+1, m] := M[k+1, m] * j od od fi;
M end:
RiordanSquare(1 + 2*x/(1 + x + sqrt(1 - 2*x - 3*x^2)), 8);

RiordanSquare[gf_, len_] := Module[{T}, T[n_, k_] := T[n, k] = If[k == 0, SeriesCoefficient[gf, {x, 0, n}], Sum[T[j, k-1] T[n-j, 0], {j, k-1, n-1}]]; Table[T[n, k], {n, 0, len-1}, {k, 0, n}]];
M = RiordanSquare[1 + 2x/(1 + x + Sqrt[1 - 2x - 3x^2]), 12];
M // Flatten (* Jean-François Alcover, Nov 24 2018 *)

# uses[riordan_array from A256893]
def riordan_square(gf, len, exp=false):
    return riordan_array(gf, gf, len, exp)
riordan_square(1 + 2*x/(1 + x + sqrt(1 - 2*x - 3*x^2)), 10)
# Alternatively, given a list S:
def riordan_square_array(S):
    N = len(S)
    M = matrix(ZZ, N, N)
    for n in (0..N-1): M[n, 0] = S[n]
    for k in (1..N-1):
        for m in (k..N-1):
            M[m, k] = sum(M[j, k-1]*S[m-j] for j in (k-1..m-1))
    return M
riordan_square_array([1, 1, 0, 1, 1, 3, 6, 15, 36]) # Peter Luschny, Apr 03 2020

Given a generating function g and an positive integer N compute the Taylor expansion at the origin t(k) = [x^k] g(x) for k in [0...N-1] and set T(n, 0) = t(n) for n in [0...N-1]. Then compute T(m, k) = Sum_{j in [k-1...m-1]} T(j, k - 1) t(m - j) for k in [1...N-1] and for m in [k...N-1]. The resulting (0, 0)-based lower triangular array is the Riordan square generated by g.

A152035 Expansion of g.f. (1-2x^2)/(1-2x-2*x^2).

Original entry on oeis.org

1, 2, 4, 12, 32, 88, 240, 656, 1792, 4896, 13376, 36544, 99840, 272768, 745216, 2035968, 5562368, 15196672, 41518080, 113429504, 309895168, 846649344, 2313089024, 6319476736, 17265131520, 47169216512, 128868696064, 352075825152, 961889042432, 2627929735168, 7179637555200, 19615134580736

Offset: 0

Roger L. Bagula, Nov 20 2008

Essentially same as A028860. - Philippe Deléham, Sep 21 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2).

Cf. A028860. Row sums of A322942.

[1] cat [n le 2 select 2^n else 2*(Self(n-1) +Self(n-2)): n in [1..30]]; // G. C. Greubel, Sep 20 2023

a := proc(n) option remember;
`if`(n < 3, [1, 2, 4][n+1], 2*(a(n-1) + a(n-2))) end:
seq(a(n), n=0..31); # Peter Luschny, Jan 03 2019

f[n_] = 2^n*Product[(1 + 2*Cos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}]; Table[FullSimplify[ExpandAll[f[n]]], {n, 0, 15}]
CoefficientList[Series[(1-2x^2)/(1-2x-2x^2),{x,0,40}],x] (* Harvey P. Dale, Sep 23 2014 *)
LinearRecurrence[{2,2},{1,2,4},40] (* Harvey P. Dale, May 12 2023 *)

@CachedFunction
def a(n): # a = A152035
    if n<3: return (1,2,4)[n]
    else: return 2*(a(n-1) + a(n-2))
[a(n) for n in range(31)] # G. C. Greubel, Sep 20 2023

a(n) = 2*(a(n-1) + a(n-2)) for n >= 3. - Peter Luschny, Jan 03 2019

Edited by N. J. A. Sloane, Apr 11 2009, based on comments from Philippe Deléham and R. J. Mathar

More terms from Philippe Deléham, Sep 21 2009

A331319 a(n) = [x^n](x - 2x^3)/(1 - 2x*(x + 1))^2.

Original entry on oeis.org

0, 1, 4, 14, 48, 156, 496, 1544, 4736, 14352, 43072, 128224, 379136, 1114560, 3260160, 9494656, 27545600, 79642880, 229573632, 659951104, 1892478976, 5414755328, 15461117952, 44064835584, 125371383808, 356137570304, 1010187124736, 2861518086144, 8095486246912

Offset: 0

Peter Luschny, Jan 14 2020

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,0,-8,-4).

Cf. A322942 (Jacobsthal triangle), A331320, A331321.

gf := (x - 2*x^3)/(1 - 2*x*(x + 1))^2: ser := series(gf, x, 32):
seq(coeff(ser, x, n), n=0..28);

LinearRecurrence[ {4, 0, -8, -4}, {0, 1, 4, 14}, 28]

concat(0, Vec(x*(1 - 2*x^2) / (1 - 2*x - 2*x^2)^2 + O(x^30))) \\ Colin Barker, Jan 14 2020

a(n) = Sum_{k=0..n} A322942(n, k)*k.
a(n) = 2*((n^2 - n - 2)*a(n-2) + (n^2 - 2*n - 4)*a(n-1))/(n^2 - 3*n).
a(n) = n! [x^n] (1/9)*exp(x)*(sqrt(3)*(3*x+2)*sinh(sqrt(3)*x)+3*x*cosh(sqrt(3)*x)).
From Colin Barker, Jan 14 2020: (Start)
a(n) = ((1-sqrt(3))^n*(-2*sqrt(3) + 3*n) + (1+sqrt(3))^n*(2*sqrt(3) + 3*n)) / 18.
a(n) = 4*a(n-1) - 8*a(n-3) - 4*a(n-4) for n>3.
(End)

A331320 a(n) = [x^n] ((x + 1)(2x - 1)(2x^2 - 1))/(2x^2 + 2x - 1)^2.

Original entry on oeis.org

1, 3, 8, 26, 80, 244, 736, 2200, 6528, 19248, 56448, 164768, 478976, 1387328, 4005376, 11530624, 33107968, 94839552, 271091712, 773380608, 2202374144, 6261404672, 17774206976, 50384312320, 142636515328, 403306786816, 1139055820800, 3213593911296, 9057375289344

Offset: 0

Peter Luschny, Jan 14 2020

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,0,-8,-4).

Cf. A322942 (Jacobsthal triangle), A331319, A331321.

a := proc(n) option remember; if n < 3 then return [1, 3, 8][n+1] fi;
(12*(n - 3)*a(n-3) + (14*n - 6)*a(n-2) + (70 - 4*n)*a(n-1))/(n+19) end:
seq(a(n), n=0..28);
# Alternative:
gf := ((x + 1)*(2*x - 1)*(2*x^2 - 1))/(2*x^2 + 2*x - 1)^2:
ser := series(gf, x, 32): seq(coeff(ser, x, n), n=0..28);

LinearRecurrence[{4,0,-8,-4},{1,3,8,26,80},40] (* Harvey P. Dale, Jun 14 2025 *)

Vec((1 + x)*(1 - 2*x)*(1 - 2*x^2) / (1 - 2*x - 2*x^2)^2 + O(x^30)) \\ Colin Barker, Jan 14 2020

a(n) = Sum_{k=0..n} A322942(n,k)*(k+1).
a(n) = (12*(n - 3)*a(n-3) + (14*n - 6)*a(n-2) + (70 - 4*n)*a(n-1))/(n + 19).
Let h(k) = (1+k)*exp((1+k)*x)*(3*x+12-4*k)/18 then
a(n) = n!*[x^n](h(sqrt(3)) + h(-sqrt(3)) + 1).
From Colin Barker, Jan 14 2020: (Start)
a(n) = 4*a(n-1) - 8*a(n-3) - 4*a(n-4) for n>4.
a(n) = (-8*sqrt(3)*((1-sqrt(3))^n - (1+sqrt(3))^n) + 3*((1-sqrt(3))^n + (1+sqrt(3))^n)*n) / 18 for n>0.
(End)

A322941 Coefficients of orthogonal polynomials p(n, x) where p(n, 0) is A026150 with 1 prepended. Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 4, 7, 4, 1, 10, 22, 17, 6, 1, 28, 68, 64, 31, 8, 1, 76, 208, 230, 138, 49, 10, 1, 208, 628, 796, 568, 252, 71, 12, 1, 568, 1880, 2680, 2208, 1170, 414, 97, 14, 1, 1552, 5584, 8832, 8232, 5052, 2140, 632, 127, 16, 1, 4240, 16480, 28608, 29712, 20676, 10160, 3598, 914, 161, 18, 1

Offset: 0

Peter Luschny, Jan 06 2019

			The first few polynomials are:
[0] p(0, x) = 1;
[1] p(1, x) = x + 1;
[2] p(2, x) = x^2 +  2*x + 1;
[3] p(3, x) = x^3 +  4*x^2 +  7*x + 4;
[4] p(4, x) = x^4 +  6*x^3 + 17*x^2 +  22*x + 10;
[5] p(5, x) = x^5 +  8*x^4 + 31*x^3 +  64*x^2 +  68*x + 28;
[6] p(6, x) = x^6 + 10*x^5 + 49*x^4 + 138*x^3 + 230*x^2 + 208*x + 76;
The triangle starts:
[0]    1;
[1]    1,    1;
[2]    1,    2,    1;
[3]    4,    7,    4,    1;
[4]   10,   22,   17,    6,    1;
[5]   28,   68,   64,   31,    8,    1;
[6]   76,  208,  230,  138,   49,   10,   1;
[7]  208,  628,  796,  568,  252,   71,  12,   1;
[8]  568, 1880, 2680, 2208, 1170,  414,  97,  14,  1;
[9] 1552, 5584, 8832, 8232, 5052, 2140, 632, 127, 16, 1;

Row sums are A322940, alternating row sums are A000007.

Cf. A026150, A322942.

p := proc(n) option remember;
`if`(n < 3, [1, x+1, x^2 + 2*x + 1][n+1], (x+2)*p(n-1) + 2*p(n-2));
sort(expand(%)) end: seq(print(p(n)), n=0..11); # Computes the polynomials.
seq(seq(coeff(p(n), x, k), k=0..n), n=0..10);

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

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

A329918 Coefficients of orthogonal polynomials related to the Jacobsthal numbers A152046, triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 0, 4, 0, 1, 0, 4, 0, 6, 0, 1, 0, 0, 12, 0, 8, 0, 1, 0, 8, 0, 24, 0, 10, 0, 1, 0, 0, 32, 0, 40, 0, 12, 0, 1, 0, 16, 0, 80, 0, 60, 0, 14, 0, 1, 0, 0, 80, 0, 160, 0, 84, 0, 16, 0, 1, 0, 32, 0, 240, 0, 280, 0, 112, 0, 18, 0, 1

Offset: 0

Peter Luschny, Nov 28 2019

			Triangle starts:
  [0] 1;
  [1] 0,  1;
  [2] 0,  0,  1;
  [3] 0,  2,  0,  1;
  [4] 0,  0,  4,  0,  1;
  [5] 0,  4,  0,  6,  0,  1;
  [6] 0,  0, 12,  0,  8,  0,  1;
  [7] 0,  8,  0, 24,  0, 10,  0,  1;
  [8] 0,  0, 32,  0, 40,  0, 12,  0, 1;
  [9] 0, 16,  0, 80,  0, 60,  0, 14, 0, 1;
The first few polynomials:
  p(0,x) = 1;
  p(1,x) = x;
  p(2,x) = x^2;
  p(3,x) = 2*x + x^3;
  p(4,x) = 4*x^2 + x^4;
  p(5,x) = 4*x + 6*x^3 + x^5;
  p(6,x) = 12*x^2 + 8*x^4 + x^6;

Row sums are A001045 starting with 1, which is A152046. These are in signed form also the alternating row sums. Diagonal sums are aerated A133494.

Cf. A110509, A113953, A114192, A167431, A322942.

using Nemo # Returns row n.
function A329918(row)
    R, x = PolynomialRing(ZZ, "x")
    function p(n)
        n < 3 && return x^n
        x*p(n-1) + 2*p(n-2)
    end
    p = p(row)
    [coeff(p, k) for k in 0:row]
end
for row in 0:9 println(A329918(row)) end # prints triangle

T := (n, k) -> `if`((n+k)::odd, 0, 2^((n-k)/2)*binomial((n+k)/2-1, (n-k)/2)):
seq(seq(T(n, k), k=0..n), n=0..11);

p(n) = x*p(n-1) + 2*p(n-2) for n >= 3; p(0) = 1, p(1) = x, p(2) = x^2.
T(n, k) = [x^k] p(n).
T(n, k) = 2^((n-k)/2)*binomial((n+k)/2-1, (n-k)/2) if n+k is even otherwise 0.

A323210 a(n) = 9*J(n)^2 where J(n) are the Jacobsthal numbers A001045 with J(0) = 1.

Original entry on oeis.org

1, 9, 9, 81, 225, 1089, 3969, 16641, 65025, 263169, 1046529, 4198401, 16769025, 67125249, 268402689, 1073807361, 4294836225, 17180131329, 68718952449, 274878955521, 1099509530625, 4398050705409, 17592177655809, 70368760954881, 281474943156225, 1125899973951489

Offset: 0

Peter Luschny, Jan 09 2019

Colin Barker conjectures that A208556 is a shifted version of this sequence.

Index entries for linear recurrences with constant coefficients, signature (3,6,-8).

Cf. A001045, A062510, A139818, A208556, A322942, A323232.

gf := (8*x^3 - 24*x^2 + 6*x + 1)/((4*x - 1)*(2*x + 1)*(x - 1)):
ser := series(gf,x,32): seq(coeff(ser,x,n), n=0..25);

LinearRecurrence[{3, 6, -8}, {1, 9, 9, 81}, 25]

# Demonstrates the product formula.
CC = ComplexField(200)
def t(n,k): return CC(3)*cos(CC(pi*k/n)) - CC(i)*sin(CC(pi*k/n))
def T(n,k): return t(n,k)*(t(n,k).conjugate())
def a(n): return prod(T(n,k) for k in (1..n))
print([a(n).real().round() for n in (0..29)])

a(n) = Product_{k=1..n} T(n, k) where T(n, k) = t(n,k)*conjugate(t(n,k)) and t(n,k) = 3*cos(Pi*k/n) - i*sin(Pi*k/n), i is the imaginary unit.
a(n) = [x^n] (8*x^3 - 24*x^2 + 6*x + 1)/((4*x - 1)*(2*x + 1)*(x - 1)).
a(n) = n! [x^n] (1 + exp(x) - 2*exp(-2*x) + exp(4*x)).
a(n) = 3*a(n-1) + 6*a(n-2) - 8*a(n-3) for n >= 4.
A062510(n) = sqrt(a(n)) for n > 0.
a(n) = 4^n-2*(-2)^n+1, n>0. - R. J. Mathar, Mar 06 2022

cp's OEIS Frontend

A323232 a(n) = 2^n*J(n, 1/2) where J(n, x) are the Jacobsthal polynomials as defined in A322942.

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A330794 Inverse of the Jacobsthal triangle (A322942). Triangle read by rows, T(n, k) for 0 <= k <= n.

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A321620 The Riordan square of the Riordan numbers, triangle read by rows, T(n, k) for 0 <= k <= n.

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A152035 Expansion of g.f. (1-2*x^2)/(1-2*x-2*x^2).

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A331319 a(n) = [x^n](x - 2*x^3)/(1 - 2*x*(x + 1))^2.

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A331320 a(n) = [x^n] ((x + 1)*(2*x - 1)*(2*x^2 - 1))/(2*x^2 + 2*x - 1)^2.

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A322941 Coefficients of orthogonal polynomials p(n, x) where p(n, 0) is A026150 with 1 prepended. Triangle read by rows, T(n, k) for 0 <= k <= n.

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A152035 Expansion of g.f. (1-2x^2)/(1-2x-2*x^2).

A331319 a(n) = [x^n](x - 2x^3)/(1 - 2x*(x + 1))^2.

A331320 a(n) = [x^n] ((x + 1)(2x - 1)(2x^2 - 1))/(2x^2 + 2x - 1)^2.