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a(n) = [x^n] (1 - 2*x - sqrt((1 - 3*x)/(1 + x)))/(2*x^3).
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%I #12 Jun 27 2022 05:24:06

%S 1,1,3,6,15,36,91,232,603,1585,4213,11298,30537,83097,227475,625992,

%T 1730787,4805595,13393689,37458330,105089229,295673994,834086421,

%U 2358641376,6684761125,18985057351,54022715451,154000562758,439742222071,1257643249140,3602118427251

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

%F D-finite with recurrence a(n) = (2*a(n - 1) + 3*a(n - 2))*(n + 1)/(n + 3) for n >= 3.

%F a(n) = (-1)^n*hypergeom([1/2, -2 - n], [2], 4].

%F a(n) ~ (3^(n + 7/2)*(16*n + 11))/(128*sqrt(Pi)*(n + 2)^(5/2)).

%p gf := (1 - 2*x - sqrt((1 - 3*x)/(1 + x)))/(2*x^3): ser := series(gf, x, 36):

%p seq(coeff(ser, x, n), n = 0..30);

%p a := proc(n) option remember; `if`(n < 3, [1, 1, 3][n + 1],

%p ((2*a(n - 1) + 3*a(n - 2))*(n + 1))/(n + 3)) end: seq(a(n), n=0..30);

%t a[n_] := (-1)^n*HypergeometricPFQ[{1/2, -2 - n}, {2}, 4]

%t Table[a[n], {n, 0, 30}]

%o (Python)

%o def rnum():

%o a, b, n = 1, 3, 3

%o yield 1

%o yield 1

%o while True:

%o yield b

%o n += 1

%o a, b = b, (n*(3*a + 2*b))//(n + 2)

%o A342912 = rnum()

%o print([next(A342912) for _ in range(31)])

%Y The diagonal sums of the Motzkin triangle A064189 (with the Motzkin numbers A001006 as first column), the row sums of A020474, and a shifted version of the Riordan numbers A005043.

%K nonn,easy

%O 0,3

%A _Peter Luschny_, Apr 18 2021