A119369 -id:A119369 - cp's OEIS frontend

A119373 A lower diagonal of pendular trinomial triangle A119369.

1, 2, 6, 20, 70, 253, 938, 3546, 13617, 52967, 208255, 826315, 3304456, 13304924, 53891402, 219442686, 897772983, 3688451380, 15211545938, 62950542636, 261329456566, 1087985751336, 4541524025769, 19003488710465, 79696345430789

Offset: 0

Paul D. Hanna, May 17 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A119369, A119370, A119371, A119372, A119374, A119375, A119376.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( 8*(1+x)/( ((1+x^2) +Sqrt((1+x^2)^2 -4*x*(1+x)))^2*(1+4*x+x^2 +Sqrt((1+4*x+x^2)^2 -4*x*(1+x)*(3+2*x))) ) )); // G. C. Greubel, Mar 16 2021

CoefficientList[Series[8*(1+x)/( ((1+x^2) + Sqrt[(1+x^2)^2 -4*x*(1+x)])^2*(1 + 4*x +x^2 +Sqrt[(1+4*x+x^2)^2 -4*x*(1+x)*(3+2*x)])), {x,0,30}], x] (* G. C. Greubel, Mar 16 2021 *)

{a(n)=polcoeff(8*(1+x)/((1+x^2)+sqrt((1+x^2)^2-4*x*(1+x)+x*O(x^n)))^2 /(1+4*x+x^2 + sqrt((1+4*x+x^2)^2 - 4*x*(1+x)*(3+2*x)+x*O(x^n))),n)}

def A119373_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( 8*(1+x)/( ((1+x^2) +sqrt((1+x^2)^2 -4*x*(1+x)))^2*(1+4*x+x^2 +sqrt((1+4*x+x^2)^2 -4*x*(1+x)*(3+2*x))) ) ).list()
A119373_list(30) # G. C. Greubel, Mar 16 2021

G.f.: A(x) = B(x)^2/(1+x - x*B(x)) = B(x)^2*G(x) = B(x)*H(x), where B(x) is g.f. of A119370, G(x) is g.f. of A119371 and H(x) is g.f. of A119372.

G.f.: 8*(1+x)/( ((1+x^2) +sqrt((1+x^2)^2 -4*x*(1+x)))^2*(1+4*x+x^2 +sqrt((1+4*x+x^2)^2 -4*x*(1+x)*(3+2*x))) ).

A119374 A lower diagonal of pendular trinomial triangle A119369.

Original entry on oeis.org

1, 3, 10, 36, 133, 501, 1918, 7440, 29180, 115522, 461044, 1852938, 7492846, 30464306, 124461782, 510696350, 2103708187, 8696498477, 36066269640, 150015248758, 625664295594, 2615929689642, 10962436020878, 46037427169060

Offset: 0

Paul D. Hanna, May 17 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A119369, A119370, A119371, A119372, A119373, A119375, A119376.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( 16*(1+x)/( ((1+x^2) +Sqrt((1+x^2)^2-4*x*(1+x)))^3*(1+4*x+x^2 +Sqrt((1+4*x+x^2)^2 - 4*x*(1+x)*(3+2*x))) ) )); // G. C. Greubel, Mar 16 2021

CoefficientList[Series[16*(1+x)/( ((1+x^2) +Sqrt[(1+x^2)^2 -4*x*(1+x)])^3*(1+4*x +x^2 +Sqrt[(1+4*x+x^2)^2 -4*x*(1+x)*(3+2*x)])), {x,0,30}], x] (* G. C. Greubel, Mar 16 2021 *)

{a(n)=polcoeff(16*(1+x)/((1+x^2)+sqrt((1+x^2)^2-4*x*(1+x)+x*O(x^n)))^3 /(1+4*x+x^2 + sqrt((1+4*x+x^2)^2 - 4*x*(1+x)*(3+2*x)+x*O(x^n))),n)}

def A119374_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( 16*(1+x)/( ((1+x^2) +sqrt((1+x^2)^2-4*x*(1+x)))^3*(1+4*x+x^2 +sqrt((1+4*x+x^2)^2 - 4*x*(1+x)*(3+2*x))) ) ).list()
A119374_list(30) # G. C. Greubel, Mar 16 2021

G.f.: A(x) = B(x)^3/(1+x - x*B(x)) = B(x)^3*G(x) = B(x)^2*H(x) = B(x)*I(x), where B(x) is g.f. of A119370, G(x) is g.f. of A119371, H(x) is g.f. of A119372 and I(x) is g.f. of A119373.

G.f.: 16*(1+x)/( ((1+x^2) +sqrt((1+x^2)^2-4*x*(1+x)))^3*(1+4*x+x^2 +sqrt((1+4*x+x^2)^2 - 4*x*(1+x)*(3+2*x))) ).

A119375 Diagonal above the central terms of pendular trinomial triangle A119369, ignoring leading zeros.

Original entry on oeis.org

1, 3, 11, 40, 149, 564, 2166, 8420, 33074, 131085, 523599, 2105727, 8519469, 34652696, 141621164, 581266730, 2394961851, 9902433681, 41074316737, 170869972460, 712729001716, 2980264528670, 12490379959184, 52458339164169

Offset: 0

Paul D. Hanna, May 17 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A119369, A119370, A119371, A119372, A119373, A119374, A119376.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( (1-2*x-x^2 - Sqrt(1-4*x-2*x^2+x^4))/( 1+2*x^3+x^4 +(1+x)^2*Sqrt(1-4*x-2*x^2+x^4) ) )); // G. C. Greubel, Mar 16 2021

CoefficientList[Series[(1-2*x-x^2 -Sqrt[1-4*x-2*x^2+x^4])/(x^2*(1+2*x^3+x^4 +(1+x)^2*Sqrt[1-4*x-2*x^2+x^4])), {x,0,30}], x] (* G. C. Greubel, Mar 16 2021 *)

{a(n)=polcoeff(2/((1+x^2)+sqrt((1+x^2)^2-4*x*(1+x)+x^3*O(x^n)))* (2*(1+x)/(1+4*x+x^2 + sqrt((1+4*x+x^2)^2-4*x*(1+x)*(3+2*x)+x^3*O(x^n)))-1)/x^2,n)}

def A119375_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( (1-2*x-x^2 -sqrt(1-4*x-2*x^2+x^4))/( x^2*(1+2*x^3+x^4 +(1+x)^2*sqrt(1-4*x-2*x^2+x^4)) ) ).list()
A119375_list(30) # G. C. Greubel, Mar 16 2021

G.f.: A(x) = B(x)*(G(x) - 1)/x^2 = B(x)*(B(x) - 1)/(x+x^2 - x^2*B(x)), where B(x) is g.f. of A119370 and G(x) is g.f. of A119371 (central terms of A119369).

G.f.: (1-2*x-x^2 -sqrt(1-4*x-2*x^2+x^4))/( x^2*(1+2*x^3+x^4 +(1+x)^2*sqrt(1-4*x-2*x^2+x^4)) ). - G. C. Greubel, Mar 16 2021

A119376 Second diagonal above the central terms of pendular trinomial triangle A119369, ignoring leading zeros.

Original entry on oeis.org

1, 4, 16, 63, 248, 980, 3894, 15563, 62555, 252789, 1026623, 4188390, 17159382, 70570380, 291253664, 1205935204, 5008047097, 20854723702, 87064706122, 364334839028, 1527943938306, 6420911995109, 27033938458595

Offset: 0

Paul D. Hanna, May 17 2006

nonn

Equals convolution of A119370 and A119375, which is the prior diagonal above the central terms of triangle A119369.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A119369, A119370, A119371, A119372, A119373, A119374, A119375.

R:=PowerSeriesRing(Rationals(), 30);
f:= func< x | Sqrt(1-4*x-2*x^2+x^4) >;
Coefficients(R!( 2*(1-2*x-x^2 -f(x))/( x^2*(1+2*x^3+x^4 +(1+x)^2*f(x))*(1+x^2 +f(x)) ) )); // G. C. Greubel, Mar 17 2021

f[x_]:= Sqrt[1-4*x-2*x^2+x^4];
CoefficientList[Series[2*(1-2*x-x^2 -f[x])/(x^2*(1+2*x^3+x^4 +(1+x)^2*f[x])*(1+x^2 +f[x])), {x,0,30}], x] (* G. C. Greubel, Mar 17 2021 *)

{a(n)=polcoeff(4/((1+x^2)+sqrt((1+x^2)^2-4*x*(1+x)+x^3*O(x^n)))^2* (2*(1+x)/(1+4*x+x^2 + sqrt((1+4*x+x^2)^2-4*x*(1+x)*(3+2*x)+x^3*O(x^n)))-1)/x^2,n)}

def f(x): return sqrt(1-4*x-2*x^2+x^4)
def A119376_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( 2*(1-2*x-x^2 -f(x))/( x^2*(1+2*x^3+x^4 +(1+x)^2*f(x))*(1+x^2 +f(x)) ) ).list()
A119376_list(30) # G. C. Greubel, Mar 17 2021

G.f.: A(x) = B(x)^2*(G(x) - 1)/x^2 = B(x)^2*(B(x) - 1)/(x+x^2 - x^2*B(x)), where B(x) is g.f. of A119370 and G(x) is g.f. of A119371 (central terms of A119369).

G.f.: 2*(1-2*x-x^2-f(x))/( x^2*(1+2*x^3+x^4+(1+x)^2*f(x))*(1+x^2+f(x)) ) where f(x) = sqrt(1-4*x-2*x^2+x^4). - G. C. Greubel, Mar 17 2021

A119370 G.f. satisfies A(x) = 1 + xA(x)^2 + x^2(A(x)^2 - A(x)).

Original entry on oeis.org

1, 1, 2, 6, 19, 64, 225, 816, 3031, 11473, 44096, 171631, 675130, 2679728, 10719237, 43168826, 174885089, 712222799, 2914150406, 11973792218, 49385167369, 204386777160, 848530495383, 3532844222611, 14747626307436, 61712139464939

Offset: 0

Paul D. Hanna, May 16 2006

Equals base sequence of pendular trinomial triangle A119369; iterated convolutions of this sequence with the central terms (A119371) generates all diagonals of A119369. For example: A119372 = A119370 * A119371; A119373 = A119370^2 * A119371.

Diagonal sums of number array A133336. - Philippe Deléham, Nov 09 2009

			A(x) = 1 + x + 2*x^2 + 6*x^3 + 19*x^4 + 64*x^5 + 225*x^6 + 816*x^7 +...
x*A(x)^2 = x + 2*x^2 + 5*x^3 + 16*x^4 + 54*x^5 + 190*x^6 + 690*x^7 +...
x^2*( A(x)^2 - A(x) ) = 1*x^3 + 3*x^4 + 10*x^5 + 35*x^6 + 126*x^7 +...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Stoyan Dimitrov, On permutation patterns with constrained gap sizes, arXiv:2002.12322 [math.CO], 2020.
Hannah Golab, Pattern avoidance in Cayley permutations, Master's Thesis, Northern Arizona Univ. (2024). See p. 41.

Cf. A104547, A119369, A119371, A119372, A119373, A119374, A119375, A119376.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( (1+x^2 -Sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)) )); // G. C. Greubel, Mar 17 2021

m:= 30;
S:= series( (1+x^2 -sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)), x, m+1);
seq(coeff(S, x, j), j = 0..m); # G. C. Greubel, Mar 17 2021

CoefficientList[Series[((1+x^2)-Sqrt[(1+x^2)^2-4*x*(1+x)])/(2*x*(1+x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 11 2013 *)

{a(n)=polcoeff(2/((1+x^2)+sqrt((1+x^2)^2-4*x*(1+x)+x*O(x^n))),n)}

def A119370_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( (1+x^2 -sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)) ).list()
A119370_list(30) # G. C. Greubel, Mar 17 2021

G.f.: A(x) = ((1+x^2) - sqrt( (1+x^2)^2 - 4*x*(1+x) ))/(2*x*(1+x)). Equals the inverse binomial transform of A104547.
Recurrence: (n+1)*a(n) = 3*(n-1)*a(n-1) + 6*(n-1)*a(n-2) + 2*(n-2)*a(n-3) - (n-5)*a(n-4) - (n-5)*a(n-5). - Vaclav Kotesovec, Sep 11 2013
a(n) ~ sqrt(-z^2-3*z+1)*(4+2*z-z^3)^(n+1)*(-z^3+z^2+z+3) / (8*sqrt(Pi) * n^(3/2)), where z = 1/(2*sqrt(3/(4+(280-24*sqrt(129))^(1/3) + 2*(35 + 3*sqrt(129))^(1/3)))) - 1/2*sqrt(8/3-1/3*(280-24*sqrt(129))^(1/3) - 2/3*(35+3*sqrt(129))^(1/3) + 8*sqrt(3/(4+(280-24*sqrt(129))^(1/3) + 2*(35 + 3*sqrt(129))^(1/3)))) = 0.225270426... is the root of the equation 1-2*z^2+z^4-4*z=0. - Vaclav Kotesovec, Sep 11 2013
G.f.: 1/G(0) where G(k) = 1 - q/(1 - (q + q^2) / G(k+1) ). - Joerg Arndt, Dec 06 2014
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k-1,k) * binomial(2*n-3*k+1,n-2*k)/(2*n-3*k+1). - Seiichi Manyama, Aug 28 2023
Conjecture: A(x) = 1 + x*exp(Sum_{n >= 1} g(n, x)*x^n/n), where g(n, x) = Sum_{k = 0..n} binomial(n, k)^2*(1 + x)^k. Cf. A105633 and A167638. - Peter Bala, Sep 10 2024

A122445 Pendular trinomial triangle, read by rows of 2n+1 terms (n>=0), defined by the recurrence: if 0 < k < n, T(n,k) = T(n-1,k) + 2*T(n,2n-1-k); otherwise, if n-1 < k < 2n-1, T(n,k) = T(n-1,k) + T(n,2n-2-k); with T(n,0) = T(n+1,2n) = 1 and T(n+1,2n+1) = T(n+1,2n+2) = 0.

Original entry on oeis.org

1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 2, 3, 2, 1, 0, 0, 1, 3, 6, 10, 8, 3, 1, 0, 0, 1, 4, 10, 22, 36, 28, 12, 4, 1, 0, 0, 1, 5, 15, 39, 83, 135, 107, 47, 17, 5, 1, 0, 0, 1, 6, 21, 62, 155, 324, 525, 418, 189, 72, 23, 6, 1, 0, 0, 1, 7, 28, 92, 259, 629, 1298, 2094, 1676, 773, 305, 104, 30, 7, 1, 0, 0

Offset: 0

Paul D. Hanna, Sep 07 2006

The diagonals may be generated by iterated convolutions of a base sequence B with the sequence C of central terms. The g.f. B(x) of the base sequence satisfies: B = 1 + x*B^2 + 2x^2*(B^2 - B); the g.f. C(x) of the central terms satisfies: C(x) = 1/(1+x - xB(x)).

			To obtain row 4, pendular sums of row 3 are carried out as follows.
  [1, 2, 3,  2, 1, 0, 0]: given row 3;
  [1, _, _, __, _, _, _]: start with T(4,0) = T(3,0) = 1;
  [1, _, _, __, _, _, 1]: T(4,6) = T(4,0) + 2*T(3,6) = 1 + 2*0 = 1;
  [1, 3, _, __, _, _, 1]: T(4,1) = T(4,6) + 1*T(3,1) = 1 + 1*2 = 3;
  [1, 3, _, __, _, 3, 1]: T(4,5) = T(4,1) + 2*T(3,5) = 3 + 2*0 = 3;
  [1, 3, 6, __, _, 3, 1]: T(4,2) = T(4,5) + 1*T(3,2) = 3 + 1*3 = 6;
  [1, 3, 6, __, 8, 3, 1]: T(4,4) = T(4,2) + 2*T(3,4) = 6 + 2*1 = 8;
  [1, 3, 6, 10, 8, 3, 1]: T(4,3) = T(4,4) + 1*T(3,3) = 8 + 1*2 = 10;
  [1, 3, 6, 10, 8, 3, 1,0,0]: complete row 4 by appending two zeros.
Triangle begins:
  1;
  1, 0,  0;
  1, 1,  1,  0,   0;
  1, 2,  3,  2,   1,   0,   0;
  1, 3,  6, 10,   8,   3,   1,   0,   0;
  1, 4, 10, 22,  36,  28,  12,   4,   1,  0,  0;
  1, 5, 15, 39,  83, 135, 107,  47,  17,  5,  1, 0, 0;
  1, 6, 21, 62, 155, 324, 525, 418, 189, 72, 23, 6, 1, 0, 0;
Central terms are:
  C = A122447 = [1, 0, 1, 2, 8, 28, 107, 418, 1676, 6848, ...].
Lower diagonals start:
  D1 = A122448 = [1, 1, 3, 10, 36, 135, 525, 2094, 8524, ...];
  D2 = A122449 = [1, 2, 6, 22, 83, 324, 1298, 5302, 22002, ...].
Diagonals above central terms (ignoring leading zeros) start:
  U1 = A122450 = [1, 3, 12, 47, 189, 773, 3208, 13478, 57222, ...];
  U2 = A122451 = [1, 4, 17, 72, 305, 1300, 5576, 24068, 104510, ...].
There exists the base sequence:
  B = A122446 = [1, 1, 2, 7, 24, 88, 336, 1321, 5316, 21788, ...]
which generates all diagonals by convolutions with central terms:
  D2 = B * D1 = B^2 * C
  U2 = B * U1 = B^2 * C"
where C" = [1, 2, 8, 28, 107, 418, 1676, 6848, 28418, ...]
are central terms not including the initial [1,0].

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

Cf. A122446, A122447 (central terms), A122452 (row sums).
Diagonals: A122448, A122449, A122450, A122451.
Variants: A118340, A118345, A118350, A119369.

T:= proc(n, k) option remember;
      if k=0 and n=0 then 1
    elif k<0 or k>2*(n-1) then 0
    elif n=2 and k<3 then 1
    else T(n-1, k) + `if`(kG. C. Greubel, Mar 16 2021

T[n_, k_]:= T[n, k]= If[n==0 && k==0, 1, If[k<0 || k>2*(n-1), 0, If[n==2 && k<3, 1, T[n-1, k] + If[kG. C. Greubel, Mar 16 2021 *)

{T(n,k)= if(k==0 && n==0, 1, if(k>2*n-2 || k<0, 0, if(n==2 && k<=2, 1, if(k

@CachedFunction
def T(n, k):
    if (n==0 and k==0): return 1
    elif (k<0 or k>2*(n-1)): return 0
    elif (n==2 and k<3): return 1
    else: return T(n-1, k) + ( T(n, 2*n-k-1) if kG. C. Greubel, Mar 16 2021

A119371 G.f. satisfies: A(x) = (1+x) - x(4+x)A(x) + x(3+2x)*A(x)^2.

Original entry on oeis.org

1, 0, 1, 2, 7, 23, 81, 293, 1087, 4110, 15783, 61387, 241329, 957400, 3828055, 15410651, 62410871, 254095382, 1039394147, 4269718110, 17606507789, 72852976317, 302403773303, 1258855723796, 5254253027485, 21983753239135

Offset: 0

Paul D. Hanna, May 17 2006

nonn

Equals central terms of pendular trinomial triangle A119369.

			            A(x) = 1 + x^2 + 2*x^3 + 7*x^4 + 23*x^5 + 81*x^6 ...;
   -x*(4+x)*A(x) = -4*x -x^2 -4*x^3 -9*x^4 -30*x^5 -99*x^6 - ...;
x*(3+2*x)*A(x)^2 = 3*x +2*x^2 +6*x^3 +16*x^4 +53*x^5 +180*x^6 + ...;

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A119369, A119370, A119372, A119373, A119374, A119375, A119376.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( (1+4*x+x^2 -Sqrt(1-4*x-2*x^2+x^4))/(2*x*(3+2*x)) )); // G. C. Greubel, Mar 17 2021

m:= 30;
S:= series( (1+4*x+x^2 -sqrt(1-4*x-2*x^2+x^4))/(2*x*(3+2*x)), x, m+1);
seq(coeff(S, x, j), j = 0..m); # G. C. Greubel, Mar 17 2021

CoefficientList[Series[(1+4*x+x^2-Sqrt[(1+4*x+x^2)^2-4*x*(1+x)*(3+2*x)])/(2*x*(3+2*x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 11 2013 *)

{a(n)=polcoeff((2*(1+x))/(1+4*x+x^2+sqrt((1+4*x+x^2)^2-4*x*(1+x)*(3+2*x)+x*O(x^n))),n)}

def A119371_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( (1+4*x+x^2 -sqrt(1-4*x-2*x^2+x^4))/(2*x*(3+2*x)) ).list()
A119371_list(30) # G. C. Greubel, Mar 17 2021

G.f.: A(x) = (1+4*x+x^2 - sqrt((1+4*x+x^2)^2 - 4*x*(1+x)*(3+2*x)))/(2*x*(3+2*x)).
G.f.: A(x) = 1/(1+x - x*B(x)) = (1 + x*H(x))/(1+x) = 1 + x^2*F(x)/B(x), where B(x) is g.f. of A119370, H(x) is g.f. of A119372, F(x) is g.f. of A119375.
Recurrence: 3*(n+1)*a(n) = 2*(5*n-4)*a(n-1) + 2*(7*n-8)*a(n-2) + 4*(n-2)*a(n-3) - 3*(n-5)*a(n-4) - 2*(n-5)*a(n-5). - Vaclav Kotesovec, Sep 11 2013
a(n) ~ sqrt(-z^2-3*z+1)*(4+2*z-z^3)^(n+1)*(35-8*z^3+12*z^2-2*z) /(242*sqrt(Pi)*n^(3/2)), where z = 1/(2*sqrt(3/(4+(280-24*sqrt(129))^(1/3) + 2*(35+3*sqrt(129))^(1/3)))) - 1/2*sqrt(8/3-1/3*(280-24*sqrt(129))^(1/3) - 2/3*(35+3*sqrt(129))^(1/3) + 8*sqrt(3/(4+(280-24*sqrt(129))^(1/3) + 2*(35+3*sqrt(129))^(1/3)))) = 0.225270426... is the root of the equation 1-2*z^2+z^4-4*z=0. - Vaclav Kotesovec, Sep 11 2013

A119372 G.f. satisfies: A(x) = 1 + x(1-x-x^2)A(x) + x^2(3+2x)*A(x)^2.

Original entry on oeis.org

1, 1, 3, 9, 30, 104, 374, 1380, 5197, 19893, 77170, 302716, 1198729, 4785455, 19238706, 77821522, 316506253, 1293489529, 5309112257, 21876225899, 90459484106, 375256749620, 1561259497099, 6513108751281, 27238006266620

Offset: 0

Paul D. Hanna, May 17 2006

nonn

Equals diagonal and row sums of pendular trinomial triangle A119369. Also equals convolution of A119370 and A119371 (central terms of A119369).

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A119369, A119370, A119371, A119373, A119374, A119375, A119376.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( (1-x+x^2+x^3 -(1+x)*Sqrt(1-4*x-2*x^2+x^4))/(2*x^2*(3+2*x)) )); // G. C. Greubel, Mar 17 2021

m:= 30;
S:= series( (1-x+x^2+x^3 -(1+x)*sqrt(1-4*x-2*x^2+x^4))/(2*x^2*(3+2*x)), x, m+1);
seq(coeff(S, x, j), j = 0..m); # G. C. Greubel, Mar 17 2021

CoefficientList[Series[(1-x+x^2+x^3-Sqrt[(1-x+x^2+x^3)^2-4*x^2*(3+2*x)])/(2*x^2*(3+2*x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 11 2013 *)

{a(n)=polcoeff(2/(1-x+x^2+x^3+sqrt((1-x+x^2+x^3)^2-4*x^2*(3+2*x)+x*O(x^n))),n)}

def A119372_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( (1-x+x^2+x^3 -(1+x)*sqrt(1-4*x-2*x^2+x^4))/(2*x^2*(3+2*x)) ).list()
A119372_list(30) # G. C. Greubel, Mar 17 2021

G.f.: A(x) = (1-x+x^2+x^3 - sqrt( (1-x+x^2+x^3)^2 - 4*x^2*(3+2*x)) )/(2*x^2*(3+2*x)).
G.f.: A(x) = B(x)/(1+x - x*B(x)) = B(x)*G(x), where B(x) is g.f. of A119370, G(x) is g.f. of A119371.
Recurrence: 3*(n+2)*(2*n-1)*a(n) = (20*n^2 - 6*n - 11)*a(n-1) + (28*n^2 - 18*n + 5)*a(n-2) + (8*n^2-12*n-17)*a(n-3) - 3*(2*n^2 - 9*n + 1)*a(n-4) - 2*(n-5)*(2*n+1)*a(n-5). - Vaclav Kotesovec, Sep 11 2013
a(n) ~ sqrt(-8*z^2-5*z^3+2-5*z)*(4+2*z-z^3)^n*(-18-8*z+4*z^3+z^2)*(-35+8*z^3-12*z^2+2*z)/(242*sqrt(Pi)*n^(3/2)), where z = 1/(2*sqrt(3/(4+(280-24*sqrt(129))^(1/3) + 2*(35+3*sqrt(129))^(1/3)))) - 1/2*sqrt(8/3-1/3*(280-24*sqrt(129))^(1/3) - 2/3*(35+3*sqrt(129))^(1/3) + 8*sqrt(3/(4+(280-24*sqrt(129))^(1/3) + 2*(35+3*sqrt(129))^(1/3)))) = 0.225270426... is the root of the equation 1-2*z^2+z^4-4*z=0. - Vaclav Kotesovec, Sep 11 2013

A167769 Pendular trinomial triangle (p=0), read by rows of 2n+1 terms (n>=0), defined by the recurrence : if 0

Original entry on oeis.org

1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 2, 3, 2, 1, 0, 0, 1, 3, 6, 8, 6, 3, 1, 0, 0, 1, 4, 10, 18, 24, 18, 10, 4, 1, 0, 0, 1, 5, 15, 33, 57, 75, 57, 33, 15, 5, 1, 0, 0, 1, 6, 21, 54, 111, 186, 243, 186, 111, 54, 21, 6, 1, 0, 0, 1, 7, 28, 82, 193, 379, 622, 808, 622, 379, 193, 82, 28, 7, 1, 0, 0

Offset: 0

Philippe Deléham, Nov 11 2009

See A119369 for p=1 and A122445 for p=2. The diagonals may be generated by iterated convolutions of a base sequence B (A000108(n)) with the sequence C (A000957(n+1)) of central terms.

			Triangle begins :
  1;
  1, 0,  0;
  1, 1,  1,  0,  0;
  1, 2,  3,  2,  1,  0,  0;
  1, 3,  6,  8,  6,  3,  1,  0,  0;
  1, 4, 10, 18, 24, 18, 10,  4,  1, 0, 0,
  1, 5, 15, 33, 57, 75, 57, 33, 15, 5, 1, 0, 0; ...

Kim, Ki Hang; Rogers, Douglas G.; Roush, Fred W. Similarity relations and semiorders. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 577--594, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561081 (81i:05013) - From N. J. A. Sloane, Jun 05 2012

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

Cf. A000108, A000957, A000958, A001558, A001559, A071724, A104629.

T:= proc(n, k) option remember;
      if k=0 and n=0 then 1;
    elif k<0 or k>2*(n-1) then 0;
    elif n=2 and k<3 then 1;
    elif kG. C. Greubel, Mar 17 2021

T[n_, k_]:= T[n, k]= If[k==0 && n==0, 1, If[k<0 || k>2*(n-1), 0, If[n==2 && k<3, 1, If[kG. C. Greubel, Mar 17 2021 *)

T(n, k)=if(k==0 && n==0, 1, if(k>2*n-2 || k<0, 0, if(n==2 && k<=2, 1, if(kPaul D. Hanna, Nov 12 2009

@CachedFunction
def T(n, k):
    if (k==0 and n==0): return 1
    elif (k<0 or k>2*(n-1)): return 0
    elif (n==2 and k<3): return 1
    elif (kG. C. Greubel, Mar 17 2021

Sum_{k=0..2*n} T(n,k) = A071724(n) = [n=0] + 3*binomial(2n,n-1)/(n+2) = [n=0] + n*C(n)/(n+2), where C(n) are the Catalan numbers (A000108). - G. C. Greubel, Mar 17 2021

cp's OEIS Frontend

A119373 A lower diagonal of pendular trinomial triangle A119369.

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A119374 A lower diagonal of pendular trinomial triangle A119369.

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A119375 Diagonal above the central terms of pendular trinomial triangle A119369, ignoring leading zeros.

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A119376 Second diagonal above the central terms of pendular trinomial triangle A119369, ignoring leading zeros.

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A119370 G.f. satisfies A(x) = 1 + x*A(x)^2 + x^2*(A(x)^2 - A(x)).

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A122445 Pendular trinomial triangle, read by rows of 2n+1 terms (n>=0), defined by the recurrence: if 0 < k < n, T(n,k) = T(n-1,k) + 2*T(n,2n-1-k); otherwise, if n-1 < k < 2n-1, T(n,k) = T(n-1,k) + T(n,2n-2-k); with T(n,0) = T(n+1,2n) = 1 and T(n+1,2n+1) = T(n+1,2n+2) = 0.

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A119371 G.f. satisfies: A(x) = (1+x) - x*(4+x)*A(x) + x*(3+2*x)*A(x)^2.

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A119370 G.f. satisfies A(x) = 1 + xA(x)^2 + x^2(A(x)^2 - A(x)).

A119371 G.f. satisfies: A(x) = (1+x) - x(4+x)A(x) + x(3+2x)*A(x)^2.

A119372 G.f. satisfies: A(x) = 1 + x(1-x-x^2)A(x) + x^2(3+2x)*A(x)^2.