A167022 -id:A167022 - cp's OEIS frontend

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

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.

A168051 Expansion of (1+x+sqrt(1-2x-3x^2))/2.

Original entry on oeis.org

1, 0, -1, -1, -2, -4, -9, -21, -51, -127, -323, -835, -2188, -5798, -15511, -41835, -113634, -310572, -853467, -2356779, -6536382, -18199284, -50852019, -142547559, -400763223, -1129760415, -3192727797, -9043402501, -25669818476

Offset: 0

Paul Barry, Nov 17 2009

A signed variant of the Motzkin numbers A001006. Hankel transform is A168052.

			G.f. = 1 - x^2 - x^3 - 2*x^4 - 4*x^5 - 9*x^6 - 21*x^7 - 51*x^8 - 127*x^9 + ...

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

Cf. A168049, A166587, A167022.

Cf. A001006, A005043.

a[ n_] := SeriesCoefficient[ (1 + x + Sqrt[1 - 2 x - 3 x^2]) / 2, {x, 0, n}] (* Michael Somos, Jan 25 2014 *)

{a(n) = polcoeff( (1 + x + sqrt(1 - 2*x - 3*x^2 + x * O(x^n))) / 2, n)}; /* Michael Somos, Jan 25 2014 */

D-finite with recurrence: n*a(n) -(2n-3)*a(n-1) -3*(n-3)*a(n-2)=0 if n>2. - R. J. Mathar, Dec 20 2011 [Edited by Michael Somos, Jan 25 2014]
0 = a(n) * (9*a(n+1) + 15*a(n+2) - 12*a(n+3)) + a(n+1) * (-3*a(n+1) + 10*a(n+2) - 5*a(n+3)) + a(n+2) * (a(n+2) + a(n+3)) if n>0. - Michael Somos, Jan 25 2014
G.f.: 1 + x - (x + x^2) / (1 + x - (x + x^2) / (1 + x - ...)). - Michael Somos, Mar 27 2014
Convolution inverse of A005043. - Michael Somos, Mar 27 2014
a(n) ~ -3^(n - 1/2) / (2 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jun 05 2018
From Gennady Eremin, Feb 25 2021: (Start)
For n > 1, a(n) = A167022(n) / 2.
G.f.: (1 + x + A(x)) / 2, where A(x) is the g.f. of A167022. (End)

A295112 a(n) = Sum_{k=0..n} binomial(n,2k)binomial(2k,k)/(2k-1).

Original entry on oeis.org

-1, -1, 1, 5, 13, 29, 63, 139, 317, 749, 1827, 4575, 11699, 30419, 80161, 213573, 574253, 1556077, 4244835, 11647151, 32122231, 88995879, 247573565, 691246369, 1936445619, 5441165699, 15331341373, 43308322049, 122624939677, 347957102909, 989335822559, 2818200111867

Offset: 0

Zhi-Wei Sun, Nov 14 2017

As binomial(2*k,k) = 2*(2*k-1)*A000108(k-1) for all k = 1,2,..., we see that a(n) is always an odd integer. Clearly, a(n) > 0 for all n > 1. a(n) can be viewed as an analog of Motzkin numbers, which should have some combinatorial interpretations.
Conjecture: The sequence a(n+1)/a(n) (n = 5,6,...) is strictly increasing with limit 3, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 9,10,...) is strictly decreasing to the limit 1.
See also A295113 for a conjecture involving the current sequence.

			a(3) = 5 since binomial(3,2*0)*binomial(2*0,0)/(2*0-1) + binomial(3,2*1)*binomial(2*1,1)/(2*1-1) = -1 + 3*2 = 5.

Zhi-Wei Sun, Table of n, a(n) for n = 0..400
Min Bian, Olivia X. M. Yao, Yan Zhang, and Alina F. Y. Zhao, Proof of a conjecture of Z. W. Sun, Contributions to Discrete Mathematics (2019) Vol. 14, No. 1, 214-221.
Zhi-Wei Sun, Conjectures involving arithmetical sequences, arXiv:1208.2683 [math.CO], 2012-2013; in: Number Theory: Arithmetic in Shangri-La (eds., S. Kanemitsu, H. Li and J. Liu), Proc. 6th China-Japan Seminar (Shanghai, August 15-17, 2011), World Sci., Singapore, 2013, pp. 244-258.

Cf. A000108, A001006, A167022, A295113.

a := n -> -hypergeom([-1/2, 1/2 - n/2, -n/2], [1/2, 1], 4):
seq(simplify(a(n)), n=0..31); # Peter Luschny, Nov 15 2017

W[n_]:=W[n]=Sum[Binomial[n,2k]Binomial[2k,k]/(2k-1),{k,0,n/2}]; Table[W[n],{n,0,35}]
a[n_] := -AppellF1[-n, -1/2, -1/2, 1, 2, -2]; Table[a[n], {n,0,31}] (* Peter Luschny, Nov 15 2017 *)

Via the Zeilberger algorithm we have the recurrence: (n+3)*a(n+3) = (3n+7)*a(n+2) + (n-5)*a(n+1) - 3*(n+1)*a(n) for any nonnegative integer n.
a(n) = -hypergeom([-1/2, 1/2 - n/2, -n/2], [1/2, 1], 4). - Peter Luschny, Nov 15 2017
a(n) ~ 3^(n + 3/2) / (4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Nov 15 2017
From Mélika Tebni, Sep 03 2025: (Start)
G.f.: -sqrt((1 + x)*(1 - 3*x)) / (1 - x)^2.
a(n) = -Sum_{k=0..n} Sum_{j=0..k} A167022(j). (End)

A168058 Expansion of x + sqrt(1-2x-3x^2).

Original entry on oeis.org

1, 0, -2, -2, -4, -8, -18, -42, -102, -254, -646, -1670, -4376, -11596, -31022, -83670, -227268, -621144, -1706934, -4713558, -13072764, -36398568, -101704038, -285095118, -801526446, -2259520830, -6385455594, -18086805002

Offset: 0

Paul Barry, Nov 17 2009

a(n+2) = -2*A001006(n). Hankel transform is (-1)^n*A168057(n).

Essentially the same as A167022. - R. J. Mathar, Nov 18 2009

			1 - 2*x^2 - 2*x^3 - 4*x^4 - 8*x^5 - 18*x^6 - 42*x^7 - 102*x^8 - 254*x^9 - ...

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

Cf. A168055.

CoefficientList[Series[x + Sqrt[1 - 2 x - 3 x^2], {x, 0, 50}], x] (* G. C. Greubel, Jul 08 2016 *)

a(n) = 0^n - 2*Sum_{k=0..floor((n-2)/2)} C(n-2,2k)*A000108(k).

D-finite with recurrence: n*a(n) +(-2*n+3)*a(n-1) +3*(-n+3)*a(n-2)=0. - R. J. Mathar, Jan 23 2020

A308848 Expansion of e.g.f. exp(-x) / BesselI(0,2*x).

Original entry on oeis.org

1, -1, -1, 5, 7, -71, -139, 2071, 5335, -103207, -331511, 7853251, 30256381, -847377805, -3808492297, 123081031165, 632196102455, -23155450005175, -133802756269735, 5477371955388355, 35167483918412257, -1591161899246627297, -11237664710770159597, 556875003328690925825, 4290500676272573740429

Offset: 0

Ilya Gutkovskiy, Jun 28 2019

sign

E.g.f. is inverse of e.g.f. for A002426 (central trinomial coefficients).

Cf. A002426, A167022, A308847.

nmax = 24; CoefficientList[Series[Exp[-x]/BesselI[0, 2 x], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = -Sum[Binomial[n, k] 3^k Hypergeometric2F1[1/2, -k, 1, 4/3] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 24}]

my(x='x+O('x^30)); Vec(serlaplace(exp(-x) / besseli(0,2*x))) \\ Michel Marcus, Jul 02 2019

E.g.f.: 1 / Sum_{k>=0} A002426(k)*x^k/k!.

cp's OEIS Frontend

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

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A168051 Expansion of (1+x+sqrt(1-2x-3x^2))/2.

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A295112 a(n) = Sum_{k=0..n} binomial(n,2*k)*binomial(2*k,k)/(2*k-1).

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A168058 Expansion of x + sqrt(1-2x-3x^2).

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A308848 Expansion of e.g.f. exp(-x) / BesselI(0,2*x).

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A295112 a(n) = Sum_{k=0..n} binomial(n,2k)binomial(2k,k)/(2k-1).