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Triangle read by rows: the reversed x = 1+q Narayana triangle at m=2.
4

%I #52 Apr 25 2023 10:14:47

%S 1,3,1,12,8,1,55,55,15,1,273,364,156,24,1,1428,2380,1400,350,35,1,

%T 7752,15504,11628,4080,680,48,1,43263,100947,92169,41895,9975,1197,63,

%U 1,246675,657800,708400,396704,123970,21560,1960,80,1,1430715,4292145,5328180,3552120,1381380,318780,42504,3036,99,1

%N Triangle read by rows: the reversed x = 1+q Narayana triangle at m=2.

%C See Novelli-Thibon (2014) for precise definition.

%C From _Tom Copeland_, Dec 13 2022: (Start)

%C The row polynomials are the nonvanishing numerator polynomials generated in the compositional, or Lagrange, inversion in x about the origin of the odd o.g.f. Od1(x,t) = x*(t*(1-x^2)-x^2) / (1-x^2) = t*x - x^3 - x^5 - x^7 - x^9 - ... .

%C For example, from the Lagrange inversion formula (LIF), the tenth derivative in x of (x/Od1(x,t))^11 / 11! = (1/((t*(1-x^2)-x^2) / (1-x^2)))^11 / 11! at x = 0 is (t^4 + 24*t^3 + 156*t^2 + 364*t + 273) / t^16. These polynomials are also generated by the iterated derivatives ((1/(D Od1(x,t)) D)^n g(x) evaluated at x = 0 where D = d/dx.

%C An explicit generating function for the polynomials can be obtained by finding the solution of the cubic equation y - t*x - y*x^2 + (1+t)*x^3 = 0 for x in terms of y and t that satisfies y(x=0;t) = 0 = x(y=0;t).

%C The row polynomials are also the polynomials generated in the compositional inverse of O(x,t) = x / (1+(1+t)x)*(1+x)^2) = x + (-t - 3)*x^2 + (t^2 + 4 t + 6)*x^3 + (-t^3 - 5*t^2 - 10*t - 10)*x^4 + ..., containing the truncated Pascal polynomials of A104712 / A325000.

%C For example, from the LIF, the third derivative of ((1 + (1+t)*x)*(1+x)^2)^4 / 4! at x = 0 is 55 + 55*t + 15*t^2 + t^3.

%C A natural refinement of this array was provided in a letter by Isaac Newton in 1676--a set of partition polynomials for generating the o.g.f. of the compositional inverse of the generic odd o.g.f. x + u_1 x^3 + u_2 x^5 + ... in the infinite set of indeterminates u_n. (End)

%C T(n,k) is the number of noncrossing cacti with n+1 nodes and n+1-k blocks. See A361242. - _Andrew Howroyd_, Apr 13 2023

%H Michael De Vlieger, <a href="/A243662/b243662.txt">Table of n, a(n) for n = 1..11325</a> (rows 1 <= n <= 150, flattened)

%H Paul Barry, <a href="https://arxiv.org/abs/2101.06713">On the inversion of Riordan arrays</a>, arXiv:2101.06713 [math.CO], 2021.

%H J.-C. Novelli and J.-Y. Thibon, <a href="http://arxiv.org/abs/1403.5962">Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions</a>, arXiv preprint arXiv:1403.5962 [math.CO], 2014. See Fig. 10.

%F T(n,k) = (binomial(3*n+1,n) * binomial(n,k-1) * binomial(n-1,k-1)) / (binomial(3*n,k-1) * (3*n+1)) = (A001764(n) * A001263(n,k) * k) / binomial(3*n,k-1) for 1 <= k <= n (conjectured). - _Werner Schulte_, Nov 22 2018

%F T(n,k) = binomial(3*n+1-k,n-k) * binomial(n,k-1) / n for 1 <= k <= n, more generally: T_m(n,k) = binomial((m+1)*n+1-k,n-k) * binomial(n,k-1) / n for 1 <= k <= n and some fixed integer m > 1. - _Werner Schulte_, Nov 22 2018

%F G.f.: A(x,y) is the series reversion of x/((1 + x + x*y)*(1 + x)^2). - _Andrew Howroyd_, Apr 13 2023

%e Triangle begins:

%e 1;

%e 3, 1;

%e 12, 8, 1;

%e 55, 55, 15, 1;

%e 273, 364, 156, 24, 1;

%e 1428, 2380, 1400, 350, 35, 1;

%e ...

%t T[m_][n_, k_] := Binomial[(m + 1) n + 1 - k, n - k] Binomial[n, k - 1]/n;

%t Table[T[2][n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Feb 12 2019 *)

%o (PARI)

%o T(n)=[Vecrev(p) | p<-Vec(serreverse(x/((1+x+x*y)*(1+x)^2) + O(x*x^n)))]

%o { my(A=T(10)); for(i=1, #A, print(A[i])) } \\ _Andrew Howroyd_, Apr 13 2023

%Y Cf. A001764, A001263, A243663 (m=3).

%Y Row sums give A003168.

%Y Row reversed triangle is A102537.

%K nonn,tabl

%O 1,2

%A _N. J. A. Sloane_, Jun 13 2014

%E Data and Example (T(2,2) and T(5,3)) corrected and more terms added by _Werner Schulte_, Nov 22 2018