%I
%S 1,1,2,1,17,26,2,62,192,60,1,1382,7192,5097,502,2,21844,171511,217186,
%T 55196,2036,2,929569,10262046,20376780,9893440,1089330,16356,4,
%U 6404582,94582204,271154544,215114420,48673180,2567568,16376,1
%N Triangle of polynomial coefficients related to the o.g.f.s of the RES1 polynomials.
%C In A160464 we defined the ES1 matrix by ES1[2*m1,n=1] and in A094665 it was shown that the nth term of the coefficients of matrix row ES1[12*m,n] for m >= 1 can be generated with the RES1(12*m,n) polynomials.
%C We define the o.g.f.s. of these polynomials by GFRES1(z,12*m) = sum(RES1(12*m,n)*z^(n1), n=1..infinity) for m >= 1. The general expression of the o.g.f.s. is GFRES1(z,12*m) = (1)*RE(z,12*m)/(2*p(m1)*(z1)^(m)). The p(m1), m >= 1, sequence is Gould's sequence A001316.
%C The coefficients of the RE(z,12*m) polynomials lead to the triangle given above.
%C The E(z,n) = numer(sum((1)^(n+1)*k^n*z^(k1), k=1..infinity)) polynomials with n >= 1, see the Maple algorithm, lead to the Eulerian numbers A008292.
%C Some of our results are conjectures based on numerical evidence.
%H Grzegorz Rzadkowski, M Urlinska, <a href="http://arxiv.org/abs/1612.06635">A Generalization of the Eulerian Numbers</a>, arXiv preprint arXiv:1612.06635, 2016
%e The first few rows are:
%e [1]
%e [1]
%e [2, 1]
%e [17, 26, 2]
%e [62, 192, 60, 1]
%e The first few polynomials RE(z,m) are:
%e RE(z,1) = 1
%e RE(z,3) = 1
%e RE(z,5) = 2+z
%e RE(z,7) = 17+26*z+2*z^2
%e The first few GFRES1(z,m) are:
%e GFRES1(z,1) = (1/1)*(1)/(2*(z1)^1)
%e GFRES1(z,3) = (1/2)*(1)/(2*(z1)^2)
%e GFRES1(z,5) = (1/2)*(2+z)/(2*(z1)^3)
%e GFRES1(z,7) = (1/4)*(17+26*z+2*z^2)/(2*(z1)^4)
%p nmax := 8; mmax := nmax: T(0, x) := 1: for i from 1 to nmax do dgr := degree(T(i1, x), x): for na from 0 to dgr do c(na) := coeff(T(i1, x), x, na) od: T(i1, x+1) := 0: for nb from 0 to dgr do T(i1, x+1) := T(i1, x+1) + c(nb)*(x+1)^nb od: for nc from 0 to dgr do ECGP(i1, nc+1) := coeff(T(i1, x), x, nc) od: T(i, x) := expand((2*x+1)*(x+1)*T(i1, x+1)  2*x^2*T(i1, x)) od: dgr := degree (T(nmax, x), x): kmax := nmax: for k from 1 to kmax do p := k: for m from 1 to k do E(m, k) := sum((1)^(mq)*(q^k)*binomial(k+1, mq), q=1..m) od: fx(p) := (1)^(p+1) * (sum(E(r, k)*z^(kr), r=1..k))/(z1)^(p+1): GF((2*p+1)) := sort(simplify(((1)^p* 1/2^(p+1)) * sum(ECGP(k1, ks)*fx(ks), s=0..k1)), ascending): NUMGF((2*p+1)) := numer(GF((2*p+1))): for n from 1 to mmax+1 do A(k+1, n) := coeff(NUMGF((2*p+1)), z, n1) od: od: for m from 2 to mmax do A(1, m) := 0 od: A(1, 1) := 1: FT(1) := 1: for n from 1 to nmax do for m from 1 to n do FT((n)*(n1)/2+m+1) := A(n+1, m) end do end do: a := n> FT(n): seq(a(n), n = 1..(nmax+1)*(nmax)/2+1);
%t T[ n_, k_] := Coefficient[a[2 n]/2^IntegerExponent[(2 n)!, 2], x, n + k];
%t a[0] = a[1] = 1; a[ m_] := a[m] = With[{n = m  1}, x Sum[ a[k] a[n  k] Binomial[n, k], {k, 0, n}]]; Join[{1}, Flatten@Table[T[n, k], {n, 1, 8}, {k, 0, n  1}]] (* _Michael Somos_, Apr 22 2020 *)
%Y Cf. A160464, A094665 and A083061.
%Y For the Eulerian numbers E(n, k) see A008292.
%Y The p(n) sequence equals Gould's sequence A001316.
%Y The first right hand column of the triangle equals A048896.
%Y The first left hand column equals A160469.
%Y The row sums equal the absolute values of A117972.
%K easy,nonn,tabf
%O 1,3
%A _Johannes W. Meijer_, May 24 2009
%E Edited by _Johannes W. Meijer_, Sep 23 2012
