%I
%S 1,1,2,4,7,12,23,46,89,168,311,594,1194,2355,4570,8745,16532,32948,
%T 65761,129632,252697,487647,936785,1884892,3754166,7407451,14489982,
%U 28118751,54868937,110096666,219129673,432847116,848952949,1654022768,3256427202,6524228863,12983131874,25671612977,50454577444
%N List of largest row numbers of Pascallike triangles with index of asymmetry y = 1 and index of obliqueness z = 0 or z = 1.
%C Triangle with index of asymmetry y = 1 and index of obliqueness z = 0, read by rows, with recurrence for G(n, k) as follows: G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, k) = G(n+1, k1) + G(n+1, k) + G(n+2, k) for k = 1..(n+1).
%C Triangle with index of asymmetry y = 1 and index of obliqueness z = 1, read by rows, with recurrence for G(n, k) as follows: G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, k) = G(n+1, k1) + G(n+1, k2) + G(n+2, k1) for k = 2..(n+2). [Edited by _Petros Hadjicostas_, Jun 11 2019]
%C From _Petros Hadjicostas_, Jun 10 2019: (Start)
%C For the triangle with index of asymmetry y = 1 and index of obliqueness z = 0, read by rows, we have G(n, k) = A140998(n, k) for 0 <= k <= n.
%C For the triangle with index of asymmetry y = 1 and index of obliqueness z = 1, read by rows, we have G(n, k) = A140993(n+1, k+1) for n >= 0 and k >= 0.
%C Thus, except for the (unfortunate) shifting of the indices by 1, triangular arrays A140998 and A140993 are mirror images of each other.
%C (End)
%F a(n) = max(A140993(n,k), k = 1..n).  _R. J. Mathar_, Apr 28 2010
%F a(n) = max(A140998(n1, k1), k = 1..n).  _Petros Hadjicostas_, Jun 10 2019
%e Triangle with y = 1 and z = 0 (i.e., triangle A140998) begins as follows:
%e a(1) = max(1) = 1;
%e a(2) = max(1, 1) = 1;
%e a(3) = max(1, 2, 1) = 2;
%e a(4) = max(1, 4, 2, 1) = 4;
%e a(5) = max(1, 7, 5, 2, 1) = 7;
%e a(6) = max(1, 12, 11, 5, 2, 1) = 12;
%e a(7) = max(1, 20, 23, 12, 5, 2, 1) = 23;
%e a(8) = max(1, 33, 46, 28, 12, 5, 2, 1) = 46;
%e a(9) = max(1, 54, 89, 63, 29, 12, 5, 2, 1) = 89;
%e ...
%p # Here, BB is the bivariate g.f. of sequence A140993.
%p BB := proc(x, y) y*x*(1  y*x  x^2*y^2 + x^3*y^2)/((1  x)*(1  y*x)*(1  y*x  x^2*y  x^2*y^2)); end proc;
%p #
%p # Here, we find the nth row of sequence A140993 and find the maximum of the row:
%p ff := proc(n) local xx, k, yy;
%p xx := 0;
%p for k from 1 to n do
%p yy := coeftayl(coeftayl(BB(x, y), x = 0, n), y = 0, k);
%p xx := max(xx, yy); end do; xx;
%p end proc;
%p #
%p # Here, we print the maxima of the rows:
%p for i from 1 to 40 do
%p ff(i);
%p end do; # _Petros Hadjicostas_, Jun 10 2019
%Y Cf. A007318, A140993, A140998.
%K nonn,changed
%O 1,3
%A _JuriStepan Gerasimov_, Jul 11 2008
%E a(4) and offset corrected by _Gary W. Adamson_, Jul 11 2008
%E More terms from _R. J. Mathar_, Apr 28 2010
%E Name edited and more terms by _Petros Hadjicostas_, Jun 10 2019
