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A068763 Irregular triangle of the Fibonacci polynomials of A011973 multiplied diagonally by the Catalan numbers.

%I A068763 #51 Feb 26 2024 11:01:01
%S A068763 1,1,1,2,2,5,6,1,14,20,6,42,70,30,2,132,252,140,20,429,924,630,140,5,
%T A068763 1430,3432,2772,840,70,4862,12870,12012,4620,630,14,16796,48620,51480,
%U A068763 24024,4620,252,58786,184756,218790
%N A068763 Irregular triangle of the Fibonacci polynomials of A011973 multiplied diagonally by the Catalan numbers.
%C A068763 The row length sequence of this array is [1,2,2,3,3,4,4,5,5,...] = A008619(n+2), n>=0.
%C A068763 The row polynomials p(n,x) := Sum_{m=0..floor((n+1)/2)} a(n,m)*x^m produce, for x = (b-a^2)/a^2 (not 0), the two parameter family of sequences K(a,b; n) := (a^(n+1))*p(n,(b-a^2)/a^2) with g.f. K(a,b; x) := (1-sqrt(1-4*x*(a+x*(b-a^2))))/(2*x).
%C A068763 Some members are: K(1,1; n)=A000108(n) (Catalan), K(1,2; n)=A025227(n-1), K(2,1; n)=A025228(n-1), K(1,3; n)=A025229(n-1), K(3,1; n)=A025230(n-1). For a=b=2..10 the sequences K(a,a; n)/a are A068764-A068772.
%C A068763 The column sequences (without leading 0's) are: A000108 (Catalan), A000984 (central binomial), A002457, 2*A002802, 5*A020918, 14*A020920, 42*A020922, ...
%C A068763 a(n,m) is the number of ways to designate exactly m cherries over all binary trees with n internal nodes.  A cherry is an internal node whose descendants are both external nodes.  Cf. A091894 which gives the number of binary trees with m cherries. - _Geoffrey Critzer_, Jul 24 2020
%C A068763 This irregular triangle is essentially that of A011973 with its diagonals multiplied by the Catalan numbers of A000108. The diagonals of this triangle are then rows of the Pascal matrix A007318 multiplied by the Catalan numbers. - _Tom Copeland_, Dec 23 2023
%H A068763 P. Flajolet and R. Sedgewick, <a href="http://algo.inria.fr/flajolet/Publications/books.html">Analytic Combinatorics</a>, 2009; see page 213.
%H A068763 Tad White, <a href="https://arxiv.org/abs/2401.01462">Quota Trees</a>, arXiv:2401.01462 [math.CO], 2024. See p. 20.
%F A068763 a(n, m) = binomial(n+1-m, m)*C(n-m) if 0 <= m <= floor((n+1)/2), otherwise 0, with C(n) := A000108(n) (Catalan).
%F A068763 G.f. for column m=1, 2, ...: (x^(2*m-1))*C(m-1)/(1-4*x)^((2*m-1)/2); m=0: c(x), g.f. for A000108 (Catalan).
%F A068763 G.f. for row polynomials p(n, x): c(z) + x*z*c(x*(z^2)/(1-4*z))/sqrt(1-4*z) = (1-sqrt(1-4*z*(1+x*z)))/(2*z), where c(x) is the g.f. of A000108 (Catalan).
%F A068763 G.f. for triangle: (1 - sqrt(1 - 4*x (1 + y*x)))/(2*x). - _Geoffrey Critzer_, Jul 24 2020
%F A068763 The alternating row sums are {1,repeat 0} = {A000007(n)}_{n>=0}. From the second comment p(n, x=-1) = K(1,0; n), with g.f. K(1,0; x) = 1. Or from the g.f. of the triangle with y = -1. Thanks to Aaron Li for asking for a proof. - _Wolfdieter Lang_, Jan 21 2023
%F A068763 The series expansion of f(x) = (1 + 2sx - sqrt(1 + 4sx + 4d^2x^2))/(2x) at x = 0 is (s^2 - d^2) x + (2 d^2s - 2 s^3) x^2 + (d^4 - 6 d^2 s^2 + 5 s^4) x^3  + (-6 d^4 s + 20 d^2 s^3 - 14 s^5) x^4 + ..., containing the coefficients of this array. With s = (a+b)/2 and d = (a-b)/2, then f(x)/ab = g(x) = (1 + (a+b)x - sqrt((1+(a+b)x)^2 - 4abx^2))/(2abx) = x - (a + b) x^2 + (a^2 + 3 a b + b^2) x^3 - (a^3 + 6 a^2 b + 6 a b^2 + b^3) x^4 + ..., containing the Narayana polynomials of A001263, which can be simply transformed into A033282. The compositional inverse about the origin of g(x) is g^(-1)(x) = x/((1-ax)(1-bx)) = x/((1-(s+d)x)(1-(s-d)x)) = x + (a + b) x^2 + (a^2 + a b + b^2) x^3 + (a^3 + a^2 b + a b^2 + b^3) x^4 + ..., containing the complete homogeneous symmetric polynomials h_n(a,b) = (a^n - b^n)/(a-b), which are the polynomials of A034867 when expressed in s and d, e.g., ((s + d)^7 - (s - d)^7)/(2 d) = d^6 + 21 d^4 s^2 + 35 d^2 s^4 + 7 s^6. A133437 and A134264 for compositional inversion of o.g.f.s can be used to relate the sets of polynomials above. - _Tom Copeland_, Nov 28 2023
%e A068763 The irregular triangle begins:
%e A068763    n\m    0     1     2     3    4   5
%e A068763    0:     1
%e A068763    1:     1     1
%e A068763    2:     2     2
%e A068763    3:     5     6     1
%e A068763    4:    14    20     6
%e A068763    5:    42    70    30     2
%e A068763    6:   132   252   140    20
%e A068763    7:   429   924   630   140    5
%e A068763    8:  1430  3432  2772   840   70
%e A068763    9:  4862 12870 12012  4620  630  14
%e A068763   10: 16796 48620 51480 24024 4620 252
%e A068763   ...
%e A068763 p(3,x) = 5 + 6*x + x^2.
%t A068763 nn = 10; b[z_] := (1 - Sqrt[1 - 4 z])/(2 z);Map[Select[#, # > 0 &] &,
%t A068763 CoefficientList[Series[v b[v z] /. v -> (1 + u z ), {z, 0, nn}], {z, u}]] // Grid (* _Geoffrey Critzer_, Jul 24 2020 *)
%Y A068763 Cf. A025227(n-1) (row sums).
%Y A068763 Cf. A000007(n) (alternating row sums).
%Y A068763 Cf. A000108, A001263, A007318, A011973, A033832, A034867, A133437 and A134264.
%K A068763 nonn,easy,tabf
%O A068763 0,4
%A A068763 _Wolfdieter Lang_, Mar 04 2002
%E A068763 Title changed by _Tom Copeland_, Dec 23 2023