OFFSET
1,6
COMMENTS
LINKS
Paul D. Hanna, Table of n, a(n) for n = 1..1275 (first 50 rows).
FORMULA
G.f. A(x,y) = Sum_{n>=1} Sum_{k=0..n} T(n,k) * x^n*y^k satisfies the following formulas.
(1) A(x,y) = A(x^3 + 3*x*y*A(x,y)^3, y) / A(x^2 + 2*x*y*A(x,y)^2, y).
(2) A(x,y=1) = C(x) = C(x^3 + 3*x*C(x)^3) / C(x^2 + 2*x*C(x)^2), where C(x) = x + C(x)^2 is the g.f. of the Catalan numbers (A000108).
(3) T(n+1,n) = A001764(n) for n >= 0, with g.f. D(x) = 1 + x*D(x)^3.
EXAMPLE
G.f. A(x,y) = x + y*x^2 + (3*y^2 - y)*x^3 + (12*y^3 - 8*y^2 + y)*x^4 + (55*y^4 - 49*y^3 + 7*y^2 + y)*x^5 + (273*y^5 - 296*y^4 + 56*y^3 + 9*y^2)*x^6 + (1428*y^6 - 1815*y^5 + 498*y^4 + 14*y^3 + 9*y^2 - 2*y)*x^7 + (7752*y^7 - 11284*y^6 + 4181*y^5 - 288*y^4 + 91*y^3 - 23*y^2)*x^8 + (43263*y^8 - 70924*y^7 + 33168*y^6 - 4487*y^5 + 522*y^4 - 108*y^3 - 3*y^2 - y)*x^9 + (246675*y^9 - 449616*y^8 + 253590*y^7 - 49239*y^6 + 3971*y^5 - 579*y^4 + 82*y^3 - 23*y^2 + y)*x^10 + ...
where A(x,y) = A(x^3 + 3*x*y*A(x,y)^3, y) / A(x^2 + 2*x*y*A(x,y)^2, y).
TRIANGLE.
Triangle of coefficients T(n,k) of x^n*y^k in A(x,y) begins
1;
0, 1;
0, -1, 3;
0, 1, -8, 12;
0, 1, 7, -49, 55;
0, 0, 9, 56, -296, 273;
0, -2, 9, 14, 498, -1815, 1428;
0, 0, -23, 91, -288, 4181, -11284, 7752;
0, -1, -3, -108, 522, -4487, 33168, -70924, 43263;
0, 1, -23, 82, -579, 3971, -49239, 253590, -449616, 246675;
0, 0, 5, -373, 2419, -6510, 46017, -478291, 1892593, -2869779, 1430715;
0, 0, -2, -65, -3746, 28523, -74367, 554792, -4334344, 13891755, -18418400, 8414640; ...
PROG
(PARI) {T(n) = my(A=[0, 1], Ax=x); for(i=1, n, A=concat(A, 0); Ax=Ser(A);
A[#A] = polcoeff( subst(Ax, x, x^3 + 3*y*x*Ax^3 ) - Ax*subst(Ax, x, x^2 + 2*y*x*Ax^2 ), #A+1)); A[n+1]}
\\ Print the rows of the triangle
my(Rown); for(n=1, 12, Rown = T(n); for(k=0, n-1, print1(polcoef(Rown, k), ", ")); print(""))
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Paul D. Hanna, Jul 14 2025
STATUS
approved