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A355346
G.f.: A(x,y) = Sum_{n=-oo..+oo} (x*y)^(n*(n+1)/2) * C(x)^(2*n-1), where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
1
2, -4, 2, -1, -4, 0, -3, 7, 0, 2, -8, 5, 0, -4, 0, -23, 14, 0, 23, 0, 0, -70, 41, 0, 21, 0, 0, 2, -222, 127, 0, 90, 0, 0, -4, 0, -726, 409, 0, 297, 0, 0, 47, 0, 0, -2431, 1355, 0, 1001, 0, 0, 45, 0, 0, 0, -8294, 4587, 0, 3431, 0, 0, 284, 0, 0, 0, 2, -28730, 15795, 0, 11927, 0, 0, 1001, 0, 0, 0, -4, 0, -100776, 55146, 0, 41955, 0, 0, 3640, 0, 0, 0, 79, 0, 0, -357238, 194752, 0, 149072, 0, 0, 13260, 0, 0, 0, 77, 0, 0, 0
OFFSET
0,1
FORMULA
G.f. A(x,y) = Sum_{n>=0} x^n * Sum_{k=0..n} T(n,k) * y^k may be obtained from the following expressions; here, C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
(1) A(x,y) = Sum_{n=-oo..+oo} (x*y)^(n*(n+1)/2) * C(x)^(2*n-1).
(2) A(x,y) = Sum_{n>=0} (x*y)^(n*(n+1)/2) * (C(x)^(2*n-1) + 1/C(x)^(2*n+3)).
(3) A(x,y) = 1/C(x)^3 * Product_{n>=1} (1 + (x*y)^(n-1)*C(x)^2) * (1 + (x*y)^n/C(x)^2) * (1-(x*y)^n), by the Jacobi triple product identity.
EXAMPLE
G.f.: A(x,y) = 2 + (2*y - 4)*x + (-4*y - 1)*x^2 + (2*y^3 + 7*y - 3)*x^3 + (-4*y^3 + 5*y - 8)*x^4 + (23*y^3 + 14*y - 23)*x^5 + (2*y^6 + 21*y^3 + 41*y - 70)*x^6 + (-4*y^6 + 90*y^3 + 127*y - 222)*x^7 + (47*y^6 + 297*y^3 + 409*y - 726)*x^8 + (45*y^6 + 1001*y^3 + 1355*y - 2431)*x^9 + (2*y^10 + 284*y^6 + 3431*y^3 + 4587*y - 8294)*x^10 + ...
such that
A(x,y) = ... + (x*y)^6/C(x)^9 + (x*y)^3/C(x)^7 + (x*y)/C(x)^5 + 1/C(x)^3 + 1/C(x) + (x*y)*C(x) + (x*y)^3*C(x)^3 + (x*y)^6*C(x)^5 + (x*y)^10*C(x)^7 + (x*y)^15*C(x)^9 + ... + (x*y)^(n*(n+1)/2) * C(x)^(2*n-1) + ...
also
A(x,y) = 1/C(x)^3 * (1 + C(x)^2)*(1 + x*y/C(x)^2)*(1-x) * (1 + x*y*C(x)^2)*(1 + (x*y)^2/C(x)^2)*(1-x^2) * (1 + (x*y)^2*C(x)^2)*(1 + (x*y)^3/C(x)^2)*(1-(x*y)^3) * (1 + (x*y)^3*C(x)^2)*(1 + (x*y)^4/C(x)^2)*(1-(x*y)^4) * ... * (1 + (x*y)^(n-1)*C(x)^2)*(1 + (x*y)^n/C(x)^2)*(1-(x*y)^n) * ...
where C(x) = 1 + x*C(x)^2 begins
C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + 1430*x^8 + ... + A000108(n)*x^n + ...
This triangle of coefficients T(n,k) of x^n*y^k in A(x,y), for k = 0..n in row n, n >= 0, begins:
2;
-4, 2;
-1, -4, 0;
-3, 7, 0, 2;
-8, 5, 0, -4, 0;
-23, 14, 0, 23, 0, 0;
-70, 41, 0, 21, 0, 0, 2;
-222, 127, 0, 90, 0, 0, -4, 0;
-726, 409, 0, 297, 0, 0, 47, 0, 0;
-2431, 1355, 0, 1001, 0, 0, 45, 0, 0, 0;
-8294, 4587, 0, 3431, 0, 0, 284, 0, 0, 0, 2;
-28730, 15795, 0, 11927, 0, 0, 1001, 0, 0, 0, -4, 0;
-100776, 55146, 0, 41955, 0, 0, 3640, 0, 0, 0, 79, 0, 0;
-357238, 194752, 0, 149072, 0, 0, 13260, 0, 0, 0, 77, 0, 0, 0;
-1277788, 694450, 0, 534251, 0, 0, 48450, 0, 0, 0, 692, 0, 0, 0, 0;
-4605980, 2496790, 0, 1928992, 0, 0, 177649, 0, 0, 0, 2537, 0, 0, 0, 0, 2;
...
the row sums of which yield A355345:
[2, -2, -5, 6, -7, 14, -6, -9, 27, -30, 10, -11, 44, -77, 55, -10, -13, 65, -156, 182, -91, ...].
The row sums in turn form the antidiagonals of the rectangular table given by:
n = 0: [ 2, -5, 14, -30, 55, -91, 140, ...];
n = 1: [ -2, -7, 27, -77, 182, -378, 714, ...];
n = 2: [ 6, -9, 44, -156, 450, -1122, 2508, ...];
n = 3: [ -6, -11, 65, -275, 935, -2717, 7007, ...];
n = 4: [ 10, -13, 90, -442, 1729, -5733, 16744, ...];
n = 5: [-10, -15, 119, -665, 2940, -10948, 35700, ...];
n = 6: [ 14, -17, 152, -952, 4692, -19380, 69768, ...];
n = 7: [-14, -19, 189, -1311, 7125, -32319, 127281, ...];
...
in which row n has g.f.: (-1)^n*(2*n+1) + (1-x)/(1+x)^(2*n+4) for n >= 0.
PROG
(PARI) {T(n, k) = my(A, C = serreverse(x-x^2 +x^2*O(x^n))/x, M = sqrtint(2*n+9));
A = sum(m=0, n+2, (x*y)^(m*(m+1)/2) * (C^(2*m-1) + 1/C^(2*m+3))); polcoeff(polcoeff(A, n, x), k, y)}
for(n=0, 16, for(k=0, n, print1(T(n, k), ", ")); print(""))
CROSSREFS
Sequence in context: A079045 A021417 A368517 * A364608 A105791 A116515
KEYWORD
sign,tabl
AUTHOR
Paul D. Hanna, Jul 25 2022
STATUS
approved