OFFSET
0,5
COMMENTS
The term T(n,k) is found in row n and column k of this triangle, and can be used to derive the following sequences.
A355351(n) = Sum_{k=0..n} T(n,k) for n >= 0 (row sums).
A355352(n) = Sum_{k=0..n} T(n,k) * 2^k for n >= 0.
A355353(n) = Sum_{k=0..n} T(n,k) * 3^k for n >= 0.
A355354(n) = Sum_{k=0..n} T(n,k) * 4^k for n >= 0.
A355355(n) = Sum_{k=0..n} T(n,k) * 5^k for n >= 0.
A355356(n) = Sum_{k=0..floor(n/2)} T(n-k,k) for n >= 0 (antidiagonal sums).
A355357(n) = Sum_{k=0..floor(n/2)} T(n-k,n-2*k) for n >= 0.
A354658(n) = T(2*n,n) for n >= 0 (central terms of this triangle).
Conjectures:
(C.1) Column 1 equals A000716, the number of partitions into parts of 3 kinds;
(C.2) Column 2 equals A023005, the number of partitions into parts of 6 kinds.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..5150
FORMULA
G.f. A(x) = Sum_{n>=0} x^n * Sum_{k=0..n} T(n,k)*y^k satisfies:
(1) x*y = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n+1)/2) * A(x,y)^n.
(2) x*y*P(x) = Product_{n>=1} (1 - x^n*A(x,y)) * (1 - x^(n-1)/A(x,y)), where P(x) = Product_{n>=1} 1/(1 - x^n) is the partition function (A000041), due to the Jacobi triple product identity.
EXAMPLE
G.f.: A(x,y) = 1 + x*y + x^2*(3*y + y^2) + x^3*(9*y + 6*y^2 + y^3) + x^4*(22*y + 27*y^2 + 10*y^3 + y^4) + x^5*(51*y + 98*y^2 + 66*y^3 + 15*y^4 + y^5) + x^6*(108*y + 315*y^2 + 340*y^3 + 135*y^4 + 21*y^5 + y^6) + x^7*(221*y + 918*y^2 + 1495*y^3 + 910*y^4 + 246*y^5 + 28*y^6 + y^7) + x^8*(429*y + 2492*y^2 + 5838*y^3 + 5070*y^4 + 2086*y^5 + 413*y^6 + 36*y^7 + y^8) + x^9*(810*y + 6372*y^2 + 20805*y^3 + 24543*y^4 + 14280*y^5 + 4284*y^6 + 652*y^7 + 45*y^8 + y^9) + x^10*(1479*y + 15525*y^2 + 68816*y^3 + 106535*y^4 + 83559*y^5 + 35168*y^6 + 8100*y^7 + 981*y^8 + 55*y^9 + y^10) + ...
where
x*y = ... - x^10/A(x,y)^5 + x^6/A(x,y)^4 - x^3/A(x,y)^3 + x/A(x,y)^2 - 1/A(x,y) + 1 - x*A(x,y) + x^3*A(x,y)^2 - x^6*A(x,y)^3 + x^10*A(x,y)^4 -+ ... + (-1)^n * x^(n*(n+1)/2) * A(x,y)^n + ...
also, given P(x) is the partition function (A000041),
x*y*P(x) = (1 - x*A(x,y))*(1 - 1/A(x,y)) * (1 - x^2*A(x,y))*(1 - x/A(x,y)) * (1 - x^3*A(x,y))*(1 - x^2/A(x,y)) * (1 - x^4*A(x,y))*(1 - x^3/A(x,y)) * ... * (1 - x^n*A(x,y))*(1 - x^(n-1)/A(x,y)) * ...
TRIANGLE.
The triangle of coefficients T(n,k) of x^n*y^k in A(x,y), for k = 0..n in row n, begins:
n=0: [1];
n=1: [0, 1];
n=2: [0, 3, 1];
n=3: [0, 9, 6, 1];
n=4: [0, 22, 27, 10, 1];
n=5: [0, 51, 98, 66, 15, 1];
n=6: [0, 108, 315, 340, 135, 21, 1];
n=7: [0, 221, 918, 1495, 910, 246, 28, 1];
n=8: [0, 429, 2492, 5838, 5070, 2086, 413, 36, 1];
n=9: [0, 810, 6372, 20805, 24543, 14280, 4284, 652, 45, 1];
n=10: [0, 1479, 15525, 68816, 106535, 83559, 35168, 8100, 981, 55, 1];
n=11: [0, 2640, 36280, 213945, 423390, 432930, 243208, 78282, 14355, 1420, 66, 1];
n=12: [0, 4599, 81816, 630890, 1563705, 2033244, 1472261, 629280, 160965, 24145, 1991, 78, 1];
...
in which column 1 appears to equal A000716, the coefficients in P(x)^3,
and column 2 appears to equal A023005, the coefficients in P(x)^6,
where P(x) is the partition function and begins
P(x) = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 7*x^5 + 11*x^6 + 15*x^7 + 22*x^8 + 30*x^9 + 42*x^10 + ... + A000041(n)*x^n + ...
Also, the power series expansions of P(x)^3 and P(x)^6 begin
P(x)^3 = 1 + 3*x + 9*x^2 + 22*x^3 + 51*x^4 + 108*x^5 + 221*x^6 + 429*x^7 + 810*x^8 + 1479*x^9 + 2640*x^10 + ... + A000716(n)*x^n + ...
P(x)^6 = 1 + 6*x + 27*x^2 + 98*x^3 + 315*x^4 + 918*x^5 + 2492*x^6 + 6372*x^7 + 15525*x^8 + 36280*x^9 + 81816*x^10 + ... + A023005(n)*x^n + ...
PROG
(PARI) {T(n, k) = my(A=[1, y], t); for(i=1, n, A=concat(A, 0); t = ceil(sqrt(2*(#A)+9));
A[#A] = -polcoeff( sum(m=-t, t, (-1)^m*x^(m*(m+1)/2)*Ser(A)^m ), #A-1)); polcoeff(A[n+1], k, y)}
for(n=0, 12, for(k=0, n, print1( T(n, k), ", ")); print(""))
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Paul D. Hanna, Jun 29 2022
STATUS
approved