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A111184
Triangle T(n,k), 0<=k<=n, read by rows, given by [0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938.
2
1, 0, 1, 0, 2, 1, 0, 6, 6, 1, 0, 24, 34, 12, 1, 0, 120, 210, 110, 20, 1, 0, 720, 1452, 974, 270, 30, 1, 0, 5040, 11256, 8946, 3248, 560, 42, 1, 0, 40320, 97296, 87504, 38338, 8792, 1036, 56, 1
OFFSET
0,5
LINKS
Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.
R. Cori, Indecomposable permutations, hypermaps and labeled Dyck paths, J. Comb. Theory A 116 (2009) 1326-1343, end of Section 1.2.2.
FORMULA
O.g.f. satisfies: A(x,y) = (1 + x^2*A'(x,y)) / (1+x - x*y - x*A(x,y)), where A'(x,y) = d/dx A(x,y). - Paul D. Hanna, Jul 31 2011
O.g.f. satisfies: A(x,y) = 1 - x * d/dx log(1+x - x*y - x*A(x,y)). - Paul D. Hanna, Jul 30 2011
Sum_{k=0..n} T(n, k) = A003319(n+1).
Sum_{k=0..n} T(n, k)*2^(n-k) = A004208(n).
From Mikhail Kurkov, Jul 15 2025: (Start)
Conjecture 1: Sum_{k=1..n} T(n,k)*q^(k-1) = A111528(q,n) for n > 0, q >= 0.
Conjecture 2: Sum_{k=1..n} T(n,k)*(-1/q)^(k-1) = R(n,q)/q^(n-1) for n > 0, q > 0 where log(1 + x + q*x*[Sum_{k>=1} R(k,q)*x^k]) = Sum_{k>=1} R(k,q)/k*x^k.
Conjecture 3: n-th row polynomial is x*v_n for n > 0 where we start with vector v of fixed length m with elements v_i = 1, and for i=1..m-1, j=i+1..m apply A := v_i (at the beginning of each cycle for i) and also apply A := A + v_j, v_j := (j-i+x-1)*v_j + A.
Here last conjecture provides fast and simple algorithm, that allow to compute sums in the previous conjectures by substituting x = q and x = -1/q respectively. (End)
EXAMPLE
Rows begin:
1;
0, 1;
0, 2, 1;
0, 6, 6, 1;
0, 24, 34, 12, 1;
0, 120, 210, 110, 20, 1;
0, 720, 1452, 974, 270, 30, 1;
0, 5040, 11256, 8946, 3248, 560, 42, 1;
0, 40320, 97296, 87504, 38338, 8792, 1036, 56, 1;
...
MATHEMATICA
DELTA[r_, s_, m_] := Module[{p, q, t, x, y}, q[k_] := x r[[k+1]] + y s[[k+1]]; p[0, _] = 1; p[_, -1] = 0; p[n_ /; n >= 1, k_ /; k >= 0] := p[n, k] = p[n, k-1] + q[k] p[n-1, k+1] // Expand; t[n_, k_] := Coefficient[p[n, 0], x^(n-k) y^k]; t[0, 0] = p[0, 0]; Table[t[n, k], {n, 0, m}, {k, 0, n}]];
DELTA[LinearRecurrence[{1, 1, -1}, {0, 2, 1}, 10], Mod[Range[10], 2], 10] // Flatten (* Jean-François Alcover, Jul 27 2018 *)
PROG
(PARI) {T(n, k)=local(A=1+x*y); for(i=1, n, A=1-x*deriv(log(1+x-x*y-x*A +x*O(x^n)))); polcoeff(polcoeff(A, n, x), k, y)} /* Paul D. Hanna */
(PARI) {T(n, k)=local(A=1+x*y); for(i=1, n, A=(1 + x^2*A')/(1 + x - x*y - x*A +x*O(x^n))); polcoeff(polcoeff(A, n, x), k, y)} /* Paul D. Hanna */
/* Print 10 Rows of the triangle: */
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
STATUS
approved