OFFSET
1,5
COMMENTS
Equivalently, T(n,k) = (1-3*(n-k))*T(n-1,k-1) + (4-3*k)*T(n-1,k) for 1 < k < n.
The triangle is symmetric: T(n,k) = T(n,n+1-k).
Row sums S_n = Sum_{k=1..n} T(n,k) satisfy the exact ratio S_n/S_{n-1} = -(3*n-8) for n >= 4, hence S_n = S_3*Product_{i=4..n} (-(3*i-8)), with S_1 = 1, S_2 = 2, S_3 = -2.
The row polynomial is R_n(y) = Sum_{k=1..n} T(n,k)*y^(k-1).
For even n, R_n(-1) = 0, i.e., Sum_{k=1..n} (-1)^(k-1)*T(n,k) = 0, so odd- and even-indexed entries contribute equally to the row sum.
For odd n >= 3, R_n(-1) takes the values 6, 0, 1296, -155520, 35271936, ... (at n = 3, 5, 7, 9, 11).
Conjecture: T(n,k) >= 0 for all even n and all 1 <= k <= n. Verified for even n <= 400. Since all entries are nonneg for even n, the roots of R_n(y) lie in the annulus determined by consecutive coefficient ratios (Enestrom-Kakeya theorem).
Under the parameter map m -> -m-2, this triangle (m = -3) pairs with the m = 1 case (classical Eulerian numbers, A008292), sharing |m+1| = 2.
LINKS
Xin Chen and Yudian Chen, Supporting data, figures, and reproducible code for a parametric Pascal triangle family (m = -3, -4, -5), 2026.
FORMULA
T(n,1) = T(n,n) = 1 for n >= 1.
For 1 < k < n: T(n,k) = (m*(n-k)+1)*T(n-1,k-1) + (m*(k-1)+1)*T(n-1,k) with m = -3.
Row sums: S_1 = 1, S_2 = 2, S_3 = -2, and for n >= 4: S_n/S_{n-1} = -(3*n-8).
EXAMPLE
Triangle begins:
n=1: 1;
n=2: 1, 1;
n=3: 1, -4, 1;
n=4: 1, 3, 3, 1;
n=5: 1, -14, -30, -14, 1;
n=6: 1, 17, 262, 262, 17, 1;
n=7: 1, -48, -1497, -4192, -1497, -48, 1;
n=8: 1, 79, 8157, 50003, 50003, 8157, 79, 1;
n=9: 1, -178, -42128, -514222, -1100066, -514222, -42128, -178, 1;
n=10: 1, 333, 214200, 4829952, 19299834, 19299834, 4829952, 214200, 333, 1;
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Xin Chen and Yudian Chen, Jan 18 2026
STATUS
approved
