OFFSET
1,5
COMMENTS
Equivalently, T(n,k) = (1-5*(n-k))*T(n-1,k-1) + (6-5*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} = -(5*n-12) for n >= 4, hence S_n = S_3*Product_{i=4..n} (-(5*i-12)), with S_1 = 1, S_2 = 2, S_3 = -6.
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, R_n(-1) is nonzero; first values (at n = 3, 5, 7, 9): 10, -200, 40000, -15200000.
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 = -5) pairs with the m = 3 case (A142458), sharing |m+1| = 4.
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 = -5.
Row sums: S_1 = 1, S_2 = 2, S_3 = -6, and for n >= 4: S_n/S_{n-1} = -(5*n-12).
EXAMPLE
Triangle begins:
n=1: 1;
n=2: 1, 1;
n=3: 1, -8, 1;
n=4: 1, 23, 23, 1;
n=5: 1, -106, -414, -106, 1;
n=6: 1, 405, 5210, 5210, 405, 1;
n=7: 1, -1644, -54585, -145880, -54585, -1644, 1;
n=8: 1, 6547, 530721, 3079435, 3079435, 530721, 6547, 1;
n=9: 1, -26222, -4966352, -55849394, -117018530, -55849394, -4966352, -26222, 1;
n=10: 1, 104849, 45588716, 925915724, 3563737526, 3563737526, 925915724, 45588716, 104849, 1;
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Xin Chen and Yudian Chen, Feb 01 2026
STATUS
approved
