%I #16 Aug 20 2021 17:46:40

%S 1,1,1,1,-2,1,1,7,3,1,1,-20,22,-4,1,1,61,90,50,5,1,1,-182,511,-260,95,

%T -6,1,1,547,2373,2331,595,161,7,1,1,-1640,12412,-13944,7686,-1176,252,

%U -8,1,1,4921,60420,110020,55230,20622,2100,372,9,1,1,-14762,307021,-709720,607090,-171612,47922,-3480,525,-10,1

%N Triangle, read by rows, defined by recurrence: T(n,k) = T(n-1,k-1) + (-1)^k * (2 * k + 1) * T(n-1,k) for 0 < k < n with initial values T(n,0) = T(n,n) = 1 for n >= 0 and T(i,j) = 0 if j < 0 or j > i.

%F G.f. of column k >= 0: col(t,k) = Sum_{n >= k} T(n,k) * t^n = t^k / (Product_{i=0..k} (1 - (-1)^i * (2 * i + 1) * t)), i.e., col(t,k) = col(t,k-1) * t / (1 - (-1)^k * (2 * k + 1) * t) for k > 0.

%F Matrix inverse M = T^(-1) has row polynomials p(n,x) = Sum_{k=0..n} M(n,k) * x^k = Product_{i=1..n} (x + (-1)^i * (2 * i - 1)) for n >= 0 and empty product 1, i.e., p(n,x) = p(n-1,x) * (x + (-1)^n * (2 * n - 1)) for n > 0 with initial value p(0,x) = 1.

%F Conjecture: E.g.f. of column k >= 0: Sum_{n >= k} T(n,k) * t^n / (n!) = (Sum_{i=0..k} (-1)^(i * (i + 1 ) / 2) * binomial(k,floor((k - i) / 2)) * exp((-1)^i * (2 * i + 1) * t)) * (-1)^(k * (k - 1) / 2) / (4^k * (k!)), i.e., T(n,k) = (Sum_{i=0..k} (-1)^(i * (i + 1) / 2) * binomial(k,floor((k - i) / 2)) * ((-1)^i * (2 * i + 1))^n) * (-1)^(k * (k - 1) / 2) / (4^k * (k!)) for 0 <= k <= n.

%F Conjecture: E.g.f. of column k >= 0: Sum_{n >= k} T(n,k) * t^n / (n!) = exp(t) * (exp(4*t) - 1)^k / (4^k * (k!) * exp(4*t*floor((k+1)/2))), i.e., T(n,k) = (Sum_{i=0..k} (-1)^i * binomial(k,i) *(1 + 4*i - 4*floor((k+1)/2))^n) * (-1)^k / (4^k * (k!)) for 0 <= k <= n. Proved by Burkhard Hackmann and Werner Schulte (distinction of two cases: odd k, even k). - _Werner Schulte_, Aug 03 2021

%e The triangle T(n,k) for 0 <= k <= n starts:

%e n\k : 0 1 2 3 4 5 6 7 8 9

%e =============================================================

%e 0 : 1

%e 1 : 1 1

%e 2 : 1 -2 1

%e 3 : 1 7 3 1

%e 4 : 1 -20 22 -4 1

%e 5 : 1 61 90 50 5 1

%e 6 : 1 -182 511 -260 95 -6 1

%e 7 : 1 547 2373 2331 595 161 7 1

%e 8 : 1 -1640 12412 -13944 7686 -1176 252 -8 1

%e 9 : 1 4921 60420 110020 55230 20622 2100 372 9 1

%e etc.

%o (Python)

%o from functools import cache

%o @cache

%o def T(n, k):

%o if k == 0 or k == n: return 1

%o return T(n-1, k-1) + (-1)**k*(2*k + 1)*T(n-1, k)

%o for n in range(10):

%o print([T(n, k) for k in range(n+1)]) # _Peter Luschny_, Jul 22 2021

%Y Cf. A000012 (column 0 and main diagonal), A014983 (column 1), A181983 (1st subdiagonal), A002412 (2nd subdiagonal), A264851 (3rd subdiagonal without signs).

%Y Cf. A051159.

%K sign,easy,tabl

%O 0,5

%A _Werner Schulte_, Jul 04 2021