1, -1, 3, -1, -15, 10, -1, 105, -105, 10, 15, -1, -945, 1260, -280, -210, 35, 21, -1, 10395, -17325, 6300, 3150, -280, -1260, -378, 35, 56, 28, -1, -135135, 270270, -138600, -51975, 15400, 34650, 6930, -2100, -1575, -2520, -630, 126, 84, 36, -1
Offset: 1
Examples and checks:
1) Let u_1 = -1 and u_n = 1 for n>1,
then f(t) = exp(u.*t) - u(0) = exp(t)-2t-1
and g(t) = [e.g.f. of signed A000311];
therefore, the row sums of unsigned [C(j,k)] are A000311 =
(0,1,1,4,26,236,2752,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),4!*P(4,1),...).
2) Let u_1 = -1 and u_n = (n-1)! for n>1,
then f(t) = -log(1-t)-2t
and g(t) = [e.g.f. of signed (0,A032188)]
with (0,A032188) = (0,1,1,5,41,469,6889,...) = (0,-P(1,1),2!*P(2,1),-3!P(3,1),...).
3) Let u_1 = -1 and u_n = (-1)^n (n-2)! for n>1, then
f(t) = (1+t) log(1+t) - 2t
and g(t) = [e.g.f. of signed (0,A074059)]
with (0,A074059) = (0,1,1,2,7,34,213,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),...).
4) Let u_1 = 1, u_2 = -1 and u_n = 0 for n>2,
then f(t) = t(1-t/2)
and g(t) = [e.g.f. of (0,A001147)] = 1 - (1-2t)^(1/2)
with (0,A001147) = (0,1,1,3,15,105,945...) =(0,P(1,1),2!*P(2,1),3!*P(3,1),...).
5) Let u_1 = 1, u_2 = -2 and u_n = 0 for n>2,
then f(t)= t(1-t)
and g(t) = t * [o.g.f. of A000108] = [1 - (1-4t)^(1/2)] / 2
with (0,A000108) = (0,1,1,2,5,14,42,...) = (0,P(1,1),P(2,1),P(3,1),...).
.
From _Peter Luschny_, Feb 19 2021: (Start)
Triangle starts:
[1] 1;
[2] -1;
[3] 3, -1;
[4] -15, 10, -1;
[5] 105, -105, [10, 15], -1;
[6] -945, 1260, [-280, -210], [35, 21], -1;
[7] 10395, -17325, [6300, 3150], [-280, -1260, -378], [35, 56, 28], -1;
[8] -135135, 270270, [-138600, -51975], [15400, 34650, 6930], [-2100, -1575, -2520, -630], [126, 84, 36], -1
The coefficients can be seen as a refinement of the Ward numbers: Let R(n, k) = Sum T(n, k), where the sum collects adjacent terms with equal sign, as indicated by the square brackets in the table, then R(n+1, k+1) = (-1)^(n-k)*W(n, k), where W(n, k) are the Ward numbers A181996, for n >= 0 and 0 <= k <= n-1. (End)
Comments