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A143987 Eigentriangle of (A007318)^(-1); row sums = A014182, exp(1-x-exp(-x)).

%I A143987 #8 Dec 29 2023 10:23:00
%S A143987 1,-1,1,1,2,0,-1,3,0,-1,1,-4,0,4,1,-1,5,0,-10,-5,2,1,-6,0,20,15,-12,
%T A143987 -9,-1,7,0,-35,-35,42,63,9,1,-8,0,56,70,-112,-252,-72,50,-1,9,0,-84,
%U A143987 -126,252,756,324,-450,-267,1,-10,0,120,210,-504,-1890,-1080,2250,2670,413
%N A143987 Eigentriangle of (A007318)^(-1); row sums = A014182, exp(1-x-exp(-x)).
%C A143987 Sum of n-th row terms = rightmost term of next row. Row sums = A014182: (1, 0, -1, 1, 2, -9, 9, 50, -267,...).
%C A143987 Right border = A014182 shifted: (1, 1, 0, -1, 1, 2, -9,...).
%F A143987 (A007318^(-1) * (A014182 * 0^(n-k))) 0<=k<=n
%F A143987 A007318^(-1) = the inverse of Pascal's triangle.
%F A143987 Given A014182: (1, 0, -1, 1, 2, -9, 9,...) = expansion of exp(1-x-exp(-x)), we preface A014182 with a "1" getting (1, 1, 0, -1, 1, 2, -9,...).
%F A143987 Then diagonalize it as an infinite lower triangular matrix R =
%F A143987 1;
%F A143987 0, 1;
%F A143987 0, 0, 0;
%F A143987 0, 0, 0, -1;
%F A143987 0, 0, 0, 0, 1;
%F A143987 ...
%F A143987 Finally, take the inverse binomial transform of triangle R, getting A143987.
%F A143987 Given the inverse of Pascal's triangle by rows, we apply termwise products of equal numbers of terms in the sequence: (1, 1, 0, -1, 1, 2, -9, 9,...).
%e A143987 First few rows of the triangle =
%e A143987 1;
%e A143987 -1, 1;
%e A143987 1, -2, 0;
%e A143987 -1, 3, 0, -1;
%e A143987 1, -4, 0, 4, 1;
%e A143987 -1, 5, 0, -10, -5, 2;
%e A143987 1, -6, 0, 20, 15, -12, -9;
%e A143987 -1, 7, 0, -35, -35, 42, 63, 9;
%e A143987 1, -8, 0, 56, 70, -112, -252, 72, 50;
%e A143987 ...
%e A143987 Example: row 4 = (1, -4, 0, 4, 1) = termwise products of (1, -4, 6, -4, 1) and (1, 1, 0, -1, 1).= (1*1, -4*1, 6*0, -4*-1, 1*1).
%Y A143987 Cf. A007318, A014182.
%K A143987 tabl,sign
%O A143987 0,5
%A A143987 _Gary W. Adamson_, Sep 07 2008