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Antidiagonal expansion of the polynomials: f(x,n) = 1/(exp(t) - Sum_{i=1..n} t^i/i!).
0

%I #8 Dec 08 2016 09:03:23

%S 1,1,0,1,0,-3,1,0,0,-4,1,0,0,-4,25,1,0,0,0,-5,114,1,0,0,0,-5,-6,-287,

%T 1,0,0,0,0,-6,133,-4152,1,0,0,0,0,-6,-7,552,-1647,1,0,0,0,0,0,-7,-8,

%U 1629,192230,1,0,0,0,0,0,-7,-8,621,-12610,807961,1,0,0,0,0,0,0,-8,-9,2510,-128579,-10164804,1,0,0,0,0,0,0,-8,-9,-10,7381

%N Antidiagonal expansion of the polynomials: f(x,n) = 1/(exp(t) - Sum_{i=1..n} t^i/i!).

%C Row sums are:

%C {1, 2, 3, 20, 125, 804, 4501, 36896, 362673, 3831560, 40591001, 467518248, 6106124713, 87661533764, 1323052370025}.

%C Triangle sequence rows last terms are:

%C Table[n!*a[[1]][[n]], {n, 1, 15}]

%C {1, 0, -3, -4, 25,114, -287, -4152, -1647, 192230, 807961, -10164804, -111209111, 454840554, 14657978385}

%F f(x,n) = 1/(exp(t) - Sum_{i=1..n} t^i/i!); t(n,m) = Expansion(f(x,n)); t_out(n,m) = m!*t(n-m+1,m).

%e {1},

%e {1, 0},

%e {1, 0, -3},

%e {1, 0, 0, -4},

%e {1, 0, 0, -4, 25},

%e {1, 0, 0, 0, -5, 114},

%e {1, 0, 0, 0, -5, -6, -287},

%e {1, 0, 0, 0, 0, -6, 133, -4152},

%e {1, 0, 0, 0, 0, -6, -7, 552, -1647},

%e {1, 0, 0, 0, 0, 0, -7, -8,1629, 192230},

%e {1, 0, 0, 0, 0, 0, -7, -8, 621, -12610, 807961},

%e {1, 0, 0, 0, 0, 0, 0, -8, -9, 2510, -128579, -10164804},

%e {1, 0, 0, 0, 0, 0, 0, -8, -9, -10, 7381, -725484, -111209111},

%e {1, 0, 0, 0, 0,0, 0, 0, -9, -10, 2761, 18996, 1522651, 454840554},

%e {1, 0, 0, 0, 0, 0, 0,0, -9, -10, -11, 11076, -404989, 54082014, 14657978385}

%t Clear[f, b, a, g, h, n, t]; f[t_, n_] = 1/(Exp[t] - Sum[t^i/i!, {i, 1, n}]); a = Table[Table[SeriesCoefficient[Series[f[t, m], {t, 0, 30}], n], {n, 0, 30}], {m, 1, 31}]; b = Table[Table[m!*a[[n - m + 1]][[m]], {m, 1, n }], {n, 1, 15}]; Flatten[b]

%Y Cf. A089148.

%K uned,sign

%O 1,6

%A _Roger L. Bagula_ and _Gary W. Adamson_, Oct 06 2008