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Triangle in which row n consists of the coefficients in Sum_{m=0..n} x^m * Product_{k=m+1..n} (1-k*x), as read by rows.
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%I #33 Apr 25 2018 19:25:48

%S 1,1,0,1,-2,1,1,-5,7,-2,1,-9,27,-30,9,1,-14,72,-165,159,-44,1,-20,156,

%T -597,1149,-998,265,1,-27,296,-1689,5328,-9041,7251,-1854,1,-35,512,

%U -4057,18840,-51665,79579,-59862,14833,1,-44,827,-8665,55353,-221225,544564,-776073,553591,-133496

%N Triangle in which row n consists of the coefficients in Sum_{m=0..n} x^m * Product_{k=m+1..n} (1-k*x), as read by rows.

%C If m=n, we have Sum_{k=0..n} A008277(n, k) = A000110(n) = Sum_{j=0..n} T(n,j)*A008277(2n-j,n) where A000110(n) is the n-th Bell number. - _Robert A. Russell_, Apr 08 2018

%F Right border equals A000166, the subfactorial numbers.

%F Row sums equal A000166 (shift right 1 place).

%F Row sums of unsigned terms yields A002467(n) = n! - A000166(n).

%F Sum_{k=0..n} A008277(m, k) = Sum_{j=0..n} T(n, j)*A008277(m+n-j, n) where A008277(m, k) are Stirling subset numbers. - _Robert A. Russell_, Mar 30 2018

%F T(n,0) = 1.

%F For k>0, T(n,k) = [k==n] + [k<n]*T(n-1,k) - n*T(n-1,k-1). - _Robert A. Russell_, Apr 25 2018

%e Triangle begins:

%e 1;

%e 1, 0;

%e 1, -2, 1;

%e 1, -5, 7, -2;

%e 1, -9, 27, -30, 9;

%e 1, -14, 72, -165, 159, -44;

%e 1, -20, 156, -597, 1149, -998, 265;

%e 1, -27, 296, -1689, 5328, -9041, 7251, -1854;

%e 1, -35, 512, -4057, 18840, -51665, 79579, -59862, 14833;

%e 1, -44, 827, -8665, 55353, -221225, 544564, -776073, 553591, -133496;

%e 1, -54, 1267, -16935, 142003, -774755, 2756814, -6221713, 8314321, -5669406, 1334961; ...

%e Generating method for row n:

%e n=0: 1 = 1;

%e n=1: 1 + 0*x = (1-x) * ( 1 + x/(1-x) );

%e n=2: 1 - 2*x + x^2 = (1-x)*(1-2*x) * ( 1 + x/(1-x) + x^2/((1-x)*(1-2*x)) );

%e n=3: 1 - 5*x + 7*x^2 - 2*x^3 = (1-x)*(1-2*x)*(1-3*x) * ( 1 + x/(1-x) + x^2/((1-x)*(1-2*x)) + x^3/((1-x)*(1-2*x)*(1-3*x)) );

%e n=4: 1 - 9*x + 27*x^2 - 30*x^3 + 9*x^4 = (1-x)*(1-2*x)*(1-3*x)*(1-4*x) * ( 1 + x/(1-x) + x^2/((1-x)*(1-2*x)) + x^3/((1-x)*(1-2*x)*(1-3*x)) + x^4/((1-x)*(1-2*x)*(1-3*x)*(1-4*x)) ); ...

%e Compare the row g.f.s to the o.g.f. of Bell numbers (A000110):

%e B(x) = 1 + x/(1-x) + x^2/((1-x)*(1-2*x)) + x^3/((1-x)*(1-2*x)*(1-3*x)) + x^4/((1-x)*(1-2*x)*(1-3*x)*(1-4*x)) +...

%e Central terms of triangle begin:

%e [1, -2, 27, -597, 18840, -774755, 39320575, -2375828028, 166592007731, -13300276081039, 1191315248017730, ...].

%t Table[LinearSolve[Table[StirlingS2[m+j, n], {m, 0, n}, {j, n, 0, -1}],

%t Table[Sum[StirlingS2[m, j], {j, 0, n}], {m, 0, n}]], {n, 0, 20}]

%t // Flatten (* _Robert A. Russell_, Mar 30 2018 *)

%t Table[PadRight[CoefficientList[Sum[x^m*Product[1-j*x, {j, m+1, n}],

%t {m, 0, n}], x], n+1], {n, 0, 20}] // Flatten (* _Robert A. Russell_, Apr 08 2018 *)

%t T[n_, 0] := T[n,0] = 1;

%t T[n_, k_] := T[n,k] = If[k<n, T[n-1, k], 1] - n T[n-1, k-1];

%t Table[T[n, k], {n, 0, 15}, {k, 0, n}] // Flatten (* _Robert A. Russell_, Apr 25 2018 *)

%o (PARI) {T(n,k)=polcoeff(sum(m=0,n, x^m*prod(j=m+1,n,1-j*x)), k)}

%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

%Y Cf. A000166, A002467, A000110.

%K sign,tabl

%O 0,5

%A _Paul D. Hanna_, Oct 16 2014