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A166341 Triangle T(n, k) = coefficients of ( t(n, x) ) where t(n, x) = (1-x)^(n+1)*p(n, x)/x, p(n, x) = x*D( p(n-1, x) ), with p(1, x) = x/(1-x)^2, p(2, x) = x*(1+x)/(1-x)^3, and p(3, x) = x*(1+10*x+x^2)/(1-x)^4, read by rows. 5

%I

%S 1,1,1,1,10,1,1,23,23,1,1,50,138,50,1,1,105,614,614,105,1,1,216,2367,

%T 4912,2367,216,1,1,439,8397,31483,31483,8397,439,1,1,886,28264,176314,

%U 314830,176314,28264,886,1,1,1781,91880,903104,2632034,2632034,903104,91880,1781,1

%N Triangle T(n, k) = coefficients of ( t(n, x) ) where t(n, x) = (1-x)^(n+1)*p(n, x)/x, p(n, x) = x*D( p(n-1, x) ), with p(1, x) = x/(1-x)^2, p(2, x) = x*(1+x)/(1-x)^3, and p(3, x) = x*(1+10*x+x^2)/(1-x)^4, read by rows.

%D Douglas C. Montgomery and Lynwood A. Johnson, Forecasting and Time Series Analysis, MaGraw-Hill, New York, 1976, page 91

%H G. C. Greubel, <a href="/A166341/b166341.txt">Rows n = 1..50 of the triangle, flattened</a>

%F T(n, k) = coefficients of ( t(n, x) ) where t(n, x) = (1-x)^(n+1)*p(n, x)/x, p(n, x) = x*D( p(n-1, x) ), with p(1, x) = x/(1-x)^2, p(2, x) = x*(1+x)/(1-x)^3, and p(3, x) = x*(1+10*x+x^2)/(1-x)^4.

%F From _G. C. Greubel_, Mar 11 2022: (Start)

%F T(n, k) = t(n-1, k) - t(n-1, k-1), T(n,1) = 1, where t(n, k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n+1, k-j)*b(n, j), b(n, k) = k^(n-2)*A007588(k), b(n, 0) = 1, and b(1, k) = 1.

%F T(n, n-k) = T(n, k). (End)

%e Triangle begins as:

%e 1;

%e 1, 1;

%e 1, 10, 1;

%e 1, 23, 23, 1;

%e 1, 50, 138, 50, 1;

%e 1, 105, 614, 614, 105, 1;

%e 1, 216, 2367, 4912, 2367, 216, 1;

%e 1, 439, 8397, 31483, 31483, 8397, 439, 1;

%e 1, 886, 28264, 176314, 314830, 176314, 28264, 886, 1;

%e 1, 1781, 91880, 903104, 2632034, 2632034, 903104, 91880, 1781, 1;

%t (* First program *)

%t p[x_, 1]:= x/(1-x)^2;

%t p[x_, 2]:= x*(1+x)/(1-x)^3;

%t p[x_, 3]:= x*(1+10*x+x^2)/(1-x)^4;

%t p[x_, n_]:= p[x, n]= x*D[p[x, n-1], x]

%t Table[CoefficientList[(1-x)^(n+1)*p[x, n]/x, x], {n,12}]//Flatten

%t (* Second program *)

%t b[n_, k_, m_]:= If[n<2, 1, If[k==0, 0, k^(n-1)*((m+3)*k^2 - m)/3]];

%t t[n_, k_, m_]:= t[n,k,m]= Sum[(-1)^(k-j)*Binomial[n+1, k-j]*b[n,j,m], {j,0,k}];

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

%t Table[T[n,k,3], {n,12}, {k,n}]//Flatten (* _G. C. Greubel_, Mar 11 2022 *)

%o (Sage)

%o def b(n,k,m):

%o if (n<2): return 1

%o elif (k==0): return 0

%o else: return k^(n-1)*((m+3)*k^2 - m)/3

%o @CachedFunction

%o def t(n,k,m): return sum( (-1)^(k-j)*binomial(n+1, k-j)*b(n,j,m) for j in (0..k) )

%o def A166341(n,k): return 1 if (k==1) else t(n-1,k,3) - t(n-1,k-1,3)

%o flatten([[A166341(n,k) for k in (1..n)] for n in (1..12)]) # _G. C. Greubel_, Mar 11 2022

%Y Cf. A166340, A166343, A166344, A166345, A166346, A166349.

%Y Cf. A007588, A123125.

%K nonn,tabl

%O 1,5

%A _Roger L. Bagula_, Oct 12 2009

%E Edited by _G. C. Greubel_, Mar 11 2022

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Last modified May 17 23:07 EDT 2022. Contains 353779 sequences. (Running on oeis4.)