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%I #7 Jun 16 2016 23:27:41
%S 1,1,2,2,6,2,6,24,4,8,24,120,0,48,24,120,720,-120,384,72,144,720,5040,
%T -1680,3696,-432,1296,720,5040,40320,-20160,40320,-15840,17280,2880,
%U 5760,40320,362880,-241920,483840,-311040,288000,-46080,69120,40320
%N A triangle related to the GF(z) formulas of the rows of the ED1 array A167546.
%C The GF(z) formulas given below correspond to the first ten rows of the ED1 array A167546. The polynomials in their numerators lead to the triangle given above.
%e Row 1: GF(z) = 1/(1-z).
%e Row 2: GF(z) = (1 + 2*z)/(1-z)^2.
%e Row 3: GF(z) = (2 + 6*z + 2*z^2)/(1-z)^3.
%e Row 4: GF(z) = (6 + 24*z + 4*z^2 + 8*z^3)/(1-z)^4.
%e Row 5: GF(z) = (24 + 120*z + 0*z^2 + 48*z^3 + 24*z^4)/(1-z)^5.
%e Row 6: GF(z) = (120 + 720*z - 120*z^2 + 384*z^3 + 72*z^4 + 144*z^5)/ (1-z)^6.
%e Row 7: GF(z) = (720 + 5040*z - 1680*z^2 + 3696*z^3 - 432*z^4 + 1296*z^5 + 720*z^6)/(1-z)^7.
%e Row 8: GF(z) = (5040 + 40320*z - 20160*z^2 + 40320*z^3 - 15840*z^4 + 17280*z^5 + 2880*z^6 + 5760*z^7)/(1-z)^8.
%e Row 9: GF(z) = (40320 +362880*z -241920*z^2 + 483840*z^3 - 311040*z^4 + 288000*z^5 - 46080*z^6 + 69120*z^7 + 40320*z^8)/(1-z)^9.
%e Row 10: GF(z) = (362880 +3628800*z -3024000*z^2 +6289920*z^3 -5495040*z^4 + 5276160*z^5 - 2131200*z^6 + 1382400*z^7 + 201600*z^8 + 403200*z^9)/(1-z)^10;
%Y A167546 is the ED1 array.
%Y A000142, A000142 (n=>2) and 120*A062148 (with three extra terms at the beginning of the sequence) equal the first three left hand triangle columns.
%Y A098557(n) and A098557(n)*A064455(n) equal the first two right hand triangle columns.
%Y A007680 equals the row sums.
%K sign,tabl
%O 1,3
%A _Johannes W. Meijer_, Nov 10 2009
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