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%I #14 Feb 19 2015 16:26:51
%S 1,-2,1,1,-3,1,0,4,-4,1,-1,-4,8,-5,1,2,3,-12,13,-6,1,-3,-1,15,-25,19,
%T -7,1,4,-2,-16,40,-44,26,-8,1,-5,6,14,-56,84,-70,34,-9,1,6,-11,-8,70,
%U -140,154,-104,43,-10,1,-7,17,-3,-78,210,-294,258,-147,53,-11,1,8,-24,20,75,-288,504,-552,405,-200,64,-12,1
%N Inverse Riordan triangle of A106513: Riordan ((1 - 2*x^2 )/(1 + x), x/(1+x)).
%C Columns k=0..3 give A248157, A248158, A248159, A248160.
%C Row sums have o.g.f. (1 - 2*x)/(1 + x): [1, -1, repeat(-1, 1)].
%C Alternating row sums are (-1)^n*A083318(n).
%H Wolfdieter Lang, <a href="/A248156/a248156.pdf">First 13 rows of the triangle</a>
%F O.g.f. row polynomials R(n,x) = sum(T(n,k)*x^k, k=0..n): (1 - 2*z)/((1 + z)*(1 + (1-x)*z)).
%F O.g.f. column m: x^m*(1 - 2*x^2)/(1 + x)^(m+2), m >= 0.
%F The A-sequence is [1, -1], implying the recurrence T(n,k) = T(n-1,k-1) - T(n-1,k), n >= k > = 1. The Z-sequence is -[2, 3, 7, 17, 41, 99, 239, 577, 1393, ...] = A248161, implying the recurrence T(n,0) = sum(T(n-1,k)*Z(k),k=0..n-1). See the W. Lang link under A006232 for Riordan A- and Z-sequences.
%F The standard recurrence for the sequence for column k=0 is T(0,0) = 1 and T(n,0) = -2*T(n-1,0) - T(n-2,0), n >= 3, with T(1,0) = -2 and T(2,0) = 1.
%e The triangle T(n,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9
%e 0: 1
%e 1: -2 1
%e 2: 1 -3 1
%e 3: 0 4 -4 1
%e 4: -1 -4 8 -5 1
%e 5: 2 3 -12 13 -6 1
%e 6: -3 -1 15 -25 19 -7 1
%e 7: 4 -2 -16 40 -44 26 -8 1
%e 8: -5 6 14 -56 84 -70 34 -9 1
%e 9: 6 -11 -8 70 -140 154 -104 43 -10 1
%e ...
%e For more rows see the link.
%e Recurrence from A-sequence: -12 = T(5,2) = T(4,1) - T(4,2) = -4 - 8.
%e Recurrence from the Z-sequence: 2 = T(5,0) = -(2*(-1) + 3*(-4) + 7*8 + 17*(-5) + 41*1) = 2.
%e Standard recurrence for T(n,0): 0 = T(3,0) = -2*T(2,0) - T(1,0) = -2*1 -(-2).
%t T[n_, k_] := SeriesCoefficient[x^k*(1 - 2*x^2)/(1 + x)^(k + 2), {x, 0, n}]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Oct 09 2014 *)
%Y Cf. A106513, A083318, A248157, A248158, A248159, A248160, A248161.
%K sign,easy,tabl
%O 0,2
%A _Wolfdieter Lang_, Oct 05 2014