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A248156 Inverse Riordan triangle of A106513: Riordan ((1 - 2*x^2 )/(1 + x), x/(1+x)).

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%I A248156 #22 May 28 2025 01:03:30
%S A248156 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 A248156 -7,1,4,-2,-16,40,-44,26,-8,1,-5,6,14,-56,84,-70,34,-9,1,6,-11,-8,70,
%U A248156 -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 A248156 Inverse Riordan triangle of A106513: Riordan ((1 - 2*x^2 )/(1 + x), x/(1+x)).
%C A248156 Row sums have o.g.f. (1 - 2*x)/(1 + x): [1, -1, repeat(-1, 1)].
%H A248156 G. C. Greubel, <a href="/A248156/b248156.txt">Rows n = 0..50 of the triangle, flattened</a>
%H A248156 Wolfdieter Lang, <a href="/A248156/a248156.pdf">First 13 rows of the triangle</a>
%F A248156 O.g.f. row polynomials R(n,x) = Sum_{k=0..n} T(n,k)*x^k = [(-z)^n] (1 - 2*z^2)/( (1 + z)*(1 + (1-x)*z)).
%F A248156 O.g.f. column m: x^m*(1 - 2*x^2)/(1 + x)^(m+2), m >= 0.
%F A248156 The A-sequence is [1, -1], implying the recurrence T(n,k) = T(n-1, k-1) - T(n-1, k), n >= k > = 1.
%F A248156 The Z-sequence is -[2, 3, 7, 17, 41, 99, 239, 577, 1393, ...] = A248161, implying the recurrence T(n, 0) = Sum_{k=0..n-1} T(n-1,k)*Z(k). See the W. Lang link under A006232 for Riordan A- and Z-sequences.
%F A248156 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.
%F A248156 From _G. C. Greubel_, May 27 2025: (Start)
%F A248156 Sum_{k=0..n} T(n, k) = (-1)^(n+1) + 2*([n=0] - [n=1]).
%F A248156 Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = the repeated pattern of [1, -2, 0, 3, -4, 2]. (End)
%e A248156 The triangle T(n,k) begins:
%e A248156   n\k  0   1   2   3    4    5    6    7    8    9
%e A248156   0:   1
%e A248156   1:  -2   1
%e A248156   2:   1  -3   1
%e A248156   3:   0   4  -4   1
%e A248156   4:  -1  -4   8  -5    1
%e A248156   5:   2   3 -12  13   -6    1
%e A248156   6:  -3  -1  15 -25   19   -7    1
%e A248156   7:   4  -2 -16  40  -44   26   -8    1
%e A248156   8:  -5   6  14 -56   84  -70   34   -9    1
%e A248156   9:   6 -11  -8  70 -140  154 -104   43  -10    1
%e A248156   ...
%e A248156 For more rows see the link.
%e A248156 Recurrence from A-sequence: T(5,2) = T(4,1) - T(4,2) = -4 - 8 = -12.
%e A248156 Recurrence from the Z-sequence: T(5,0) = -(2*(-1) + 3*(-4) + 7*8 + 17*(-5) + 41*1) = 2.
%e A248156 Standard recurrence for T(n,0): T(3,0) = -2*T(2,0) - T(1,0) = -2*1 - (-2) = 0.
%t A248156 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 *)
%t A248156 T[n_, k_]:= T[n, k]= If[k==n,1, If[k==0,(-1)^n*(3-n), T[n-1,k-1]-T[n-1,k]]];
%t A248156 Table[T[n,k], {n,0,25}, {k,0,n}]//Flatten (* _G. C. Greubel_, May 27 2025 *)
%o A248156 (Magma)
%o A248156 function T(n,k) // T = A248156
%o A248156   if k eq n then return 1;
%o A248156   elif k eq 0 then return (-1)^n*(3-n);
%o A248156   else return T(n-1,k-1) - T(n-1,k);
%o A248156   end if;
%o A248156 end function;
%o A248156 [T(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, May 27 2025
%o A248156 (SageMath)
%o A248156 def T(n,k): # T = A248156
%o A248156     if (k==n): return 1
%o A248156     elif (k==0): return (-1)^n*(3-n)
%o A248156     else: return T(n-1,k-1) - T(n-1,k)
%o A248156 print(flatten([[T(n,k) for k in range(n+1)] for n in range(13)])) # _G. C. Greubel_, May 27 2025
%Y A248156 Cf. A006232, A083318, A106513, A248157, A248158, A248159, A248160, A248161.
%Y A248156 Columns: A248157 (k=0), A248158 (k=1), A248159 (k=2), A248160 (k=3).
%Y A248156 Diagonals: A000012 (k=n), A022958(n+3) (k=n-1), -A034856(n-1) (k=n-2), A000297(n-4) (k=n-3), A014309(n-3) (k=n-4).
%Y A248156 Sums: (-1)^n*A001611(n) (diagonal), (-1)^n*A083318(n) (alternating sign row).
%K A248156 sign,easy,tabl
%O A248156 0,2
%A A248156 _Wolfdieter Lang_, Oct 05 2014