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A248156
Inverse Riordan triangle of A106513: Riordan ((1 - 2*x^2 )/(1 + x), x/(1+x)).
6
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, -7, 1, 4, -2, -16, 40, -44, 26, -8, 1, -5, 6, 14, -56, 84, -70, 34, -9, 1, 6, -11, -8, 70, -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
OFFSET
0,2
COMMENTS
Columns k=0..3 give A248157, A248158, A248159, A248160.
Row sums have o.g.f. (1 - 2*x)/(1 + x): [1, -1, repeat(-1, 1)].
Alternating row sums are (-1)^n*A083318(n).
FORMULA
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)).
O.g.f. column m: x^m*(1 - 2*x^2)/(1 + x)^(m+2), m >= 0.
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.
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.
EXAMPLE
The triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9
0: 1
1: -2 1
2: 1 -3 1
3: 0 4 -4 1
4: -1 -4 8 -5 1
5: 2 3 -12 13 -6 1
6: -3 -1 15 -25 19 -7 1
7: 4 -2 -16 40 -44 26 -8 1
8: -5 6 14 -56 84 -70 34 -9 1
9: 6 -11 -8 70 -140 154 -104 43 -10 1
...
For more rows see the link.
Recurrence from A-sequence: -12 = T(5,2) = T(4,1) - T(4,2) = -4 - 8.
Recurrence from the Z-sequence: 2 = T(5,0) = -(2*(-1) + 3*(-4) + 7*8 + 17*(-5) + 41*1) = 2.
Standard recurrence for T(n,0): 0 = T(3,0) = -2*T(2,0) - T(1,0) = -2*1 -(-2).
MATHEMATICA
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 *)
KEYWORD
sign,easy,tabl
AUTHOR
Wolfdieter Lang, Oct 05 2014
STATUS
approved