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A158454
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Riordan array (1/(1-x^2), x/(1+x)^2).
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15
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1, 0, 1, 1, -2, 1, 0, 4, -4, 1, 1, -6, 11, -6, 1, 0, 9, -24, 22, -8, 1, 1, -12, 46, -62, 37, -10, 1, 0, 16, -80, 148, -128, 56, -12, 1, 1, -20, 130, -314, 367, -230, 79, -14, 1, 0, 25, -200, 610, -920, 771, -376, 106, -16, 1, 1, -30, 295, -1106, 2083, -2232, 1444, -574, 137, -18, 1
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OFFSET
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0,5
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COMMENTS
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Coefficient table of the square of Chebyshev S-polynomials. For the S-polynomials see A049310, and for a proof see the array A181878, where the odd numbered rows are shifted by one to the left. - Wolfdieter Lang, Dec 15 2010
Image of the Catalan numbers A000108 by this matrix is the all 1's sequence.
Image of the central binomial numbers A000984 by this matrix is the counting numbers A000027.
Inverse array is the Riordan array (1-x^2*c(x)^4, xc(x)^2), where c(x) is the g.f. of A000108.
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k belong to the class of Boas-Buck polynomials. Hence they satisfy the Boas-Buck identity: (E_x - n*1)*R(n, x) = -Sum_{p=0..n-1} ((1 - (-1)^p)*1 + 2*(-1)^(p+1)*E_x) R(n-1-p, x) for n >= 0. See the Boas-Buck comments and references in A046521. The ensuing recurrence for the column sequences is given in the formula section. - Wolfdieter Lang, Aug 10 2017
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REFERENCES
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Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
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LINKS
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FORMULA
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Number triangle T(n, k) = Sum_{j=0..n} (-1)^(j-k)*binomial(k+j, 2*k) = Sum_{j=0..n-k} (-1)^(n-k-j)*binomial(n+k-j, 2*k).
O.g.f. column k with leading zeros (Riordan array, see NAME): (1/(1-x^2))*(x/(1+x)^2)^k, k >= 0. - Wolfdieter Lang, Dec 15 2010
T(n, k) = (-1)^(n-k)*Sum_{j=0..floor(n/2)} binomial(n+k-1-2*j, 2*k-1), 0 <= k <= n, else 0. From the o.g.f. for column k after convolution. - Wolfdieter Lang, Dec 17 2010
O.g.f. row polynomials (rising powers in y):
((1+x)/(1-x))/(1+(2-y)*x+x^2) = Sum_{n>=0} (S(n,sqrt(y))^2*x^n, with Chebyshev S-polynomials from A049310. - Wolfdieter Lang, Dec 15 2010
Recurrences from the A- and Z-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.
T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), n >= 1.
T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), n >= k >= 1.
Here Z(0)=0 and Z(j) = A000108(j), j >= 1, (o.g.f. -1 + c(x), with the Catalan o.g.f. c(x)), and A(j) = A115141(j) = [1,-2,-1,-2,-5,-14,...], j >= 0, with o.g.f. 1/c(x)^2. - Wolfdieter Lang, Dec 20 2010
Boas-Buck recurrence for column k: R(n, k) = (1/(n-k))*Sum_{p=k..n-1}((-1)^(n-p)*(2*k+1) + 1) * R(p, k), for n > k >= 0, with input R(k, k) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017
G.f.: (1 + x)/((1 - x)*(1 + x)^2 - t*x*(1 - x)). - G. C. Greubel, Dec 15 2018
T(n, k) = (-1)^(n - k)*binomial(k + n - 1, 2*k-1)*hypergeom([1, (k - n)/2, (1 + k - n)/2], [(1 - k - n)/2, (2 - k - n)/2], 1) for k >= 1 . - Peter Luschny, Aug 20 2022
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EXAMPLE
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The triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10...
0: 1
1: 0 1
2: 1 -2 1
3: 0 4 -4 1
4: 1 -6 11 -6 1
5: 0 9 -24 22 -8 1
6: 1 -12 46 -62 37 -10 1
7: 0 16 -80 148 -128 56 -12 1
8: 1 -20 130 -314 367 -230 79 -14 1
9: 0 25 -200 610 -920 771 -376 106 -16 1
10: 1 -30 295 -1106 2083 -2232 1444 -574 137 -18 1
Recurrences (from A- and Z-sequences):
1 = T(6,0) = 0*0 + 1*9 +2*(-24) + 5*22 + 14*(-8)+ 42*1.
-80 = T(7,2) = 1*(-12) -2*(46) -1*(-62) -2*37 -5*(-10) -14*1. - Wolfdieter Lang, Dec 20 2010
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MAPLE
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A158454 := proc(n, k) (-1)^(n+k)*add(binomial(n+k-1-2*j, 2*k-1), j=0..floor(n/2)) ; end proc;
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MATHEMATICA
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nmax = 10; t[n_, k_] := (-1)^(n+k)* Sum[Binomial[n+k-1-2*j, 2*k-1], {j, 0, Floor[n/2]}]; t[n_?EvenQ, 0] = 1; Flatten[ Table[ t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Nov 08 2011, after Maple *)
With[{m = 15}, CoefficientList[CoefficientList[Series[(1+x)/((1-x)*(1 + x)^2 -t*x*(1-x)), {x, 0, m}, {t, 0, m}], x], t]]//Flatten (* G. C. Greubel, Dec 15 2018 *)
T[n_, 0] := Boole[EvenQ[n]]; T[n_, k_] := (-1)^(n - k) Binomial[k+n-1, 2*k-1] HypergeometricPFQ[{1, (k - n)/2, (1 + k - n)/2}, {(1 - k - n)/2, (2 - k - n)/2}, 1]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // TableForm (* Peter Luschny, Aug 20 2022 *)
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PROG
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(PARI) {T(n, k) = sum(j=0, n, (-1)^(j-k)*binomial(k+j, 2*k))};
for(n=0, 10, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Dec 15 2018
(Magma) [[(&+[(-1)^(j-k)*Binomial(k+j, 2*k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Dec 15 2018
(Sage) [[sum((-1)^(j-k)*binomial(k+j, 2*k) for j in range(n+1)) for k in range(n+1)] for n in range(10)] # G. C. Greubel, Dec 15 2018
(GAP) T:=Flat(List([0..10], n->List([0..n], k->Sum([0..n], j-> (-1)^(j-k)*Binomial(k+j, 2*k))))); # G. C. Greubel, Dec 15 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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