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%I #52 Apr 10 2021 16:05:02
%S 1,1,1,2,3,1,2,7,5,1,3,13,16,7,1,3,22,40,29,9,1,4,34,86,91,46,11,1,4,
%T 50,166,239,174,67,13,1,5,70,296,553,541,297,92,15,1,5,95,496,1163,
%U 1461,1068,468,121,17,1,6,125,791,2269,3544,3300,1912,695,154,19,1
%N Riordan array (1/((1-x)(1-x^2)), x/(1-x)^2). Triangle read by rows, T(n,k) for 0 <= k <= n.
%C Diagonal sums are the Jacobsthal numbers A001045.
%C Transforms r^n into the symmetric third-order sequence with g.f. 1/(1-(r+1)x-(r+1)x^2+x^3), see the formulas.
%C From _Wolfdieter Lang_, Oct 22 2019: (Start)
%C The signed triangle t(n, k) = (-1)^(n-k)*T(n, k) appears in the expansion [n+2, 2]_q / q^n = Sum_{k=0} t(n, k)*y^(2*k), with y = q^(1/2) + q^(-1/2), where [n+2, 2]_q are q-binomial coefficients (see A008967, but with a different offset). The formula is [n+2, 2]_q / q^n = S(n+1, y)*S(n, y)/y with Chebyshev S polynomials (A049310). This is a polynomial in y^2 but not in q after replacement of the given y = y(q).
%C The A-sequence for this Riordan triangle is A(n) = (-1)^n*A115141(n) with o.g.f A(x) = 1 + x*(1 + c(-x)), with c(x) generating A000108 (Catalan).
%C The Z-sequence is z(n) = (-1)^(n+1)*A071724(n), for n >= 1 and z(0) = 1. The o.g.f. is Z(x) = 1 + x*c(-x)^3. See A071724 for a link on A- and Z-sequences, and their use for the recurrence. (End)
%C T(n,k) is the number of tilings of a (2*n+1)-board (a 1 X (2*n+1) rectangular board) using 2*k+1 squares and 2*(n-k) (1,1)-fences. A (1,1)-fence is a tile composed of two squares separated by a gap of width 1. - _Michael A. Allen_, Mar 20 2021
%H G. C. Greubel, <a href="/A158909/b158909.txt">Rows n = 0..100 of the triangle, flattened</a>
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Barry4/barry64.html">Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays</a>, JIS 12 (2009) 09.8.6.
%H Kenneth Edwards and Michael A. Allen, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL24/Allen/edwards2.html">New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile</a>, JIS 24 (2021) 21.3.8.
%F Sum_{k=0..n} T(n,k) = Fibonacci(n+1)*Fibonacci(n+2) = A001654(n+1).
%F From _Johannes W. Meijer_, Jul 20 2011: (Start)
%F T(n, k) = Sum_{i=0..n-k} (-1)^(i+n-k) * binomial(i+2*k+1, i).
%F T(n, k) = A035317(n+k, n-k) = A092879(n, n-k).
%F Sum_{k=0..n} T(n, k)*r^k = coeftayl(1/(1-(r+1)*x-(r+1)*x^2+x^3), x=0, n). [Barry] (End)
%F T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k) + T(n-2, k-1) - T(n-3, k), T(0, 0) = 1, T(n, k) = 0 if k<0 or if k>n. - _Philippe Deléham_, Nov 12 2013
%F From _Wolfdieter Lang_, Oct 22 2019: (Start)
%F O.g.f. for the row polynomials (that is for the triangle): G(z, x) = 1/((1 + z)*(1 - (x + 2)*z + z^2)), and
%F O.g.f. for column k: x^k/((1+x)*(1-x)^(2*(k+1)) (Riordan property). (End)
%F T(n, k) = binomial(k + n + 2, n - k + 1)*hypergeom([1, k + n + 3], [n - k + 2], -1) + (-1)^(n - k)/4^(k + 1). - _Peter Luschny_, Oct 31 2019
%F From _Michael A. Allen_, Mar 20 2021: (Start)
%F T(n,k) = A335964(2*n+1,n-k).
%F T(n,k) = T(n-2,k) + binomial(n+k,2*k). (End)
%e From _Wolfdieter Lang_, Oct 22 2019: (Start)
%e The triangle T(n, k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e ----------------------------------------------------
%e 0: 1
%e 1: 1 1
%e 2: 2 3 1
%e 3: 2 7 5 1
%e 4: 3 13 16 7 1
%e 5: 3 22 40 29 9 1
%e 6: 4 34 86 91 46 11 1
%e 7: 4 50 166 239 174 67 13 1
%e 8: 5 70 296 553 541 297 92 15 1
%e 9: 5 95 496 1163 1461 1068 468 121 17 1
%e 10: 6 125 791 2269 3544 3300 1912 695 154 19 1
%e ...
%e ----------------------------------------------------------------------------
%e Recurrence: T(5, 2) = 16 + 13 + 5 + 7 - 1 = 40, and T(5, 0) = 3 + 2 - 2 = 3. [using _Philippe Deléham_'s Nov 12 2013 recurrence]
%e Recurrence from A-sequence [1, 2, -1, 2, -5, ...]: T(5, 2) = 1*13 + 2*16 - 1*7 + 2*1 = 40.
%e Recurrence from Z-sequence [1, 1, -3, 9, -28, ...]: T(5, 0) = 1*3 + 1*13 - 3*16 + 9*7 - 28*1 = 3. (End)
%p T := (n,k) -> binomial(k+n+2, n-k+1)*hypergeom([1, k+n+3], [n-k+2], -1) + (-1)^(n-k)/4^(k+1):
%p seq(seq(simplify(T(n,k)), k=0..n), n=0..9); # _Peter Luschny_, Oct 31 2019
%t Table[Sum[(-1)^(j+n-k)*Binomial[j+2*k+1, j], {j,0,n-k}], {n,0,12}, {k,0,n}] // Flatten (* _G. C. Greubel_, Mar 18 2021 *)
%o (Sage) flatten([[sum((-1)^(j+n-k)*binomial(j+2*k+1, j) for j in (0..n-k)) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Mar 18 2021
%o (Magma) [(&+[(-1)^(j+n-k)*Binomial(2*k+j+1, j): j in [0..n-k]]): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Mar 18 2021
%Y Cf. A000045, A001654 (row sums), A008967, A035317, A049310, A092879, A335964.
%K easy,nonn,tabl
%O 0,4
%A _Paul Barry_, Mar 30 2009
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