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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A158909 Riordan array (1/((1-x)(1-x^2)), x/(1-x)^2). Triangle read by rows, T(n,k) for 0 <= k <= n.

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%I A158909 #56 Oct 06 2024 09:16:43
%S A158909 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 A158909 50,166,239,174,67,13,1,5,70,296,553,541,297,92,15,1,5,95,496,1163,
%U A158909 1461,1068,468,121,17,1,6,125,791,2269,3544,3300,1912,695,154,19,1
%N A158909 Riordan array (1/((1-x)(1-x^2)), x/(1-x)^2). Triangle read by rows, T(n,k) for 0 <= k <= n.
%C A158909 Diagonal sums are the Jacobsthal numbers A001045.
%C A158909 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 A158909 From _Wolfdieter Lang_, Oct 22 2019: (Start)
%C A158909 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 A158909 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 A158909 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 A158909 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 A158909 G. C. Greubel, <a href="/A158909/b158909.txt">Rows n = 0..100 of the triangle, flattened</a>
%H A158909 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 A158909 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 A158909 Sum_{k=0..n} T(n,k) = Fibonacci(n+1)*Fibonacci(n+2) = A001654(n+1).
%F A158909 From _Johannes W. Meijer_, Jul 20 2011: (Start)
%F A158909 T(n, k) = Sum_{i=0..n-k} (-1)^(i+n-k) * binomial(i+2*k+1, i).
%F A158909 T(n, k) = A035317(n+k, n-k) = A092879(n, n-k).
%F A158909 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 A158909 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 A158909 From _Wolfdieter Lang_, Oct 22 2019: (Start)
%F A158909 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 A158909 O.g.f. for column k: x^k/((1+x)*(1-x)^(2*(k+1))) (Riordan property). (End)
%F A158909 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 A158909 From _Michael A. Allen_, Mar 20 2021: (Start)
%F A158909 T(n,k) = A335964(2*n+1,n-k).
%F A158909 T(n,k) = T(n-2,k) + binomial(n+k,2*k). (End)
%e A158909 From _Wolfdieter Lang_, Oct 22 2019: (Start)
%e A158909 The triangle T(n, k) begins:
%e A158909   n\k  0   1   2    3    4    5    6   7   8  9 10 ...
%e A158909   ----------------------------------------------------
%e A158909   0:   1
%e A158909   1:   1   1
%e A158909   2:   2   3   1
%e A158909   3:   2   7   5    1
%e A158909   4:   3  13  16    7    1
%e A158909   5:   3  22  40   29    9    1
%e A158909   6:   4  34  86   91   46   11    1
%e A158909   7:   4  50 166  239  174   67   13   1
%e A158909   8:   5  70 296  553  541  297   92  15   1
%e A158909   9:   5  95 496 1163 1461 1068  468 121  17  1
%e A158909   10:  6 125 791 2269 3544 3300 1912 695 154 19  1
%e A158909   ...
%e A158909 ----------------------------------------------------------------------------
%e A158909 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 A158909 Recurrence from A-sequence [1, 2, -1, 2, -5, ...]: T(5, 2) = 1*13 + 2*16 - 1*7 + 2*1 = 40.
%e A158909 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 A158909 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 A158909 seq(seq(simplify(T(n,k)), k=0..n), n=0..9); # _Peter Luschny_, Oct 31 2019
%t A158909 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 A158909 (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 A158909 (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 A158909 Cf. A000045, A001654 (row sums), A008967, A035317, A049310, A092879, A335964.
%K A158909 easy,nonn,tabl
%O A158909 0,4
%A A158909 _Paul Barry_, Mar 30 2009