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A158454 Riordan array (1/(1-x^2), x/(1+x)^2). 15

%I #52 May 05 2023 11:26:54

%S 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,

%T -10,1,0,16,-80,148,-128,56,-12,1,1,-20,130,-314,367,-230,79,-14,1,0,

%U 25,-200,610,-920,771,-376,106,-16,1,1,-30,295,-1106,2083,-2232,1444,-574,137,-18,1

%N Riordan array (1/(1-x^2), x/(1+x)^2).

%C 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

%C Image of the Catalan numbers A000108 by this matrix is the all 1's sequence.

%C Image of the central binomial numbers A000984 by this matrix is the counting numbers A000027.

%C Inverse array is the Riordan array (1-x^2*c(x)^4, xc(x)^2), where c(x) is the g.f. of A000108.

%C 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

%D Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.

%H G. C. Greubel, <a href="/A158454/b158454.txt">Rows n=0..100 of triangle, flattened</a>

%H Paul Barry and A. Hennessey, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Barry1/barry83.html">Notes on a Family of Riordan Arrays and Associated Integer Hankel Transforms </a>, JIS 12 (2009) 09.5.3.

%H Jia Huang, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL26/Huang/huang8.html">Partially Palindromic Compositions</a>, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 17.

%H Richard J. Mathar, <a href="https://vixra.org/abs/2208.0111">Recurrence for the Atkinson-Steenwijk Integrals for Resistors in the Infinite Triangular Lattice</a>, viXra:2208.0111, Aug. 2022.

%F 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).

%F 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

%F 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

%F O.g.f. row polynomials (rising powers in y):

%F ((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

%F Recurrences from the A- and Z-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.

%F T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), n >= 1.

%F T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), n >= k >= 1.

%F 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

%F T(n, k) = Sum_{m=0..n} A129818(m, k), 0 <= k <= n. - _Wolfdieter Lang_, Dec 15 2010

%F 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

%F G.f.: (1 + x)/((1 - x)*(1 + x)^2 - t*x*(1 - x)). - _G. C. Greubel_, Dec 15 2018

%F 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

%e The triangle T(n,k) begins:

%e n\k 0 1 2 3 4 5 6 7 8 9 10...

%e 0: 1

%e 1: 0 1

%e 2: 1 -2 1

%e 3: 0 4 -4 1

%e 4: 1 -6 11 -6 1

%e 5: 0 9 -24 22 -8 1

%e 6: 1 -12 46 -62 37 -10 1

%e 7: 0 16 -80 148 -128 56 -12 1

%e 8: 1 -20 130 -314 367 -230 79 -14 1

%e 9: 0 25 -200 610 -920 771 -376 106 -16 1

%e 10: 1 -30 295 -1106 2083 -2232 1444 -574 137 -18 1

%e ... Reformatted and extended by _Wolfdieter Lang_, Nov 24 2012

%e Recurrences (from A- and Z-sequences):

%e 1 = T(6,0) = 0*0 + 1*9 +2*(-24) + 5*22 + 14*(-8)+ 42*1.

%e -80 = T(7,2) = 1*(-12) -2*(46) -1*(-62) -2*37 -5*(-10) -14*1. - _Wolfdieter Lang_, Dec 20 2010

%p A158454 := proc(n,k) (-1)^(n+k)*add(binomial(n+k-1-2*j,2*k-1),j=0..floor(n/2)) ; end proc;

%p seq(seq(A158454(n,k),k=0..n),n=0..10) ; # _R. J. Mathar_, Dec 17 2010

%t 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 *)

%t 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 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 *)

%o (PARI) {T(n,k) = sum(j=0,n, (-1)^(j-k)*binomial(k+j, 2*k))};

%o for(n=0, 10, for(k=0,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Dec 15 2018

%o (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

%o (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

%o (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

%Y Cf. A000027, A000108, A000984, A006232, A049310, A046521 (Boas-Buck info), A115141, A181878.

%Y From _Wolfdieter Lang_, Aug 10 2017: (Start)

%Y Row sums A011655(n+1), alternating row sums A007598(n+1)*(-1)^(n+1).

%Y Column sequences k=0..5: A059841, A002620(n+2)*(-1)^(n), A001752(n)*(-1)^n, A001769(n)*(-1)^n, A001780(n)*(-1)^n, A001786(n)*(-1)^n. (End)

%K easy,sign,tabl

%O 0,5

%A _Paul Barry_, Mar 19 2009

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