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%I #75 Aug 29 2022 11:14:33
%S 1,1,-1,1,-1,-1,1,-1,-2,1,1,-1,-3,2,1,1,-1,-4,3,3,-1,1,-1,-5,4,6,-3,
%T -1,1,-1,-6,5,10,-6,-4,1,1,-1,-7,6,15,-10,-10,4,1,1,-1,-8,7,21,-15,
%U -20,10,5,-1,1,-1,-9,8,28,-21,-35,20,15,-5,-1,1,-1,-10,9,36,-28,-56,35,35,-15,-6,1,1,-1,-11,10,45,-36,-84,56,70
%N Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].
%C Matrix inverse of A124645.
%C Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:
%C L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));
%C Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);
%C Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);
%C abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);
%C T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;
%C T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;
%C T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;
%C T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;
%C T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;
%C T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;
%C T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;
%C T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;
%C T(n,n) = L(n,0) = (-1)^floor((n+1)/2);
%C L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);
%C L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;
%C L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;
%C L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;
%C L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;
%C L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;
%C L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;
%C L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;
%C L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;
%C L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;
%C L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;
%C L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;
%C L(n,13) = A085260(n);
%C L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;
%C L(n,n) = A108366(n); L(n,-n) = A108367(n).
%C Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - _Paul D. Hanna_, Jun 12 2005
%C From _L. Edson Jeffery_, Mar 12 2011: (Start)
%C Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form
%C G_N=A_{N,1}=
%C (0 1 0 ... 0)
%C (1 0 1 0 ... 0)
%C (0 1 0 1 0 ... 0)
%C ...
%C (0 ... 0 1 0 1)
%C (0 ... 0 1 1),
%C with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,
%C G_7=A_{7,1}=
%C (0 1 0)
%C (1 0 1)
%C (0 1 1).
%C We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)
%C The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - _Johannes W. Meijer_, Aug 08 2011
%C The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - _Gary W. Adamson_, Feb 07 2012
%D Friedrich L. Bauer, 'De Moivre und Lagrange: Cosinus eines rationalen Vielfachen von Pi', Informatik Spektrum 28 (Springer, 2005).
%D Jay Kappraff, S. Jablan, G. Adamson, & R. Sazdonovich: "Golden Fields, Generalized Fibonacci Sequences, & Chaotic Matrices"; FORMA, Vol 19, No 4, (2005).
%H Reinhard Zumkeller, <a href="/A108299/b108299.txt">Rows n = 0..150 of triangle, flattened</a>
%H Henry W. Gould, <a href="http://www.fq.math.ca/Scanned/3-4/gould.pdf">A Variant of Pascal's Triangle</a>, <a href="http://www.fq.math.ca/Scanned/4-1/corrections2.pdf">Corrections</a>, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, p. 257-271.
%H L. Edson Jeffery, <a href="/wiki/User:L._Edson_Jeffery/Unit-Primitive_Matrices">Unit-primitive matrices</a>.
%H Ju, Hyeong-Kwan <a href="https://doi.org/10.5831/HMJ.2017.39.4.665">On the sequence generated by a certain type of matrices.</a> Honam Math. J. 39, No. 4, 665-675 (2017).
%H Michelle Rudolph-Lilith, <a href="http://arxiv.org/abs/1508.07894">On the Product Representation of Number Sequences, with Application to the Fibonacci Family</a>, arXiv preprint arXiv:1508.07894 [math.NT], 2015.
%H Frank Ruskey and Carla Savage, <a href="https://ajc.maths.uq.edu.au/pdf/10/ocr-ajc-v10-p85.pdf">Gray codes for set partitions and restricted growth tails</a>, Australasian Journal of Combinatorics, Volume 10(1994), pp. 85-96. See Table 1 p. 95.
%F T(n,k) = binomial(n-floor((k+1)/2),floor(k/2))*(-1)^floor((k+1)/2).
%F T(n+1, k) = if sign(T(n, k-1))=sign(T(n, k)) then T(n, k-1)+T(n, k) else -T(n, k-1) for 0 < k < n, T(n, 0) = 1, T(n, n) = (-1)^floor((n+1)/2).
%F G.f.: A(x, y) = (1 - x*y)/(1 - x + x^2*y^2). - _Paul D. Hanna_, Jun 12 2005
%F The generating polynomial (in z) of row n >= 0 is (u^(2*n+1) + v^(2*n+1))/(u + v), where u and v are defined by u^2 + v^2 = 1 and u*v = z. - _Emeric Deutsch_, Jun 16 2011
%F From _Johannes W. Meijer_, Aug 08 2011: (Start)
%F abs(T(n,k)) = A065941(n,k) = abs(A187660(n,n-k));
%F T(n,n-k) = A130777(n,k); abs(T(n,n-k)) = A046854(n,k) = abs(A066170(n,k)). (End)
%e Triangle begins:
%e 1;
%e 1, -1;
%e 1, -1, -1;
%e 1, -1, -2, 1;
%e 1, -1, -3, 2, 1;
%e 1, -1, -4, 3, 3, -1;
%e 1, -1, -5, 4, 6, -3, -1;
%e 1, -1, -6, 5, 10, -6, -4, 1;
%e 1, -1, -7, 6, 15, -10, -10, 4, 1;
%e 1, -1, -8, 7, 21, -15, -20, 10, 5, -1;
%e 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1;
%e 1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1;
%e ...
%p A108299 := proc(n,k): binomial(n-floor((k+1)/2), floor(k/2))*(-1)^floor((k+1)/2) end: seq(seq(A108299 (n,k), k=0..n), n=0..11); # _Johannes W. Meijer_, Aug 08 2011
%t t[n_, k_?EvenQ] := I^k*Binomial[n-k/2, k/2]; t[n_, k_?OddQ] := -I^(k-1)*Binomial[n+(1-k)/2-1, (k-1)/2]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, May 16 2013 *)
%o (PARI) {T(n,k)=polcoeff(polcoeff((1-x*y)/(1-x+x^2*y^2+x^2*O(x^n)),n,x)+y*O(y^k),k,y)} (Hanna)
%o (Haskell)
%o a108299 n k = a108299_tabl !! n !! k
%o a108299_row n = a108299_tabl !! n
%o a108299_tabl = [1] : iterate (\row ->
%o zipWith (+) (zipWith (*) ([0] ++ row) a033999_list)
%o (zipWith (*) (row ++ [0]) a059841_list)) [1,-1]
%o -- _Reinhard Zumkeller_, May 06 2012
%Y Cf. A049310, A039961, A124645 (matrix inverse).
%Y Triangle sums (see the comments): A193884 (Kn11), A154955 (Kn21), A087960 (Kn22), A000007 (Kn3), A010892 (Fi1), A134668 (Fi2), A078031 (Ca2), A193669 (Gi1), A001519 (Gi3), A193885 (Ze1), A050935 (Ze3). - _Johannes W. Meijer_, Aug 08 2011
%Y Cf. A003558.
%Y Cf. A033999, A059841.
%K sign,tabl
%O 0,9
%A _Reinhard Zumkeller_, Jun 01 2005
%E Corrected and edited by _Philippe Deléham_, Oct 20 2008
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