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%I #64 Sep 19 2023 10:51:21
%S 1,-1,1,-1,-1,1,1,-2,-1,1,1,2,-3,-1,1,-1,3,3,-4,-1,1,-1,-3,6,4,-5,-1,
%T 1,1,-4,-6,10,5,-6,-1,1,1,4,-10,-10,15,6,-7,-1,1,-1,5,10,-20,-15,21,7,
%U -8,-1,1,-1,-5,15,20,-35,-21,28,8,-9,-1,1,1,-6,-15,35,35,-56,-28,36,9,-10,-1,1
%N Coefficients of first difference of Chebyshev S polynomials.
%C Inverse of triangle in A061554.
%C Signed version of A046854.
%C From _Paul Barry_, May 21 2009: (Start)
%C Riordan array ((1-x)/(1+x^2),x/(1+x^2)).
%C This triangle is the coefficient triangle for the Hankel transforms of the family of generalized Catalan numbers that satisfy a(n;r)=r*a(n-1;r)+sum{k=1..n-2, a(k)*a(n-1-k;r)}, a(0;r)=a(1;r)=1. The Hankel transform of a(n;r) is h(n)=sum{k=0..n, T(n,k)*r^k} with g.f. (1-x)/(1-r*x+x^2). These sequences include A086246, A000108, A002212. (End)
%C From _Wolfdieter Lang_, Jun 11 2011: (Start)
%C The Riordan array ((1+x)/(1+x^2),x/(1+x^2)) with entries Phat(n,k)= ((-1)^(n-k))*T(n,k) and o.g.f. Phat(x,z)=(1+z)/(1-x*z+z^2) for the row polynomials Phat(n,x) is related to Chebyshev C and S polynomials as follows.
%C Phat(n,x) = (R(n+1,x)-R(n,x))/(x+2) = S(2*n,sqrt(2+x))
%C with R(n,x)=C_n(x) in the Abramowitz and Stegun notation, p. 778, 22.5.11. See A049310 for the S polynomials. Proof from the o.g.f.s.
%C Recurrence for the row polynomials Phat(n,x):
%C Phat(n,x) = x*Phat(n-1,x) - Phat(n-2,x) for n>=1; Phat(-1,x)=-1, Phat(0,x)=1.
%C The A-sequence for this Riordan array Phat (see the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices) is given by 1, 0, -1, 0, -1, 0, -2, 0, -5,.., starting with 1 and interlacing the negated A000108 with zeros (o.g.f. 1/c(x^2) = 1-c(x^2)*x^2, with the o.g.f. c(x) of A000108).
%C The Z-sequence has o.g.f. sqrt((1-2*x)/(1+2*x)), and it is given by A063886(n)*(-1)^n.
%C The A-sequence of the Riordan array T(n,k) is identical with the one for the Riordan array Phat, and the Z-sequence is -A063886(n).
%C (End)
%C The row polynomials P(n,x) are the characteristic polynomials of the adjacency matrices of the graphs which look like P_n (n vertices (nodes), n-1 lines (edges)), but vertex no. 1 has a loop. - _Wolfdieter Lang_, Nov 17 2011
%C From _Wolfdieter Lang_, Dec 14 2013: (Start)
%C The zeros of P(n,x) are x(n,j) = -2*cos(2*Pi*j/(2*n+1)), j=1..n. From P(n,x) = (-1)^n*S(2*n,sqrt(2-x)) (see, e.g., the Lemma 6 of the W. Lang link).
%C The discriminants of the P-polynomials are given in A052750. (End)
%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964. Tenth printing, Wiley, 2002 (also electronically available).
%H Hyeong-Kwan Ju, <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), Theorem 2.16.
%H Wolfdieter Lang, <a href="http://arxiv.org/abs/1210.1018">The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon</a>, arXiv:1210.1018 [math.GR], 2012-2017; see Definition 1, Lemma 6 and Remark 4.
%H P. Steinbach, <a href="http://www.jstor.org/stable/2691048">Golden fields: a case for the heptagon</a>, Math. Mag. 70 (1997), p. 22-31 (formula 5).
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%F Number triangle T(n,k) = (-1)^C(n-k+1,2)*C(floor((n+k)/2),k). - _Paul Barry_, May 21 2009
%F From _Wolfdieter Lang_, Jun 11 2011: (Start)
%F Row polynomials: P(n,x) = sum(k=0..n, T(n,k)*x^k) = R(2*n+1,sqrt(2+x)) / sqrt(2+x), with Chebyshev polynomials R with coefficients given in A127672 (scaled T-polynomials).
%F R(n,x) is called C_n(x) in Abramowitz and Stegun's handbook, p. 778, 22.5.11.
%F P(n,x) = S(n,x)-S(n-1,x), n>=0, S(-1,x)=0, with the Chebyshev S-polynomials (see the coefficient triangle A049310).
%F O.g.f. for row polynomials: P(x,z):= sum(n>=0, P(n,x)*z^n ) = (1-z)/(1-x*z+z^2).
%F (from the o.g.f. for R(2*n+1,x), n>=0, computed from the o.g.f. for the R-polynomials (2-x*z)/(1-x*z+z^2) (see A127672))
%F Proof of the Chebyshev connection from the o.g.f. for Riordan array property of this triangle (see the P. Barry comment above).
%F For the A- and Z-sequences of this Riordan array see a comment above. (End)
%F abs(T(n,k)) = A046854(n,k) = abs(A066170(n,k)) T(n,n-k) = A108299(n,k); abs(T(n,n-k)) = A065941(n,k). - _Johannes W. Meijer_, Aug 08 2011
%F From _Wolfdieter Lang_, Jul 31 2014: (Start)
%F Similar to the triangles A157751, A244419 and A180070 one can give for the row polynomials P(n,x) besides the usual three term recurrence another one needing only one recurrence step. This uses also a negative argument, namely P(n,x) = (-1)^(n-1)*(-1 + x/2)*P(n-1,-x) + (x/2)*P(n-1,x), n >= 1, P(0,x) = 1. Proof by computing the o.g.f. and comparing with the known one. This entails the alternative triangle recurrence T(n,k) = (-1)^(n-k)*T(n-1,k) + (1/2)*(1 + (-1)^(n-k))*T(n-1,k-1), n >= m >= 1, T(n,k) = 0 if n < k and T(n,0) = (-1)^floor((n+1)/2) = A057077(n+1). [P(n,x) recurrence corrected Aug 03 2014]
%F (End)
%e The triangle T(n,k) begins:
%e n\k 0 1 1 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
%e 0: 1
%e 1: -1 1
%e 2: -1 -1 1
%e 3: 1 -2 -1 1
%e 4: 1 2 -3 -1 1
%e 5: -1 3 3 -4 -1 1
%e 6: -1 -3 6 4 -5 -1 1
%e 7: 1 -4 -6 10 5 -6 -1 1
%e 8: 1 4 -10 -10 15 6 -7 -1 1
%e 9: -1 5 10 -20 -15 21 7 -8 -1 1
%e 10: -1 -5 15 20 -35 -21 28 8 -9 -1 1
%e 11: 1 -6 -15 35 35 -56 -28 36 9 -10 -1 1
%e 12: 1 6 -21 -35 70 56 -84 -36 45 10 -11 -1 1
%e 13: -1 7 21 -56 -70 126 84 -120 -45 55 11 -12 -1 1
%e 14: -1 -7 28 56 -126 -126 210 120 -165 -55 66 12 -13 -1 1
%e 15: 1 -8 -28 84 126 -252 -210 330 165 -220 -66 78 13 -14 -1 1
%e ... reformatted and extended - _Wolfdieter Lang_, Jul 31 2014.
%e ---------------------------------------------------------------------------
%e From _Paul Barry_, May 21 2009: (Start)
%e Production matrix is
%e -1, 1,
%e -2, 0, 1,
%e -2, -1, 0, 1,
%e -4, 0, -1, 0, 1,
%e -6, -1, 0, -1, 0, 1,
%e -12, 0, -1, 0, -1, 0, 1,
%e -20, -2, 0, -1, 0, -1, 0, 1,
%e -40, 0, -2, 0, -1, 0, -1, 0, 1,
%e -70, -5, 0, -2, 0, -1, 0, -1, 0, 1 (End)
%e Row polynomials as first difference of S polynomials:
%e P(3,x) = S(3,x) - S(2,x) = (x^3 - 2*x) - (x^2 -1) = 1 - 2*x - x^2 +x^3.
%e Alternative triangle recurrence (see a comment above): T(6,2) = T(5,2) + T(5,1) = 3 + 3 = 6. T(6,3) = -T(5,3) + 0*T(5,1) = -(-4) = 4. - _Wolfdieter Lang_, Jul 31 2014
%p A130777 := proc(n,k): (-1)^binomial(n-k+1,2)*binomial(floor((n+k)/2),k) end: seq(seq(A130777(n,k), k=0..n), n=0..11); # _Johannes W. Meijer_, Aug 08 2011
%t T[n_, k_] := (-1)^Binomial[n - k + 1, 2]*Binomial[Floor[(n + k)/2], k];
%t Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Nov 14 2017, from Maple *)
%o (Sage)
%o @CachedFunction
%o def A130777(n,k):
%o if n< 0: return 0
%o if n==0: return 1 if k == 0 else 0
%o h = A130777(n-1,k) if n==1 else 0
%o return A130777(n-1,k-1) - A130777(n-2,k) - h
%o for n in (0..9): [A130777(n,k) for k in (0..n)] # _Peter Luschny_, Nov 20 2012
%Y Cf. A066170, A046854, A057077 (first column).
%Y Row sums: A010892(n+1); repeat(1,0,-1,-1,0,1). Alternating row sums: A061347(n+2); repeat(1,-2,1).
%K sign,tabl,easy
%O 0,8
%A _Philippe Deléham_, Jul 14 2007
%E New name and Chebyshev comments by _Wolfdieter Lang_, Jun 11 2010
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