

A049310


Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).


488



1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 1, 0, 3, 0, 1, 0, 3, 0, 4, 0, 1, 1, 0, 6, 0, 5, 0, 1, 0, 4, 0, 10, 0, 6, 0, 1, 1, 0, 10, 0, 15, 0, 7, 0, 1, 0, 5, 0, 20, 0, 21, 0, 8, 0, 1, 1, 0, 15, 0, 35, 0, 28, 0, 9, 0, 1, 0, 6, 0, 35, 0, 56, 0, 36, 0, 10, 0, 1, 1, 0, 21, 0, 70, 0, 84, 0
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OFFSET

0,8


COMMENTS

G.f. for row polynomials S(n,x) (signed triangle): 1/(1x*z+z^2). Unsigned triangle a(n,m) has Fibonacci polynomials F(n+1,x) as row polynomials with g.f. 1/(1x*zz^2). a(n,m) triangle has rows of Pascal's triangle A007318 in the evennumbered diagonals (oddnumbered ones have only 0's).
Row sums (unsigned triangle) A000045(n+1) (Fibonacci). Row sums (signed triangle) S(n,1) sequence = periodic(1,1,0,1,1,0) = A010892.
S(n,x) is the characteristic polynomial of the adjacency matrix of the npath.  Michael Somos, Jun 24 2002
S(n,x) is also the matching polynomial of the npath.  Eric W. Weisstein, Apr 10 2017
T(n,k) = number of compositions of n+1 into k+1 odd parts. Example: T(7,3) = 10 because we have (1,1,3,3), (1,3,1,3), (1,3,3,1), (3,1,1,3), (3,1,3,1), (3,3,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1) and (5,1,1,1).  Emeric Deutsch, Apr 09 2005
S(n,x)= R(n,x) + S(n2,x), n >= 2, S(1,x)=0, S(0,x)=1, R(n,x):=2*T(n,x/2) = Sum_{m=0..n} A127672(n,m)*x^m (monic integer Chebyshev TPolynomials). This is the rewritten socalled trace of the transfer matrix formula for the Tpolynomials.  Wolfdieter Lang, Dec 02 2010
In a regular Ngon inscribed in a unit circle, the side length is d(N,1) = 2*sin(Pi/N). The length ratio R(N,k):=d(N,k)/d(N,1) for the (k1)th diagonal, with k from {2,3,...,floor(N/2)}, N >= 4, equals S(k1,x) = sin(k*Pi/N)/sin(Pi/N) with x=rho(N):=R(N,2) = 2*cos(Pi/N). Example: N=7 (heptagon), rho=R(7,2), sigma:=R(N,3) = S(2,rho) = rho^2  1. Motivated by the quoted paper by P. Steinbach.  Wolfdieter Lang, Dec 02 2010
In q or basic analysis, qnumbers are [n]_q :=S(n1,q+1/q) = (q^n(1/q)^n})/(q1/q), with the row polynomials S(n,x), n >= 0.
The zeros of the row polynomials S(n1,x) are (from those of Chebyshev Upolynomials):
x(n1;k) = + t(k,rho(n)), k = 1..ceiling((n1)/2), n >= 2, with t(n,x) the row polynomials of A127672 and rho(n):= 2*cos(Pi/n). The simple vanishing zero for even n appears here as +0 and 0.
Factorization of the row polynomials S(n1,x), x >= 1, in terms of the minimal polynomials of cos(2 Pi/2), called Psi(n,x), with coefficients given by A181875/A181876:
S(n1,x) = (2^(n1))*Product_{n>=1}(Psi(d,x/2), 2 < d  2n).
(From the rewritten eq. (3) of the Watkins and Zeitlin reference, given under A181872.) [See the W. Lang ArXiv link, Proposition 9, eq. (62).  Wolfdieter Lang, Apr 14 2018]
(End)
This is an example for a subclass of Riordan convolution arrays (lower triangular matrices) called Bell arrays. See the L. W. Shapiro et al. reference under A007318. If a Riordan array is named (G(z),F(z)) with F(z)=z*Fhat(z), the o.g.f. for the row polynomials is G(z)/(1x*z*Fhat(z)), and it becomes a Bell array if G(z)=Fhat(z). For the present Bell type triangle G(z)=1/(1+z^2) (see the o.g.f. comment above). This leads to the o.g.f. for the column no. k, k >= 0, x^k/(1+x^2)^(k+1) (see the formula section), the one for the row sums and for the alternating row sums (see comments above). The Riordan (Bell) A and Zsequences (defined in a W. Lang link under A006232, with references) have o.g.f.s 1x*c(x^2) and x*c(x^2), with the o.g.f. of the Catalan numbers A000108. Together they lead to a recurrence given in the formula section.  Wolfdieter Lang, Nov 04 2011
The determinant of the N x N matrix S(N,[x[1], ..., x[N]]) with elements S(m1,x[n]), for n, m = 1, 2, ..., N, and for any x[n], is identical with the determinant of V(N,[x[1], ..., x[N]]) with elements x[n]^(m1) (a Vandermondian, which equals Product_{1 <= i < j<= N} (x[j]  x[i]). This is a special instance of a theorem valid for any N >= 1 and any monic polynomial system p(m,x), m>=0, with p(0,x) = 1. For this theorem see the VeinDale reference, p. 59. Thanks to L. Edson Jeffery for an email asking for a proof of the nonsingularity of the matrix S(N,[x[1], ...., x[N]]) if and only if the x[j], j = 1..N, are pairwise distinct.  Wolfdieter Lang, Aug 26 2013
These S polynomials also appear in the context of modular forms. The rescaled Hecke operator T*_n = n^((1k)/2)*T_n acting on modular forms of weight k satisfies T*_(p^n) = S(n, T*_p), for each prime p and positive integer n. See the KoecherKrieg reference, p. 223.  Wolfdieter Lang, Jan 22 2016
For a shifted o.g.f. (mod signs), its compositional inverse, and connections to Motzkin and Fibonacci polynomials, noncrossing partitions and other combinatorial structures, see A097610.  Tom Copeland, Jan 23 2016
Solutions of the Diophantine equation u^2 + v^2  k*u*v = 1 for integer k given by (u(k,n), v(k,n)) = (S(n,k), S(n1,k)) because of the CassiniSimson identity: S(n,x)^2  S(n+1,x)*S(n1, x) = 1, after use of the Srecurrence. Note that S(n, x) = S(n2, x), n >= 1, and the periodicity of some S(n, k) sequences.
Hence another way to obtain the row polynomials would be to take powers of the matrix [x, 1; 1,0]: S(n, x) = ([x, 1; 1, 0])^n)[1,1], n >= 0.
See also a Feb 01 2016 comment on A115139 for a wellknown S(n, x) sum formula.
Then we have with the present T triangle
A039834(n) = i^(n+1)*T(n1, k) where i is the imaginary unit and n >= 0.
A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the
A181545(n) = Sum_{i=0..n+1} abs(T(n,i)^3),
A181546(n) = Sum_{i=0..n+1} T(n,i)^4,
A181547(n) = Sum_{i=0..n+1} abs(T(n,i)^5).
S(n, 0) = A056594(n), and for k = 1..10 the sequences S(n1, k) with offset n = 0 are A128834, A001477, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.
(End)
For more on the Diophantine equation presented by Kul, see the Ismail paper.  Tom Copeland, Jan 31 2016
The o.g.f. for the Legendre polynomials L(n,x) is 1 / sqrt(1 2x*z + z^2), and squaring it gives the o.g.f. of U(n,x), A053117, so Sum_{k=0..n} L(k,x/2) L(nk,x/2) = S(n,x). This gives S(n,x) = L(n/2,x/2)^2 + 2*Sum_{k=0..n/21} L(k,x/2) L(nk,x/2) for n even and S(n,x) = 2*Sum_{k=0..(n1)/2} L(k,x/2) L(nk,x/2) for odd n. For a connection to elliptic curves and modular forms, see A053117. For the normalized Legendre polynomials, see A100258. For other properties and relations to other polynomials, see Allouche et al.  Tom Copeland, Feb 04 2016
LG(x,h1,h2) = log(1  h1*x + h2*x^2) = Sum_{n>0} F(n,h1,h2,0,..,0) x^n/n is a log series generator of the bivariate row polynomials of A127672 with A127672(0,0) = 0 and where F(n,b1,b2,..,bn) are the Faber polynomials of A263916. Exp(LG(x,h1,h2)) = 1 / (1  h1*x + h2*x^2 ) is the o.g.f. of the bivariate row polynomials of this entry.  Tom Copeland, Feb 15 2016 (Instances of the bivariate o.g.f. for this entry are on pp. 5 and 18 of Sunada.  Tom Copeland, Jan 18 2021)
For distinct odd primes p and q the Legendre symbol can be written as Legendre(q,p) = Product_{k=1..P} S(q1, 2*cos(2*Pi*k/p)), with P = (p1)/2. See the Lemmermeyer reference, eq. (8.1) on p. 236. Using the zeros of S(q1, x) (see above) one has S(q1, x) = Product_{l=1..Q} (x^2  (2*cos(Pi*l/q))^2), with Q = (q1)/2. Thus S(q1, 2*cos(2*Pi*k/p)) = ((4)^Q)*Product_{l=1..Q} (sin^2(2*Pi*k/p)  sin^2(Pi*l/q)) = ((4)^Q)*Product_{m=1..Q} (sin^2(2*Pi*k/p)  sin^2(2*Pi*m/q)). For the proof of the last equality see a W. Lang comment on the triangle A057059 for n = Q and an obvious function f. This leads to Eisenstein's proof of the quadratic reciprocity law Legendre(q,p) = ((1)^(P*Q)) * Legendre(p,q), See the Lemmermeyer reference, pp. 236237.  Wolfdieter Lang, Aug 28 2016
For connections to generalized Fibonacci polynomials, compare their generating function on p. 5 of the Amdeberhan et al. link with the o.g.f. given above for the bivariate row polynomials of this entry.  Tom Copeland, Jan 08 2017
The formula for Ramanujan's tau function (see A000594) for prime powers is tau(p^k) = p^(11*k/2)*S(k, p^(11/2)*tau(p)) for k >= 1, and p = A000040(n), n >= 1. See the Hardy reference, p. 164, eqs. (10.3.4) and (10.3.6) rewritten in terms of S.  Wolfdieter Lang, Jan 27 2017
The number of zeros Z(n) of the S(n, x) polynomials in the open interval (1,+1) is 2*b(n) for even n >= 0 and 1 + 2*b(n) for odd n >= 1, where b(n) = floor(n/2)  floor((n+1)/3). This b(n) is the number of integers k in the interval (n+1)/3 < k <= floor(n/2). See a comment on the zeros of S(n, x) above, and b(n) = A008615(n2), n >= 0. The numbers Z(n) have been proposed (with a conjecture related to A008611) by Michel Lagneau, Mar 2017, as the number of zeros of Fibonacci polynomials on the imaginary axis (I,+I), with I=sqrt(1). They are Z(n) = A008611(n1), n >= 0, with A008611(1) = 0. Also Z(n) = A194960(n4), n >= 0. Proof using the A008611 version. A194960 follows from this.
In general the number of zeros Z(a;n) of S(n, x) for n >= 0 in the open interval (a,+a) for a from the interval (0,2) (x >= 2 never has zeros, and a=0 is trivial: Z(0;n) = 0) is with b(a;n) = floor(n//2)  floor((n+1)*arccos(a/2)/Pi), as above Z(a;n) = 2*b(a;n) for even n >= 0 and 1 + 2*b(a;n) for odd n >= 1. For the closed interval [a,+a] Z(0;n) = 1 and for a from (0,1) one uses for Z(a;n) the values b(a;n) = floor(n/2)  ceiling((n+1)*arccos(a/2)/Pi) + 1. (End)
The Riordan row polynomials S(n, x) (Chebyshev S) belong to the BoasBuck class (see a comment and references in A046521), hence they satisfy the BoasBuck identity: (E_x  n*1)*S(n, x) = (E_x + 1)*Sum_{p=0..n1} (1  (1)^p)*(1)^((p+1)/2)*S(n1p, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle T(n, k) this entails a recurrence for the sequence of column k, given in the formula section.  Wolfdieter Lang, Aug 11 2017
The e.g.f. E(x,t) := Sum_{n>=0} (t^n/n!)*S(n,x) for the row polynomials is obtained via inverse Laplace transformation from the above given o.g.f. as E(x,t) = ((1/xm)*exp(t/xm)  (1/xp)*exp(t/xp) )/(xp  xm) with xp = (x + sqrt(x^24))/2 and xm = (x  sqrt(x^24))/2.  Wolfdieter Lang, Nov 08 2017
Factorization of row polynomials S(n, x), for n >= 1, in terms of C polynomials (not Chebyshev C) with coefficients given in A187360. This is obtained from the factorization into Psi polynomials (see the Jul 12 2011 comment above) but written in terms of minimal polynomials of 2*cos(2*Pi/n) with coefficients in A232624:
S(2*k, x) = Product_{2 <= d  (2*k+1)} C(d, x)*(1)^deg(d)*C(d, x), with deg(d) = A055034(d) the degree of C(d, x).
S(2*k+1, x) = Product_{2 <= d  2*(k+1)} C(d, x) * Product_{3 <= 2*d + 1  (k+1)} (1)^(deg(2*d+1))*C(2*d+1, x).
Note that (1)^(deg(2*d+1)*C(2*d+1, x)*C(2*d+1 ,x) pairs always appear.
The number of C factors of S(2*k, x), for k >= 0, is 2*(tau(2*k+1)  1) = 2*(A099774(k+1)  1) = 2*A095374(k), and for S(2*k+1, x), for k >= 0, it is tau(2*(k+1)) + tau_{odd}(k+1)  2 = A302707(k), with tau(2*k+1) = A099774(k+1), tau(n) = A000005 and tau(2*(k+1)) = A099777(k+1).
For the reverse problem, the factorization of C polynomials into S polynomials, see A255237. (End)
The S polynomials with general initial conditions S(a,b;n,x) = x*S(a,b;n1,x)  S(a,b;n2,x), for n >= 1, with S(a,b;1,x) = a and S(a,b;0,x) = b are S(a,b;n,x) = b*S(n, x)  a*S(n1, x), for n >= 1. Recall that S(2, x) = 1 and S(1, x) = 0. The o.g.f. is G(a,b;z,x) = (b  a*z)/(1  x*z + z^2).  Wolfdieter Lang, Oct 18 2019
Multisection of Spolynomials: S(m*n+k, x) = S(m+k, x)*S(n1, R(m, x))  S(k, x)*S(n2, R(m, x)), with R(n, x) = S(n, x)  S(n2, x) (see A127672), S(2, x) = 1, and S(1, x) = 0, for n >= 0, m >= 1, and k = 0, 1, ..., m1.
O.g.f. of {S(m*n+k, y)}_{n>=0}: G(m,k,y,x) = (S(k, y)  (S(k, y)*R(m, y)  S(m+k, y))*x)/(1  R(m,y)*x + x^2).
See eqs. (40) and (49), with r = x or y and s =1, of the G. Detlefs and W. Lang link at A034807. (End)
S(n, x) for complex n and complex x: S(n, x) = ((i/2)/sqrt(1  (x/2)^2))*(q(x/2)*exp(+n*log(q(x/2)))  (1/q(x/2))*exp(n*log(q(x/2)))), with q(x) = x + sqrt(1  x^2)*i. Here log(z) = z + Arg(z)*i, with Arg(z) from [Pi,+Pi) (principal branch). This satisfies the recurrence relation for S because it is derived from the Binet  de Moivre formula for S. Examples: S(n/m, 0) = cos((n/m)*Pi/4), for n >= 0 and m >= 1. S(n*i, 0) = (1/2)*(1 + exp(n*Pi))*exp((n/2)*Pi), for n >= 0. S(1+i, 2+i) = 0.6397424847... + 1.0355669490...*i. Thanks to Roberto Alfano for asking a question leading to this formula.  Wolfdieter Lang, Jun 05 2023
Lim_{n>oo} S(n, x)/S(n1, x) = r(x) = (x  sqrt(x^2 4))/2, for x >= 2. For x = +2, this limit is +1.  Wolfdieter Lang, Nov 15 2023


REFERENCES

G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, AMS Chelsea Publishing, Providence, Rhode Island, 2002, p. 164.
Max Koecher and Aloys Krieg, Elliptische Funktionen und Modulformen, 2. Auflage, Springer, 2007, p. 223.
Franz Lemmermeyer, Reciprocity Laws. From Euler to Eisenstein, Springer, 2000.
D. S. Mitrinovic, Analytic Inequalities, SpringerVerlag, 1970; p. 232, Sect. 3.3.38.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990, pp. 60  61.
R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer, 1999.


LINKS

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy], Table 22.8, p.797.


FORMULA

T(n,k) := 0 if n < k or n+k odd, otherwise ((1)^((n+k)/2+k))*binomial((n+k)/2, k); T(n, k) = T(n2, k)+T(n1, k1), T(n, 1) := 0 =: T(1, k), T(0, 0)=1, T(n, k)= 0 if n < k or n+k odd; g.f. kth column: (1 / (1 + x^2)^(k + 1)) * x^k.  Michael Somos, Jun 24 2002
T(n,k) = binomial((n+k)/2, (nk)/2)*cos(Pi*(nk)/2)*(1+(1)^(nk))/2.  Paul Barry, Aug 28 2005
Recurrence for the (unsigned) Fibonacci polynomials: F(1)=1, F(2)=x; for n > 2, F(n) = x*F(n1) + F(n2).
The Riordan A and Zsequences, given in a comment above, lead together to the recurrence:
T(n,k) = 0 if n < k, if k=0 then T(0,0)=1 and
T(n,0)= Sum_{i=0..floor((n1)/2)} C(i)*T(n1,2*i+1), otherwise T(n,k) = T(n1,k1)  Sum_{i=1..floor((nk)/2)} C(i)*T(n1,k1+2*i), with the Catalan numbers C(n)=A000108(n).
(End)
The row polynomials satisfy also S(n,x) = 2*(T(n+2, x/2)  T(n, x/2))/(x^24) with the Chebyshev Tpolynomials. Proof: Use the trace formula 2*T(n, x/2) = S(n, x)  S(n2, x) (see the Dec 02 2010 comment above) and the Srecurrence several times. This is a formula which expresses the S in terms of the Tpolynomials.  Wolfdieter Lang, Aug 07 2014
The nonvanishing, unsigned subdiagonals Diag_(2n) contain the elements D(n,k) = Sum_{j=0..k} D(n1,j) = (k+1) (k+2) ... (k+n) / n! = binomial(n+k,n), so the o.g.f. for the subdiagonal is (1x)^((n+1)). E.g., Diag_4 contains D(2,3) = D(1,0) + D(1,1) + D(1,2) + D(1,3) = 1 + 2 + 3 + 4 = 10 = binomial(5,2). Diag_4 is shifted A000217; Diag_6, shifted A000292: Diag_8, shifted A000332; and Diag_10, A000389.
The nonvanishing antidiagonals are signed rows of the Pascal triangle A007318.
For a reversed, unsigned version with the zeros removed, see A011973. (End)
The BoasBuck recurrence (see a comment above) for the sequence of column k is: S(n, k) = ((k+1)/(nk))*Sum_{p=0..n1k} (1  (1)^p)*(1)^((p+1)/2) * S(n1p, k), for n > k >= 0 and input S(k, k) = 1.  Wolfdieter Lang, Aug 11 2017
The mth row consecutive nonzero entries in order are (1)^c*(c+b)!/c!b! with c = m/2, m/21, ..., 0 and b = m2c if m is even and with c = (m1)/2, (m1)/21, ..., 0 with b = m2c if m is odd. For the 8th row starting at a(36) the 5 consecutive nonzero entries in order are 1,10,15,7,1 given by c = 4,3,2,1,0 and b = 0,2,4,6,8.  Richard Turk, Aug 20 2017
O.g.f.: exp( Sum_{n >= 0} 2*T(n,x/2)*t^n/n ) = 1 + x*t + (1 + x^2)*t^2 + (2*x + x^3)*t^3 + (1  3*x^2 + x^4)*t^4 + ..., where T(n,x) denotes the nth Chebyshev polynomial of the first kind.  Peter Bala, Aug 15 2022


EXAMPLE

The triangle T(n, k) begins
n\k 0 1 2 3 4 5 6 7 8 9 10 11
0: 1
1: 0 1
2: 1 0 1
3: 0 2 0 1
4: 1 0 3 0 1
5: 0 3 0 4 0 1
6: 1 0 6 0 5 0 1
7: 0 4 0 10 0 6 0 1
8: 1 0 10 0 15 0 7 0 1
9: 0 5 0 20 0 21 0 8 0 1
10: 1 0 15 0 35 0 28 0 9 0 1
11: 0 6 0 35 0 56 0 36 0 10 0 1
For more rows see the link.
E.g., fourth row {0,2,0,1} corresponds to polynomial S(3,x)= 2*x + x^3.
Zeros of S(3,x) with rho(4)= 2*cos(Pi/4) = sqrt(2):
+ t(1,sqrt(2)) = + sqrt(2) and
+ t(2,sqrt(2)) = + 0.
Factorization of S(3,x) in terms of Psi polynomials:
S(3,x) = (2^3)*Psi(4,x/2)*Psi(8,x/2) = x*(x^22).
(End)
A and Z sequence recurrence:
T(4,0) =  (C(0)*T(3,1) + C(1)*T(3,3)) = (2 + 1) = +1,
T(5,3) = 3  1*1 = 4.
(End)
BoasBuck recurrence for column k = 2, n = 6: S(6, 2) = (3/4)*(0  2* S(4 ,2) + 0 + 2*S(2, 2)) = (3/4)*(2*(3) + 2) = 6.  Wolfdieter Lang, Aug 11 2017
Factorization into C polynomials (see the Apr 12 2018 comment):
S(4, x) = 1  3*x^2 + x^4 = (1 + x + x^2)*(1  x + x^2) = (C(5, x)) * C(5, x); the number of factors is 2 = 2*A095374(2).
S(5, x) = 3*x  4*x^3 + x^5 = x*(1 + x)*(1 + x)*(3 + x^2) = C(2, x)*C(3, x)*(C(3, x))*C(6, x); the number of factors is 4 = A302707(2). (End)


MAPLE

A049310 := proc(n, k): binomial((n+k)/2, (nk)/2)*cos(Pi*(nk)/2)*(1+(1)^(nk))/2 end: seq(seq(A049310(n, k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011
# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n > ifelse(irem(n, 2) = 0, 0, (1)^iquo(n1, 2))); # Peter Luschny, Oct 06 2022


MATHEMATICA

t[n_, k_] /; EvenQ[n+k] = ((1)^((n+k)/2+k))*Binomial[(n+k)/2, k]; t[n_, k_] /; OddQ[n+k] = 0; Flatten[Table[t[n, k], {n, 0, 12}, {k, 0, n}]][[;; 86]] (* JeanFrançois Alcover, Jul 05 2011 *)
Table[Coefficient[(I)^n Fibonacci[n + 1,  I x], x, k], {n, 0, 10}, {k, 0, n}] //Flatten (* Clark Kimberling, Aug 02 2011; corrected by Eric W. Weisstein, Apr 06 2017 *)
CoefficientList[ChebyshevU[Range[0, 10], x/2], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)
CoefficientList[Table[(I)^n Fibonacci[n + 1, I x], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)


PROG

(PARI) {T(n, k) = if( k<0  k>n  (n + k)%2, 0, (1)^((n + k)/2 + k) * binomial((n + k)/2, k))} /* Michael Somos, Jun 24 2002 */
(SageMath)
@CachedFunction
if n< 0: return 0
if n==0: return 1 if k == 0 else 0
(Magma)
A049310:= func< n, k  ((n+k) mod 2) eq 0 select (1)^(Floor((n+k)/2)+k)*Binomial(Floor((n+k)/2), k) else 0 >;


CROSSREFS

Cf. A000005, A000217, A000292, A000332, A000389, A001227, A007318, A008611, A008615, A101455, A010892, A011973, A053112 (without zeros), A053117, A053119 (reflection), A053121 (inverse triangle), A055034, A097610, A099774, A099777, A100258, A112552 (first column clipped), A127672, A168561 (absolute values), A187360. A194960, A232624, A255237.
Triangles of coefficients of Chebyshev's S(n,x+k) for k = 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5: A207824, A207823, A125662, A078812, A101950, A049310, A104562, A053122, A207815, A159764, A123967.


KEYWORD



AUTHOR



STATUS

approved



