A056594 -id:A056594 - cp's OEIS frontend

A049310 Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).

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

Offset: 0

Wolfdieter Lang

G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with g.f. 1/(1-x*z-z^2). |a(n,m)| triangle has rows of Pascal's triangle A007318 in the even-numbered diagonals (odd-numbered 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.
Alternating row sums A049347(n) = S(n,-1) = periodic(1,-1,0). - Wolfdieter Lang, Nov 04 2011
S(n,x) is the characteristic polynomial of the adjacency matrix of the n-path. - Michael Somos, Jun 24 2002
S(n,x) is also the matching polynomial of the n-path. - 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(n-2,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 T-Polynomials). This is the rewritten so-called trace of the transfer matrix formula for the T-polynomials. - Wolfdieter Lang, Dec 02 2010
In a regular N-gon 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 (k-1)-th diagonal, with k from {2,3,...,floor(N/2)}, N >= 4, equals S(k-1,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
From Wolfdieter Lang, Jul 12 2011: (Start)
In q- or basic analysis, q-numbers are [n]_q := S(n-1,q+1/q) = (q^n-(1/q)^n)/(q-1/q), with the row polynomials S(n,x), n >= 0.
The zeros of the row polynomials S(n-1,x) are (from those of Chebyshev U-polynomials):
  x(n-1;k) = +- t(k,rho(n)), k = 1..ceiling((n-1)/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(n-1,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(n-1,x) = (2^(n-1))*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)
The discriminants of the S(n,x) polynomials are found in A127670. - Wolfdieter Lang, Aug 03 2011
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)/(1-x*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 Z-sequences (defined in a W. Lang link under A006232, with references) have o.g.f.s 1-x*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(m-1,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]^(m-1) (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 Vein-Dale reference, p. 59. Thanks to L. Edson Jeffery for an email asking for a proof of the non-singularity 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^((1-k)/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 Koecher-Krieg 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, non-crossing partitions and other combinatorial structures, see A097610. - Tom Copeland, Jan 23 2016
From M. Sinan Kul, Jan 30 2016; edited by Wolfdieter Lang, Jan 31 2016 and Feb 01 2016: (Start)
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(n-1,k)) because of the Cassini-Simson identity: S(n,x)^2 - S(n+1,x)*S(n-1, x) = 1, after use of the S-recurrence. Note that S(-n, x) = -S(-n-2, 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 well-known S(n, x) sum formula.
Then we have with the present T triangle
  A039834(n) = -i^(n+1)*T(n-1, k) where i is the imaginary unit and n >= 0.
  A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the Philippe Deléham, Nov 21 2005 formula),
  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(n-1, 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(n-k,x/2) = S(n,x). This gives S(n,x) = L(n/2,x/2)^2 + 2*Sum_{k=0..n/2-1} L(k,x/2) L(n-k,x/2) for n even and S(n,x) = 2*Sum_{k=0..(n-1)/2} L(k,x/2) L(n-k,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(q-1, 2*cos(2*Pi*k/p)), with P = (p-1)/2. See the Lemmermeyer reference, eq. (8.1) on p. 236. Using the zeros of S(q-1, x) (see above) one has S(q-1, x) = Product_{l=1..Q} (x^2 - (2*cos(Pi*l/q))^2), with Q = (q-1)/2. Thus S(q-1, 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. 236-237. - 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
From Wolfdieter Lang, May 08 2017: (Start)
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(n-2), n >= 0. The numbers Z(n) have been proposed (with a conjecture related to A008611) by Michel Lagneau, as the number of zeros of Fibonacci polynomials on the imaginary axis (-I,+I), with I=sqrt(-1). They are Z(n) = A008611(n-1), n >= 0, with A008611(-1) = 0. Also Z(n) = A194960(n-4), 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 Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*S(n, x) = (E_x + 1)*Sum_{p=0..n-1} (1 - (-1)^p)*(-1)^((p+1)/2)*S(n-1-p, 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^2-4))/2 and xm = (x - sqrt(x^2-4))/2. - Wolfdieter Lang, Nov 08 2017
From Wolfdieter Lang, Apr 12 2018: (Start)
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;n-1,x) - S(a,b;n-2,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(n-1, 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
Also the convolution triangle of A101455. - Peter Luschny, Oct 06 2022
From Wolfdieter Lang, Apr 26 2023: (Start)
Multi-section of S-polynomials: S(m*n+k, x) = S(m+k, x)*S(n-1, R(m, x)) - S(k, x)*S(n-2, R(m, x)), with R(n, x) = S(n, x) - S(n-2, x) (see A127672), S(-2, x) = -1, and S(-1, x) = 0, for n >= 0, m >= 1, and k = 0, 1, ..., m-1.
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(n-1, x) = r(x) = (x - sqrt(x^2 -4))/2, for |x| >= 2. For x = +-2, this limit is +-1. - Wolfdieter Lang, Nov 15 2023

			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
  ... Reformatted and extended by _Wolfdieter Lang_, Oct 24 2012
For more rows see the link.
E.g., fourth row {0,-2,0,1} corresponds to polynomial S(3,x)= -2*x + x^3.
From _Wolfdieter Lang_, Jul 12 2011: (Start)
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^2-2).
(End)
From _Wolfdieter Lang_, Nov 04 2011: (Start)
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)
Boas-Buck 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
From _Wolfdieter Lang_, Apr 12 2018: (Start)
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)

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Index entries for sequences related to Chebyshev polynomials.
Index entries for "core" sequences

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.

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 >;
[A049310(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

A049310 := proc(n,k): binomial((n+k)/2,(n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/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(n-1, 2))); # Peter Luschny, Oct 06 2022

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]] (* Jean-François 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 *)

{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 */

@CachedFunction
def A049310(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    return A049310(n-1,k-1) - A049310(n-2,k)
for n in (0..9): [A049310(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

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(n-2, k)+T(n-1, k-1), T(n, -1) := 0 =: T(-1, k), T(0, 0)=1, T(n, k)= 0 if n < k or n+k odd; g.f. k-th column: (1 / (1 + x^2)^(k + 1)) * x^k. - Michael Somos, Jun 24 2002
T(n,k) = binomial((n+k)/2, (n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2. - Paul Barry, Aug 28 2005
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Nov 21 2005
Recurrence for the (unsigned) Fibonacci polynomials: F(1)=1, F(2)=x; for n > 2, F(n) = x*F(n-1) + F(n-2).
From Wolfdieter Lang, Nov 04 2011: (Start)
The Riordan A- and Z-sequences, 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((n-1)/2)} C(i)*T(n-1,2*i+1), otherwise T(n,k) = T(n-1,k-1) - Sum_{i=1..floor((n-k)/2)} C(i)*T(n-1,k-1+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^2-4) with the Chebyshev T-polynomials. Proof: Use the trace formula 2*T(n, x/2) = S(n, x) - S(n-2, x) (see the Dec 02 2010 comment above) and the S-recurrence several times. This is a formula which expresses the S- in terms of the T-polynomials. - Wolfdieter Lang, Aug 07 2014
From Tom Copeland, Dec 06 2015: (Start)
The non-vanishing, unsigned subdiagonals Diag_(2n) contain the elements D(n,k) = Sum_{j=0..k} D(n-1,j) = (k+1) (k+2) ... (k+n) / n! = binomial(n+k,n), so the o.g.f. for the subdiagonal is (1-x)^(-(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 non-vanishing antidiagonals are signed rows of the Pascal triangle A007318.
For a reversed, unsigned version with the zeros removed, see A011973. (End)
The Boas-Buck recurrence (see a comment above) for the sequence of column k is: S(n, k) = ((k+1)/(n-k))*Sum_{p=0..n-1-k} (1 - (-1)^p)*(-1)^((p+1)/2) * S(n-1-p, k), for n > k >= 0 and input S(k, k) = 1. - Wolfdieter Lang, Aug 11 2017
The m-th row consecutive nonzero entries in order are (-1)^c*(c+b)!/c!b! with c = m/2, m/2-1, ..., 0 and b = m-2c if m is even and with c = (m-1)/2, (m-1)/2-1, ..., 0 with b = m-2c 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 n-th Chebyshev polynomial of the first kind. - Peter Bala, Aug 15 2022

A002193 Decimal expansion of square root of 2.

Original entry on oeis.org

1, 4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5, 0, 4, 8, 8, 0, 1, 6, 8, 8, 7, 2, 4, 2, 0, 9, 6, 9, 8, 0, 7, 8, 5, 6, 9, 6, 7, 1, 8, 7, 5, 3, 7, 6, 9, 4, 8, 0, 7, 3, 1, 7, 6, 6, 7, 9, 7, 3, 7, 9, 9, 0, 7, 3, 2, 4, 7, 8, 4, 6, 2, 1, 0, 7, 0, 3, 8, 8, 5, 0, 3, 8, 7, 5, 3, 4, 3, 2, 7, 6, 4, 1, 5, 7

Offset: 1

N. J. A. Sloane

Sometimes called Pythagoras's constant.
Its continued fraction expansion is [1; 2, 2, 2, ...] (see A040000). - Arkadiusz Wesolowski, Mar 10 2012
The discovery of irrational numbers is attributed to Hippasus of Metapontum, who may have proved that sqrt(2) is not a rational number; thus sqrt(2) is often regarded as the earliest known irrational number. - Clark Kimberling, Oct 12 2017
From Clark Kimberling, Oct 12 2017: (Start)
In the first million digits,
  0 occurs  99814 times;
  1 occurs  99925 times;
  2 occurs 100436 times;
  3 occurs 100190 times;
  4 occurs 100024 times;
  5 occurs 100155 times;
  6 occurs  99886 times;
  7 occurs 100008 times;
  8 occurs 100441 times;
  9 occurs 100121 times. (End)
Diameter of a sphere whose surface area equals 2*Pi. More generally, the square root of x is also the diameter of a sphere whose surface area equals x*Pi. - Omar E. Pol, Nov 10 2018
Sqrt(2) = 1 + area of region bounded by y = sin x, y = cos x, and x = 0. - Clark Kimberling, Jul 03 2020
Also aspect ratio of the ISO 216 standard for paper sizes. - Stefano Spezia, Feb 24 2021
The standard deviation of a roll of a 5-sided die. - Mohammed Yaseen, Feb 23 2023
From Michal Paulovic, Mar 22 2023: (Start)
The length of a unit square diagonal.
The infinite tetration (power tower) sqrt(2)^(sqrt(2)^(sqrt(2)^(...))) equals 2 from the identity (x^(1/x))^((x^(1/x))^((x^(1/x))^(...))) = x where 1/e <= x <= e. (End)

			1.41421356237309504880168872420969807856967187537694807317667...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 182.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, Section 1.1.
David Flannery, The Square Root of 2, Copernicus Books Springer-Praxis Pub. 2006.
Martin Gardner, Gardner's Workout, Chapter 2 "The Square Root of 2=1.414213562373095..." pp. 9-19 A. K. Peters MA 2002.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.4 Irrational Numbers and §4.4 Powers and Roots, pp. 84, 145.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 64-67.
B. Rittaud, Le fabuleux destin de sqrt(2), Le Pommier, Paris 2006.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 60, page 605.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, pp. 34-35.

Harry J. Smith, Table of n, a(n) for n = 1..20000
D. and J. Ensley, Review of "The Square Root of 2" by D. Flannery.
Steven R. Finch, Errata and Addenda to Mathematical Constants, arXiv:2001.00578 [math.HO], 2020.
M. F. Jones, 22900D approximations to the square roots of the primes less than 100, Math. Comp., 22 (1968), 234-235.
I. Khavkine, PlanetMath.org, square root of 2 is irrational.
Jason Kimberley, Index of expansions of sqrt(d) in base b.
C. E. Larson, (Avoiding) Proof by Contradiction: sqrt(2) is Not Rational, arXiv:2005.03878 [math.HO], 2020.
Robert Nemiroff and Jerry Bonnell, The Square Root of Two to 1 Million Digits.
Robert Nemiroff and Jerry Bonnell, The Square Root of Two to 5 million digits.
Robert Nemiroff and Jerry Bonnell, The first 10 million digits of the square root of 2.
Simon Plouffe, Plouffe's Inverter, The square root of 2 to 10 million digits.
Simon Plouffe, Generalized expansion of real constants.
M. Ripa and G. Morelli, Retro-analytical Reasoning IQ tests for the High Range, 2013.
Vladimir Ivanovich Smirnov, A course of higher mathematics, vol. 1 , Pergamon Press, 1964, p. 3.
Horace S. Uhler, Many-Figure Approximations To Sqrt(2), And Distribution Of Digits In Sqrt(2) And 1/Sqrt(2), Proc. Nat. Acad. Sci. U. S. A. 37, (1951). 63-67.
Eric Weisstein's World of Mathematics, Pythagoras's Constant.
Eric Weisstein's World of Mathematics, Square Root.
Index entries for algebraic numbers, degree 2.

Cf. A020807, A010503, A001790, A005187.

Cf. A004539 (binary version).

-- After Michael B. Porter's PARI program.
a002193 n = a002193_list !! (n-1)
a002193_list = w 2 0 where
w x r = dig : w (100 * (x - (20 * r + dig) * dig)) (10 * r + dig)
where dig = head (dropWhile (\d -> (20 * r + d) * d < x) [0..]) - 1
-- Reinhard Zumkeller, Nov 22 2013

Digits:=100; evalf(sqrt(2)); # Wesley Ivan Hurt, Dec 04 2013

RealDigits[N[2^(1/2), 128]] (* Vladimir Joseph Stephan Orlovsky, Dec 25 2008 *)

fpprec: 100$ ev(bfloat(sqrt(2))); /* Martin Ettl, Oct 17 2012 */

default(realprecision, 20080); x=sqrt(2); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b002193.txt", n, " ", d)); \\ Harry J. Smith, Apr 21 2009

r=0; x=2; /* Digit-by-digit method */
for(digits=1,100,{d=0;while((20*r+d)*d <= x,d++);
d--; /* while loop overshoots correct digit */
print(d);x=100*(x-(20*r+d)*d);r=10*r+d}) \\ Michael B. Porter, Oct 20 2009

\\ Works in v2.15.0; n = 100 decimal places
my(n=100); digits(floor(10^n*quadgen(8))) \\ Michal Paulovic, Mar 22 2023

Sqrt(2) = 14 * Sum_{n >= 0} (A001790(n)/2^A005187(floor(n/2)) * 10^(-2n-1)) where A001790(n) are numerators in expansion of 1/sqrt(1-x) and the denominators in expansion of 1/sqrt(1-x) are 2^A005187(n). 14 = 2*7, see A010503 (expansion of 1/sqrt(2)). - Gerald McGarvey, Jan 01 2005
Limit_{n -> +oo} (1/n)*(Sum_{k = 1..n} frac(sqrt(1+zeta(k+1)))) = 1/(1+sqrt(2)). - Yalcin Aktar, Jul 14 2005
sqrt(2) = 2 + n*A167199(n-1)/A167199(n) as n -> infinity (conjecture). - Mats Granvik, Oct 30 2009
sqrt(2) = limit as n goes to infinity of A179807(n+1)/A179807(n) - 1. - Mats Granvik, Feb 15 2011
sqrt(2) = Product_{l=0..k-1} 2*cos((2*l+1)*Pi/(4*k)) = (Product_{l=0..k-1} R(2*l+1,rho(4*k)) - 1), identical for k >= 1, with the row polynomials R(n, x) from A127672 and rho(4*k) := 2*cos(Pi/(4*k)) is the length ratio (smallest diagonal)/side in a regular (4*k)-gon. From the product formula given in a Oct 21 2013 formula contribution to A056594, with n -> 2*k, using cos(Pi-alpha) = - cos(alpha) to obtain 2 for the square of the present product. - Wolfdieter Lang, Oct 22 2013
If x = sqrt(2), 1/log(x - 1) + 1/log(x + 1) = 0. - Kritsada Moomuang, Jul 10 2020
From Amiram Eldar, Jul 25 2020: (Start)
Equals Product_{k>=0} (1 + (-1)^k/(2*k + 1)).
Equals Sum_{k>=0} binomial(2*k,k)/8^k. (End)
Equals i^(1/2) + i^(-1/2). - Gary W. Adamson, Jul 11 2022
Equals (sqrt(2) + (sqrt(2) + (sqrt(2) + ...)^(1/3))^(1/3))^(1/3). - Michal Paulovic, Mar 22 2023
Equals 1 + Sum_{k>=1} (-1)^(k-1)/(2^(2*k)*(2*k - 1))*binomial(2*k,k) [Newton]. - Stefano Spezia, Oct 15 2024
From Antonio Graciá Llorente, Dec 19 2024: (Start)
Equals Sum_{k>=0} 2*k*binomial(2*k,k)/8^k.
Equals Product_{k>=2}  k/sqrt(k^2 + 1).
Equals Product_{k>=0} (6*k + 3)/((6*k + 3) - (-1)^k).
Equals Product_{k>=1} (2*k + 1)/((2*k + 1) + (-1)^k).
Equals Product_{k>=0} ((4*k + 3)*(4*k + 1 + (-1)^k))/((4*k + 1)*(4*k + 3 + (-1)^k)). (End)
Equals hypergeom([1/2, 1/2], [1/2], 1/2). - Stefano Spezia, Jan 05 2025

A033999 a(n) = (-1)^n.

Original entry on oeis.org

1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1

Offset: 0

Vasiliy Danilov (danilovv(AT)usa.net), Jun 15 1998

(-1)^(n+1) = signed area of parallelogram with vertices (0,0), U=(F(n),F(n+1)), V=(F(n+1),F(n+2)), where F = A000045 (Fibonacci numbers). The area of every such parallelogram is 1. The signed area is -1 if and only if F(n+1)^2 > F(n)*F(n+2), or, equivalently, n is even, or, equivalently, the vector U is "above" V, indicating that U and V "cross" as n -> n+1. - Clark Kimberling, Sep 09 2013
Periodic with period length 2. - Ray Chandler, Apr 03 2017
From Bernard Schott, May 11 2022: (Start)
Cesàro mean theorem: When a(n) has a limit (finite or infinite) in the usual sense, then c(n) = (a(1)+...+a(n))/n has the same Cesàro limit, but the converse is false. This sequence is a counterexample in the case of a finite Cesàro limit (see A237420 for counterexample with an infinite Cesàro limit).
This sequence is not convergent in the usual sense because a(2n) = 1 while a(2n+1) = -1; the successive arithmetic means c(n) of the first n terms of the sequence are 1/1, 0/2, 1/3, 0/4, 1/5, 0/6, ... so c(2n) = 1/(2n+1) and c(2n+1) = 0, hence the Cesàro limit is 0 because c(n) -> 0 when n -> oo.
In fact, when sequence a(n) is "Period k: [a1, a2, ..., ak]", then the Cesàro limit c of this sequence is (a1+a2+...+ak)/k.
Note that the converse of the theorem is true iff a(n) is monotonic (End).

			G.f. = 1 - x + x^2 - x^3 + x^4 - x^5 + x^6 - x^7 + x^8 - x^9 + x^10 - x^11 + x^12 + ...

J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.

Antti Karttunen, Table of n, a(n) for n = 0..10000
Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.
F. Javier de Vega, An extension of Furstenberg's theorem of the infinitude of primes, arXiv:2003.13378 [math.NT], 2020.
S. K. Ghosal and J. K. Mandal, Stirling Transform Based Color Image Authentication, Procedia Technology, 2013 Volume 10, 2013, Pages 95-104.
Tanya Khovanova, Recursive Sequences
Mathematics Stack Exchange, Convergence of series implies convergence of Cesàro mean, 2013.
László Németh, The trinomial transform triangle, J. Int. Seqs., Vol. 21 (2018), Article 18.7.3. Also arXiv:1807.07109 [math.NT], 2018.
ProofWiki, Cesàro mean.
Michael Somos, Rational Function Multiplicative Coefficients
Eric Weisstein's World of Mathematics, Inverse Tangent
Eric Weisstein's World of Mathematics, Stirling Transform
Wikipedia, Ernesto Cesàro.
Wikipedia, Grandi's series
Wikipedia, +/-1-sequence
Wikipedia, Dirichlet eta function
Wikipédia, Lemme de Cesàro (in French).
Index entries for linear recurrences with constant coefficients, signature (-1).
Index to sequences related to inverse of cyclotomic polynomials

Cf. A056594, A059841, A063007, A122803.
About Cesàro mean theorem: A114112, A237420.
Cf. A072691 (abs. val. Dgf at s=2), A197070 (abs. val. Dgf at s=3), A267315 (abs. val. Dgf at s=4).

a033999 = (1 -) . (* 2) . (`mod` 2)
a033999_list = cycle [1,-1] -- Reinhard Zumkeller, May 06 2012, Jan 02 2012

[(-1)^n : n in [0..100]]; // Wesley Ivan Hurt, Nov 19 2014

A033999 := n->(-1)^n: seq(A033999(n), n=0..100);

Table[(-1)^n, {n, 0, 88}] (* Alonso del Arte, Nov 30 2009 *)
PadRight[{}, 89, {1, -1}] (* Arkadiusz Wesolowski, Sep 16 2012 *)

a(n)=1-2*(n%2) /* Jaume Oliver Lafont, Mar 20 2009 */

def A033999(n): return -1 if n % 2 else 1 # Chai Wah Wu, May 24 2022

G.f.: 1/(1+x).
E.g.f.: exp(-x).
Linear recurrence: a(0)=1, a(n)=-a(n-1) for n>0. - Jaume Oliver Lafont, Mar 20 2009
Sum_{k=0..n} a(k) = A059841(n). - Jaume Oliver Lafont, Nov 21 2009
Sum_{k>=0} a(k)/(k+1) = log(2). - Jaume Oliver Lafont, Mar 30 2010
Euler transform of length 2 sequence [ -1, 1]. - Michael Somos, Mar 21 2011
Moebius transform is length 2 sequence [ -1, 2]. - Michael Somos, Mar 21 2011
a(n) = -b(n) where b(n) = multiplicative with b(2^e) = -1 if e>0, b(p^e) = 1 if p>2. - Michael Somos, Mar 21 2011
a(n) = a(-n) = a(n + 2) = cos(n * Pi). a(n) = c_2(n) if n>1 where c_k(n) is Ramanujan's sum. - Michael Somos, Mar 21 2011
a(n) = (1/2)*Product_{k=0..2*n-1} 2*cos((2*k+1)*Pi/(4*n)), n >= 1. See the product given in the Oct 21 2013 formula comment in A056594, and replace there n -> 2*n. - Wolfdieter Lang, Oct 23 2013
D.g.f.: (2^(1-s)-1)*zeta(s) = -eta(s) (the Dirichlet eta function). - Ralf Stephan, Mar 27 2015
From Ilya Gutkovskiy, Aug 17 2016: (Start)
a(n) = T_n(-1), where T_n(x) are the Chebyshev polynomials of the first kind.
Binomial transform of A122803. (End)
a(n) = exp(i*Pi*n) = exp(-i*Pi*n). - Carauleanu Marc, Sep 15 2016
a(n) = Sum_{k=0..n} (-1)^k*A063007(n, k), n >= 0. - Wolfdieter Lang, Sep 13 2016

A008288 Square array of Delannoy numbers D(i,j) (i >= 0, j >= 0) read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 13, 7, 1, 1, 9, 25, 25, 9, 1, 1, 11, 41, 63, 41, 11, 1, 1, 13, 61, 129, 129, 61, 13, 1, 1, 15, 85, 231, 321, 231, 85, 15, 1, 1, 17, 113, 377, 681, 681, 377, 113, 17, 1, 1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1, 1, 21, 181, 833, 2241, 3653, 3653

Offset: 0

N. J. A. Sloane

In the Formula section, some contributors use T(n,k) = D(n-k, k) (for 0 <= k <= n), which is the triangular version of the square array (D(n,k): n,k >= 0). Conversely, D(n,k) = T(n+k,k) for n,k >= 0. - Petros Hadjicostas, Aug 05 2020
Also called the tribonacci triangle [Alladi and Hoggatt (1977)]. - N. J. A. Sloane, Mar 23 2014
D(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0), (0,1), (1,1). - Joerg Arndt, Jul 01 2011 [Corrected by N. J. A. Sloane, May 30 2020]
Or, triangle read by rows of coefficients of polynomials P[n](x) defined by P[0] = 1, P[1] = x+1; for n >= 2, P[n] = (x+1)*P[n-1] + x*P[n-2].
D(n, k) is the number of k-matchings of a comb-like graph with n+k teeth. Example: D(1, 3) = 7 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has seven 3-matchings: four triples of three teeth and the three triples {Aa, Bb, CD}, {Aa, Dd, BC}, {Cc, Dd, AB}. Also D(3, 1)=7, the 1-matchings of the same graph being the seven edges: {AB}, {BC}, {CD}, {Aa}, {Bb}, {Cc}, {Dd}. - Emeric Deutsch, Jul 01 2002
Sum of n-th antidiagonal of the array D is A000129(n+1). - Reinhard Zumkeller, Dec 03 2004 [Edited by Petros Hadjicostas, Aug 05 2020 so that the counting of antidiagonals of D starts at n = 0. That is, the sum of the terms in the n-th row of the triangles T is  A000129(n+1).]
The A-sequence for this Riordan type triangle (see one of Paul Barry's comments under Formula) is A112478 and the Z-sequence the trivial: {1, 0, 0, 0, ...}. See the W. Lang link under A006232 for Sheffer a- and z-sequences where also Riordan A- and Z-sequences are explained. This leads to the recurrence for the triangle given below. - Wolfdieter Lang, Jan 21 2008
The triangle or chess sums, see A180662 for their definitions, link the Delannoy numbers with twelve different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 and Kn15 have been added. It is remarkable that all knight sums are related to the tribonacci numbers, that is, A000073 and A001590, but none of the others. - Johannes W. Meijer, Sep 22 2010
This sequence, A008288, is jointly generated with A035607 as an array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 09 2012
Row n, for n > 0, of Roger L. Bagula's triangle in the Example section shows the coefficients of the polynomial u(n) = c(0) + c(1)*x + ... + c(n)*x^n which is the numerator of the n-th convergent of the continued fraction [k, k, k, ...], where k = sqrt(x) + 1/sqrt(x); see A230000. - Clark Kimberling, Nov 13 2013
In an n-dimensional hypercube lattice, D(n,k) gives the number of nodes situated at a Minkowski (Manhattan) distance of k from a given node. In cellular automata theory, the cells at Manhattan distance k are called the von Neumann neighborhood of radius k. For k=1, see A005843. - Dmitry Zaitsev, Dec 10 2015
These numbers appear as the coefficients of series relating spherical and bispherical harmonics, in the solutions of Laplace's equation in 3D. [Majic 2019, Eq. 22] - Matt Majic, Nov 24 2019
From Peter Bala, Feb 19 2020: (Start)
The following remarks assume an offset of 1 in the row and column indices of the triangle.
The sequence of row polynomials T(n,x), beginning with T(1,x) = x, T(2,x) = x + x^2, T(3,x) = x + 3*x^2 + x^3, ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd(T(n,x), T(m,x)) = T(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the sequence (T(n,x): n >= 1) is a divisibility sequence in the polynomial ring Z[x]; that is, if n divides m then T(n,x) divides T(m,x) in Z[x].
Let S(x) = 1 + 2*x + 6*x^2 + 22*x^3 + ... denote the o.g.f. for the large Schröder numbers A006318. The power series (x*S(x))^n, n = 2, 3, 4, ..., can be expressed as a linear combination with polynomial coefficients of S(x) and 1: (x*S(x))^n = T(n-1,-x) - T(n,-x)*S(x). The result can be extended to negative integer n if we define T(0,x) = 0 and T(-n,x) = (-1)^(n+1) * T(n,x)/x^n. Cf. A115139.
[In the previous two paragraphs, D(n,x) was replaced with T(n,x) because the contributor is referring to the rows of the triangle T(n,k), not the rows of the array D(n,k). - Petros Hadjicostas, Aug 05 2020] (End)
Named after the French amateur mathematician Henri-Auguste Delannoy (1833-1915). - Amiram Eldar, Apr 15 2021
D(i,j) = D(j,i). With this and Dmitry Zaitsev's Dec 10 2015 comment, D(i,j) can be considered the number of points at L1 distance <= i in Z^j or the number of points at L1 distance <= j in Z^i from any given point. The rows and columns of D(i,j) are the crystal ball sequences on cubic lattices. See the first example below. The n-th term in the k-th crystal ball sequence can be considered the number of points at distance <= n from any point in a k-dimensional cubic lattice, or the number of points at distance <= k from any point in an n-dimensional cubic lattice. - Shel Kaphan, Jan 01 2023 and Jan 07 2023
Dimensions of hom spaces Hom(R^{(i)}, R^{(j)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023

			The square array D(i,j) (i >= 0, j >= 0) begins:
  1, 1,  1,   1,   1,   1,    1,    1,    1,    1, ... = A000012
  1, 3,  5,   7,   9,  11,   13,   15,   17,   19, ... = A005408
  1, 5, 13,  25,  41,  61,   85,  113,  145,  181, ... = A001844
  1, 7, 25,  63, 129, 231,  377,  575,  833, 1159, ... = A001845
  1, 9, 41, 129, 321, 681, 1289, 2241, 3649, 5641, ... = A001846
  ...
For D(2,5) = 61, which is seen above in the row labeled A001844, we calculate the sum (9 + 11 + 41) of the 3 nearest terms above and/or to the left. - _Peter Munn_, Jan 01 2023
D(2,5) = 61 can also be obtained from the row labeled A005408 using a recurrence mentioned in the formula section:  D(2,5) = D(1,5) + 2*Sum_{k=0..4} D(1,k), so D(2,5) = 11 + 2*(1+3+5+7+9) = 11 + 2*25. - _Shel Kaphan_, Jan 01 2023
As a triangular array (on its side) this begins:
   0,   0,   0,   0,   1,   0,  11,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   1,   0,   3,   0,  13,   0,  63,   0, ...
   0,   1,   0,   5,   0,  25,   0, 129, ...
   0,   0,   1,   0,   7,   0,  41,   0, ...
   0,   0,   0,   1,   0,   9,   0,  61, ...
   0,   0,   0,   0,   1,   0,  11,   0, ...
   [Edited by _Shel Kaphan_, Jan 01 2023]
From _Roger L. Bagula_, Dec 09 2008: (Start)
As a triangle T(n,k) (with rows n >= 0 and columns k = 0..n), this begins:
   1;
   1,  1;
   1,  3,   1;
   1,  5,   5,   1;
   1,  7,  13,   7,    1;
   1,  9,  25,  25,    9,    1;
   1, 11,  41,  63,   41,   11,    1;
   1, 13,  61, 129,  129,   61,   13,   1;
   1, 15,  85, 231,  321,  231,   85,  15,   1;
   1, 17, 113, 377,  681,  681,  377, 113,  17,  1;
   1, 19, 145, 575, 1289, 1683, 1289, 575, 145, 19, 1;
   ... (End)
Triangle T(n,k) recurrence: 63 = T(6,3) = 25 + 13 + 25 = T(5,2) + T(4,2) + T(5,3).
Triangle T(n,k) recurrence with A-sequence A112478: 63 = T(6,3) = 1*25 + 2*25 - 2*9 + 6*1 (T entries from row n = 5 only). [Here the formula T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j) is used with n = 6 and k = 3; i.e., T(6,3) = Sum_{j=0..3} A111478(j) * T(5, 2+j). - _Petros Hadjicostas_, Aug 05 2020]
From _Philippe Deléham_, Mar 29 2012: (Start)
Subtriangle of the triangle given by (1, 0, 1, -1, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   1,  0;
   1,  1,  0;
   1,  3,  1,  0;
   1,  5,  5,  1,  0;
   1,  7, 13,  7,  1,  0;
   1,  9, 25, 25,  9,  1, 0;
   1, 11, 41, 63, 41, 11, 1, 0;
   ...
Subtriangle of the triangle given by (0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) where DELTA is the operator defined in A084938:
   1;
   0, 1;
   0, 1,  1;
   0, 1,  3,  1;
   0, 1,  5,  5,  1;
   0, 1,  7, 13,  7,  1;
   0, 1,  9, 25, 25,  9,  1;
   0, 1, 11, 41, 63, 41, 11, 1;
   ... (End)

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Sums of antidiagonals: A000129 (Pell numbers).
Main diagonal: A001850 (central Delannoy numbers), which has further information and references.
A002002, A026002, and A190666 are +-k-diagonals for k=1, 2, 3 resp. - Shel Kaphan, Jan 01 2023
Rows 0..10: A000012, A005408, A001844, A001845, A001846, A001847, A001848, A001849, A008417, A008419, A008421.
See also A027618.
Cf. A059446.
Has same main diagonal as A064861. Different from A100936.
Cf. A101164, A101167, A128966, A131887, A131935.
Cf. A035607, A108625, A142979, A142992, A143007.
Read mod small primes: A211312, A211313, A211314, A211315.
Triangle sums (see the comments): A000129 (Row1); A056594 (Row2); A000073 (Kn11 & Kn21); A089068 (Kn12 & Kn22); A180668 (Kn13 & Kn23); A180669 (Kn14 & Kn24); A180670 (Kn15 & Kn25); A099463 (Kn3 & Kn4); A116404 (Fi1 & Fi2); A006498 (Ca1 & Ca2); A006498(3*n) (Ca3 & Ca4); A079972 (Gi1 & Gi2); A079972(4*n) (Gi3 & Gi4); A079973(3*n) (Ze1 & Ze2); A079973(2*n) (Ze3 & Ze4).
Cf. A102413, A128966. (D(n,1)) = A005843. Cf. A115139.

a008288 n k = a008288_tabl !! n !! k
a008288_row n = a008288_tabl !! n
a008288_tabl = map fst $ iterate
    (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                       zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
-- Reinhard Zumkeller, Jul 21 2013

A008288 := proc(n, k) option remember; if k = 0 then 1 elif n=k then 1 else procname(n-1, k-1) + procname(n-2, k-1) + procname(n-1, k) end if; end proc: seq(seq(A008288(n,k),k=0..n), n=0..10); # triangular indices n and k
P[0]:=1; P[1]:=x+1; for n from 2 to 12 do P[n]:=expand((x+1)*P[n-1]+x*P[n-2]); lprint(P[n]); lprint(seriestolist(series(P[n],x,200))); end do:

(* Next, A008288 jointly generated with A035607 *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A008288 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A035607 *)
(* Clark Kimberling, Mar 09 2012 *)
d[n_, k_] := Binomial[n+k, k]*Hypergeometric2F1[-k, -n, -n-k, -1]; A008288 = Flatten[Table[d[n-k, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Apr 05 2012, after 3rd formula *)

from functools import cache
@cache
def delannoy_row(n: int) -> list[int]:
    if n == 0: return [1]
    if n == 1: return [1, 1]
    rov = delannoy_row(n - 2)
    row = delannoy_row(n - 1) + [1]
    for k in range(n - 1, 0, -1):
        row[k] += row[k - 1] + rov[k - 1]
    return row
for n in range(10): print(delannoy_row(n))  # Peter Luschny, Jul 30 2023

for k in range(8):  # seen as an array, read row by row
    a = lambda n: hypergeometric([-n, -k], [1], 2)
    print([simplify(a(n)) for n in range(11)]) # Peter Luschny, Nov 19 2014

D(n, 0) = 1 = D(0, n) for n >= 0; D(n, k) = D(n, k-1) + D(n-1, k-1) + D(n-1, k).
Bivariate o.g.f.: Sum_{n >= 0, k >= 0} D(n, k)*x^n*y^k = 1/(1 - x - y - x*y).
D(n, k) = Sum_{d = 0..min(n,k)} binomial(k, d)*binomial(n+k-d, k) = Sum_{d=0..min(n,k)} 2^d*binomial(n, d)*binomial(k, d). [Edited by Petros Hadjicostas, Aug 05 2020]
Seen as a triangle read by rows: T(n, 0) = T(n, n) = 1 for n >= 0 and T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-1, k), 0 < k < n and n > 1. - Reinhard Zumkeller, Dec 03 2004
Read as a number triangle, this is the Riordan array (1/(1-x), x(1+x)/(1-x)) with T(n, k) = Sum_{j=0..n-k} C(n-k, j) * C(k, j) * 2^j. - Paul Barry, Jul 18 2005
T(n,k) = Sum_{j=0..n-k} C(k,j)*C(n-j,k). - Paul Barry, May 21 2006
Let y^k(n) be the number of Khalimsky-continuous functions f from [0,n-1] to Z such that f(0) = 0 and f(n-1) = k. Then y^k(n) = D(i,j) for i = (1/2)*(n-1-k) and j = (1/2)*(n-1+k) where n-1+k belongs to 2Z. - Shiva Samieinia (shiva(AT)math.su.se), Oct 08 2007
Recurrence for triangle from A-sequence (see the Wolfdieter Lang comment above): T(n,k) = Sum_{j=0..n-k} A112478(j) * T(n-1, k-1+j), n >= 1, k >= 1. [For k > n, the sum is empty, in which case T(n,k) = 0.]
From Peter Bala, Jul 17 2008: (Start)
The n-th row of the square array is the crystal ball sequence for the product lattice A_1 x ... x A_1 (n copies). A035607 is the table of the associated coordination sequences for these lattices.
The polynomial p_n(x) := Sum {k = 0..n} 2^k * C(n,k) * C(x,k) = Sum_{k = 0..n} C(n,k) * C(x+k,n), whose values [p_n(0), p_n(1), p_n(2), ... ] give the n-th row of the square array, is the Ehrhart polynomial of the n-dimensional cross polytope (the hyperoctahedron) [Bump et al. (2000), Theorem 6].
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The reciprocity law p_n(m) = p_m(n) reflects the symmetry of the table.
The polynomial p_n(x) is the unique polynomial solution of the difference equation (x+1)*f(x+1) - x*f(x-1) = (2*n+1)*f(x), normalized so that f(0) = 1.
These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis [Bump et al. (2000), Theorem 4]. The o.g.f. for the p_n(x) is (1 + t)^x/(1 - t)^(x + 1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ... .
The square array of Delannoy numbers has a close connection with the constant log(2). The entries in the n-th row of the array occur in the series acceleration formula log(2) = (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n) + (-1)^n * Sum_{k>=1} (-1)^(k+1)/(k*D(n,k-1)*D(n,k)). [T(n,k) was replaced with D(n,k) in the formula to agree with the beginning of the paragraph. - Petros Hadjicostas, Aug 05 2020]
For example, the fourth row of the table (n = 3) gives the series log(2) = 1 - 1/2 + 1/3 - 1/(1*1*7) + 1/(2*7*25) - 1/(3*25*63) + 1/(4*63*129) - ... . See A142979 for further details.
Also the main diagonal entries (the central Delannoy numbers) give the series acceleration formula Sum_{n>=1} 1/(n*D(n-1,n-1)*D(n,n)) = (1/2)*log(2), a result due to Burnside. [T(n,n) was replaced here with D(n,n) to agree with the previous paragraphs. - Petros Hadjicostas, Aug 05 2020]
Similar relations hold between log(2) and the crystal ball sequences of the C_n lattices A142992. For corresponding results for the constants zeta(2) and zeta(3), involving the crystal ball sequences for root lattices of type A_n and A_n x A_n, see A108625 and A143007 respectively. (End)
From Peter Bala, Oct 28 2008: (Start)
Hilbert transform of Pascal's triangle A007318 (see A145905 for the definition of this term).
D(n+a,n) = P_n(a,0;3) for all integer a such that a >= -n, where P_n(a,0;x) is the Jacobi polynomial with parameters (a,0) [Hetyei]. The related formula A(n,k) = P_k(0,n-k;3) defines the table of asymmetric Delannoy numbers, essentially A049600. (End)
Seen as a triangle read by rows: T(n, k) = Hyper2F1([k-n, -k], [1], 2). - Peter Luschny, Aug 02 2014, Oct 13 2024.
From Peter Bala, Jun 25 2015: (Start)
O.g.f. for triangle T(n,k): A(z,t) = 1/(1 - (1 + t)*z - t*z^2) = 1 + (1 + t)*z + (1 + 3*t + t^2)*z^2 + (1 + 5*t + 5*t^2 + t^3)*z^3 + ....
1 + z*d/dz(A(z,t))/A(z,t) is the o.g.f. for A102413. (End)
E.g.f. for the n-th subdiagonal of T(n,k), n >= 0, equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(2*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 4*x + 4*x^2/2) = 1 + 5*x + 13*x^2/2! + 25*x^3/3! + 41*x^4/4! + 61*x^5/5! + .... - Peter Bala, Mar 05 2017 [The n-th subdiagonal of triangle T(n,k) is the n-th row of array D(n,k).]
Let a_i(n) be multiplicative with a_i(p^e) = D(i, e), p prime and e >= 0, then Sum_{n > 0} a_i(n)/n^s = (zeta(s))^(2*i+1)/(zeta(2*s))^i for i >= 0. - Werner Schulte, Feb 14 2018
Seen as a triangle read by rows: T(n,k) = Sum_{i=0..k} binomial(n-i, i) * binomial(n-2*i, k-i) for 0 <= k <= n. - Werner Schulte, Jan 09 2019
Univariate generating function: Sum_{k >= 0} D(n,k)*z^k = (1 + z)^n/(1 - z)^(n+1). [Dziemianczuk (2013), Eq. 5.3] - Matt Majic, Nov 24 2019
(n+1)*D(n+1,k) = (2*k+1)*D(n,k) + n*D(n-1,k). [Majic (2019), Eq. 22] - Matt Majic, Nov 24 2019
For i, j >= 1, D(i,j) = D(i,j-1) + 2*Sum_{k=0..i-1} D(k,j-1), or, because D(i,j) = D(j,i), D(i,j) = D(i-1,j) + 2*Sum_{k=0..j-1} D(i-1,k). - Shel Kaphan, Jan 01 2023
Sum_{k=0..n} T(n,k)^2 = A026933(n). - R. J. Mathar, Nov 07 2023
Let S(x) = (1 - x - (1 - 6*x + x^2)^(1/2))/(2*x) denote the g.f. of the sequence of large Schröder numbers A006318. Read as a lower triangular array, the signed n-th row polynomial R(n, -x) = 1/sqrt(1 - 6*x + x^2) *( 1/S(x)^(n+1) + (x*S(x))^(n+1) ). For example, R(4, -x) = 1 - 7*x + 13*x^2 - 7*x^3 + x^4 = 1/sqrt(1 - 6*x + x^2) * ( 1/S(x)^5 + (x*S(x))^5 ). Cf. A102413. - Peter Bala, Aug 01 2024

Expanded description from Clark Kimberling, Jun 15 1997
Additional references from Sylviane R. Schwer (schwer(AT)lipn.univ-paris13.fr), Nov 28 2001
Changed the notation to make the formulas more precise. - N. J. A. Sloane, Jul 01 2002

cp's OEIS Frontend

A075419 Convolution of A073145 with A056594.

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A075495 Convolution of A075298 with A056594.

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A159334 Transform of A056594 by the T_{1,1} transformation (see link).

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A159339 Transform of A056594 by the T_{1,0} transformation (see link).

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A159343 Transform of A056594 by the T_{0,1} transformation (see link).

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A159350 Transform of A056594 by the T_{0,0} transformation (see link).

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A049310 Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).

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