A004187 -id:A004187 - 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

A001906 F(2n) = bisection of Fibonacci sequence: a(n) = 3*a(n-1) - a(n-2).

Original entry on oeis.org

0, 1, 3, 8, 21, 55, 144, 377, 987, 2584, 6765, 17711, 46368, 121393, 317811, 832040, 2178309, 5702887, 14930352, 39088169, 102334155, 267914296, 701408733, 1836311903, 4807526976, 12586269025, 32951280099, 86267571272, 225851433717, 591286729879, 1548008755920

Offset: 0

N. J. A. Sloane

Apart from initial term, same as A088305.
Second column of array A102310 and of A028412.
Numbers k such that 5*k^2 + 4 is a square. - Gregory V. Richardson, Oct 13 2002
Apart from initial terms, also Pisot sequences E(3,8), P(3,8), T(3,8). See A008776 for definitions of Pisot sequences.
Binomial transform of A000045. - Paul Barry, Apr 11 2003
Number of walks of length 2n+1 in the path graph P_4 from one end to the other one. Example: a(2)=3 because in the path ABCD we have ABABCD, ABCBCD and ABCDCD. - Emeric Deutsch, Apr 02 2004
Simplest example of a second-order recurrence with the sixth term a square.
Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 3. - Lekraj Beedassy, Jun 11 2004
a(n) (for n > 0) is the smallest positive integer that cannot be created by summing at most n values chosen among the previous terms (with repeats allowed). - Andrew Weimholt, Jul 20 2004
All nonnegative integer solutions of Pell equation b(n)^2 - 5*a(n)^2 = +4 together with b(n) = A005248(n), n >= 0. - Wolfdieter Lang, Aug 31 2004
a(n+1) is a Chebyshev transform of 3^n (A000244), where the sequence with g.f. G(x) is sent to the sequence with g.f. (1/(1+x^2))G(x/(1+x^2)). - Paul Barry, Oct 25 2004
a(n) is the number of distinct products of matrices A, B, C, in (A+B+C)^n where commutator [A,B] = 0 but C does not commute with A or B. - Paul D. Hanna and Max Alekseyev, Feb 01 2006
Number of binary words with exactly k-1 strictly increasing runs. Example: a(3)=F(6)=8 because we have 0|0,1|0,1|1,0|01,01|0,1|01,01|1 and 01|01. Column sums of A119900. - Emeric Deutsch, Jul 23 2006
See Table 1 on page 411 of Lukovits and Janezic paper. - Parthasarathy Nambi, Aug 22 2006
Inverse: With phi = (sqrt(5) + 1)/2, log_phi((sqrt(5) a(n) + sqrt(5 a(n)^2 + 4))/2) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007
[1,3,8,21,55,144,...] is the Hankel transform of [1,1,4,17,75,339,1558,...](see A026378). - Philippe Deléham, Apr 13 2007
The Diophantine equation a(n) = m has a solution (for m >= 1) if and only if floor(arcsinh(sqrt(5)*m/2)/log(phi)) <> floor(arccosh(sqrt(5)*m/2)/log(phi)) where phi is the golden ratio. An equivalent condition is A130259(m) = A130260(m). - Hieronymus Fischer, May 25 2007
a(n+1) = AB^(n)(1), n >= 0, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 1=`1`, 3=`10`, 8=`100`, 21=`1000`, ..., in Wythoff code.
Equals row sums of triangles A140069, A140736 and A140737. - Gary W. Adamson, May 25 2008
a(n) is also the number of idempotent order-preserving partial transformations (of an n-element chain) of width n (width(alpha) = max(Im(alpha))). Equivalently, it is the number of idempotent order-preserving full transformations (of an n-element chain). - Abdullahi Umar, Sep 08 2008
a(n) is the number of ways that a string of 0,1 and 2 of size (n-1) can be arranged with no 12-pairs. - Udita Katugampola, Sep 24 2008
Starting with offset 1 = row sums of triangle A175011. - Gary W. Adamson, Apr 03 2010
As a fraction: 1/71 = 0.01408450... or 1/9701 = 0.0001030821.... - Mark Dols, May 18 2010
Sum of the products of the elements in the compositions of n (example for n=3: the compositions are 1+1+1, 1+2, 2+1, and 3; a(3) = 1*1*1 + 1*2 + 2*1 + 3 = 8). - Dylon Hamilton, Jun 20 2010, Geoffrey Critzer, Joerg Arndt, Dec 06 2010
a(n) relates to regular polygons with even numbers of edges such that Product_{k=1..(n-2)/2} (1 + 4*cos^2 k*Pi/n) = even-indexed Fibonacci numbers with a(n) relating to the 2*n-gons. The constants as products = roots to even-indexed rows of triangle A152063. For example: a(5) = 55 satisfies the product formula relating to the 10-gon. - Gary W. Adamson, Aug 15 2010
Alternatively, product of roots to x^4 - 12x^3 + 51x^2 - 90x + 55, (10th row of triangle A152063) = (4.618...)*(3.618...)*(2.381...)*(1.381...) = 55. - Gary W. Adamson, Aug 15 2010
a(n) is the number of generalized compositions of n when there are i different types of i, (i=1,2,...). - Milan Janjic, Aug 26 2010
Starting with "1" = row sums of triangle A180339, and eigensequence of triangle A137710. - Gary W. Adamson, Aug 28 2010
a(2) = 3 is the only prime.
Number of nonisomorphic graded posets with 0 and uniform hasse graph of rank n > 0, with exactly 2 elements of each rank level above 0. (Uniform used in the sense of Retakh, Serconek, and Wilson.  Graded used in Stanley's sense that every maximal chain has the same length n.) - David Nacin, Feb 13 2012
Pisano period lengths: 1, 3, 4, 3, 10, 12, 8, 6, 12, 30, 5, 12, 14, 24, 20, 12, 18, 12, 9, 30, ... - R. J. Mathar, Aug 10 2012
Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy + 1. - Michel Lagneau, Feb 01 2014
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2}. - Milan Janjic, Jan 25 2015
With a(0) = 0, for n > 1, a(n) is the smallest number not already in the sequence such that a(n)^2 - a(n-1)^2 is a Fibonacci number. - Derek Orr, Jun 08 2015
Let T be the tree generated by these rules: 0 is in T, and if p is in T, then p + 1 is in T and x*p is in T and y*p is in T. The n-th generation of T consists of A001906(n) polynomials, for n >= 0. - Clark Kimberling, Nov 24 2015
For n > 0, a(n) = exactly the maximum area of a quadrilateral with sides in order of lengths F(n), F(n), L(n), and L(n) with L(n)=A000032(n). - J. M. Bergot, Jan 20 2016
a(n) = twice the area of a triangle with vertices at (L(n+1), L(n+2)), (F(n+1), F(n+1)), and (L(n+2), L(n+1)), with L(n)=A000032(n). - J. M. Bergot, Apr 20 2016
Except for the initial 0, this is the p-INVERT of (1,1,1,1,1,...) for p(S) = 1 - S - S^2; see A291000. - Clark Kimberling, Aug 24 2017
a(n+1) is the number of spanning trees of the graph T_n, where T_n is a sequence of n triangles, where adjacent triangles share an edge. - Kevin Long, May 07 2018
a(n) is the number of ways to partition [n] such that each block is a run of consecutive numbers, and each block has a fixed point, e.g., for n=3, 12|3 with 1 and 3 as fixed points is valid, but 13|2 is not valid as 1 and 3 do not form a run. Consequently, a(n) also counts the spanning trees of the graph given by taking a path with n vertices and adding another vertex adjacent to all of them. - Kevin Long, May 11 2018
From Wolfdieter Lang, May 31 2018: (Start)
The preceding comment can be paraphrased as follows. a(n) is the row sum of the array A305309 for n >= 1. The array A305309(n, k) gives the sum of the products of the block lengths of the set partition of [n] := {1, 2, ..., n} with A048996(n, k) blocks of consecutive numbers, corresponding to the compositions obtained from the k-th partition of n in Abramowitz-Stegun order. See the comments and examples at A305309.
{a(n)} also gives the infinite sequence of nonnegative numbers k for which k * ||k*phi|| < 1/sqrt(5), where the irrational number phi = A001622 (golden section), and ||x|| is the absolute value of the difference between x and the nearest integer. See, e.g., the Havil reference, pp. 171-172. (End)
This Chebyshev sequence a(n) = S(n-1, 3) (see a formula below) is related to the bisection of Fibonacci sequences {F(a,b;n)}{n>=0} with input F(a,b;0) = a and F(a,b;1) = b, by F(a,b;2*k) = (a+b)*S(k-1, 3) - a*S(k-2, 3) and F(a,b;2*k+1) = b*S(k, 3) + (a-b)*S(k-1, 3), for k >= 0, and S(-2, 3) = -1. Proof via the o.g.f.s GFeven(a,b,t) = (a - t*(2*a-b))/(1 - 3*t + t^2) and GFodd(a,b,t) = (b + t*(a-b))/(1 - 3*t + t^2). The special case a = 0, b = 1 gives back F(2*k) = S(k-1, 3) = a(k). - _Wolfdieter Lang, Jun 07 2019
a(n) is the number of tilings of two n X 1 rectangles joined orthogonally at a common end-square (so to have 2n-1 squares in a right-angle V shape) with only 1 X 1 and 2 X 1 tiles. This is a consequence of F(2n) = F(n+1)*F(n) + F(n)*F(n-1). - Nathaniel Gregg, Oct 10 2021
These are the denominators of the upper convergents to the golden ratio, tau; they are also the numerators of the lower convergents (viz. 1/1 < 3/2 < 8/5 < 21/13 < ... < tau < ... 13/8 < 5/3 < 2/1). - Clark Kimberling, Jan 02 2022
For n > 1, a(n) is the smallest Fibonacci number of unit equilateral triangle tiles needed to make an isosceles trapezoid of height F(n) triangles. - Kiran Ananthpur Bacche, Sep 01 2024

			G.f. = x + 3*x^2 + 8*x^3 + 21*x^4 + 55*x^5 + 144*x^6 + 377*x^7 + 987*x^8 + ...
a(3) = 8 because there are exactly 8 idempotent order-preserving full transformations on a 3-element chain, namely: (1,2,3)->(1,1,1),(1,2,3)->(2,2,2),(1,2,3)->(3,3,3),(1,2,3)->(1,1,3),(1,2,3)->(2,2,3),(1,2,3)->(1,2,2),(1,2,3)->(1,3,3),(1,2,3)->(1,2,3)-mappings are coordinate-wise. - _Abdullahi Umar_, Sep 08 2008

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N. J. A. Sloane, Transforms
Michael Somos, In the Elliptic Realm.
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Murray Tannock, Equivalence classes of mesh patterns with a dominating pattern, MSc Thesis, Reykjavik Univ., May 2016. See Appendix B2.
Eric Weisstein's World of Mathematics, Fibonacci Hyperbolic Functions.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Math. 46 (2013), 15-27.
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009) 310-329, MR2555042
Index entries for "core" sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-1).
Index entries for two-way infinite sequences

Fibonacci A000045 = union of this sequence and A001519.
Cf. A000041, A000032, A001622, A048996, A052529, A055991, A078812, A305309, A058038 (pairwise products), A153386.
Inverse sequences A130259 and A130260.
Cf. A033888, A033890, A249450, A344212.

a001906 n = a001906_list !! n
a001906_list =
   0 : 1 : zipWith (-) (map (* 3) $ tail a001906_list) a001906_list
-- Reinhard Zumkeller, Oct 03 2011

[Fibonacci(2*n): n in [0..30]]; // Vincenzo Librandi, Sep 10 2014

with(combstruct): SeqSeqSeqL := [T, {T=Sequence(S, card > 0), S=Sequence(U, card > 1), U=Sequence(Z, card >0)}, unlabeled]: seq(count(SeqSeqSeqL, size=n+1), n=0..28); # Zerinvary Lajos, Apr 04 2009
H := (n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -4):
a := n -> `if`(n = 0, 0, H(2*n, 1, 1/2)):
seq(simplify(a(n)), n=0..30); # Peter Luschny, Sep 03 2019
A001906 := proc(n)
    combinat[fibonacci](2*n) ;
end proc:
seq(A001906(n),n=0..20) ; # R. J. Mathar, Jan 11 2024

f[n_] := Fibonacci[2n]; Array[f, 28, 0] (* or *)
LinearRecurrence[{3, -1}, {0, 1}, 28] (* Robert G. Wilson v, Jul 13 2011 *)
Take[Fibonacci[Range[0,60]],{1,-1,2}] (* Harvey P. Dale, May 23 2012 *)
Table[ ChebyshevU[n-1, 3/2], {n, 0, 30}] (* Jean-François Alcover, Jan 25 2013, after Michael Somos *)
CoefficientList[Series[(x)/(1 - 3x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Sep 10 2014 *)

makelist(fib(2*n),n,0,30); /* Martin Ettl, Oct 21 2012 */

numlib::fibonacci(2*n) $ n = 0..35; // Zerinvary Lajos, May 09 2008

{a(n) = fibonacci(2*n)}; /* Michael Somos, Dec 06 2002 */

{a(n) = subst( poltchebi(n+1)*4 - poltchebi(n)*6, x, 3/2)/5}; /* Michael Somos, Dec 06 2002 */

{a(n) = polchebyshev( n-1, 2, 3/2)}; /* Michael Somos Jun 18 2011 */

Vec(x/(1-3*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Oct 24 2012

def a(n, adict={0:0, 1:1}):
    if n in adict:
        return adict[n]
    adict[n]=3*a(n-1) - a(n-2)
    return adict[n] # David Nacin, Mar 04 2012

[lucas_number1(n,3,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

[fibonacci(2*n) for n in range(0, 28)] # Zerinvary Lajos, May 15 2009

G.f.: x / (1 - 3*x + x^2). -  Simon Plouffe in his 1992 dissertation
a(n) = 3*a(n-1) - a(n-2) = A000045(2*n).
a(n) = -a(-n).
a(n) = A060921(n-1, 0), n >= 1.
a(n) = sqrt((A005248(n)^2 - 4)/5).
a(n) = A007598(n) - A007598(n-2), n > 1.
a(n) = (ap^n - am^n)/(ap-am), with ap := (3+sqrt(5))/2, am := (3-sqrt(5))/2.
Invert transform of natural numbers: a(n) = Sum_{k=1..n} k*a(n-k), a(0) = 1. - Vladeta Jovovic, Apr 27 2001
a(n) = S(n-1, 3) with S(n, x) = U(n, x/2) Chebyshev's polynomials of the 2nd kind, see A049310.
a(n) = Sum_{k=0..n} binomial(n, k)*F(k). - Benoit Cloitre, Sep 03 2002
Limit_{n->infinity} a(n)/a(n-1) = 1 + phi = (3 + sqrt(5))/2. This sequence includes all of the elements of A033888 combined with A033890.
a(0)=0, a(1)=1, a(2)=3, a(n)*a(n-2) + 1 = a(n-1)^2. - Benoit Cloitre, Dec 06 2002
a(n) = n + Sum_{k=0..n-1} Sum_{i=0..k} a(i) = n + A054452(n). - Benoit Cloitre, Jan 26 2003
a(n) = Sum_{k=1..n} binomial(n+k-1, n-k). - Vladeta Jovovic, Mar 23 2003
E.g.f.: (2/sqrt(5))*exp(3*x/2)*sinh(sqrt(5)*x/2). - Paul Barry, Apr 11 2003
Second diagonal of array defined by T(i, 1) = T(1, j) = 1, T(i, j) = Max(T(i-1, j) + T(i-1, j-1); T(i-1, j-1) + T(i, j-1)). - Benoit Cloitre, Aug 05 2003
a(n) = F(n)*L(n) = A000045(n)*A000032(n). - Lekraj Beedassy, Nov 17 2003
F(2n+2) = 1, 3, 8, ... is the binomial transform of F(n+2). - Paul Barry, Apr 24 2004
Partial sums of A001519(n). - Lekraj Beedassy, Jun 11 2004
a(n) = Sum_{i=0..n-1} binomial(2*n-1-i, i)*5^(n-i-1)*(-1)^i. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
a(n) = Sum_{k=0..n} binomial(n+k, n-k-1) = Sum_{k=0..n} binomial(n+k, 2k+1).
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*3^(n-2*k). - Paul Barry, Oct 25 2004
a(n) = (n*L(n) - F(n))/5 = Sum_{k=0..n-1} (-1)^n*L(2*n-2*k-1).
The i-th term of the sequence is the entry (1, 2) in the i-th power of the 2 X 2 matrix M = ((1, 1), (1, 2)). - Simone Severini, Oct 15 2005
Computation suggests that this sequence is the Hankel transform of A005807. The Hankel transform of {a(n)} is Det[{{a(1), ..., a(n)}, {a(2), ..., a(n+1)}, ..., {a(n), ..., a(2n-1)}}]. - John W. Layman, Jul 21 2000
a(n+1) = (A005248(n+1) - A001519(n))/2. - Creighton Dement, Aug 15 2004
a(n+1) = Sum_{i=0..n} Sum_{j=0..n} binomial(n-i, j)*binomial(n-j, i). - N. J. A. Sloane, Feb 20 2005
a(n) = (2/sqrt(5))*sinh(2*n*psi), where psi:=log(phi) and phi=(1+sqrt(5))/2. - Hieronymus Fischer, Apr 24 2007
a(n) = ((phi+1)^n - A001519(n))/phi with phi=(1+sqrt(5))/2. - Reinhard Zumkeller, Nov 22 2007
Row sums of triangle A135871. - Gary W. Adamson, Dec 02 2007
a(n)^2 = Sum_{k=1..n} a(2*k-1). This is a property of any sequence S(n) such that S(n) = B*S(n-1) - S(n-2) with S(0) = 0 and S(1) = 1 including {0,1,2,3,...} where B = 2. - Kenneth J Ramsey, Mar 23 2008
a(n) = 1/sqrt(5)*(phi^(2*n+2) - phi^(-2*n-2)), where phi = (1+sqrt(5))/2, the golden ratio. - Udita Katugampola (SIU), Sep 24 2008
If p[i] = i and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i<=j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n) = det(A). - Milan Janjic, May 02 2010
If p[i] = Stirling2(i,2) and if A is the Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i<=j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, May 08 2010
a(n) = F(2*n+10) mod F(2*n+5).
a(n) = 1 + a(n-1) + Sum_{i=1..n-1} a(i), with a(0)=0. - Gary W. Adamson, Feb 19 2011
a(n) is equal to the permanent of the (n-1) X (n-1) Hessenberg matrix with 3's along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
a(n), n > 1 is equal to the determinant of an (n-x) X (n-1) tridiagonal matrix with 3's in the main diagonal, 1's in the super and subdiagonals, and the rest 0's. - Gary W. Adamson, Jun 27 2011
a(n) = b such that Integral_{x=0..Pi/2} sin(n*x)/(3/2-cos(x)) dx = c + b*log(3). - Francesco Daddi, Aug 01 2011
a(n+1) = Sum_{k=0..n} A101950(n,k)*2^k. - Philippe Deléham, Feb 10 2012
G.f.: A(x) = x/(1-3*x+x^2) = G(0)/sqrt(5); where G(k)= 1 -(a^k)/(1 - b*x/(b*x - 2*(a^k)/G(k+1))), a = (7-3*sqrt(5))/2, b = 3+sqrt(5), if |x|<(3-sqrt(5))/2 = 0.3819660...; (continued fraction 3 kind, 3-step ). - Sergei N. Gladkovskii, Jun 25 2012
a(n) = 2^n*b(n;1/2) = -b(n;-1), where b(n;d), n=0,1,...,d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also Witula's et al. papers). - Roman Witula, Jul 12 2012
Product_{n>=1} (1 + 1/a(n)) = 1 + sqrt(5). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = (1/6)*(1 + sqrt(5)). - Peter Bala, Dec 23 2012
G.f.: x/(1-2*x) + x^2/(1-2*x)/(Q(0)-x) where Q(k) = 1 - x/(x*k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Feb 23 2013
G.f.: G(0)/2 - 1, where G(k) = 1 + 1/( 1 - x/(x + (1-x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013
G.f.: x*G(0)/(2-3*x), where G(k) = 1 + 1/( 1 - x*(5*k-9)/(x*(5*k-4) - 6/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 17 2013
Sum_{n>=1} 1/(a(n) + 1/a(n)) = 1. Compare with A001519, A049660 and A049670. - Peter Bala, Nov 29 2013
a(n) = U(n-1,3/2) where U(n-1,x) is Chebyshev polynomial of the second kind. - Milan Janjic, Jan 25 2015
The o.g.f. A(x) satisfies A(x) + A(-x) + 6*A(x)*A(-x) = 0. The o.g.f. for A004187 equals -A(sqrt(x))*A(-sqrt(x)). - Peter Bala, Apr 02 2015
For n > 1, a(n) = (3*F(n+1)^2 + 2*F(n-2)*F(n+1) - F(n-2)^2)/4. - J. M. Bergot, Feb 16 2016
For n > 3, a(n) = floor(MA) - 4 for n even and floor(MA) + 5 for n odd.  MA is the maximum area of a quadrilateral with lengths of sides in order L(n), L(n), F(n-3), F(n+3), with L(n)=A000032(n). The ratio of the longer diagonal to the shorter approaches 5/3. - J. M. Bergot, Feb 16 2016
a(n+1) = Sum_{j=0..n} Sum_{k=0..j} binomial(n-j,k)*binomial(j,k)*2^(j-k). - Tony Foster III, Sep 18 2017
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} C(k+i,k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = Sum_{k=1..A000041(n)} A305309(n, k), n >= 1. Also row sums of triangle A078812.- Wolfdieter Lang, May 31 2018
a(n) = H(2*n, 1, 1/2) for n > 0 where H(n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -4). - Peter Luschny, Sep 03 2019
Sum_{n>=1} 1/a(n) = A153386. - Amiram Eldar, Oct 04 2020
a(n) = A249450(n) + 2. - Leo Tavares, Oct 10 2021
a(n) = -2/(sqrt(5)*tan(2*arctan(phi^(2*n)))), where phi = A001622 is the golden ratio. - Diego Rattaggi, Nov 21 2021
a(n) = sinh(2*n*arcsinh(1/2))/sqrt(5/4). - Peter Luschny, May 21 2022
From Amiram Eldar, Dec 02 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 1 + 1/sqrt(5) (A344212).
Product_{n>=2} (1 + (-1)^n/a(n)) = (5/6) * (1 + 1/sqrt(5)). (End)
a(n) = Sum_{k>=0} Fibonacci(2*n*k)/(Lucas(2*n)^(k+1)). - Diego Rattaggi, Jan 12 2025
Sum_{n>=0} a(n)/3^n = 3. - Diego Rattaggi, Jan 20 2025

cp's OEIS Frontend

A340099 Odd composite integers m such that A004187(m-J(m,45)) == 0 (mod m) and gcd(m,45)=1, where J(m,45) is the Jacobi symbol.

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A338011 Odd composite integers m such that A004187(m)^2 == 1 (mod m).

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A340124 Odd composite integers m such that A004187(2*m-J(m,45)) == J(m,45) (mod m) and gcd(m,45)=1, where J(m,45) is the Jacobi symbol.

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A337781 Odd composite integers m such that U(m)^2 == 1 (mod m) and V(m) == 7 (mod m), where U(m)=A004187(m) and V(m)=A056854(m) are the m-th generalized Lucas and Pell-Lucas numbers of parameters a=7 and b=1, respectively.

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A337782 Even composite integers m such that U(m)^2 == 1 (mod m) and V(m) == 7 (mod m), where U(m)=A004187(m) and V(m)=A056854(m) are the m-th generalized Lucas and Pell-Lucas numbers of parameters a=7 and b=1, respectively.

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A340241 Odd composite integers m such that A004187(3*m-J(m,45)) == 7*J(m,45) (mod m) and gcd(m,45)=1, where J(m,45) is the Jacobi symbol.

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A337783 Even composite integers m such that U(m)^2 == 1 (mod m), where U(m)=A004187(m) is the m-th generalized Lucas number of parameters a=7 and b=1.

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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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A340241 Odd composite integers m such that A004187(3m-J(m,45)) == 7J(m,45) (mod m) and gcd(m,45)=1, where J(m,45) is the Jacobi symbol.