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A217476 Coefficient triangle for the square of the monic integer Chebyshev T-polynomials A127672.

Original entry on oeis.org

4, 0, 1, 4, -4, 1, 0, 9, -6, 1, 4, -16, 20, -8, 1, 0, 25, -50, 35, -10, 1, 4, -36, 105, -112, 54, -12, 1, 0, 49, -196, 294, -210, 77, -14, 1, 4, -64, 336, -672, 660, -352, 104, -16, 1, 0, 81, -540, 1386, -1782, 1287, -546, 135, -18, 1, 4, -100, 825, -2640, 4290, -4004, 2275, -800, 170, -20, 1
Offset: 0

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Author

Wolfdieter Lang, Oct 17 2012

Keywords

Comments

The monic integer T-polynomials, called R(n,x) (in Abramowitz-Stegun C(n,x)), with their coefficient triangle given in A127672, when squared, become polynomials in y=x^2:
R(n,x)^2 = sum(T(n,k)*y^k,m=0..n).
R(n,x)^2 = 2 + R(2*n,x). From the bisection of the R-(or T-)polynomials, the even part. Directly from the R(m*n,x)=R(m,R(n,x)) property for m=2.
The o.g.f. is G(z,y) := sum((R(n,sqrt(y))^2)*z^n ,n=0..infinity) = (4 + (4 - 3*y)*z + y*z^2)/((1 +(2-y)*z + z^2)*(1-z)). From the bisection.
The o.g.f.s of the columns k>=1 are x^k*(1-x)/(1+x)^(2*k+1),
and for k=0 the o.g.f. is 4/(1-x^2).
Hetmaniok et al. (2015) refer to these as "modified Chebyshev" polynomials. - N. J. A. Sloane, Sep 13 2016

Examples

			The triangle begins:
n\k 0    1    2      3     4      5     6     7    8   9  10
0:  4
1:  0    1
2:  4   -4    1
3:  0    9   -6      1
4:  4  -16   20     -8     1
5:  0   25  -50     35   -10      1
6:  4  -36  105   -112    54    -12     1
7:  0   49 -196    294  -210     77   -14     1
8:  4  -64  336   -672   660   -352   104   -16    1
9:  0   81 -540   1386 -1782   1287  -546   135  -18   1
10: 4 -100  825  -2640  4290  -4004  2275  -800  170 -20   1
...
n=2:  R(2,x) = -2 + y, R(2,x)^2 = 4 -4*y + y^2, with y=x^2.
n=3:  R(3,x) = 3*x - x^3, R(3,x)^2 = 9*y - 6*y^2 +y^3, with y=x^2.
T(4,1) = 8*(-1)^3*binomial(5,3)/5 = -16.
T(4,0) = 2 + 8*(-1)^4*binomial(4,4)/4 = 4.
T(n,1) = (-1)^(n-1)*2*n*(n+1)!/((n-1)!*2!*(n+1)) = -((-1)^n)*n^2 = A162395(n), n >= 1.
T(n,2) = (-1)^n*A002415(n), n >= 0.
T(n,3) = -(-1)^n*A040977(n-3), n >= 3.
T(n,4) = (-1)^n*A053347(n-4), n >= 4.
T(n,5) = -(-1)^n*A054334(n-5), n >= 5.
		

References

  • E Hetmaniok, P Lorenc, S Damian, et al., Periodic orbits of boundary logistic map and new kind of modified Chebyshev polynomials in R. Witula, D. Slota, W. Holubowski (eds.), Monograph on the Occasion of 100th Birthday Anniversary of Zygmunt Zahorski. Wydawnictwo Politechniki Slaskiej, Gliwice 2015, pp. 325-343.

Crossrefs

Cf. A127672, A158454 (square of S-polynomials), A128495 (sum of square of S-polynomials).

Formula

T(n,k) = [x^(2*k)]R(n,x)^2, with R(n,x) the monic integer version of the Chebyshev T(n,x) polynomial.
T(n,k) = 0 if n=1. ([k=0] means 1 if k=0 else 0).

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

Original entry on oeis.org

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

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Author

Keywords

Comments

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

Examples

			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)
		

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, Springer-Verlag, 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.

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.

Programs

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

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(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

A000384 Hexagonal numbers: a(n) = n*(2*n-1).

Original entry on oeis.org

0, 1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, 496, 561, 630, 703, 780, 861, 946, 1035, 1128, 1225, 1326, 1431, 1540, 1653, 1770, 1891, 2016, 2145, 2278, 2415, 2556, 2701, 2850, 3003, 3160, 3321, 3486, 3655, 3828, 4005, 4186, 4371, 4560
Offset: 0

Views

Author

Keywords

Comments

Number of edges in the join of two complete graphs, each of order n, K_n * K_n. - Roberto E. Martinez II, Jan 07 2002
The power series expansion of the entropy function H(x) = (1+x)log(1+x) + (1-x)log(1-x) has 1/a_i as the coefficient of x^(2i) (the odd terms being zero). - Tommaso Toffoli (tt(AT)bu.edu), May 06 2002
Partial sums of A016813 (4n+1). Also with offset = 0, a(n) = (2n+1)(n+1) = A005408 * A000027 = 2n^2 + 3n + 1, i.e., a(0) = 1. - Jeremy Gardiner, Sep 29 2002
Sequence also gives the greatest semiperimeter of primitive Pythagorean triangles having inradius n-1. Such a triangle has consecutive longer sides, with short leg 2n-1, hypotenuse a(n) - (n-1) = A001844(n), and area (n-1)*a(n) = 6*A000330(n-1). - Lekraj Beedassy, Apr 23 2003
Number of divisors of 12^(n-1), i.e., A000005(A001021(n-1)). - Henry Bottomley, Oct 22 2001
More generally, if p1 and p2 are two arbitrarily chosen distinct primes then a(n) is the number of divisors of (p1^2*p2)^(n-1) or equivalently of any member of A054753^(n-1). - Ant King, Aug 29 2011
Number of standard tableaux of shape (2n-1,1,1) (n>=1). - Emeric Deutsch, May 30 2004
It is well known that for n>0, A014105(n) [0,3,10,21,...] is the first of 2n+1 consecutive integers such that the sum of the squares of the first n+1 such integers is equal to the sum of the squares of the last n; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2.
Less well known is that for n>1, a(n) [0,1,6,15,28,...] is the first of 2n consecutive integers such that sum of the squares of the first n such integers is equal to the sum of the squares of the last n-1 plus n^2; e.g., 15^2 + 16^2 + 17^2 = 19^2 + 20^2 + 3^2. - Charlie Marion, Dec 16 2006
a(n) is also a perfect number A000396 when n is an even superperfect number A061652. - Omar E. Pol, Sep 05 2008
Sequence found by reading the line from 0, in the direction 0, 6, ... and the line from 1, in the direction 1, 15, ..., in the square spiral whose vertices are the generalized hexagonal numbers A000217. - Omar E. Pol, Jan 09 2009
For n>=1, 1/a(n) = Sum_{k=0..2*n-1} ((-1)^(k+1)*binomial(2*n-1,k)*binomial(2*n-1+k,k)*H(k)/(k+1)) with H(k) harmonic number of order k.
The number of possible distinct colorings of any 2 colors chosen from n colors of a square divided into quadrants. - Paul Cleary, Dec 21 2010
Central terms of the triangle in A051173. - Reinhard Zumkeller, Apr 23 2011
For n>0, a(n-1) is the number of triples (w,x,y) with all terms in {0,...,n} and max(|w-x|,|x-y|) = |w-y|. - Clark Kimberling, Jun 12 2012
a(n) is the number of positions of one domino in an even pyramidal board with base 2n. - César Eliud Lozada, Sep 26 2012
Partial sums give A002412. - Omar E. Pol, Jan 12 2013
Let a triangle have T(0,0) = 0 and T(r,c) = |r^2 - c^2|. The sum of the differences of the terms in row(n) and row(n-1) is a(n). - J. M. Bergot, Jun 17 2013
With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for A176230, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 11 2013
a(n) is the number of length 2n binary sequences that have exactly two 1's. a(2) = 6 because we have: {0,0,1,1}, {0,1,0,1}, {0,1,1,0}, {1,0,0,1}, {1,0,1,0}, {1,1,0,0}. The ordinary generating function with interpolated zeros is: (x^2 + 3*x^4)/(1-x^2)^3. - Geoffrey Critzer, Jan 02 2014
For n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n^(2*m) is a multiple of k + n is given by k = 2*n^(2*m) - n. - Derek Orr, Sep 04 2014
Binomial transform of (0, 1, 4, 0, 0, 0, ...) and second partial sum of (0, 1, 4, 4, 4, ...). - Gary W. Adamson, Oct 05 2015
a(n) also gives the dimension of the simple Lie algebras D_n, for n >= 4. - Wolfdieter Lang, Oct 21 2015
For n > 0, a(n) equals the number of compositions of n+11 into n parts avoiding parts 2, 3, 4. - Milan Janjic, Jan 07 2016
Also the number of minimum dominating sets and maximal irredundant sets in the n-cocktail party graph. - Eric W. Weisstein, Jun 29 and Aug 17 2017
As Beedassy's formula shows, this Hexagonal number sequence is the odd bisection of the Triangle number sequence. Both of these sequences are figurative number sequences. For A000384, a(n) can be found by multiplying its triangle number by its hexagonal number. For example let's use the number 153. 153 is said to be the 17th triangle number but is also said to be the 9th hexagonal number. Triangle(17) Hexagonal(9). 17*9=153. Because the Hexagonal number sequence is a subset of the Triangle number sequence, the Hexagonal number sequence will always have both a triangle number and a hexagonal number. n* (2*n-1) because (2*n-1) renders the triangle number. - Bruce J. Nicholson, Nov 05 2017
Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central valley and the largest Dyck path has a central peak, n >= 1. Thus all hexagonal numbers > 0 have middle divisors. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
k^a(n-1) mod n = 1 for prime n and k=2..n-1. - Joseph M. Shunia, Feb 10 2019
Consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z: a(n+1) gives the semiperimeter of related triangles; A005408, A046092 and A001844 give the X, Y and Z values. - Ralf Steiner, Feb 25 2020
See A002939(n) = 2*a(n) for the corresponding perimeters. - M. F. Hasler, Mar 09 2020
It appears that these are the numbers k with the property that the smallest subpart in the symmetric representation of sigma(k) is 1. - Omar E. Pol, Aug 28 2021
The above conjecture is true. See A280851 for a proof. - Hartmut F. W. Hoft, Feb 02 2022
The n-th hexagonal number equals the sum of the n consecutive integers with the same parity starting at n; for example, 1, 2+4, 3+5+7, 4+6+8+10, etc. In general, the n-th 2k-gonal number is the sum of the n consecutive integers with the same parity starting at (k-2)*n - (k-3). When k = 1 and 2, this result generates the positive integers, A000027, and the squares, A000290, respectively. - Charlie Marion, Mar 02 2022
Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)} such that |A|=|B|=k and A+B={0,1,2,...,2*a(n)}} = 2*n. - Michael Chu, Mar 09 2022

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 38.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 53-54, 129-130, 132.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 21.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 122-123.

Crossrefs

a(n)= A093561(n+1, 2), (4, 1)-Pascal column.
a(n) = A100345(n, n-1) for n>0.
Cf. A002939 (twice a(n): sums of Pythagorean triples (X, Y, Z=Y+1)).
Cf. A280851.

Programs

  • Haskell
    a000384 n = n * (2 * n - 1)
    a000384_list = scanl (+) 0 a016813_list
    -- Reinhard Zumkeller, Dec 16 2012
    
  • Maple
    A000384:=n->n*(2*n-1); seq(A000384(k), k=0..100); # Wesley Ivan Hurt, Sep 27 2013
  • Mathematica
    Table[n*(2 n - 1), {n, 0, 100}] (* Wesley Ivan Hurt, Sep 27 2013 *)
    LinearRecurrence[{3, -3, 1}, {0, 1, 6}, 50] (* Harvey P. Dale, Sep 10 2015 *)
    Join[{0}, Accumulate[Range[1, 312, 4]]] (* Harvey P. Dale, Mar 26 2016 *)
    (* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[6], n], {n, 0, 48}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
    PolygonalNumber[6, Range[0, 20]] (* Eric W. Weisstein, Aug 17 2017 *)
    CoefficientList[Series[x*(1 + 3*x)/(1 - x)^3 , {x, 0, 100}], x] (* Stefano Spezia, Sep 02 2018 *)
  • PARI
    a(n)=n*(2*n-1)
    
  • PARI
    a(n) = binomial(2*n,2) \\ Altug Alkan, Oct 06 2015
    
  • Python
    # Intended to compute the initial segment of the sequence, not isolated terms.
    def aList():
         x, y = 1, 1
         yield 0
         while True:
             yield x
             x, y = x + y + 4, y + 4
    A000384 = aList()
    print([next(A000384) for i in range(49)]) # Peter Luschny, Aug 04 2019

Formula

a(n) = Sum_{k=1..n} tan^2((k - 1/2)*Pi/(2n)). - Ignacio Larrosa Cañestro, Apr 17 2001
E.g.f.: exp(x)*(x+2x^2). - Paul Barry, Jun 09 2003
G.f.: x*(1+3*x)/(1-x)^3. - Simon Plouffe in his 1992 dissertation, dropping the initial zero
a(n) = A000217(2*n-1) = A014105(-n).
a(n) = 4*A000217(n-1) + n. - Lekraj Beedassy, Jun 03 2004
a(n) = right term of M^n * [1,0,0], where M = the 3 X 3 matrix [1,0,0; 1,1,0; 1,4,1]. Example: a(5) = 45 since M^5 *[1,0,0] = [1,5,45]. - Gary W. Adamson, Dec 24 2006
Row sums of triangle A131914. - Gary W. Adamson, Jul 27 2007
Row sums of n-th row, triangle A134234 starting (1, 6, 15, 28, ...). - Gary W. Adamson, Oct 14 2007
Starting with offset 1, = binomial transform of [1, 5, 4, 0, 0, 0, ...]. Also, A004736 * [1, 4, 4, 4, ...]. - Gary W. Adamson, Oct 25 2007
a(n)^2 + (a(n)+1)^2 + ... + (a(n)+n-1)^2 = (a(n)+n+1)^2 + ... + (a(n)+2n-1)^2 + n^2; e.g., 6^2 + 7^2 = 9^2 + 2^2; 28^2 + 29^2 + 30^2 + 31^2 = 33^2 + 34^2 + 35^2 + 4^2. - Charlie Marion, Nov 10 2007
a(n) = binomial(n+1,2) + 3*binomial(n,2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=0, a(1)=1, a(2)=6. - Jaume Oliver Lafont, Dec 02 2008
a(n) = T(n) + 3*T(n-1), where T(n) is the n-th triangular number. - Vincenzo Librandi, Nov 10 2010
a(n) = a(n-1) + 4*n - 3 (with a(0)=0). - Vincenzo Librandi, Nov 20 2010
a(n) = A007606(A000290(n)). - Reinhard Zumkeller, Feb 12 2011
a(n) = 2*a(n-1) - a(n-2) + 4. - Ant King, Aug 26 2011
a(n+1) = A045896(2*n). - Reinhard Zumkeller, Dec 12 2011
a(2^n) = 2^(2n+1) - 2^n. - Ivan N. Ianakiev, Apr 13 2013
a(n) = binomial(2*n,2). - Gary Detlefs, Jul 28 2013
a(n+1) = A128918(2*n+1). - Reinhard Zumkeller, Oct 13 2013
a(4*a(n)+7*n+1) = a(4*a(n)+7*n) + a(4*n+1). - Vladimir Shevelev, Jan 24 2014
Sum_{n>=1} 1/a(n) = 2*log(2) = 1.38629436111989...= A016627. - Vaclav Kotesovec, Apr 27 2016
Sum_{n>=1} (-1)^n/a(n) = log(2) - Pi/2. - Vaclav Kotesovec, Apr 20 2018
a(n+1) = trinomial(2*n+1, 2) = trinomial(2*n+1, 4*n), for n >= 0, with the trinomial irregular triangle A027907. a(n+1) = (n+1)*(2*n+1) = (1/Pi)*Integral_{x=0..2} (1/sqrt(4 - x^2))*(x^2 - 1)^(2*n+1)*R(4*n-2, x) with the R polynomial coefficients given in A127672. [Comtet, p. 77, the integral formula for q=3, n -> 2*n+1, k = 2, rewritten with x = 2*cos(phi)]. - Wolfdieter Lang, Apr 19 2018
Sum_{n>=1} 1/(a(n))^2 = 2*Pi^2/3-8*log(2) = 1.0345588... = 10*A182448 - A257872. - R. J. Mathar, Sep 12 2019
a(n) = (A005408(n-1) + A046092(n-1) + A001844(n-1))/2. - Ralf Steiner, Feb 27 2020
Product_{n>=2} (1 - 1/a(n)) = 2/3. - Amiram Eldar, Jan 21 2021
a(n) = floor(Sum_{k=(n-1)^2..n^2} sqrt(k)), for n >= 1. - Amrit Awasthi, Jun 13 2021
a(n+1) = A084265(2*n), n>=0. - Hartmut F. W. Hoft, Feb 02 2022
a(n) = A000290(n) + A002378(n-1). - Charles Kusniec, Sep 11 2022

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

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

Views

Author

Keywords

Comments

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)

Examples

			1.41421356237309504880168872420969807856967187537694807317667...
		

References

  • 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.

Crossrefs

Cf. A004539 (binary version).

Programs

  • Haskell
    -- 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
  • Maple
    Digits:=100; evalf(sqrt(2)); # Wesley Ivan Hurt, Dec 04 2013
  • Mathematica
    RealDigits[N[2^(1/2), 128]] (* Vladimir Joseph Stephan Orlovsky, Dec 25 2008 *)
  • Maxima
    fpprec: 100$ ev(bfloat(sqrt(2))); /* Martin Ettl, Oct 17 2012 */
    
  • PARI
    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
    
  • PARI
    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
    
  • PARI
    \\ Works in v2.15.0; n = 100 decimal places
    my(n=100); digits(floor(10^n*quadgen(8))) \\ Michal Paulovic, Mar 22 2023
    

Formula

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

A000096 a(n) = n*(n+3)/2.

Original entry on oeis.org

0, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377, 1430, 1484, 1539, 1595, 1652, 1710, 1769
Offset: 0

Views

Author

Keywords

Comments

For n >= 1, a(n) is the maximal number of pieces that can be obtained by cutting an annulus with n cuts. See illustration. - Robert G. Wilson v
n(n-3)/2 (n >= 3) is the number of diagonals of an n-gon. - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr)
n(n-3)/2 (n >= 4) is the degree of the third-smallest irreducible presentation of the symmetric group S_n (cf. James and Kerber, Appendix 1).
a(n) is also the multiplicity of the eigenvalue (-2) of the triangle graph Delta(n+1). (See p. 19 in Biggs.) - Felix Goldberg (felixg(AT)tx.technion.ac.il), Nov 25 2001
For n > 3, a(n-3) = dimension of the traveling salesman polytope T(n). - Benoit Cloitre, Aug 18 2002
Also counts quasi-dominoes (quasi-2-ominoes) on an n X n board. Cf. A094170-A094172. - Jon Wild, May 07 2004
Coefficient of x^2 in (1 + x + 2*x^2)^n. - Michael Somos, May 26 2004
a(n) is the number of "prime" n-dimensional polyominoes. A "prime" n-polyomino cannot be formed by connecting any other n-polyominoes except for the n-monomino and the n-monomino is not prime. E.g., for n=1, the 1-monomino is the line of length 1 and the only "prime" 1-polyominoes are the lines of length 2 and 3. This refers to "free" n-dimensional polyominoes, i.e., that can be rotated along any axis. - Bryan Jacobs (bryanjj(AT)gmail.com), Apr 01 2005
Solutions to the quadratic equation q(m, r) = (-3 +- sqrt(9 + 8(m - r))) / 2, where m - r is included in a(n). Let t(m) = the triangular number (A000217) less than some number k and r = k - t(m). If k is neither prime nor a power of two and m - r is included in A000096, then m - q(m, r) will produce a value that shares a divisor with k. - Andrew S. Plewe, Jun 18 2005
Sum_{k=2..n+1} 4/(k*(k+1)*(k-1)) = ((n+3)*n)/((n+2)*(n+1)). Numerator(Sum_{k=2..n+1} 4/(k*(k+1)*(k-1))) = (n+3)*n/2. - Alexander Adamchuk, Apr 11 2006
Number of rooted trees with n+3 nodes of valence 1, no nodes of valence 2 and exactly two other nodes. I.e., number of planted trees with n+2 leaves and exactly two branch points. - Theo Johnson-Freyd (theojf(AT)berkeley.edu), Jun 10 2007
If X is an n-set and Y a fixed 2-subset of X then a(n-2) is equal to the number of (n-2)-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
For n >= 1, a(n) is the number of distinct shuffles of the identity permutation on n+1 letters with the identity permutation on 2 letters (12). - Camillia Smith Barnes, Oct 04 2008
If s(n) is a sequence defined as s(1) = x, s(n) = kn + s(n-1) + p for n > 1, then s(n) = a(n-1)*k + (n-1)*p + x. - Gary Detlefs, Mar 04 2010
The only primes are a(1) = 2 and a(2) = 5. - Reinhard Zumkeller, Jul 18 2011
a(n) = m such that the (m+1)-th triangular number minus the m-th triangular number is the (n+1)-th triangular number: (m+1)(m+2)/2 - m(m+1)/2 = (n+1)(n+2)/2. - Zak Seidov, Jan 22 2012
For n >= 1, number of different values that Sum_{k=1..n} c(k)*k can take where the c(k) are 0 or 1. - Joerg Arndt, Jun 24 2012
On an n X n chessboard (n >= 2), the number of possible checkmate positions in the case of king and rook versus a lone king is 0, 16, 40, 72, 112, 160, 216, 280, 352, ..., which is 8*a(n-2). For a 4 X 4 board the number is 40. The number of positions possible was counted including all mirror images and rotations for all four sides of the board. - Jose Abutal, Nov 19 2013
If k = a(i-1) or k = a(i+1) and n = k + a(i), then C(n, k-1), C(n, k), C(n, k+1) are three consecutive binomial coefficients in arithmetic progression and these are all the solutions. There are no four consecutive binomial coefficients in arithmetic progression. - Michael Somos, Nov 11 2015
a(n-1) is also the number of independent components of a symmetric traceless tensor of rank 2 and dimension n >= 1. - Wolfdieter Lang, Dec 10 2015
Numbers k such that 8k + 9 is a square. - Juri-Stepan Gerasimov, Apr 05 2016
Let phi_(D,rho) be the average value of a generic degree D monic polynomial f when evaluated at the roots of the rho-th derivative of f, expressed as a polynomial in the averaged symmetric polynomials in the roots of f. [See the Wojnar et al. link] The "last" term of phi_(D,rho) is a multiple of the product of all roots of f; the coefficient is expressible as a polynomial h_D(N) in N:=D-rho. These polynomials are of the form h_D(N)= ((-1)^D/(D-1)!)*(D-N)*N^chi*g_D(N) where chi = (1 if D is odd, 0 if D is even) and g_D(N) is a monic polynomial of degree (D-2-chi). Then a(n) are the negated coefficients of the next to the highest order term in the polynomials N^chi*g_D(N), starting at D=3. - Gregory Gerard Wojnar, Jul 19 2017
For n >= 2, a(n) is the number of summations required to solve the linear regression of n variables (n-1 independent variables and 1 dependent variable). - Felipe Pedraza-Oropeza, Dec 07 2017
For n >= 2, a(n) is the number of sums required to solve the linear regression of n variables: 5 for two variables (sums of X, Y, X^2, Y^2, X*Y), 9 for 3 variables (sums of X1, X2, Y1, X1^2, X1*X2, X1*Y, X2^2, X2*Y, Y^2), and so on. - Felipe Pedraza-Oropeza, Jan 11 2018
a(n) is the area of a triangle with vertices at (n, n+1), ((n+1)*(n+2)/2, (n+2)*(n+3)/2), ((n+2)^2, (n+3)^2). - J. M. Bergot, Jan 25 2018
Number of terms less than 10^k: 1, 4, 13, 44, 140, 446, 1413, 4471, 14141, 44720, 141420, 447213, ... - Muniru A Asiru, Jan 25 2018
a(n) is also the number of irredundant sets in the (n+1)-path complement graph for n > 2. - Eric W. Weisstein, Apr 11 2018
a(n) is also the largest number k such that the largest Dyck path of the symmetric representation of sigma(k) has exactly n peaks, n >= 1. (Cf. A237593.) - Omar E. Pol, Sep 04 2018
For n > 0, a(n) is the number of facets of associahedra. Cf. A033282 and A126216 and their refinements A111785 and A133437 for related combinatorial and analytic constructs. See p. 40 of Hanson and Sha for a relation to projective spaces and string theory. - Tom Copeland, Jan 03 2021
For n > 0, a(n) is the number of bipartite graphs with 2n or 2n+1 edges, no isolated vertices, and a stable set of cardinality 2. - Christian Barrientos, Jun 13 2022
For n >= 2, a(n-2) is the number of permutations in S_n which are the product of two different transpositions of adjacent points. - Zbigniew Wojciechowski, Mar 31 2023
a(n) represents the optimal stop-number to achieve the highest running score for the Greedy Pig game with an (n-1)-sided die with a loss on a 1. The total at which one should stop is a(s-1), e.g. for a 6-sided die, one should pass the die at 20. See Sparks and Haran. - Nicholas Stefan Georgescu, Jun 09 2024

Examples

			G.f. = 2*x + 5*x^2 + 9*x^3 + 14*x^4 + 20*x^5 + 27*x^6 + 35*x^7 + 44*x^8 + 54*x^9 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), Table 22.7, p. 797.
  • Norman Biggs, Algebraic Graph Theory, 2nd ed. Cambridge University Press, 1993.
  • G. James and A. Kerber, The Representation Theory of the Symmetric Group, Encyclopedia of Maths. and its Appls., Vol. 16, Addison-Wesley, 1981, Reading, MA, U.S.A.
  • D. G. Kendall et al., Shape and Shape Theory, Wiley, 1999; see p. 4.
  • 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).

Crossrefs

Complement of A007401. Column 2 of A145324. Column of triangle A014473, first skew subdiagonal of A033282, a diagonal of A079508.
Occurs as a diagonal in A074079/A074080, i.e., A074079(n+3, n) = A000096(n-1) for all n >= 2. Also A074092(n) = 2^n * A000096(n-1) after n >= 2.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488.
Similar sequences are listed in A316466.

Programs

Formula

G.f.: A(x) = x*(2-x)/(1-x)^3. a(n) = binomial(n+1, n-1) + binomial(n, n-1).
Connection with triangular numbers: a(n) = A000217(n+1) - 1.
a(n) = a(n-1) + n + 1. - Bryan Jacobs (bryanjj(AT)gmail.com), Apr 01 2005
a(n) = 2*t(n) - t(n-1) where t() are the triangular numbers, e.g., a(5) = 2*t(5) - t(4) = 2*15 - 10 = 20. - Jon Perry, Jul 23 2003
a(-3-n) = a(n). - Michael Somos, May 26 2004
2*a(n) = A008778(n) - A105163(n). - Creighton Dement, Apr 15 2005
a(n) = C(3+n, 2) - C(3+n, 1). - Zerinvary Lajos, Dec 09 2005
a(n) = A067550(n+1) / A067550(n). - Alexander Adamchuk, May 20 2006
a(n) = A126890(n,1) for n > 0. - Reinhard Zumkeller, Dec 30 2006
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Paul Curtz, Jan 02 2008
Starting (2, 5, 9, 14, ...) = binomial transform of (2, 3, 1, 0, 0, 0, ...). - Gary W. Adamson, Jul 03 2008
For n >= 0, a(n+2) = b(n+1) - b(n), where b(n) is the sequence A005586. - K.V.Iyer, Apr 27 2009
A002262(a(n)) = n. - Reinhard Zumkeller, May 20 2009
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n>=1, a(n-1)=coeff(charpoly(A,x),x^(n-2)). - Milan Janjic, Jul 08 2010
a(n) = Sum_{k=1..n} (k+1)!/k!. - Gary Detlefs, Aug 03 2010
a(n) = n(n+1)/2 + n = A000217(n) + n. - Zak Seidov, Jan 22 2012
E.g.f.: F(x) = 1/2*x*exp(x)*(x+4) satisfies the differential equation F''(x) - 2*F'(x) + F(x) = exp(x). - Peter Bala, Mar 14 2012
a(n) = binomial(n+3, 2) - (n+3). - Robert G. Wilson v, Mar 15 2012
a(n) = A181971(n+1, 2) for n > 0. - Reinhard Zumkeller, Jul 09 2012
a(n) = A214292(n+2, 1). - Reinhard Zumkeller, Jul 12 2012
G.f.: -U(0) where U(k) = 1 - 1/((1-x)^2 - x*(1-x)^4/(x*(1-x)^2 - 1/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 27 2012
A023532(a(n)) = 0. - Reinhard Zumkeller, Dec 04 2012
a(n) = A014132(n,n) for n > 0. - Reinhard Zumkeller, Dec 12 2012
a(n-1) = (1/n!)*Sum_{j=0..n} binomial(n,j)*(-1)^(n-j)*j^n*(j-1). - Vladimir Kruchinin, Jun 06 2013
a(n) = 2n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
a(n) = Sum_{i=2..n+1} i. - Wesley Ivan Hurt, Jun 28 2013
Sum_{n>0} 1/a(n) = 11/9. - Enrique Pérez Herrero, Nov 26 2013
a(n) = Sum_{i=1..n} (n - i + 2). - Wesley Ivan Hurt, Mar 31 2014
A023531(a(n)) = 1. - Reinhard Zumkeller, Feb 14 2015
For n > 0: a(n) = A101881(2*n-1). - Reinhard Zumkeller, Feb 20 2015
a(n) + a(n-1) = A008865(n+1) for all n in Z. - Michael Somos, Nov 11 2015
a(n+1) = A127672(4+n, n), n >= 0, where A127672 gives the coefficients of the Chebyshev C polynomials. See the Abramowitz-Stegun reference. - Wolfdieter Lang, Dec 10 2015
a(n) = (n+1)^2 - A000124(n). - Anton Zakharov, Jun 29 2016
Dirichlet g.f.: (zeta(s-2) + 3*zeta(s-1))/2. - Ilya Gutkovskiy, Jun 30 2016
a(n) = 2*A000290(n+3) - 3*A000217(n+3). - J. M. Bergot, Apr 04 2018
a(n) = Stirling2(n+2, n+1) - 1. - Peter Luschny, Jan 05 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/3 - 5/9. - Amiram Eldar, Jan 10 2021
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = 3.
Product_{n>=1} (1 - 1/a(n)) = 3*cos(sqrt(17)*Pi/2)/(4*Pi). (End)
Product_{n>=0} a(4*n+1)*a(4*n+4)/(a(4*n+2)*a(4*n+3)) = Pi/6. - Michael Jodl, Apr 05 2025

A014105 Second hexagonal numbers: a(n) = n*(2*n + 1).

Original entry on oeis.org

0, 3, 10, 21, 36, 55, 78, 105, 136, 171, 210, 253, 300, 351, 406, 465, 528, 595, 666, 741, 820, 903, 990, 1081, 1176, 1275, 1378, 1485, 1596, 1711, 1830, 1953, 2080, 2211, 2346, 2485, 2628, 2775, 2926, 3081, 3240, 3403, 3570, 3741, 3916, 4095, 4278
Offset: 0

Views

Author

N. J. A. Sloane, Jun 14 1998

Keywords

Comments

Note that when starting from a(n)^2, equality holds between series of first n+1 and next n consecutive squares: a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2*n)^2; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Henry Bottomley, Jan 22 2001; with typos fixed by Zak Seidov, Sep 10 2015
a(n) = sum of second set of n consecutive even numbers - sum of the first set of n consecutive odd numbers: a(1) = 4-1, a(3) = (8+10+12) - (1+3+5) = 21. - Amarnath Murthy, Nov 07 2002
Partial sums of odd numbers 3 mod 4, that is, 3, 3+7, 3+7+11, ... See A001107. - Jon Perry, Dec 18 2004
If Y is a fixed 3-subset of a (2n+1)-set X then a(n) is the number of (2n-1)-subsets of X intersecting Y. - Milan Janjic, Oct 28 2007
More generally (see the first comment), for n > 0, let b(n,k) = a(n) + k*(4*n + 1). Then b(n,k)^2 + (b(n,k) + 1)^2 + ... + (b(n,k) + n)^2 = (b(n,k) + n + 1 + 2*k)^2 + ... + (b(n,k) + 2*n + 2*k)^2 + k^2; e.g., if n = 3 and k = 2, then b(n,k) = 47 and 47^2 + ... + 50^2 = 55^2 + ... + 57^2 + 2^2. - Charlie Marion, Jan 01 2011
Sequence found by reading the line from 0, in the direction 0, 10, ..., and the line from 3, in the direction 3, 21, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Nov 09 2011
a(n) is the number of positions of a domino in a pyramidal board with base 2n+1. - César Eliud Lozada, Sep 26 2012
Differences of row sums of two consecutive rows of triangle A120070, i.e., first differences of A016061. - J. M. Bergot, Jun 14 2013 [In other words, the partial sums of this sequence give A016061. - Leo Tavares, Nov 23 2021]
a(n)*Pi is the total length of half circle spiral after n rotations. See illustration in links. - Kival Ngaokrajang, Nov 05 2013
For corresponding sums in first comment by Henry Bottomley, see A059255. - Zak Seidov, Sep 10 2015
a(n) also gives the dimension of the simple Lie algebras B_n (n >= 2) and C_n (n >= 3). - Wolfdieter Lang, Oct 21 2015
With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for unsigned A130757, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 13 2015
Partial sums of squares with alternating signs, ending in an even term: a(n) = 0^2 - 1^2 +- ... + (2*n)^2, cf. Example & Formula from Berselli, 2013. - M. F. Hasler, Jul 03 2018
Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central peak and the largest Dyck path has a central valley, n > 0. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
a(n) is the area of a triangle with vertices at (0,0), (2*n+1, 2*n), and ((2*n+1)^2, 4*n^2). - Art Baker, Dec 12 2018
This sequence is the largest subsequence of A000217 such that gcd(a(n), 2*n) = a(n) mod (2*n) = n, n > 0 up to a given value of n. It is the interleave of A033585 (a(n) is even) and A033567 (a(n) is odd). - Torlach Rush, Sep 09 2019
A generalization of Hasler's Comment (Jul 03 2018) follows. Let P(k,n) be the n-th k-gonal number. Then for k > 1, partial sums of {P(k,n)} with alternating signs, ending in an even term, = n*((k-2)*n + 1). - Charlie Marion, Mar 02 2021
Let U_n(H) = {A in M_n(H): A*A^H = I_n} be the group of n X n unitary matrices over the quaternions (A^H is the conjugate transpose of A. Note that over the quaternions we still have A*A^H = I_n <=> A^H*A = I_n by mapping A and A^H to (2n) X (2n) complex matrices), then a(n) is the dimension of its Lie algebra u_n(H) = {A in M_n(H): A + A^H = 0} as a real vector space. A basis is given by {(E_{st}-E_{ts}), i*(E_{st}+E_{ts}), j*(E_{st}+E_{ts}), k*(E_{st}+E_{ts}): 1 <= s < t <= n} U {i*E_{tt}, j*E_{tt}, k*E_{tt}: t = 1..n}, where E_{st} is the matrix with all entries zero except that its (st)-entry is 1. - Jianing Song, Apr 05 2021

Examples

			For n=6, a(6) = 0^2 - 1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2 - 7^2 + 8^2 - 9^2 + 10^2 - 11^2 + 12^2 = 78. - _Bruno Berselli_, Aug 29 2013
		

References

  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)

Crossrefs

Second column of array A094416.
Equals A033586(n) divided by 4.
See Comments of A132124.
Second n-gonal numbers: A005449, A147875, A045944, A179986, A033954, A062728, A135705.
Row sums in triangle A253580.

Programs

Formula

a(n) = 3*Sum_{k=1..n} tan^2(k*Pi/(2*(n + 1))). - Ignacio Larrosa Cañestro, Apr 17 2001
a(n)^2 = n*(a(n) + 1 + a(n) + 2 + ... + a(n) + 2*n); e.g., 10^2 = 2*(11 + 12 + 13 + 14). - Charlie Marion, Jun 15 2003
From N. J. A. Sloane, Sep 13 2003: (Start)
G.f.: x*(3 + x)/(1 - x)^3.
E.g.f.: exp(x)*(3*x + 2*x^2).
a(n) = A000217(2*n) = A000384(-n). (End)
a(n) = A084849(n) - 1; A100035(a(n) + 1) = 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = A126890(n, k) + A126890(n, n-k), 0 <= k <= n. - Reinhard Zumkeller, Dec 30 2006
a(2*n) = A033585(n); a(3*n) = A144314(n). - Reinhard Zumkeller, Sep 17 2008
a(n) = a(n-1) + 4*n - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = Sum_{k=0.2*n} (-1)^k*k^2. - Bruno Berselli, Aug 29 2013
a(n) = A242342(2*n + 1). - Reinhard Zumkeller, May 11 2014
a(n) = Sum_{k=0..2} C(n-2+k, n-2) * C(n+2-k, n), for n > 1. - J. M. Bergot, Jun 14 2014
a(n) = floor(Sum_{j=(n^2 + 1)..((n+1)^2 - 1)} sqrt(j)). Fractional portion of each sum converges to 1/6 as n -> infinity. See A247112 for a similar summation sequence on j^(3/2) and references to other such sequences. - Richard R. Forberg, Dec 02 2014
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3, with a(0) = 0, a(1) = 3, and a(2) = 10. - Harvey P. Dale, Feb 10 2015
Sum_{n >= 1} 1/a(n) = 2*(1 - log(2)) = 0.61370563888010938... (A188859). - Vaclav Kotesovec, Apr 27 2016
From Wolfdieter Lang, Apr 27 2018: (Start)
a(n) = trinomial(2*n, 2) = trinomial(2*n, 2*(2*n-1)), for n >= 1, with the trinomial irregular triangle A027907; i.e., trinomial(n,k) = A027907(n,k).
a(n) = (1/Pi) * Integral_{x=0..2} (1/sqrt(4 - x^2)) * (x^2 - 1)^(2*n) * R(4*(n-1), x), for n >= 0, with the R polynomial coefficients given in A127672, and R(-m, x) = R(m, x). [See Comtet, p. 77, the integral formula for q = 3, n -> 2*n, k = 2, rewritten with x = 2*cos(phi).] (End)
a(n) = A002943(n)/2. - Ralf Steiner, Jul 23 2019
a(n) = A000290(n) + A002378(n). - Torlach Rush, Nov 02 2020
a(n) = A003215(n) - A000290(n+1). See Squared Hexagons illustration. Leo Tavares, Nov 23 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/2 + log(2) - 2. - Amiram Eldar, Nov 28 2021

Extensions

Link added and minor errors corrected by Johannes W. Meijer, Feb 04 2010

A056594 Period 4: repeat [1,0,-1,0]; expansion of 1/(1 + x^2).

Original entry on oeis.org

1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0
Offset: 0

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Author

Wolfdieter Lang, Aug 04 2000

Keywords

Comments

G.f. is inverse of cyclotomic(4,x). Unsigned: A000035(n+1).
Real part of i^n and imaginary part of i^(n+1), i=sqrt(-1). - Reinhard Zumkeller, Jul 22 2007
The BINOMIAL transform generates A009116(n); the inverse BINOMIAL transform generates (-1)^n*A009116(n). - R. J. Mathar, Apr 07 2008
a(n-1), n >= 1, is the nontrivial Dirichlet character modulo 4, called Chi_2(4;n) (the trivial one is Chi_1(4;n) given by periodic(1,0) = A000035(n)). See the Apostol reference, p. 139, the k = 4, phi(k) = 2 table. - Wolfdieter Lang, Jun 21 2011
a(n-1), n >= 1, is the character of the Dirichlet beta function. - Daniel Forgues, Sep 15 2012
a(n-1), n >= 1, is also the (strongly) multiplicative function h(n) of Theorem 5.12, p. 150, of the Niven-Zuckerman reference. See the formula section. This function h(n) can be employed to count the integer solutions to n = x^2 + y^2. See A002654 for a comment with the formula. - Wolfdieter Lang, Apr 19 2013
This sequence is duplicated in A101455 but with offset 1. - Gary Detlefs, Oct 04 2013
For n >= 2 this gives the determinant of the bipartite graph with 2*n nodes and the adjacency matrix A(n) with elements A(n;1,2) = 1 = A(n;n,n-1), and for 1 < i < n A(n;i,i+1) = 1 = A(n;i,i-1), otherwise 0. - Wolfdieter Lang, Jun 25 2023

Examples

			With a(n-1) = h(n) of Niven-Zuckerman: a(62) = h(63) = h(3^2*7^1) = (-1)^(2*1)*(-1)^(1*3) = -1 = h(3)^2*h(7) = a(2)^2*a(6) = (-1)^2*(-1) = -1. - _Wolfdieter Lang_, Apr 19 2013
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1986.
  • I. S. Gradstein and I. M. Ryshik, Tables of series, products, and integrals, Volume 1, Verlag Harri Deutsch, 1981.
  • Ivan Niven and Herbert S. Zuckerman, An Introduction to the Theory of Numbers, New York: John Wiley (1980), p. 150.
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 32, equation 32:6:1 at page 300.

Crossrefs

Cf. A049310, A074661, A131852, A002654, A146559 (binomial transform).

Programs

  • Magma
    &cat[ [1, 0, -1, 0]: n in [0..23] ]; // Bruno Berselli, Feb 08 2011
    
  • Maple
    A056594 := n->(1-irem(n,2))*(-1)^iquo(n,2); # Peter Luschny, Jul 27 2011
  • Mathematica
    CoefficientList[Series[1/(1 + x^2), {x, 0, 50}], x]
    a[n_]:= KroneckerSymbol[-4,n+1];Table[a[n],{n,0,93}] (* Thanks to Jean-François Alcover. - Wolfdieter Lang, May 31 2013 *)
    CoefficientList[Series[1/Cyclotomic[4, x], {x, 0, 100}], x] (* Vincenzo Librandi, Apr 03 2014 *)
  • Maxima
    A056594(n) := block(
            [1,0,-1,0][1+mod(n,4)]
    )$ /* R. J. Mathar, Mar 19 2012 */
    
  • PARI
    {a(n) = real( I^n )}
    
  • PARI
    {a(n) = kronecker(-4, n+1) }
    
  • Python
    def A056594(n): return (1,0,-1,0)[n&3] # Chai Wah Wu, Sep 23 2023

Formula

G.f.: 1/(1+x^2).
E.g.f.: cos(x).
a(n) = (1/2)*((-i)^n + i^n), where i = sqrt(-1). - Mitch Harris, Apr 19 2005
a(n) = (1/2)*((-1)^(n+floor(n/2)) + (-1)^floor(n/2)).
Recurrence: a(n)=a(n-4), a(0)=1, a(1)=0, a(2)=-1, a(3)=0.
a(n) = T(n, 0) = A053120(n, 0); T(n, x) Chebyshev polynomials of the first kind. - Wolfdieter Lang, Aug 21 2009
a(n) = S(n, 0) = A049310(n, 0); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind.
Sum_{k>=0} a(k)/(k+1) = Pi/4. - Jaume Oliver Lafont, Mar 30 2010
a(n) = Sum_{k=0..n} A101950(n,k)*(-1)^k. - Philippe Deléham, Feb 10 2012
a(n) = (1/2)*(1 + (-1)^n)*(-1)^(n/2). - Bruno Berselli, Mar 13 2012
a(0) = 1, a(n-1) = 0 if n is even, a(n-1) = Product_{j=1..m} (-1)^(e_j*(p_j-1)/2) if the odd n-1 = p_1^(e_1)*p_2^(e_2)*...*p_m^(e_m) with distinct odd primes p_j, j=1..m. See the function h(n) of Theorem 5.12 of the Niven-Zuckerman reference. - Wolfdieter Lang, Apr 19 2013
a(n) = (-4/(n+1)), n >= 0, where (k/n) is the Kronecker symbol. See the Eric Weisstein and Wikipedia links. Thanks to Wesley Ivan Hurt. - Wolfdieter Lang, May 31 2013
a(n) = R(n,0)/2 with the row polynomials R of A127672. This follows from the product of the zeros of R, and the formula Product_{k=0..n-1} 2*cos((2*k+1)*Pi/(2*n)) = (1 + (-1)^n)*(-1)^(n/2), n >= 1 (see the Gradstein and Ryshik reference, p. 63, 1.396 4., with x = sqrt(-1)). - Wolfdieter Lang, Oct 21 2013
a(n) = Sum_{k=0..n} i^(k*(k+1)), where i=sqrt(-1). - Bruno Berselli, Mar 11 2015
Dirichlet g.f. of a(n) shifted right: L(chi_2(4),s) = beta(s) = (1-2^(-s))*(d.g.f. of A034947), see comments by Lang and Forgues. - Ralf Stephan, Mar 27 2015
a(n) = cos(3*n*Pi/2). - Ridouane Oudra, Sep 29 2024

A025192 a(0)=1; a(n) = 2*3^(n-1) for n >= 1.

Original entry on oeis.org

1, 2, 6, 18, 54, 162, 486, 1458, 4374, 13122, 39366, 118098, 354294, 1062882, 3188646, 9565938, 28697814, 86093442, 258280326, 774840978, 2324522934, 6973568802, 20920706406, 62762119218, 188286357654, 564859072962, 1694577218886, 5083731656658, 15251194969974
Offset: 0

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Author

Keywords

Comments

Warning: there is considerable overlap between this entry and the essentially identical A008776.
Shifts one place left when plus-convolved (PLUSCONV) with itself. a(n) = 2*Sum_{i=0..n-1} a(i). - Antti Karttunen, May 15 2001
Let M = { 0, 1, ..., 2^n-1 } be the set of all n-bit numbers. Consider two operations on this set: "sum modulo 2^n" (+) and "bitwise exclusive or" (XOR). The results of these operations are correlated.
To give a numerical measure, consider the equations over M: u = x + y, v = x XOR y and ask for how many pairs (u,v) is there a solution? The answer is exactly a(n) = 2*3^(n-1) for n >= 1. The fraction a(n)/4^n of such pairs vanishes as n goes to infinity. - Max Alekseyev, Feb 26 2003
Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+2, s(0) = 3, s(2n+2) = 3. - Herbert Kociemba, Jun 10 2004
Number of compositions of n into parts of two kinds. For a string of n objects, before the first, choose first kind or second kind; before each subsequent object, choose continue, first kind, or second kind. For example, compositions of 3 are 3; 2,1; 1,2; and 1,1,1. Using parts of two kinds, these produce respectively 2, 4, 4 and 8 compositions, 2+4+4+8 = 18. - Franklin T. Adams-Watters, Aug 18 2006
In the compositions the kinds of parts are ordered inside a run of identical parts, see example. Replacing "ordered" by "unordered" gives A052945. - Joerg Arndt, Apr 28 2013
Number of permutations of {1, 2, ..., n+1} such that no term is more than 2 larger than its predecessor. For example, a(3) = 18 because all permutations of {1, 2, 3, 4} are valid except 1423, 1432, 2143, 3142, 2314, 3214, in which 1 is followed by 4. Proof: removing (n + 1) gives a still-valid sequence. For n >= 2, can insert (n + 1) either at the beginning or immediately following n or immediately following (n - 1), but nowhere else. Thus the number of such permutations triples when we increase the sequence length by 1. - Joel B. Lewis, Nov 14 2006
Antidiagonal sums of square array A081277. - Philippe Deléham, Dec 04 2006
Equals row sums of triangle A160760. - Gary W. Adamson, May 25 2009
Let M = a triangle with (1, 2, 4, 8, ...) as the left border and all other columns = (0, 1, 2, 4, 8, ...). A025192 = lim_{n->oo} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
Number of nonisomorphic graded posets with 0 and uniform hasse graph of rank n with no 3-element antichain. ("Uniform" used in the sense of Retakh, Serconek and Wilson. By "graded" we mean that all maximal chains have the same length n.) - David Nacin, Feb 13 2012
Equals partial sums of A003946 prefaced with a 1: (1, 1, 4, 12, 36, 108, ...). - Gary W. Adamson, Feb 15 2012
Number of vertices (or sides) of the (n-1)-th iteration of a Gosper island. - Arkadiusz Wesolowski, Feb 07 2013
Row sums of triangle in A035002. - Jon Perry, May 30 2013
a(n) counts walks (closed) on the graph G(1-vertex; 1-loop, 1-loop, 2-loop, 2-loop, 3-loop, 3-loop, ...). - David Neil McGrath, Jan 01 2015
From Tom Copeland, Dec 03 2015: (Start)
For n > 0, a(n) are the traces of the even powers of the adjacency matrix M of the simple Lie algebra B_3, tr(M^(2n)) where M = Matrix(row 1; row 2; row 3) = Matrix[0,1,0; 1,0,2; 0,1,0], same as the traces of Matrix[0,2,0; 1,0,1; 0,1,0] (cf. Damianou). The traces of the odd powers vanish.
The characteristic polynomial of M equals determinant(x*I - M) = x^3 - 3x = A127672(3,x), so 1 - 3*x^2 = det(I - x M) = exp(-Sum_{n>=1} tr(M^n) x^n / n), implying Sum_{n>=1} a(n+1) x^(2n) / (2n) = -log(1 - 3*x^2), giving a logarithmic generating function for the aerated sequence, excluding a(0) and a(1).
a(n+1) = tr(M^(2n)), where tr(M^n) = 3^(n/2) + (-1)^n * 3^(n/2) = 2^n*(cos(Pi/6)^n + cos(5*Pi/6)^n) = n-th power sum of the eigenvalues of M = n-th power sum of the zeros of the characteristic polynomial.
The relation det(I - x M) = exp(-Sum_{n>=1} tr(M^n) x^n / n) = Sum_{n>=0} P_n(-tr(M), -tr(M^2), ..., -tr(M^n)) x^n/n! = exp(P.(-tr(M), -tr(M^2), ...)x), where P_n(x(1), ..., x(n)) are the partition polynomials of A036039 implies that with x(2n) = -tr(M^(2n)) = -a(n+1) for n > 0 and x(n) = 0 otherwise, the partition polynomials evaluate to zero except for P_2(x(1), x(2)) = P_2(0,-6) = -6.
Because of the inverse relation between the partition polynomials of A036039 and the Faber polynomials F_k(b1,b2,...,bk) of A263916, F_k(0,-3,0,0,...) = tr(M^k) gives aerated a(n), excluding n=0,1. E.g., F_2(0,-3) = -2(-3) = 6, F_4(0,-3,0,0) = 2 (-3)^2 = 18, and F_6(0,-3,0,0,0,0) = -2(-3)^3 = 54. (Cf. A265185.)
(End)
Number of permutations of length n > 0 avoiding the partially ordered pattern (POP) {1>2, 1>3, 1>4} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is the largest. - Sergey Kitaev, Dec 08 2020
For n > 0, a(n) is the number of 3-colorings of the grid graph P_2 X P_(n-1). More generally, for q > 1, the number of q-colorings of the grid graph P_2 X P_n is given by q*(q - 1)*((q - 1)*(q - 2) + 1)^(n - 1). - Sela Fried, Sep 25 2023
For n > 1, a(n) is the largest solution to the equation phi(x) = a(n-1). - M. Farrokhi D. G., Oct 25 2023
Number of dotted compositions of degree n. - Diego Arcis, Feb 01 2024

Examples

			There are a(3)=18 compositions of 3 into 2 kinds of parts. Here p:s stands for "part p of sort s":
01:  [ 1:0  1:0  1:0  ]
02:  [ 1:0  1:0  1:1  ]
03:  [ 1:0  1:1  1:0  ]
04:  [ 1:0  1:1  1:1  ]
05:  [ 1:0  2:0  ]
06:  [ 1:0  2:1  ]
07:  [ 1:1  1:0  1:0  ]
08:  [ 1:1  1:0  1:1  ]
09:  [ 1:1  1:1  1:0  ]
10:  [ 1:1  1:1  1:1  ]
11:  [ 1:1  2:0  ]
12:  [ 1:1  2:1  ]
13:  [ 2:0  1:0  ]
14:  [ 2:0  1:1  ]
15:  [ 2:1  1:0  ]
16:  [ 2:1  1:1  ]
17:  [ 3:0  ]
18:  [ 3:1  ]
- _Joerg Arndt_, Apr 28 2013
G.f. = 1 + 2*x + 6*x^2 + 18*x^3 + 54*x^4 + 162*x^5 + 486*x^6 + 1458*x^7 + ...
		

References

  • Richard P. Stanley, Enumerative combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.

Crossrefs

First differences of 3^n (A000244). Other self-convolved sequences: A000108, A007460, A007461, A007462, A007463, A007464, A061922.
Apart from initial term, same as A008776.

Programs

  • Haskell
    a025192 0 = 1
    a025192 n = 2 * 3 ^ (n -1)
    a025192_list = 1 : iterate (* 3) 2  -- Reinhard Zumkeller, Nov 27 2012
  • Maple
    A025192 := proc(n): if n=0 then 1 else 2*3^(n-1) fi: end: seq(A025192(n),n=0..26);
  • Mathematica
    Join[{1},2*3^(Range[30]-1)]  (* Harvey P. Dale, Mar 22 2011 *)
  • PARI
    a(n)=max(1,2*3^(n-1)) \\ Charles R Greathouse IV, Jul 25 2011
    
  • PARI
    Vec((1-x)/(1-3*x) + O(x^100)) \\ Altug Alkan, Dec 05 2015
    
  • Python
    [1]+[2*3**(n-1) for n in range(1,30)] # David Nacin, Mar 04 2012
    

Formula

G.f.: (1-x)/(1-3*x).
E.g.f.: (2*exp(3*x) + exp(0))/3. - Paul Barry, Apr 20 2003
a(n) = phi(3^n) = A000010(A000244(n)). - Labos Elemer, Apr 14 2003
a(0) = 1, a(n) = Sum_{k=0..n-1} (a(k) + a(n-k-1)). - Benoit Cloitre, Jun 24 2003
a(n) = A002326((3^n-1)/2). - Vladimir Shevelev, May 26 2008
a(1) = 2, a(n) = 3*a(n-1). - Vincenzo Librandi, Jan 01 2011
a(n) = lcm(a(n-1), Sum_{k=1..n-1} a(k)) for n >= 3. - David W. Wilson, Sep 27 2011
a(n) = ((2*n-1)*a(n-1) + (3*n-6)*a(n-2))/(n-1); a(0)=1, a(1)=2. - Sergei N. Gladkovskii, Jul 16 2012
From Sergei N. Gladkovskii, Jul 17 2012: (Start)
For the e.g.f. E(x) = (2/3)*exp(3*x) + exp(0)/3 we have
E(x) = 2*G(0)/3 where G(k) = 1 + k!/(3*(9*x)^k - 3*(9*x)^(2*k+1)/((9*x)^(k+1) + (k+1)!/G(k+1))); (continued fraction, 3rd kind, 3-step).
E(x) = 1+2*x/(G(0)-3*x) where G(k) = 3*x + 1 + k - 3*x*(k+1)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). (End)
a(n) = A114283(0,0). - Reinhard Zumkeller, Nov 27 2012
G.f.: 1 + ((1/2)/G(0) - 1)/x where G(k) = 1 - 2^k/(2 - 4*x/(2*x - 2^k/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 22 2012
G.f.: 1 + x*W(0), where W(k) = 1 + 1/(1 - x*(2*k+3)/(x*(2*k+4) + 1/W(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 28 2013
G.f.: 1 / (1 - 2*x / (1 - x)). - Michael Somos, Apr 03 2014
Construct the power matrix T(n,j) = [A(n)^*j]*[S(n)^*(j-1)] where A(n)=(2,2,2,...) and S(n)=(0,1,0,0,...). (* is convolution operation.) Then a(n) = Sum_{j=1..n} T(n,j). - David Neil McGrath, Jan 01 2015
G.f.: 1 + 2*x/(1 + 2*x)*( 1 + 5*x/(1 + 5*x)*( 1 + 8*x/(1 + 8*x)*( 1 + 11*x/(1 + 11*x)*( 1 + .... - Peter Bala, May 27 2017
Sum_{n>=0} 1/a(n) = 7/4. - Bernard Schott, Oct 02 2021
From Amiram Eldar, May 08 2023: (Start)
Sum_{n>=0} (-1)^n/a(n) = 5/8.
Product_{n>=1} (1 - 1/a(n)) = A132019. (End)

Extensions

Additional comments from Barry E. Williams, May 27 2000
a(22) corrected by T. D. Noe, Feb 08 2008
Maple programs simplified by Johannes W. Meijer, Jun 02 2011

A187360 Coefficient array for minimal polynomials of 2*cos(Pi/n) (rising powers of x).

Original entry on oeis.org

2, 1, 0, 1, -1, 1, -2, 0, 1, -1, -1, 1, -3, 0, 1, 1, -2, -1, 1, 2, 0, -4, 0, 1, -1, -3, 0, 1, 5, 0, -5, 0, 1, -1, 3, 3, -4, -1, 1, 1, 0, -4, 0, 1, -1, -3, 6, 4, -5, -1, 1, -7, 0, 14, 0, -7, 0, 1, 1, -4, -4, 1, 1, 2, 0, -16, 0, 20, 0, -8, 0, 1, 1, 4, -10, -10, 15, 6, -7, -1, 1
Offset: 1

Views

Author

Wolfdieter Lang, Jul 14 2011

Keywords

Comments

The degree delta(n) of the monic integer row polynomial, call it C(n,x), is A055034(n).
This minimal polynomial of the algebraic number 2*cos(Pi/n), n>=1, is given by
C(n,x) = sum(a(n,m)*x^m,m=0..A055034(n)) = (2^delta(n))*Psi(2n,x/2), with Psi(n,x) the minimal polynomial of cos(2Pi/n), with rational coefficient array A181875/A181876. There also references and links are found. See especially Watkins and Zeitlin for Psi(n,x).
The zeros of C(n,x), n>=2, are 2*cos(Pi k/n), with k=1,...,n-1 and gcd(k,2n)=1. For n=1 the zero is -2. Alternatively, these zeros are 2*cos(Pi(2l+1)/n), with l=0,...,floor((n-2)/2) and gcd(2l+1,n)=1. For n=1 take l=0.
The first column looks like the differently signed A020513(n),n>=1.
The polynomial for row n=2^m, m>=1, coincides with the row polynomial R(2^(m-1),x) of A127672. See the factorization of these R-polynomials (also known as Chebyshev C-polynomials) given there. - Wolfdieter Lang, Sep 15 2011
From Wolfdieter Lang, Nov 04 2013: (Start)
The norm N(rho(n)) of rho(n) = 2*cos(Pi/n), n >= 1, in the algebraic number field Q(rho(n)) is given by (-1)^delta(n)* C(n, 0), with C(n, x) of degree delta(n) = A055034(n). If N(rho(n)) equals +1 or -1 then 1/rho(n), which is an element of Q(rho(n)), is in fact an integer in this number field. For the 1/rho(n) formula in terms of the C coefficients see A230079. Thus 1/rho(n) is a Q(rho(n))-integer if and only if C(n, 0) is +1 or -1 , and this happens if and only if n is from the set {A230078(k), k >= 2}.
The negation says that, for n a positive integer, 1/rho(n) is not a Q(rho(n))-integer if and only if n is 1 or of the form 2*p^m, m >= 0 and p a prime, which are the numbers of A138929 including 1.
The proof uses for case (i): n = 2*m+1, m >= 1, the fact that C(2*m+1, 0)^2 = (product( 2*cos(Pi*(2*l+1)/(2*m+1)), l=0 .. m-1 and gcd(2*l+1, 2*m+1) = 1))^2 = (product(2*cos(Pi*k/(2*m+1)), k=1..L and gcd(k, 2*m+1) = 1))^2 = cyclotomic(2*m +1, -1). See the linked Q(rho(n)) paper, eq. (31), for a product formula for cyclotomic(n, -1). With the prime factorization of 2*m+1, and the fact that only the squarefree kernel of 2*m+1 enters (see an Oct 29 2013 comment on A013595), one finds, form the formula for cyclotomic(p1*p2*...*pk, x) involving the Moebius function, cyclotomic(2*m +1, -1) = +1, m >= 1. C(1, 0) = +2. For case (ii): n even, one has C(2^m, 0) = 0, -2, +2, for m = 1 , 2, >=3, respectively (see eq. (39) of the linked Q(rho(n)) paper). For odd prime p: (-1)^((p-1)/2)*C(2*p^m, 0) = cyclotomic(2*p^m, -1) = cyclotomic(2*p, -1) = cyclotomic(p, +1) = p, for m >= 1. For more than one odd prime in the squarefree kernel of n = 2*m, m >= 1, one finds C(2*m, 0) = +1 from cyclotomic(2*p1*...*pk, -1) = +1, for k >= 2. (End)
For the conversion of the C-polynomials into sums of Chebyshev's S-polynomials (A049310) see A255237. - Wolfdieter Lang, Mar 16 2015

Examples

			n=1:  2, 1;
n=2:  0, 1;
n=3: -1, 1;
n=4: -2, 0, 1;
n=5: -1,-1, 1;
n=6: -3, 0, 1;
n=7:  1,-2,-1, 1;
n=8:  2, 0,-4, 0, 1;
n=9: -1,-3, 0, 1;
n=10: 5, 0,-5, 0, 1;
...
C(2,x) = R(1,x), C(4,x) = R(2,x), C(8,x) = R(4,x),... with R(n,x) from A127672. - _Wolfdieter Lang_, Sep 15 2011
		

Crossrefs

Cf. A192003 (row sums). A192004 (alternating row sums).

Programs

  • Maple
    f:= proc(n) local P,z,j;
       P:= factor(evala(Norm(z-convert(2*cos(Pi/n),RootOf))));
       if type(P,`^`) then P:= op(1,P) fi;
       seq(coeff(P,z,j),j=0..degree(P));
    end proc:
    seq(f(n),n=1..20); # Robert Israel, Aug 04 2015
  • Mathematica
    Flatten[ CoefficientList[ Table[ MinimalPolynomial[2*Cos[Pi/n], x], {n, 1, 17}], x]] (* Jean-François Alcover, Sep 26 2011 *)
  • PARI
    halftot(n)=if(n<=2, 1, eulerphi(n)/2); \\ A023022
    default(realprecision, 110);
    row(n) = Vecrev(algdep(2*cos(2*Pi/n), halftot(n))); \\ Michel Marcus, Sep 19 2023

Formula

a(n,m) = [x^m] C(n,x), n>=1, m=0..A055034(n), with the minimal (monic and integer) polynomial C(n,x) of 2*cos(Pi/n). See the comment above.

A007070 a(n) = 4*a(n-1) - 2*a(n-2) with a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 14, 48, 164, 560, 1912, 6528, 22288, 76096, 259808, 887040, 3028544, 10340096, 35303296, 120532992, 411525376, 1405035520, 4797091328, 16378294272, 55918994432, 190919389184, 651839567872, 2225519493120, 7598398836736, 25942556360704, 88573427769344, 302408598355968
Offset: 0

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Keywords

Comments

Joe Keane (jgk(AT)jgk.org) observes that this sequence (beginning at 4) is "size of raises in pot-limit poker, one blind, maximum raising."
It appears that this sequence is the BinomialMean transform of A002315 - see A075271. - John W. Layman, Oct 02 2002
Number of (s(0), s(1), ..., s(2n+3)) such that 0 < s(i) < 8 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+3, s(0) = 1, s(2n+3) = 4. - Herbert Kociemba, Jun 11 2004
a(n) = number of distinct matrix products in (A+B+C+D)^n where commutators [A,B]=[C,D]=0 but neither A nor B commutes with C or D. - Paul D. Hanna and Joshua Zucker, Feb 01 2006
The n-th term of the sequence is the entry (1,2) in the n-th power of the matrix M=[1,-1;-1,3]. - Simone Severini, Feb 15 2006
Hankel transform of this sequence is [1,-2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
A204089 convolved with A000225, e.g., a(4) = 164 = (1*31 + 1*15 + 4*7 + 14*3 + 48*1) = (31 + 15 + 28 + 42 + 48). - Gary W. Adamson, Dec 23 2008
Equals INVERT transform of A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 03 2009
For n>=1, a(n-1) is the number of generalized compositions of n when there are 2^i-1 different types of the part i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Binomial transform of A078057. - R. J. Mathar, Mar 28 2011
Pisano period lengths: 1, 1, 8, 1, 24, 8, 6, 1, 24, 24, 120, 8, 168, 6, 24, 1, 8, 24, 360, 24, ... . - R. J. Mathar, Aug 10 2012
a(n) is the diagonal of array A228405. - Richard R. Forberg, Sep 02 2013
From Wolfdieter Lang, Oct 01 2013: (Start)
a(n) appears together with A106731, both interspersed with zeros, in the representation of nonnegative powers of the algebraic number rho(8) = 2*cos(Pi/8) = A179260 of degree 4, which is the length ratio of the smallest diagonal and the side in the regular octagon.
The minimal polynomial for rho(8) is C(8,x) = x^4 - 4*x^2 + 2, hence rho(8)^n = A(n+1)*1 + A(n)*rho(8) + B(n+1)*rho(8)^2 + B(n)*rho(8)^3, n >= 0, with A(2*k) = 0, k >= 0, A(1) = 1, A(2*k+1) = A106731(k-1), k >= 1, and B(2*k) = 0, k >= 0, B(1) = 0, B(2*k+1) = a(k-1), k >= 1. See also the P. Steinbach reference given under A049310. (End)
The ratio a(n)/A006012(n) converges to 1+sqrt(2). - Karl V. Keller, Jr., May 16 2015
From Tom Copeland, Dec 04 2015: (Start)
An aerated version of this sequence is given by the o.g.f. = 1 / (1 - 4 x^2 + 2 x^4) = 1 / [x^4 a_4(1/x)] = 1 / determinant(I - x M) = exp[-log(1 -4 x + 2 x^4)], where M is the adjacency matrix for the simple Lie algebra B_4 given in A265185 with the characteristic polynomial a_4(x) = x^4 - 4 x^2 + 2 = 2 T_4(x/2) = A127672(4,x), where T denotes a Chebyshev polynomial of the first kind.
A133314 relates a(n) to the reciprocal of the e.g.f. 1 - 4 x + 4 x^2/2!. (End)
a(n) is the number of vertices of the Minkowski sum of n simplices with vertices e_(2*i+1), e_(2*i+2), e_(2*i+3), e_(2*i+4) for i=0,...,n-1, where e_i is a standard basis vector. - Alejandro H. Morales, Oct 03 2022

Examples

			a(3) = 48 = 3 * 4 + 4 + 1 + 1 = 3*a(2) + a(1) + a(0) + 1.
Example for the octagon rho(8) powers: rho(8)^4  = 2 + sqrt(2) = -2*1 + 4*rho(8)^2  = A(5)*1 + A(4)*rho(8) + B(5)*rho(8)^2 + B(4)*rho(8)^3, with a(5) = A106731(1) = -2, B(5) = a(1) = 4, A(4) = 0, B(4) = 0. - _Wolfdieter Lang_, Oct 01 2013
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row sums of A059474. - David W. Wilson, Aug 14 2006
Equals 2 * A003480, n>0.
Row sums of A140071.

Programs

  • Haskell
    a007070 n = a007070_list !! n
    a007070_list = 1 : 4 : (map (* 2) $ zipWith (-)
       (tail $ map (* 2) a007070_list) a007070_list)
    -- Reinhard Zumkeller, Jan 16 2012
  • Magma
    Z:=PolynomialRing(Integers()); N:=NumberField(x^2-8); S:=[ ((4+r)^(1+n)-(4-r)^(1+n))/((2^(1+n))*r): n in [0..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Vincenzo Librandi, Mar 27 2011
    
  • Magma
    [n le 2 select 3*n-2 else 4*Self(n-1)-2*Self(n-2): n in [1..23]];  // Bruno Berselli, Mar 28 2011
    
  • Maple
    A007070 :=proc(n) option remember; if n=0 then 1 elif n=1 then 4 else 4*procname(n-1)-2*procname(n-2); fi; end:
    seq(A007070(n), n=0..30); # Wesley Ivan Hurt, Dec 06 2015
  • Mathematica
    LinearRecurrence[{4,-2}, {1,4}, 30] (* Harvey P. Dale, Sep 16 2014 *)
  • PARI
    a(n)=polcoeff(1/(1-4*x+2*x^2)+x*O(x^n),n)
    
  • PARI
    a(n)=if(n<1,1,ceil((2+sqrt(2))*a(n-1)))
    
  • Sage
    [lucas_number1(n,4,2) for n in range(1, 24)]# Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: 1/(1 - 4*x + 2*x^2).
Preceded by 0, this is the binomial transform of the Pell numbers A000129. Its e.g.f. is then exp(2*x)*sinh(sqrt(2)*x)/sqrt(2). - Paul Barry, May 09 2003
a(n) = ((2+sqrt(2))^(n+1) - (2-sqrt(2))^(n+1))/sqrt(8). - Al Hakanson (hawkuu(AT)gmail.com), Dec 27 2008, corrected Mar 28 2011
a(n) = (2 - sqrt(2))^n*(1/2 - sqrt(2)/2) + (2 + sqrt(2))^n*(1/2 + sqrt(2)/2). - Paul Barry, May 09 2003
a(n) = ceiling((2 + sqrt(2))*a(n-1)). - Benoit Cloitre, Aug 15 2003
a(n) = U(n, sqrt(2))*sqrt(2)^n. - Paul Barry, Nov 19 2003
a(n) = (1/4)*Sum_{r=1..7} sin(r*Pi/8)*sin(r*Pi/2)*(2*cos(r*Pi/8))^(2*n+3). - Herbert Kociemba, Jun 11 2004
a(n) = center term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 1 1 / 1 2 1 / 1 1 1]. M^n * [1 1 1] = [A007052(n) a(n) A007052(n)]. E.g., a(3) = 48 since M^3 * [1 1 1] = [34 48 34], where 34 = A007052(3). - Gary W. Adamson, Dec 18 2004
This is the binomial mean transform of A002307. See Spivey and Steil (2006). - Michael Z. Spivey (mspivey(AT)ups.edu), Feb 26 2006
a(2n) = Sum_{r=0..n} 2^(2n-1-r)*(4*binomial(2n-1,2r) + 3*binomial(2n-1,2r+1)) a(2n-1) = Sum_{r=0..n} 2^(2n-2-r)*(4*binomial(2n-2,2r) + 3*binomial(2n-2,2r+1)). - Jeffrey Liese, Oct 12 2006
a(n) = 3*a(n - 1) + a(n - 2) + a(n - 3) + ... + a(0) + 1. - Gary W. Adamson, Feb 18 2011
G.f.: 1/(1 - 4*x + 2*x^2) = 1/( x*(1 + U(0)) ) - 1/x where U(k)= 1 - 2^k/(1 - x/(x - 2^k/U(k+1) )); (continued fraction 3rd kind, 3-step). - Sergei N. Gladkovskii, Dec 05 2012
G.f.: A(x) = G(0)/(1-2*x) where G(k) = 1 + 2*x/(1 - 2*x - x*(1-2*x)/(x + (1-2*x)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 04 2013
G.f.: G(0)/(2*x) - 1/x, where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - (1-x)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n-1) = Sum_{k=0..n} binomial(2*n, n+k)*(k|8) where (k|8) is the Kronecker symbol. - Greg Dresden, Oct 11 2022
E.g.f.: exp(2*x)*(cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x)). - Stefano Spezia, May 20 2024
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