cp's OEIS Frontend

A transform of A000244 under the mapping g(x)->(1/(1+x^3))g(x/(1+x^3)). - Paul Barry, Oct 20 2004

b(n) := (-1)^n*a(n) appears in the formula for the nonpositive powers of rho(9) := 2*cos(Pi/9), when written in the power basis of the algebraic number field Q(rho(9)) of degree 3. See A187360 for the minimal polynomial C(9, x) of rho(9), and a link to the Q(2*cos(pi/n)) paper. 1/rho(9) = -3*1 + 0*rho(9) + 1*rho(9)^2 (see A230079, row n=5). 1/rho(9)^n = b(n)*1 + b(n-2)*rho(9) + b(n-1)*rho(9)^2, n >= 0, with b(-1) = 0 = b(-2). - Wolfdieter Lang, Nov 04 2013

The limit b(n+1)/b(n) = -a(n+1)/a(n) for n -> infinity is -tau(9) := -(1 + rho(9)) = 1/(2*cos(Pi*5/9)), approximately -2.445622407. tau(9) is known to be the length ratio (longest diagonal)/side in the regular 9-gon. This limit follows from the b(n)-recurrence and the solutions of X^3 + 3*X^2 - 1 = 0, which are given by the inverse of the known solutions of the minimal polynomial C(9, x) of rho(9) (see A187360). The other two X solutions are 1/rho(9) = -3 + rho(9)^2, approximately 0.5320888860 and 1/(2*cos(Pi*7/9)) = 1 + rho(9) - rho(9)^2, approximately -0.6527036445, and they are therefore irrelevant for this sequence. - Wolfdieter Lang, Nov 08 2013

a(n) is also the number of ternary (0,1,2) sequences of length n without a consecutive '110' because the patterns A=012 and B=110 have the same autocorrelation, i.e., AA=100=BB, in the sense of Guibas and Odlysko (1981). (A cyclic version of this sequence can be found in sequence A274018.) - Petros Hadjicostas, Sep 12 2017

rho_3 := +2*cos(5*Pi/7) is the negative zero of the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of the algebraic number rho(7) = 2*cos(Pi/7), the length ratio of the smaller diagonal and the side in the regular 7-gon (heptagon). See A187360 and a link to the arXiv paper given there, eq. (20) for the zeros of C(n, x). The positive zeros are rho(7) and rho_2 = 2*cos(3*Pi/7) shown in A160389 and A255241.

Essentially the same as A231187 and A116425. - R. J. Mathar, Mar 14 2015

This is also the decimal expansion of 2*sin(Pi/14).

rho_2 := 2*cos(3*Pi/7) and rho(7) := 2*cos(Pi/7) (see A160389) are the two positive zeros of the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of the algebraic number rho(7), the length ratio of the smaller diagonal and the side in the regular 7-gon (heptagon). See A187360 and a link to the arXiv paper given there, eq. (20) for the zeros of C(n, x). The other zero is negative, rho_3 := 2*cos(5*Pi/n). See -A255249.

Also the edge length of a regular 14-gon with unit circumradius. Such an m-gon is not constructible using a compass and a straightedge (see A004169). With an even m, in fact, it would be constructible only if the (m/2)-gon were constructible, which is not true in this case (see A272487). - Stanislav Sykora, May 01 2016

2*T(2*n,x) = Sum_{m=0..n} a(n,m)*(2*x)^(2*m).

Closely related to A284982, which has opposite signs and rows begin with 0 of alternating signs instead of +/2. - Eric W. Weisstein, Apr 07 2017

Bisection triangle of A127672 (without zero entries, even part). The odd part is ((-1)^(n-m))*A111125(n,m).

If the leading 2 is replaced by a 1 we get the essentially identical sequence A110162. - N. J. A. Sloane, Jun 09 2007

Also row n gives coefficients of characteristic polynomial of the Cartan matrix for the root system B_n (or, equally, C_n). - Roger L. Bagula, May 23 2007

From Wolfdieter Lang, Oct 04 2013: (Start)

This triangle a(n,m) is used to express the length ratio side/R given by s(4*n+2) = 2*sin(Pi/(4*n+2)) = 2*cos(2*n*Pi/(4*n+2)) in a regular (4*n+2)-gon, inscribed in a circle with radius R, in terms of rho(4*n+2) = 2*cos(Pi/4*n+2), the length ratio of (the smallest diagonal)/side (for n=2 there is no such diagonal).

s(4*n+2) = Sum_{m=0..n}a(n,m)*rho(4*n+2)^(2*m). This formula is needed to show that the total sum of all length ratios in a (4*n+2)-gon is an integer in the algebraic number field Q(rho(4*n+2)). Note that rho(4*n+2) has degree delta(4*n+2) = A055034(4*n+2). Therefore one has to take s(4*n+2) modulo C(4*n+2, x=rho(4*n+2)), the minimal polynomial of rho(4*n+2) (see A187360). Thanks go to Seppo Mustonen for asking me to look into this problem. See ((-1)^(n-m))*A111125(n,m) for the (4*n)-gon situation. (End)

Rational primes that decompose in the field Q(sqrt(-2)). - N. J. A. Sloane, Dec 25 2017

Fermat knew of the relationship between a prime being congruent to 1 or 3 mod 8 and its being the sum of a square and twice a square, and claimed to have a firm proof of this fact. These numbers are not primes in Z[sqrt(-2)], as they have x - y sqrt(-2) as a divisor. - Alonso del Arte, Dec 07 2012

Terms m in A047471 with A010051(m) = 1. - Reinhard Zumkeller, Dec 29 2012

This sequence gives the primes p which satisfy norm(rho(p)) = + 1 with rho(p) := 2*cos(Pi/p) (the length ratio (smallest diagonal)/side in the regular p-gon). The norm of an algebraic number (over Q) is the product over all zeros of its minimal polynomial. Here norm(rho(p)) = (-1)^delta(p)* C(p, 0), with the degree delta(p) = A055034(p) = (p-1)/2. For the minimal polynomial C see A187360. For p == 1 (mod 8) the norm is C(p, 0) (see a comment on 4*A005123) and for p == 3 (mod 8) the norm is -C(p, 0) (see a comment on A186297). For the primes with norm(rho(p)) = -1 see A003628. - Wolfdieter Lang, Oct 24 2013

If p is a member then it has a unique representation as x^2+2y^2 [Frei, Theorem 3]. - N. J. A. Sloane, May 30 2014

Primes that are the quarter perimeter of a Heronian triangle. Such primes are unique to the Heronian triangle (see Yiu link). - Frank M Jackson, Nov 30 2014

The graph of this sequence is made up of four linear functions: a(n_odd)=n, a(n=2+4i)=n/2, a(4+8i)=n/4, a(8i)=n/8. - Zak Seidov, Oct 30 2006. [In general, f(n) = numerator of n/(n+m) consists of linear functions n/d_i, where d_i are divisors of m (including 1 and m).]

a(n+2), n>=0, is the denominator of the harmonic mean H(n,2) = 4*n/(n+2). a(n+2) = (n+2)/gcd(n+2,8). a(n+5) = A227042(n+2, 2), n >= 0. - Wolfdieter Lang, Jul 04 2013

The sequence p(n) = a(n-4), n>=1, with a(-3) = a(3) = 3, a(-2) = a(2) = 1 and a(-1) = a(1) = 1, appears in the problem of writing 2*sin(2*Pi/n) as an integer in the algebraic number field Q(rho(q(n))), where rho(k) = 2*cos(Pi/k) and q(n) = A225975(n). One has 2*sin(2*Pi/n) = R(p(n), x) modulo C(q(n), x), with x = rho(q(n)) and the integer polynomials R and C given in A127672 and A187360, respectively. See a comment on A225975. - Wolfdieter Lang, Dec 04 2013

A204455(n) divides a(n) for n>=1. - Alexander R. Povolotsky, Apr 06 2015

A multiplicative sequence. Also, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n >= 1, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 20 2019

In general, a(n) = (2/m)*Sum_{r=1..m-1} sin(r*j*Pi/m)*sin(r*k*Pi/m)*(2*cos(r*Pi/m))^(2n+1) counts (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < m and |s(i)-s(i-1)| = 1 for i = 1,2,...,2n+1, s(0) = j, s(2n+1) = k.

a(n)/a(n-1) tends to 3.53208888...; = 2 + 2*cos(2*Pi/9) = A332438. - Gary W. Adamson, May 29 2008

From Wolfdieter Lang, Mar 27 2020: (Start)

The explicit form is written in terms of r = rho(9) = 2*cos(Pi/9) = A332437 as a Binet - de Moivre type formula a(n+2) = r^(2*(n+1))*(A(r) + Bp(r)*Cp(r)^(n+1)) + Bm(r)*Cm(r)^(n+1)), with A(r) = (1/9)*(2 + 5*r -r^2), approx. 0.87387081, Bp(r) = (1/18)*((14*r^2 - 5*r - 42)*sqrt(3*(3*r + 1)*(r - 1)) + r^2 - 5*r - 2) = (1/9)*(8 - r - 4*r^2), approx. -0.88974898, Cp(r) = (1/2)*(9*r^2 - 3*r - 26)*(3*r - 1 + sqrt(3*(3*r+1)*(r-1))) = 32 + 4*r - 11*r^2, approx. 0.66456322, Bm(r) = (1/18)*(-(14*r^2 - 5*r - 42)*sqrt(3*(3*r + 1)*(r - 1)) + r^2 - 5*r - 2) = (1/9)*(-10 -4*r + 5*r^2), approx. 0.01587816, and Cm(r) = (1/2)*(9*r^2 - 3*r - 26)*(3*r - 1 - sqrt(3*(3*r + 1)*(r - 1))) = 21 + 2*r - 7*r^2, approx. 0.03414828, for n >= 0.

Proof by partial fraction decomposition of the g.f. using the roots of 1 - 6*x + 9*x^2 - x^3 written in terms of r, which are X1(r) = 1/r^2 = 9 + r - 3*r^2, approx. 0.28311858, Xp(r) = (r/2)*(3*r - 1 + sqrt((3*(3*r+1))*(r-1))) = 1 + 2*r + r^2, approx. 8.29085937, Xm(r) = (r/2)*(3*r - 1 - sqrt((3*(3*r + 1))*(r - 1))) = -1 - 3*r + 2*r^2, approx. 0.42602205. Xp(r)*Xm(r) = r^2. The reduction with the minimal polynomial of r = rho(9), i.e., C(9, x) = x^3 - 3*x - 1 (see A187360) has been used to avoid all powers of r larger than 2. The reciprocal roots turn out to be the roots of the minimal polynomial of r^2, see A332438. 1/X1(r) = r^2, 1/Xp(r) = 2 - r, and 1/Xm(r) = 4 + r - r^2.

This proves the above stated limit of a(n+3)/a(n+2) for n to infinity, namely r^2 = A332438, approx. 3.53208889.

(End)

This algebraic number called rho(9) of degree 3 = A055034(9) has minimal polynomial C(9, x) = x^3 - 3*x - 1 (see A187360).

rho(9) gives the length ratio diagonal/side of the smallest diagonal in the regular 9-gon.

The length ratio diagonal/side of the second smallest and the third smallest (or the largest) diagonal in the regular 9-gon are rho(9)^2 - 1 = A332438 - 1 and rho(9) + 1, respectively. - Mohammed Yaseen, Oct 31 2020

This algebraic number rho(9)^2 of degree 3 is a root of its minimal polynomial x^3 - 6*x^2 + 9*x - 1.

The other two roots are x2 = (2*cos(5*Pi/9))^2 = (2*cos(4*Pi/9))^2 = (R(4,rho(9)))^2 = 2 - rho(9) = 0.120614758..., and x3 = (2*cos(7*Pi/9))^2 = (2*cos(7*Pi/9))^2 = (R(7,rho(9)))^2 = 4 + rho(9) - rho(9)^2 = 2.347296355... = A130880 + 2, with rho(9) = 2*cos(Pi/9) = A332437, the monic Chebyshev polynomials R (see A127672), and the computation is done modulo the minimal polynomial of rho(9) which is x^3 - 3*x - 1 (see A187360).

This gives the representation of these roots in the power basis of the simple field extension Q(rho(9)). See the linked W. Lang paper in A187360, sect. 4.

This number rho(9)^2 appears as limit of the quotient of consecutive numbers af various sequences, e.g., A094256 and A094829.

The algebraic number rho(9)^2 - 2 = 1.532088898... of Q(rho(9)) has minimal polynomial x^3 - 3*x + 1 over Q. The other roots are -rho(9) = -A332437 and 2 + rho(9) - rho(9)^2 = A130880. - Wolfdieter Lang, Sep 20 2022

For Modd n (not to be confused with mod n) see a comment on A203571.

For n=1 one has the Modd 1 residue class [0], the integers. The group of order 1 is the cyclic group Z_1 with the unit element 0==1 (Modd 1). [Changed by Wolfdieter Lang, Apr 04 2012]

For the non-cyclic (acyclic) values see A206552.

For these numbers n, and only for these (only the n values < 100 are shown above), there exist primitive roots Modd n. See the nonzero values of A206550 for the smallest positive ones.

For n=1 the primitive root is 0 == 1 (Modd 1), see above.

For n>=1 the multiplicative group Modd n is the Galois group Gal(Q(rho(n))/Q), with the algebraic number rho(n) := 2*cos(Pi/n) with minimal polynomial C(n,x), whose coefficients are given in A187360.