cp's OEIS Frontend

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

We follow Maple in defining Phi_0 to be x; it could equally well be taken to be 1.

From Wolfdieter Lang, Oct 29 2013: (Start)

The length of row n >= 1 of this table is phi(n) + 1 = A000010(n) + 1. Row n = 0 has here length 2.

Phi_n(x) is the minimal polynomial of omega_n := exp(i*2*Pi/n) over the rationals. Namely, Phi_n(x) = Product_{k=0..n-1, gcd(k,n)=1} (x - (omega_n)^k). See the Graham et al. reference, 4.50 a, pp. 149, 506.

Phi_n(x) = Product_{d|n} (x^d - 1)^(mu(n/d)) with the Moebius function mu(n) = A008683(n), n >= 1. See the Graham et al. reference, 4.50 b, pp. 149, 506.

Phi_n(x) = Phi_{rad(n)}(x^(n/rad(n))), n >= 2, with rad(n) = A007947(n), the squarefree kernel of n. Proof from the preceding formula, where only squarefree n/d (A005117) from the set of divisors of n enter, by mapping each factor (numerator or denominator) of the left hand side to one of the right hand side and vice versa.

(End)

Each row can be considered as the last column of the companion matrix of the cyclotomic polynomial: A000010(n) is the size of such a square matrix, last column has opposite signs and the last term (before last term of each row in A013595) equal to A008683(n). - Eric Desbiaux, Dec 14 2015

Except for the initial term a(1)=2, indices k such that A020513(k)=Phi[k](-1) is prime, where Phi is a cyclotomic polynomial.

This is illustrated by the PARI code, although it is probably more efficient to calculate a(n) as 2*A000961(n).

{ a(n)/2 ; n>1 } are also the indices for which A020500(k)=Phi[k](1) is prime.

A188666(k) = A000961(k+1) for k: a(k) <= k < a(k+1), k > 0;

A188666(a(n)) = A000961(n+1). [Reinhard Zumkeller, Apr 25 2011]

a(0) depends on the definition of the 0th cyclotomic polynomial; Maple defines it as x, but Mathematica defines it as 1. - T. D. Noe, Jul 23 2008 [a(0) = x is correct. - N. J. A. Sloane, Aug 01 2008]

A020501[2n] = A019320[n] for all odd n > 1. (Because if m > 1 is odd, then Phi_2m(x) = Phi_m(-x) as demonstrated by Bloom). - Antti Karttunen, Aug 02 2001

The complement relative to the positive integers of the present sequence is A138929.

The sequence includes all positive integers of the forms (i) odd, (ii) 2^k*p, p an odd prime and k>=2, and (iii) 2^e0*p1^e1*p2^e2 *** pk^ek, k >= 2, with odd primes p1, ..., pk, and each exponent from {e0, ..., ek} is >= 1.

For a(n) > 1 a regular a(n)-gon, with length ratio (smallest diagonal)/side rho(a(n)) = 2*cos(Pi/a(n)), the inverse of rho(a(n)), which is an element of the algebraic number field Q(rho(a(n))), is in fact a Q(rho(a(n)))-integer. For a(1)=1 rho(1) = -2, and the inverse is not a Q-integer.

Origin unknown. First encountered by this author as part of an employment-interview question at Apple Inc, in early 2016.

While holding a deck of n cards:

1. Deal the top card from the deck onto the table ('deal one').

2. Move the next card from the top of the deck to the bottom of the deck ('skip one').

3. Repeat steps 1 and 2 until all cards are on the table. This is a round.

4. Pick up the deck from the table and repeat steps 1 through 3 until the deck is in its original order.

From Robert Israel, Jul 06 2017: (Start)

a(n) <= A000793(n).

a(n) divides n!.

Conjecture: a(n) < n for infinitely many n.

Conjecture: the set of n for which the permutation is a single n-cycle, and thus a(n) = n, has nonzero density. (End)

It appears that for n = 2^k and all m > n, a(n) <= a(m). - Andrew Warren, Jul 15 2017

a(2^(k+1)) / a(2^k) = A020513(k+2) at least for 1 <= k <= 30, according to the values computed by Andrew Warren. - Andrey Zabolotskiy, Apr 02 2018

Are all terms composite? At least the first 10000 terms are.

The polynomials from A374385 (for n > 1), evaluated at x = -1.