OFFSET
1,2
COMMENTS
Arises in the approximation of 14-fold quasipatterns by 14 Fourier modes.
Let DTS(n^c) denote the set of languages accepted by a deterministic Turing machine with space n^(o(1)) and time n^(c+o(1)), and let SAT denote the Boolean satisfiability problem. Then (1) SAT is not in DTS(n^c) for any c < 2*cos(Pi/7), and (2) the Williams inference rules cannot prove that SAT is not in DTS(n^c) for any c >= 2*cos(Pi/7). These results also apply to the Boolean satisfiability problem mod m where m is in A085971 except possibly for one prime. - Charles R Greathouse IV, Jul 19 2012
rho(7):= 2*cos(Pi/7) is the length ratio (smallest diagonal)/side in the regular 7-gon (heptagon). The algebraic number field Q(rho(7)) of degree 3 is fundamental for the 7-gon. See A187360 for the minimal polynomial C(7, x) of rho(7). The other (larger) diagonal/side ratio in the heptagon is sigma(7) = -1 + rho(7)^2, approx. 2.2469796. (see the decimal expansion in A231187). sigma(7) is the limit of a(n+1)/a(n) for n->infinity for the sequences like A006054 and A077998 which can be considered as analogs of the Fibonacci sequence in the pentagon. Thus sigma(7) plays in the heptagon the role of the golden section in the pentagon. See the P. Steinbach reference. - Wolfdieter Lang, Nov 21 2013
An algebraic integer of degree 3 with minimal polynomial x^3 - x^2 - 2x + 1. - Charles R Greathouse IV, Nov 12 2014
The other two solutions of the minimal polynomial of rho(7) = 2*cos(Pi/7) are 2*cos(3*Pi/7) and 2*cos(5*Pi/7). See eq. (20) of the W. Lang link. - Wolfdieter Lang, Feb 11 2015
The constant is the square root of 3.24697... (cf. A116425). It is the fifth-longest diagonal in the regular 14-gon with unit radius, which equals 2*sin(5*Pi/14). - Gary W. Adamson, Feb 14 2022
LINKS
Harry J. Smith, Table of n, a(n) for n = 1..20000
Simon Baker, Exceptional digit frequencies and expansions in non-integer bases, arXiv:1711.10397 [math.DS], 2017. See Theorem 1.1 p. 3.
Sam Buss and Ryan Williams, Limits on alternation-trading proofs for time-space lower bounds, Electronic Colloquium on Computational Complexity 2011
Wolfdieter Lang, The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon, arXiv:1210.1018 [math.GR], Oct 03 2012.
Peter Steinbach, Golden Fields: A Case for the Heptagon, Mathematics Magazine, Vol. 70, No. 1, Feb. 1997.
Ryan Williams, Time-space tradeoffs for counting NP solutions modulo integers, Computational Complexity 17 (2008), pp. 179-219.
FORMULA
Equals 2*A073052. - Michel Marcus, Nov 21 2013
Equals (Re((-(4*7)*(1 + 3*sqrt(3)*i))^(1/3)) + 1)/3, with the real part Re, and i = sqrt(-1). - Wolfdieter Lang, Feb 24 2015
Equals i^(2/7) - i^(12/7). - Peter Luschny, Apr 04 2020
From Peter Bala, Oct 20 2021: (Start)
Equals 2 - (1 - z)*(1 - z^6)/((1 - z^3)*(1 - z^4)), where z = exp(2*Pi*i/7).
The other two zeros of the minimal polynomial x^3 - x^2 - 2*x + 1 of 2*cos(Pi/7) are given by 2 - (1 - z^3)*(1 - z^4)/((1 - z^2)*(1 - z^5)) = 2*cos(3*Pi/7) and 2 - (1 - z^2)*(1 - z^5)/((1 - z)*(1 - z^6)) = cos(5*Pi/7).
Equals Product_{n >= 0} (7*n+2)*(7*n+5)/((7*n+1)*(7*n+6)) = 1 + Product_{n >= 0} (7*n+2)*(7*n+5)/((7*n+3)*(7*n+4)) = 1/A255240.
The linear fractional mapping r -> 1/(1 - r) cyclically permutes the three zeros of the minimal polynomial x^3 - x^2 - 2*x + 1. The inverse mapping is r -> (r - 1)/r.
The quadratic mapping r -> 2 - r^2 also cyclically permutes the three zeros. The inverse mapping is r -> r^2 - r - 1. (End)
Equals i^(2/7) + i^(-2/7). - Gary W. Adamson, Feb 11 2022
EXAMPLE
1.801937735804838252472204639014890102331838324263714300107124846398864...
MAPLE
evalf(2*cos(Pi/7), 100); # Wesley Ivan Hurt, Feb 01 2017
MATHEMATICA
RealDigits[2 Cos[Pi/7], 10, 111][[1]] (* Robert G. Wilson v, Jun 11 2013 *)
PROG
(PARI) default(realprecision, 20080); x=2*cos(Pi/7); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b160389.txt", n, " ", d));
(Magma) R:= RealField(200); Reverse(Intseq(Floor(10^110*2*Cos(Pi(R)/7)))); // Marius A. Burtea, Nov 13 2019
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Harry J. Smith, May 31 2009
STATUS
approved