A047341 -id:A047341 - cp's OEIS frontend

A160389 Decimal expansion of 2*cos(Pi/7).

1, 8, 0, 1, 9, 3, 7, 7, 3, 5, 8, 0, 4, 8, 3, 8, 2, 5, 2, 4, 7, 2, 2, 0, 4, 6, 3, 9, 0, 1, 4, 8, 9, 0, 1, 0, 2, 3, 3, 1, 8, 3, 8, 3, 2, 4, 2, 6, 3, 7, 1, 4, 3, 0, 0, 1, 0, 7, 1, 2, 4, 8, 4, 6, 3, 9, 8, 8, 6, 4, 8, 4, 0, 8, 5, 5, 8, 7, 9, 9, 3, 1, 0, 0, 2, 7, 2, 2, 9, 0, 9, 4, 3, 7, 0, 2, 4, 8, 3, 0, 6, 3, 6, 6, 2

Offset: 1

Harry J. Smith, May 31 2009

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

			1.801937735804838252472204639014890102331838324263714300107124846398864...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 207.

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.
Index entries for algebraic numbers, degree 3

Cf. A039921 (continued fraction).
Cf. A047336, A047341, A116425, A255240, A255249.
Cf. A003558 (the constant is cyclic with period 3, for N = 7).

R:= RealField(200); Reverse(Intseq(Floor(10^110*2*Cos(Pi(R)/7)))); // Marius A. Burtea, Nov 13 2019

evalf(2*cos(Pi/7), 100); # Wesley Ivan Hurt, Feb 01 2017

RealDigits[2 Cos[Pi/7], 10, 111][[1]] (* Robert G. Wilson v, Jun 11 2013 *)

default(realprecision, 20080); x=2*cos(Pi/7); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b160389.txt", n, " ", d));

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) = A255241 and 2 - (1 - z^2)*(1 - z^5)/((1 - z)*(1 - z^6)) = cos(5*Pi/7) = -A362922.
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
From Amiram Eldar, Nov 22 2024: (Start)
Equals Product_{k>=1} (1 - (-1)^k/A047336(k)).
Equals 1 + cosec(3*Pi/14)/2 = 1 + Product_{k>=1} (1 + (-1)^k/A047341(k)). (End)
Equals sqrt(A116425). - Hugo Pfoertner, Nov 22 2024

A057570 Numbers of the form n*(7n+-1)/2.

Original entry on oeis.org

0, 3, 4, 13, 15, 30, 33, 54, 58, 85, 90, 123, 129, 168, 175, 220, 228, 279, 288, 345, 355, 418, 429, 498, 510, 585, 598, 679, 693, 780, 795, 888, 904, 1003, 1020, 1125, 1143, 1254, 1273, 1390, 1410, 1533, 1554, 1683, 1705, 1840, 1863, 2004

Offset: 1

N. J. A. Sloane, Oct 04 2000

Also integers of the form Sum_{k = 1..n} k/7. - Alonso del Arte, Jan 20 2012
Sequence provides all integers m such that 56*m + 1 is a square. [Bruno Berselli, Oct 07 2015]
The sequence terms occur as the exponents in the expansion of Product_{n >= 1} (1 - x^(7*n)) * (1 + x^(7*n-3)) * (1 + x^(7*n-4)) = 1 + x^3 + x^4 + x^13 + x^15 + x^30 + x^33 + .... Cf. A363801. - Peter Bala, Nov 21 2024

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A074378, A001318, A057569, A154260, A154292, A154293, A047341, A274830, A363801.

Select[Table[Plus@@Range[n]/7, {n, 0, 199}], IntegerQ] (* Alonso del Arte, Jan 20 2012 *)
CoefficientList[Series[-x (3 + x + 3 x^2) / ((1 + x)^2 (x - 1)^3), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 19 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{0,3,4,13,15},50] (* Harvey P. Dale, Sep 17 2023 *)

a(n)=(14*n*(n-1)+5*(2*n-1)*(-1)^n+5)/16 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: -x^2*(3+x+3*x^2) / ( (1+x)^2*(x-1)^3 ). - R. J. Mathar, Jan 25 2011
a(n) = +1*a(n-1)+2*a(n-2)-2*a(n-3)-1*a(n-4)+1*a(n-5). - Joerg Arndt, Jan 25 2011
a(n) = (14*n*(n-1)+5*(2*n-1)*(-1)^n+5)/16. - Bruno Berselli, Jan 25 2011
a(n)-a(n-2) = A047341(n-1) for n>2. - Bruno Berselli, Jan 25 2011
Sum_{n>=2} 1/a(n) = 14 - 2*cot(Pi/7)*Pi. - Amiram Eldar, Mar 17 2022

A164131 Numbers k such that k^2 == 2 (mod 31).

Original entry on oeis.org

8, 23, 39, 54, 70, 85, 101, 116, 132, 147, 163, 178, 194, 209, 225, 240, 256, 271, 287, 302, 318, 333, 349, 364, 380, 395, 411, 426, 442, 457, 473, 488, 504, 519, 535, 550, 566, 581, 597, 612, 628, 643, 659, 674, 690, 705, 721, 736, 752, 767, 783, 798, 814

Offset: 1

Vincenzo Librandi, Aug 11 2009

Sequences of the type n^2 == 2 (mod m) are basically defined for each m of A057126. See A047341 (m=7), A113804 (m=14), A155449 (m=17), A155450 (m=23), A158803 (m=41) etc. - R. J. Mathar, Aug 26 2009

			At n= 4, a(4)=(31-1+186)/4=54. At n=5, a(5)=(31+1+248)/4=70.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A057126, A047341, A113804, A155449, A155450, A158803.

Select[Range[850],Mod[#^2,31]==2&]  (* Harvey P. Dale, Feb 04 2011 *)

isok(k) = Mod(k, 31)^2 == 2; \\ Michel Marcus, Nov 22 2022

a(n) = a(n-1)+a(n-2)-a(n-3).
a(n) = (31+(-1)^(n-1)+62(n-1))/4.
G.f.: x*(8+15*x+8*x^2)/((1+x)*(x-1)^2). - R. J. Mathar, Aug 26 2009
a(n) = 31*(n-1)-a(n-1) with n>1, a(1)=8. - Vincenzo Librandi, Nov 30 2010
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(15*Pi/62)*Pi/31. - Amiram Eldar, Feb 28 2023

Entries checked by R. J. Mathar, Aug 26 2009

A204769 a(n) = 151*(n-1) - a(n-1) with n>1, a(1)=46.

Original entry on oeis.org

46, 105, 197, 256, 348, 407, 499, 558, 650, 709, 801, 860, 952, 1011, 1103, 1162, 1254, 1313, 1405, 1464, 1556, 1615, 1707, 1766, 1858, 1917, 2009, 2068, 2160, 2219, 2311, 2370, 2462, 2521, 2613, 2672, 2764, 2823

Offset: 1

Vincenzo Librandi, Mar 08 2012

Positive numbers k such that k^2 == 2 (mod 151), where the prime 151 == -1 (mod 8).

Equivalently, numbers k such that k == 46 or 105 (mod 151). - Bruno Berselli, Mar 08 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Sequences of the type n^2 == 2 (mod p), where p is a prime of the form 8k-1: A047341, A155450, A164131, A164135, A167533, A167534, A177044, A177046, A204766.

Sequences of the type n^2 == 2 (mod p), where p is a prime of the form 8k+1: A155449, A158803, A159007, A159008, A176010, A206525.

[(-151-33*(-1)^n+302*n)/4: n in [1..60]];

LinearRecurrence[{1,1,-1}, {46,105,197}, 40] (* or *) CoefficientList[Series[x*(46+59*x+46*x^2)/((1+x)*(x-1)^2),{x,0,33}],x] (* or *) a[1] = 46; a[n_] := a[n] = 151*(n-1) - a[n-1]; Table[a[n], {n, 1, 40}]

a(n)=(-151-33*(-1)^n+302*n)/4 \\ Charles R Greathouse IV, Oct 16 2015

G.f.: x*(46+59*x+46*x^2)/((1+x)*(x-1)^2).
a(n) = (-151-33*(-1)^n+302*n)/4.
a(n) = a(n-1) +a(n-2) -a(n-3).
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(59*Pi/302)*Pi/151. - Amiram Eldar, Feb 28 2023

A206525 a(n) = 113*(n-1) - a(n-1) with n>1, a(1)=51.

Original entry on oeis.org

51, 62, 164, 175, 277, 288, 390, 401, 503, 514, 616, 627, 729, 740, 842, 853, 955, 966, 1068, 1079, 1181, 1192, 1294, 1305, 1407, 1418, 1520, 1531, 1633, 1644, 1746, 1757, 1859, 1870, 1972, 1983, 2085, 2096, 2198, 2209

Offset: 1

Vincenzo Librandi, Mar 09 2012

Positive numbers k such that k^2 == 2 (mod 113), where the prime 113 == 1 (mod 8).

Equivalently, numbers k such that k == 51 or 62 (mod 113).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Sequences of the type n^2 == 2 (mod p), where p is a prime of the form 8k+1: A155449, A158803, A159007, A159008, A176010.

Sequences of the type n^2 == 2 (mod p), where p is a prime of the form 8k-1: A047341, A155450, A164131, A164135, A167533, A167534, A177044, A177046, A204769.

[(-113-91*(-1)^n+226*n)/4: n in [1..60]];

LinearRecurrence[{1, 1, -1}, {51, 62, 164}, 40] (* or *) CoefficientList[Series[x*(51+11*x+51*x^2)/((1+x)*(x-1)^2), {x, 0, 40}], x] (* or *) a[1] = 51; a[n_] := a[n] = 113*(n-1) - a[n-1]; Table[a[n], {n, 1, 40}]

a(n) = a(n-2) + 113.
G.f.: x*(51+11*x+51*x^2)/((1+x)*(x-1)^2).
a(n) = (-113-91*(-1)^n+226*n)/4.
a(n) = a(n-1)+a(n-2)-a(n-3).
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(11*Pi/226)*Pi/113. - Amiram Eldar, Feb 28 2023

A045432 Primes congruent to {3, 4} mod 7.

Original entry on oeis.org

3, 11, 17, 31, 53, 59, 67, 73, 101, 109, 137, 151, 157, 179, 193, 199, 227, 241, 263, 269, 277, 283, 311, 347, 353, 367, 389, 409, 431, 479, 487, 521, 557, 563, 571, 577, 599, 613, 619, 641, 647, 661, 683, 739

Offset: 1

N. J. A. Sloane

Or, primes p such that p^2 == 2 (mod 7). - Zak Seidov, May 11 2013

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A047341 (superset), A225550.

[ p: p in PrimesUpTo(1000) | p mod 7 in {3,4} ]; // Vincenzo Librandi, Aug 13 2012

Select[Prime[Range[200]],MemberQ[{3,4},Mod[#,7]]&]  (* Vincenzo Librandi, Aug 13 2012 *)
Select[Prime[Range[200]], PowerMod[#,2,7] == 2 &] (* Zak Seidov, May 11 2013 *)

A204766 a(n) = 167*(n-1)-a(n-1) with n>1, a(1)=13.

Original entry on oeis.org

13, 154, 180, 321, 347, 488, 514, 655, 681, 822, 848, 989, 1015, 1156, 1182, 1323, 1349, 1490, 1516, 1657, 1683, 1824, 1850, 1991, 2017, 2158, 2184, 2325, 2351, 2492, 2518, 2659, 2685, 2826, 2852, 2993, 3019, 3160

Offset: 1

Vincenzo Librandi, Mar 09 2012

Positive numbers k such that k^2 == 2 (mod 167), where the prime 167 == -1 (mod 8).

Equivalently, numbers k such that k == 13 or 154 (mod 167).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Sequences of the type n^2 == 2 (mod p), where p is a prime of the form 8k-1: A047341, A155450, A164131, A164135, A167533, A167534, A177044, A177046, A204769.

Sequences of the type n^2 == 2 (mod p), where p is a prime of the form 8k+1: A155449, A158803, A159007, A159008, A176010, A206525, A206526.

[(-167+115*(-1)^n+334*n)/4: n in [1..60]];

CoefficientList[Series[x*(13+141*x+13*x^2)/((1+x)*(x-1)^2), {x, 0, 40}], x] (* or *) LinearRecurrence[{1, 1, -1}, {13, 154, 180}, 40]

G.f.: x*(13+141*x+13*x^2)/((1+x)*(x-1)^2).
a(n) = (-167+115*(-1)^n+334*n)/4.
a(n) = a(n-1)+a(n-2)-a(n-3).
Sum_{n>=1} (-1)^(n+1)/a(n) = cot(13*Pi/167)*Pi/167. - Amiram Eldar, Feb 28 2023

A206526 a(n) = 137*(n-1) - a(n-1) with n>1, a(1)=31.

Original entry on oeis.org

31, 106, 168, 243, 305, 380, 442, 517, 579, 654, 716, 791, 853, 928, 990, 1065, 1127, 1202, 1264, 1339, 1401, 1476, 1538, 1613, 1675, 1750, 1812, 1887, 1949, 2024, 2086, 2161, 2223, 2298, 2360, 2435, 2497, 2572, 2634, 2709, 2771, 2846, 2908, 2983

Offset: 1

Vincenzo Librandi, Mar 09 2012

Positive numbers k such that k^2 == 2 (mod 137), where the prime 137 == 1 (mod 8).
Equivalently, numbers k such that k == 31 or 106 (mod 137).
The subsequence of primes begins: 31, 853, 1613, 1949, 2161. - Jonathan Vos Post, Mar 09 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Sequences of the type n^2 == 2 (mod p), where p is a prime of the form 8k+1: A155449, A158803, A159007, A159008, A176010, A206525.

Sequences of the type n^2 == 2 (mod p), where p is a prime of the form 8k-1: A047341, A155450, A164131, A164135, A167533, A167534, A177044, A177046, A204769.

[(-137+13*(-1)^n+274*n)/4: n in [1..60]];

[n: n in [1..3000] | n^2 mod 137 eq 2]; // Vincenzo Librandi, Mar 31 2016

LinearRecurrence[{1, 1, -1}, {31, 106, 168}, 40] (* or *) CoefficientList[Series[x*(31+75*x+31*x^2)/((1+x)*(x-1)^2), {x, 0, 50}], x] (* or *) a[1] = 31; a[n_] := a[n] = 137*(n-1) - a[n-1]; Table[a[n], {n, 1, 40}]

a(n) = a(n-2) + 137.
G.f.: x*(31+75*x+31*x^2)/((1+x)*(x-1)^2).
a(n) = (-137+13*(-1)^n+274*n)/4.
a(n) = a(n-1)+a(n-2)-a(n-3).
Sum_{n>=1} (-1)^(n+1)/a(n) = cot(31*Pi/137)*Pi/137. - Amiram Eldar, Feb 28 2023

A374584 Numbers k such that 7*k + 2 is a square.

Original entry on oeis.org

1, 2, 14, 17, 41, 46, 82, 89, 137, 146, 206, 217, 289, 302, 386, 401, 497, 514, 622, 641, 761, 782, 914, 937, 1081, 1106, 1262, 1289, 1457, 1486, 1666, 1697, 1889, 1922, 2126, 2161, 2377, 2414, 2642, 2681, 2921, 2962, 3214, 3257, 3521, 3566, 3842, 3889

Offset: 1

Juri-Stepan Gerasimov, Aug 12 2024

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

The numbers k such that (m + (9-m)*k) is a square: A000217 (m = 1), this sequence (m = 2), A003154 (m = 3), A195162 (m = 4), A028387 (m = 5), A100536 (m = 6), A059993 (m = 7), A028884 (m = 8).

Cf. A047341.

[k: k in [0..4000] | IsSquare(7*k + 2)];

((Table[7*n + {3, 4}, {n, 0, 23}] // Flatten)^2 - 2)/7 (* Amiram Eldar, Aug 12 2024 *)

a(n) = (A047341(n)^2 - 2)/7. - Amiram Eldar, Aug 12 2024

cp's OEIS Frontend

A160389 Decimal expansion of 2*cos(Pi/7).

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A057570 Numbers of the form n*(7n+-1)/2.

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A164131 Numbers k such that k^2 == 2 (mod 31).

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A204769 a(n) = 151*(n-1) - a(n-1) with n>1, a(1)=46.

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A206525 a(n) = 113*(n-1) - a(n-1) with n>1, a(1)=51.

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A045432 Primes congruent to {3, 4} mod 7.

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A204766 a(n) = 167*(n-1)-a(n-1) with n>1, a(1)=13.

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