cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A275546 a(n) = (tan(1*Pi/11))^(2*n) + (tan(2*Pi/11))^(2*n) + (tan(3*Pi/11))^(2*n) + (tan(4*Pi/11))^(2*n) + (tan(5*Pi/11))^(2*n).

Original entry on oeis.org

5, 55, 2365, 113311, 5476405, 264893255, 12813875437, 619859803695, 29985188632421, 1450508002869079, 70167091762786205, 3394273427239643839, 164195092176119969173, 7942798031108524622951, 384226104001681151724877, 18586611219134532494467151, 899111520569015285343455941, 43493755633501102693569684087, 2103973462501643822799172235773
Offset: 0

Views

Author

Kai Wang, Aug 01 2016

Keywords

Comments

(tan(1*Pi/11))^(2*n), (tan(2*Pi/11))^(2*n), (tan(3*Pi/11))^(2*n),(tan(4*Pi/11))^(2*n), (tan(5*Pi/11))^(2*n) are roots of the polynomial x^5 - 55x^4 + 330x^3 - 462x^2 + 165x - 11.
Sum_{k=1..((m-1)/2)} (tan(k*Pi/m))^(2*n) is an integer when m >= 3 is an odd integer (see AMM link); this sequence is the particular case m = 11. All terms are odd. - Bernard Schott, Apr 24 2022

Crossrefs

Similar to: A000244 (m=3), 2*A165225 (m=5), A108716 (m=7), A353410 (m=9), this sequence (m=11), A353411 (m=13).

Programs

  • PARI
    a(n)=([0,1,0,0,0;0,0,1,0,0;0,0,0,1,0;0,0,0,0,1;11,-165,462,-330,55]^n*[5;55;2365;113311;5476405])[1,1] \\ Charles R Greathouse IV, Aug 01 2016
    
  • PARI
    Vec((5-220*x+990*x^2-924*x^3+165*x^4)/(1-55*x+330*x^2-462*x^3+165*x^4-11*x^5) + O(x^20)) \\ Colin Barker, Aug 02 2016

Formula

a(-2) = 141, a(-1) = 15, a(0) = 5, a(1) = 55, a(2) = 2365.
a(n) = +55*a(n-1)-330*a(n-2)+462*a(n-3)-165*a(n-4)-11*a(n-5) for n > 2.
a(n) ~ k^n where k = 48.37415... is the largest real root of x^5 - 55x^4 + 330x^3 - 462x^2 + 165x - 11. - Charles R Greathouse IV, Aug 01 2016
G.f.: (5-220*x+990*x^2-924*x^3+165*x^4) / (1-55*x+330*x^2-462*x^3+165*x^4-11*x^5). - Colin Barker, Aug 02 2016

A228780 Power basis components of the algebraic numbers S2(n) in Q(2*cos(Pi/n)), where S2(n) is the square of the sum of the lengths of the distinct line segments (side and diagonals) in the regular n-gon.

Original entry on oeis.org

4, 3, 6, 4, 3, 4, 12, 6, -1, 4, 4, -2, -4, 6, 4, 3, 8, 4, -16, -8, 12, 6, 3, -4, -8, 4, 4, 0, 4, 10, 4, 3, 8, -8, -12, 4, 4, 28, 14, -40, -20, 12, 6, -1, 8, 12, 4, -2, -8, 28, 28, -26, -20, 6, 4, -1, -8, 16, 28, -16, -20, 4, 4, 4, 2, -12, -6, 8, 4, 3, -8, -24, 28, 44, -20, -24, 4, 4, 0, 8, 28, -4, -40, -12, 10, 4, -1, -16, -24, 0, 12, 4
Offset: 2

Views

Author

Wolfdieter Lang, Oct 01 2013

Keywords

Comments

The length of row n of this irregular array is the degree of the algebraic number rho(n):= 2*cos(Pi/n), given in A055034(n). See a Jul 19 2011 comment there.
The regular n-gon, inscribed in a circle of radius defining the length unit 1, has distinct line segments (chords) (V_0, V_j), j=1, ... , floor(n/2), with the n-gon vertices V_j, j=0, ... , n-1 distributed on the circle in the counterclockwise sense. The corresponding length ratios are denoted by L(n,j)/radius. The side length is s(n) = (V_0, V_1) = 2*sin(Pi/n), and for n >= 4 the first (the smallest) diagonal has length s(n)*rho(n), with rho(n) of degree delta(n) given above. s(2) = 2 is the ratio of the diameter of the circle. rho(2) = 0, but we use here rho(2)^0 = 1.
For n = 3: rho(3) = 1, s(3)^2 = 3. The algebraic number field Q(rho(n)) is the subject of the W. Lang link given below.
S2(n) := (sum(L(n,j)/radius, j=1, ... ,floor(n/2))^2 is seen below to be a number in the field Q(rho(n)) of degree delta(n), namely S2(n) = sum(a(n,k)*rho(n)^k, k=0..(delta(n)-1)). From the definition one has S2(n) = (s(n)*sum(S(j-1,rho(n)), j=1..floor(n/2)))^2, with the Chebyshev S-polynomials (see A049310). Due to s(n) = s(2*n)*rho(2*n), rho(2*n) = sqrt(2 + rho(n)) and an S-identity this becomes S2(n) = (s(2*n)*S(floor(n/2)-1, rho(2*n))*S(floor(n/2), rho(2*n)))^2. This can also be written as S2(n) = 4*(1 - T(2*floor(n/2), rho(2*n)/2))*(1 - T(2*(floor(n/2)+1), rho(2*n)/2))/(4-rho(2*n)^2), with Chebyshev's T-polynomials (see A053120). S2(n), written as a function of rho(n), has to be computed modulo the minimal polynomial C(n,rho(n)) of degree delta(n). These minimal polynomials are treated in A187360 (see the link to a Galois paper there, with its Table 2 and Section 3). The result is then the above given representation of S2(n) in the power basis of Q(rho(n)).
This computation was inspired by an email exchange with Seppo Mustonen. The author thanks him for sending the paper given as a link below. In this connection one should consider the even and odd n cases separately in order to find the square of the total length segments/radius in the regular n-gon, noticing that in the odd n case each distinct chord (side or diagonal) appears 2*(n/2) = n times, whereas in the even n case the longest diagonal of length 2 (in units of the radius) appears only n/2 times and the other chords appear n times.

Examples

			The table a(n,k) begins:
n\k     0    1    2    3    4    5 ...
2:      4
3:      3
4:      6    4
5:      3    4
6:     12    6
7:     -1    4    4
8:     -2   -4    6    4
9:      3    8    4
10:   -16   -8   12    6
11:     3   -4   -8    4    4
12:     0    4   10    4
13:     3    8   -8  -12    4    4
14:    28   14  -40  -20   12    6
15:    -1    8    2    4
...
n=5: S2(5) = (4-rho(5)^2)*(Sum_{j=1..2} S(j-1,rho(5)))^2 = 4 + 8*rho(5) + 3*rho(5)^2 - 2*rho(5)^3 - rho(5)^4, reduced with C(5,x) = x^2 - x - 1, with x = rho(5), using C(5,rho(5)) = 0, to eliminate all powers of rho(5) starting with power 2.
This leads to S2(5) = 3*1 + 4*rho(5). rho(5) = phi, the golden section.
The exact or approximate real values for S2(n) are, for n = 2, ..., 15: 4, 3, 11.65685426, 9.472135960, 22.39230484, 19.19566936, 36.32882142, 32.16343753, 53.49096128, 48.37415020, 73.88698896, 67.82742928, 97.52047276, 90.52313112.
		

Crossrefs

Cf. A228781, A228782 (minimal polynomials for odd and even n).

Formula

a(n,k) = [rho^k] (S2(n) modulo C(n,rho(n)), with S2(n) the square of the sum of the distinct length/radius ratios in the regular n-gon, with rho(n) = 2*cos(Pi/n) given above in a comment, and C(n,x) the minimal polynomial of rho(n) given in A187360 (see Table 2 and section 3 of the paper given in the W. Lang link below).

A275830 a(n) = (2*sqrt(7)*sin(Pi/7))^n + (-2*sqrt(7)*sin(2*Pi/7))^n + (-2*sqrt(7)*sin(4*Pi/7))^n.

Original entry on oeis.org

3, -7, 49, -196, 1029, -4802, 24010, -117649, 588245, -2941225, 14823774, -74942413, 380476866, -1936973136, 9886633715, -50563069571, 259029803333, -1328763571296, 6823754590093, -35073821767334, 180407337377834, -928487386730281, 4780794440512601, -24625601552074341, 126883328914736618
Offset: 0

Views

Author

Kai Wang, Aug 11 2016

Keywords

Comments

2*sqrt(7)*sin(Pi/7), -2*sqrt(7)*sin(2*Pi/7) and -2*sqrt(7)*sin(4*Pi/7) are roots of polynomial x^3 + 7*x^2 - 49.

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[0] == 3, a[1] == -7, a[2] == 49, a[n] == -7 a[n - 1] + 49 a[n - 3]}, a, {n, 0, 30}] (* Bruno Berselli, Aug 11 2016 *)
  • PARI
    Vec((3 + 14*x)/(1 + 7*x - 49*x^3) + O(x^30)) \\ Colin Barker, Aug 30 2016

Formula

G.f.: (3 + 14*x)/(1 + 7*x - 49*x^3). - Bruno Berselli, Aug 11 2016
a(n) = -7*a(n-1) + 49*a(n-3) with n>2, a(0)=3, a(1)=-7, a(2)=49.
a(2*n-1) = 7^n*A215493(n). - Kai Wang, May 25 2017

A353410 a(n) = (tan(1*Pi/9))^(2*n) + (tan(2*Pi/9))^(2*n) + (tan(3*Pi/9))^(2*n) + (tan(4*Pi/9))^(2*n).

Original entry on oeis.org

4, 36, 1044, 33300, 1070244, 34420356, 1107069876, 35607151476, 1145248326468, 36835122753252, 1184744167077204, 38105444942929620, 1225602095970073572, 39419576386043222340, 1267869080483029127412, 40779027899804602385460, 1311593714249667915837060, 42185362424185765127267748
Offset: 0

Views

Author

Bernard Schott, Apr 17 2022

Keywords

Comments

Sum_{k=1..(m-1)/2} (tan(k*Pi/m))^(2*n) is an integer when m >= 3 is an odd integer (see AMM link); this sequence is for the case m = 9.
Note tan(3*Pi/9) = tan(Pi/3) = sqrt(3).

Examples

			a(1) = tan^2 (Pi/9) + tan^2 (2*Pi/9) + tan^2 (3*Pi/9) + tan^2 (4*Pi/9) = 36.
		

Crossrefs

Similar with: A000244 (m=3), 2*A165225 (m=5), A108716 (m=7), this sequence (m=9), A275546 (m=11), A353411 (m=13).
Cf. A019676 (Pi/9), A019918 (tan(Pi/9)), A019938 (tan(2*Pi/9)).
Cf. A215948.

Programs

  • Mathematica
    LinearRecurrence[{36, -126, 84, -9}, {4, 36, 1044, 33300}, 18] (* Amiram Eldar, Apr 18 2022 *)

Formula

G.f.: 4*(1 - 27x + 63*x^2 - 21*x^3)/((1 - 3*x)*(1 - 33*x + 27*x^2 - 3*x^3)). - Stefano Spezia, Apr 18 2022
a(n) = A215948(n) + 3^n. - Jianing Song, Apr 19 2022

Extensions

More terms from Stefano Spezia, Apr 18 2022

A353411 a(n) = (tan(1*Pi/13))^(2*n) + (tan(2*Pi/13))^(2*n) + (tan(3*Pi/13))^(2*n) + (tan(4*Pi/13))^(2*n) + (tan(5*Pi/13))^(2*n) + (tan(6*Pi/13))^(2*n).

Original entry on oeis.org

6, 78, 4654, 312390, 21167510, 1435594238, 97371674686, 6604463476598, 447963730184230, 30384227802426030, 2060884053792801614, 139784466963241906598, 9481221017869954060214, 643086846082033986242142, 43618927438218948551328606, 2958559706907951258983758550
Offset: 0

Views

Author

Bernard Schott, Apr 19 2022

Keywords

Comments

Sum_{k=1..(m-1)/2} (tan(k*Pi/m))^(2*n) is an integer when m >= 3 is an odd integer (see AMM link); this sequence is the particular case m = 13.
All terms are even.

Examples

			a(1) = tan^2 (Pi/13) + tan^2 (2*Pi/13) + tan^2 (3*Pi/13) + tan^2 (4*Pi/13) + tan^2 (5*Pi/13) + tan^2 (6*Pi/13) = 78.
		

Crossrefs

Similar to: A000244 (m=3), 2*A165225 (m=5), A108716 (m=7), A353410 (m=9), A275546 (m=11), this sequence (m=13).

Programs

  • Mathematica
    LinearRecurrence[{78, -715, 1716, -1287, 286, -13}, {6, 78, 4654, 312390, 21167510, 1435594238}, 16] (* Amiram Eldar, Apr 19 2022 *)

Formula

G.f.: -2*(143*x^5 -1287*x^4 +2574*x^3 -1430*x^2 +195*x -3) / (13*x^6 -286*x^5 +1287*x^4 -1716*x^3 +715*x^2 -78*x +1). - Alois P. Heinz, Apr 19 2022

A217444 a(n) = A(n)*7^(-floor(n+1)/3), where A(n) = 7*A(n-1) - 14*A(n-2) + 7*A(n-3) with A(0)=0, A(1)=1, A(2)=7.

Original entry on oeis.org

0, 1, 1, 5, 22, 13, 52, 204, 113, 435, 1667, 910, 3471, 13224, 7192, 27367, 104105, 56563, 215098, 817909, 444276, 1689212, 6422529, 3488381, 13262821, 50424942, 27387681, 104126704, 395884336, 215018609, 817488295, 3108041875, 1688083894, 6417991803, 24400809980
Offset: 0

Views

Author

Roman Witula, Oct 03 2012

Keywords

Comments

The Berndt-type sequence number 18a for the argument 2Pi/7, which is closely connected with the sequence A217274. Definitions other Berndt-type sequences for the argument 2Pi/7 like A215575, A215877, A033304 in sequences from Crossrefs are given.

Crossrefs

Programs

  • Magma
    i:=35; I:=[0, 1, 7]; A:=[m le 3 select I[m] else 7*Self(m-1)-14*Self(m-2)+7*Self(m-3): m in [1..i]]; [7^(-Floor(n/3))*A[n]: n in [1..i]]; // Bruno Berselli, Oct 03 2012
    
  • Mathematica
    CoefficientList[Series[x*(1+x+5*x^2+12*x^3+3*x^4+2*x^5+x^6)/(1 - 10*x^3 + 17*x^6 - x^9), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 15 2012 *)
    LinearRecurrence[{0,0,10,0,0,-17,0,0,1}, {0, 1, 1, 5, 22, 13, 52, 204, 113}, 50] (* G. C. Greubel, Apr 23 2018 *)
  • PARI
    x='x+O('x^30); concat([0], Vec(x*(1+x+5*x^2+12*x^3+3*x^4 +2*x^5 +x^6)/(1- 10*x^3+17*x^6-x^9))) \\ G. C. Greubel, Apr 23 2018

Formula

G.f.: x*(1+x+5*x^2+12*x^3+3*x^4+2*x^5+x^6)/(1-10*x^3+17*x^6-x^9). - Bruno Berselli, Oct 03 2012
a(n) = 10*a(n-3) - 17*a(n-6) + a(n-9). - G. C. Greubel, Apr 23 2018

A228781 Irregular triangle read by rows: coefficients of minimal polynomial of a certain algebraic number S2(2*k+1) from Q(2*cos(Pi/n)) related to the regular (2*k+1)-gon, k >= 1.

Original entry on oeis.org

-3, 1, 5, -10, 1, -7, 35, -21, 1, -3, 27, -33, 1, -11, 165, -462, 330, -55, 1, 13, -286, 1287, -1716, 715, -78, 1, 1, -28, 134, -92, 1, 17, -680, 6188, -19448, 24310, -12376, 2380, -136, 1, -19, 969, -11628, 50388, -92378, 75582, -27132, 3876, -171, 1, 1, -58, 655, -1772, 1423, -186, 1
Offset: 1

Views

Author

Wolfdieter Lang, Oct 01 2013

Keywords

Comments

The row length sequence of this table is delta(2*k+1), with the degree delta(n) = A055034(n) of the algebraic number rho(n):= 2*cos(Pi/n), k >= 1.
The numbers S2(n) have been given in A228780 in the power basis of the degree delta(n) number field Q(rho(n)), with rho(n):= 2*cos(Pi/n), n >= 2. Here the odd n case, n = 2*k + 1 is considered. S2(n) is the square of the sum of the distinct length ratios side/radius or diagonal/radius with the radius of the circle in which a regular n-gon is inscribed. For two formulas for S2(n) in terms of powers of rho(n) see the comment section of A228780.
The minimal (monic) polynomial of S2(2*k+1) has degree delta(2*k+1) and is given by
p(2*k+1,x) = Product_{j=1..delta(2*k+1)} (x - S2(2*k+1)^{(j-1)} (mod C(2*k+1,delta(n))) = sum(a(k, m)*x^m, m = 0..delta(2*k+1)), where S2(2*k+1)^{(0)} = S2(2*k+1) and S2(2*k+1)^{(j-1)} is the (j-1)-th conjugate of S2(2*k+1). The conjugate of a number alpha(n) = Sum_{j=0..(delta(n)-1)} b(n, j)*rho(n)^j in Q(rho(n)) is obtained from the conjugates of rho(n), given in turn by the zeros x(n, j) of the minimal polynomial C(n, x) (see A187360 and the link to the W. Lang Galois paper, tables 2 and 3) as rho(n)^{(j-1)} = x(n, j), j = 1..delta(n), with rho(n)^{(0)} = rho(n).
The motivation to look into this problem originated from emails by Seppo Mustonen, who found experimentally polynomials which had as one zero the square of the total length/radius of all chords (sides and diagonals) in the regular n-gon. See his paper given as a link below. The author thanks Seppo Mustonen for sending his paper.
If the minimal polynomial of the algebraic number S2(n) in the n-gon with n = 2*k+1 is p(n, x) then the minimal polynomial of the square of the sum of the length of all n sides and n*(n-3)/2 diagonals is P(n, x) = n^(2*delta(n))*p(n, x/n^2).

Examples

			The irregular triangle a(k, m) begins:
n   k /m 0     1     2       3      4       5     6    7   8
3   1:  -3     1
5   2:   5   -10     1
7   3:  -7    35   -21       1
9   4:  -3    27   -33       1
11  5: -11   165  -462     330    -55      1
13  6:  13  -286  1287   -1716    715    -78      1
15  7:   1   -28   134     -92      1
17  8:  17  -680  6188  -19448  24310  -12376  2380 -136   1
...
n = 19, L = 9: -19, 969, -11628, 50388, -92378, 75582, -27132, 3876, -171, 1.
n = 21, L = 10: 1, -58, 655, -1772, 1423, -186, 1.
p(5, x) = (x - S2(5))*(x - S2(5)^{(1)}), with S2(5) = 3 + 4*rho(5), where rho(5)=phi, the golden section. C(5, x) = x^2 - x - 1 = (x - rho(5))*(x - (1-rho(5))), hence rho(5)^{(1)} = 1-rho(5), and S2(5)^{(1)} = 3 + 4*(1 - rho(5)) = 7 - 4*rho(5). Thus p(5, x) = -16*rho^2 + 21 + 16*rho -10*x + x^2 which becomes modulo C(5,rho(5)), i.e., using rho(5)^2 = rho(5) + 1, finally p(n, 5) = 5 - 10*x + x^2.
Conjecture (_Seppo Mustonen_): p(5, x) = binomial(5, 1) - binomial(5, 3)*x + binomial(5, 5)* x^2 = 5 - 10*x + x^2.
		

Crossrefs

Cf. A055034, A187360, A228780, A228782 (even case).

Formula

a(k, m) = [x^m] p(2*k+1, x), with the minimal polynomial p(2*k+1, x) of S2(2*k+1) given in the power basis in A228780. p(2*k+1, x) is given in a comment above in terms of the S2(2*k+1) and its conjugates S2(2*k+1)^{(j-1)}, j=2, ..., delta(2*k+1), where delta(n) = A055034(n).
Conjecture from Seppo Mustonen, rewritten for the p(n, x) coefficients for odd primes: p(prime(j), x) = Sum_{i=0..imax(j)} (-1)^(imax(j - i))* binomial(prime(j), 2*i+1)*x^i, with imax(j) = (prime(j)-1)/2. See the adapted eq. (5) of the S. Mustonen paper.

A228782 Irregular triangle read by rows: coefficients of minimal polynomial of a certain algebraic number S2(2*k) from Q(2*cos(Pi/(2*k))) related to the regular (2*k)-gon.

Original entry on oeis.org

-4, 1, 4, -12, 1, 36, -24, 1, 16, -96, 136, -40, 1, 16, -96, 136, -56, 1, 16, -320, 456, -80, 1, 3136, -12544, 14896, -7168, 1484, -112, 1, 256, -7168, 41216, -73472, 53344, -17472, 2576, -144, 1, 64, -1152, 5424, -6080, 2124, -168, 1, 256, -13312, 62720, -104192, 76384, -26048, 3920, -208, 1
Offset: 1

Views

Author

Wolfdieter Lang, Oct 01 2013

Keywords

Comments

The row length sequence of this table is delta(2*k), k >= 1, with the degree delta(n) = A055034(n) of the algebraic number rho(n):=2*cos(Pi/n).
The algebraic numbers S2(n) have been given in A228780 in the power basis of the degree delta(n) number field Q(rho(n)), with rho(n):=2*cos(Pi/n), n >= 2. Here the even case, n = 2*k, is considered. S2(n) is the square of the sum of the distinct length ratios side/radius and diagonal/radius with the radius of the circle in which a regular n-gon is inscribed. For two formulas for S2(n) in terms of powers of rho(n) see the comment section of A228780.
The minimal (monic) polynomial of S2(2*k) has degree delta(2*k) and is given by p(2*k,x) = Product_{j=1..delta(2*k)} (x - S2(2*k)^{(j-1)}) (mod C(2*k, rho(2*k))) = Sum_{m=0..delta(2*k)} a(k, m)*x^m, where S2(2*k)^{(0)} = S2(2*k) and S2(2*k)^{(j-1)} is the (j-1)-th conjugate of S2(2*L). For the conjugate of an algebraic number in Q(rho(n)) see a comment on A228781.
The motivation to look into this problem originated from emails by Seppo Mustonen, who found experimentally polynomials which had as one zero the square of the total length/radius of all chords (sides and diagonals) in the regular n-gon. See his paper given as a link below. The author thanks Seppo Mustonen for sending his paper.

Examples

			The irregular triangle a(k, m) begins:
  n   k / m   0      1      2      3      4       5     6    7   8
  2   1:     -4      1
  4   2:      4    -12      1
  6   3:     36    -24      1
  8   4:     16    -96    136    -40      1
  10  5:     16    -96    136    -56      1
  12  6:     16   -320    456    -80      1
  14  7:   3136 -12544  14896  -7168   1484   -112    1
  16  8:    256  -7168  41216 -73472  53344 -17472 2576 -144  1
  ...
n = 18, k = 9: 64, -1152, 5424, -6080, 2124, -168, 1;
n = 20, k = 10: 256, -13312, 62720, -104192, 76384, -26048, 3920, -208, 1.
n = 6, k = 3: p(6, x) = (x - S2(6))*(x - S2(6)^{(1)}),
with S2(6) = 12 + 6*rho(6), where rho(6) = sqrt(3). C(6, x) = x^2 - 3 = (x - rho(6))*(x - (-rho(6))), hence rho(6)^{(1)} - -rho(6) and S2(6)^{(1)} = 12 - 6*rho(6). Thus p(6, x) = 144 - 36*rho(6)^2 - 24*x + x^2, reduced with C(6, rho(6)) = 0, i.e., rho(6)^2 = 3; this becomes finally 36 - 24*x + x^2.
		

Crossrefs

Cf. A055034, A187360, A228780, A228781 (odd case).

Formula

a(k,m) = [x^m] p(2*k, x), with the minimal polynomial p(2*k, x) of S2(2*k) given in the power basis in A228780.
p(2*k, x) is given in a comment above in terms of the S2(2*k) and its conjugates S2(2*k)^{(j-1)}, j = 2, ..., delta(2*k), where delta(2*k) = A055034(2*k).

A287405 a(n) = (7*(cot(1*Pi/7))^2)^n + (7*(cot(2*Pi/7))^2)^n + (7*(cot(4*Pi/7))^2)^n.

Original entry on oeis.org

3, 35, 931, 27587, 830403, 25054435, 756187747, 22824258947, 688917131651, 20793986742179, 627637106311971, 18944339609269571, 571808137046942019, 17259221092289630307, 520945214725090792931, 15723995613526902256387, 474606601742375424297731
Offset: 0

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Author

Kai Wang, May 24 2017

Keywords

Comments

a(n) = x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of x^3 - 35*x^2 + 147*x - 49, x1 = 7*(cot(1*Pi/7))^2, x2 = 7*(cot(2*Pi/7))^2, x3 = 7*(cot(4*Pi/7))^2.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{35,-147,49},{3,35,931},30] (* Harvey P. Dale, Mar 15 2018 *)
  • PARI
    Vec((3 - 7*x)*(1 - 21*x) / (1 - 35*x + 147*x^2 - 49*x^3) + O(x^30)) \\ Colin Barker, May 26 2017
    
  • PARI
    polsym(x^3 - 35*x^2 + 147*x - 49, 20) \\ Joerg Arndt, May 26 2017

Formula

a(n) = 35*a(n-1) - 147*a(n-2) + 49*a(n-3), a(0) = 3, a(1) = 35, a(2) = 931.
Bisection of A215575: a(n) = A215575(2*n).
G.f.: (3 - 7*x)*(1 - 21*x) / (1 - 35*x + 147*x^2 - 49*x^3). - Colin Barker, May 26 2017
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