A353410 -id:A353410 - cp's OEIS frontend

A108716 a(n) = tan(Pi/14)^(-2n) + tan(3Pi/14)^(-2n) + tan(5Pi/14)^(-2n).

3, 21, 371, 7077, 135779, 2606261, 50028755, 960335173, 18434276035, 353858266965, 6792546291251, 130387472704741, 2502874814474531, 48044357383337973, 922243598852422035, 17703083191185355397

Offset: 0

Philippe Deléham, Jun 20 2005

The Berndt-type sequence number 11 for the argument 2*Pi/7 defined by the relation a(n) = t(1)^(2*n) + t(2)^(2*n) + t(4)^(2*n) = (-sqrt(7) + 4*s(1))^(2*n) + (-sqrt(7) + 4*s(2))^(2*n) + (-sqrt(7) + 4*s(4))^(2*n), where t(j) = tan(2*Pi*j/7) and s(j) = sin(2*Pi*j/7) (the respective sum with odd powers are discussed in A215794). See also A215007, A215008, A215143, A215493, A215494, A215510, A215512, A215694, A215695, A215828 and especially A215575, where a(n) = B(2n) for the function B(n) defined in the comments. - Roman Witula, Aug 23 2012
The sequence a(n+1)/a(n) is increasing and convergent to (t(2))^2 = 19,195669... (we note that the sequence A215794(n+1)/A215794(n) is decreasing and converges to the same limit). - Roman Witula, Aug 24 2012
Let L(p) be the total length of all sides and diagonals of a regular p-sided polygon inscribed in a unit circle. Then (L(p)/p)^2 = cot(Pi/(2p))^2 is the largest root of the equation: C(p,k)-C(p,2+k)*x+C(p,4+k)*x^2-C(p,6+k)*x^3+ ... +(-1)^q*x^q = 0, where k=1 if p is odd, k=0 if p is even, q = floor(p/2), and where C denotes the binomial coefficient. The complete set of roots is: x(i) = cot((2*i-1)*Pi/(2p))^2, i=1,2,...,q. Then a(n) = x(1)^n+x(2)^n+...x(q)^n for p=7. - Seppo Mustonen, Mar 25 2014
Sum_{k=1..(m-1)/2} tan^(2n) (k*Pi/m) is an integer when m >= 3 is an odd integer (see AMM link and formula); this sequence is the particular case m = 7. All terms are odd. - Bernard Schott, Apr 22 2022

Robert Israel, Table of n, a(n) for n = 0..700
Michel Bataille and Li Zhou, A Combinatorial Sum Goes on Tangent, The American Mathematical Monthly, Vol. 112, No. 7 (Aug. - Sep., 2005), Problem 11044, pp. 657-659.
Seppo Mustonen, Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations (2013).
Seppo Mustonen, Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations [Local copy]
Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2Pi/7, J. Integer Seq., 12 (2009), Article 09.8.5.
Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6
Index entries for linear recurrences with constant coefficients, signature (21,-35,7).

Similar to: A000244 (m=3), 2*A165225 (m=5), this sequence (m=7), A353410 (m=9), A275546 (m=11), A353411 (m=13).

A:= gfun:-rectoproc({-a(n+3)+21*a(n+2)-35*a(n+1)+7*a(n), a(0) = 3, a(1) = 21, a(2) = 371},a(n), remember):
seq(A(n),n=0..20); # Robert Israel, Aug 23 2015

Table[ Round[ Cot[Pi/14]^(2n) + Cot[3Pi/14]^(2n) + Cot[5Pi/14]^(2n)], {n, 0, 12}] (* Robert G. Wilson v, Jun 21 2005 *)
RecurrenceTable[{a[0]== 3, a[1]== 21, a[2]==371, a[n]== 21*a[n-1] - 35*a[n-2] + 7*a[n-3]}, a, {n,30}] (* G. C. Greubel, Aug 22 2015 *)

a(n)=round(tan(Pi/14)^(-2*n) + tan(3*Pi/14)^(-2*n) + tan(5*Pi/14)^(-2*n)); \\ Anders Hellström, Aug 22 2015

a(n) = 7^n*A(2n), where A(n) := A(n-1) + A(n-2) + A(n-3)/7, with A(0)=3, A(1)=1, and A(2)=3. - see Witula-Slota's (Section 6) and Witula's (Remark 11) papers for the proofs and details. In these papers A(n) denotes the value of the big omega function with index n for the argument 2*i/sqrt(7) (see also A215512). - Roman Witula, Aug 23 2012
Conjecture: a(n) = 21*a(n-1)-35*a(n-2)+7*a(n-3). G.f.: -(35*x^2-42*x+3) / (7*x^3-35*x^2+21*x-1). - Colin Barker, Jun 01 2013
To verify conjecture, note that the roots of 7*x^3-35*x^2+21*x-1 are tan(Pi/14)^2, tan(3*Pi/14)^2 and tan(5*Pi/14)^2. - Robert Israel, Aug 23 2015
E.g.f.: exp((tan(Pi/7))^2*x) + exp((cot(Pi/14))^2*x) + exp((cot(3*Pi/14))^2*x). - G. C. Greubel, Aug 22 2015
a(n) = A275195(2*n)/(7^n). - Kai Wang, Aug 02 2016
a(n) = (tan(1*Pi/7))^(2*n) + (tan(2*Pi/7))^(2*n) + (tan(3*Pi/7))^(2*n). - Bernard Schott, Apr 22 2022

More terms from Robert G. Wilson v, Jun 21 2005

A165225 a(0)=1, a(1)=5, a(n) = 10a(n-1) - 5a(n-2) for n > 1.

Original entry on oeis.org

1, 5, 45, 425, 4025, 38125, 361125, 3420625, 32400625, 306903125, 2907028125, 27535765625, 260822515625, 2470546328125, 23401350703125, 221660775390625, 2099601000390625, 19887706126953125, 188379056267578125

Offset: 0

Philippe Deléham, Sep 09 2009

Sum_{k=1..(m-1)/2} tan^(2n) (k*Pi/m) is an integer when m >= 3 is an odd integer (see AMM and Crux Mathematicorum links); twice this sequence is the particular case m = 5. - Bernard Schott, Apr 25 2022

Harvey P. Dale, Table of n, a(n) for n = 0..1000.
Michel Bataille and Li Zhou, A Combinatorial Sum Goes on Tangent, The American Mathematical Monthly, Vol. 112, No. 7 (Aug. - Sep., 2005), Problem 11044, pp. 657-659.
D. J. Smeenk, Problem 3452, Crux Mathematicorum, Vol. 35, No. 5 (2009), pp. 325 and 327; Solution to Problem 3452 by Roy Barbara, ibid., Vol. 36, No. 5 (2010), pp. 341-342.
Index entries for linear recurrences with constant coefficients, signature (10,-5).

Cf. A019934, A019970.

Similar with: A000244 (m=3), 2*this sequence (m=5), A108716 (m=7), A353410 (m=9), A275546 (m=11), A353411 (m=13).

LinearRecurrence[{10,-5},{1,5},30] (* Harvey P. Dale, Dec 23 2019 *)

Limit_{n->oo} a(n+1)/a(n) = 5 + 2*sqrt(5) = 9.47213595...
G.f.: (1-5x)/(1-10x+5x^2).
a(n) = ((5 - 2*sqrt(5))^n + (5 + 2*sqrt(5))^n)/2. - Klaus Brockhaus, Sep 25 2009
a(n) = (tan(Pi/5)^(2*n) + tan(2*Pi/5)^(2*n))/2 (Smeenk, 2009). - Amiram Eldar, Apr 03 2022

More terms from Klaus Brockhaus, Sep 25 2009

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

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

Kai Wang, Aug 01 2016

(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

Colin Barker, Table of n, a(n) for n = 0..550
Michel Bataille and Li Zhou, A Combinatorial Sum Goes on Tangent, The American Mathematical Monthly, Vol. 112, No. 7 (Aug. - Sep., 2005), Problem 11044, pp. 657-659.
Index entries for linear recurrences with constant coefficients, signature (55,-330,462,-165,11).

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

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

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

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

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

Original entry on oeis.org

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

Offset: 0

Bernard Schott, Apr 19 2022

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.

			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.

Michel Bataille and Li Zhou, A Combinatorial Sum Goes on Tangent, The American Mathematical Monthly, Vol. 112, No. 7 (Aug. - Sep., 2005), Problem 11044, pp. 657-659.
Index entries for linear recurrences with constant coefficients, signature (78,-715,1716,-1287,286,-13).

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

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

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

cp's OEIS Frontend

A108716 a(n) = tan(Pi/14)^(-2n) + tan(3*Pi/14)^(-2n) + tan(5*Pi/14)^(-2n).

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A165225 a(0)=1, a(1)=5, a(n) = 10*a(n-1) - 5*a(n-2) for n > 1.

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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).

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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).

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A108716 a(n) = tan(Pi/14)^(-2n) + tan(3Pi/14)^(-2n) + tan(5Pi/14)^(-2n).

A165225 a(0)=1, a(1)=5, a(n) = 10a(n-1) - 5a(n-2) for n > 1.

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

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