

A108716


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


13



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

0,1


COMMENTS

The Berndttype 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 psided 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^2C(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*i1)*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


LINKS

Robert Israel, Table of n, a(n) for n = 0..700
Seppo Mustonen, Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations (2013).
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 RamanujanType Formulas and QuasiFibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6


FORMULA

a(n) = 7^n*A(2n), where A(n) := A(n1) + A(n2) + A(n3)/7, with A(0)=3, A(1)=1, and A(2)=3.  see WitulaSlota'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(n1)35*a(n2)+7*a(n3). G.f.: (35*x^242*x+3) / (7*x^335*x^2+21*x1).  Colin Barker, Jun 01 2013
To verify conjecture, note that the roots of 7*x^335*x^2+21*x1 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


MAPLE

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


MATHEMATICA

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[n1]  35*a[n2] + 7*a[n3]}, a, {n, 30}] (* G. C. Greubel, Aug 22 2015 *)


PROG

(PARI) 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


CROSSREFS

Sequence in context: A186271 A320949 A101389 * A271570 A084620 A120603
Adjacent sequences: A108713 A108714 A108715 * A108717 A108718 A108719


KEYWORD

nonn


AUTHOR

Philippe Deléham, Jun 20 2005


EXTENSIONS

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


STATUS

approved



