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A108716
a(n) = tan(Pi/14)^(-2n) + tan(3*Pi/14)^(-2n) + tan(5*Pi/14)^(-2n).
20
3, 21, 371, 7077, 135779, 2606261, 50028755, 960335173, 18434276035, 353858266965, 6792546291251, 130387472704741, 2502874814474531, 48044357383337973, 922243598852422035, 17703083191185355397
OFFSET
0,1
COMMENTS
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
LINKS
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.
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
FORMULA
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
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[n-1] - 35*a[n-2] + 7*a[n-3]}, 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
Similar to: A000244 (m=3), 2*A165225 (m=5), this sequence (m=7), A353410 (m=9), A275546 (m=11), A353411 (m=13).
Sequence in context: A361056 A101389 A353321 * A358950 A271570 A084620
KEYWORD
nonn,easy
AUTHOR
Philippe Deléham, Jun 20 2005
EXTENSIONS
More terms from Robert G. Wilson v, Jun 21 2005
STATUS
approved