A161019 -id:A161019 - cp's OEIS frontend

A019433 Continued fraction for tan(1/10).

0, 9, 1, 28, 1, 48, 1, 68, 1, 88, 1, 108, 1, 128, 1, 148, 1, 168, 1, 188, 1, 208, 1, 228, 1, 248, 1, 268, 1, 288, 1, 308, 1, 328, 1, 348, 1, 368, 1, 388, 1, 408, 1, 428, 1, 448, 1, 468, 1, 488, 1, 508, 1, 528, 1, 548, 1, 568, 1, 588, 1, 608, 1, 628, 1, 648, 1, 668, 1, 688, 1, 708, 1, 728

Offset: 0

David W. Wilson

From Peter Bala, Oct 04 2023: (Start)
Related simple continued fractions expansions (see my comments in A019425):
tan(1/(10*k)) = [0; 10*k - 1, 1, 30*k - 2, 1, 50*k - 2, 1, 70*k - 2, 1, 90*k - 2, 1, ...] for k >= 1.
If d is a divisor of 10 with d*d' = 10 then the simple continued fraction expansion of d*tan(1/10) begins [0; d' - 1, 1, 30*d - 2, 1, 5*d' - 2, 1, 70*d - 2, 1, 9*d' - 2, 1, 110*d - 2, 1, 13*d' - 2, ...], while the simple continued fraction expansion of (1/d)*tan(1/10) begins [ 0; 10*d - 1, 1, 3*d'- 2, 1, 50*d - 2, 1, 7*d' - 2, 1, 90*d - 2, 1, 11*d' - 2, 1, 130*d - 2, ...]. (End)

			0.10033467208545054505808004... = 0 + 1/(9 + 1/(1 + 1/(28 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 14 2009

Harry J. Smith, Table of n, a(n) for n = 0..20000
G. Xiao, Contfrac
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A161019 (decimal expansion), A019425 through A019432.

LinearRecurrence[{0,2,0,-1},{0,9,1,28,1,48},80] (* or *) Join[{0,9},Riffle[NestList[20+#&,28,40],1,{1,-1,2}]] (* Harvey P. Dale, Jul 23 2023 *)

{ allocatemem(932245000); default(realprecision, 99000); x=contfrac(tan(1/10)); for (n=0, 20000, write("b019433.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 14 2009

Vec(x*(x^4-x^3+10*x^2+x+9)/((x-1)^2*(x+1)^2) + O(x^100)) \\ Colin Barker, Sep 08 2013

From Colin Barker, Sep 08 2013: (Start)
a(n) = -1/2+(3*(-1)^n)/2+5*n-5*(-1)^n*n for n>1.
a(n) = 2*a(n-2)-a(n-4) for n>5.
G.f.: x*(x^4-x^3+10*x^2+x+9) / ((x-1)^2*(x+1)^2). (End)

A161011 Decimal expansion of tan(1/2).

Original entry on oeis.org

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

Offset: 0

Harry J. Smith, Jun 13 2009

By the Lindemann-Weierstrass theorem, this constant is transcendental. - Charles R Greathouse IV, May 13 2019

			0.546302489843790513255179465780285383297551720179791246164091385932907...

Harry J. Smith, Table of n, a(n) for n = 0..20000
MathOverflow, What is the effect of adding 1/2 to a continued fraction?
Wikipedia, Lindemann-Weierstrass theorem
Index entries for transcendental numbers

Cf. A019425 (continued fraction). Cf. A049471, A161011 through A161019.

RealDigits[N[Tan[1/2],6! ]][[1]] (* Vladimir Joseph Stephan Orlovsky, Jun 13 2009 *)

default(realprecision, 20080); x=10*tan(1/2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b161011.txt", n, " ", d));

From Peter Bala, Nov 17 2019: (Start)
Related simple continued fraction expansions:
tan(1/2) = [0; 1, 1, 4, 1, 8, 1, 12, 1, 16, 1, 20, 1, ...]. See A019425.
2*tan(1/2) = [1, 10, 1, 3, 1, 26, 1, 7, 1, 42, 1, 11, 1, 58, 1, 15, 1, 74, 1, 19, 1, 90, ...]
(1/2)*tan(1/2) = [0; 3, 1, 1, 1, 18, 1, 5, 1, 34, 1, 9, 1, 50, 1, 13, 1, 66, 1, 17, 1, 82, ...].
tan(1/2)/(1 - tan(1/2)) = [1, 4, 1, 8, 1, 12, 1, 16, 1, 20, 1, 24, ...]
2*tan(1/2)/(1 - tan(1/2)) = [2, 2, 2, 4, 2, 6, 2, 8, 2, 10, 2, 12, ...]
4*tan(1/2)/(1 - tan(1/2)) = [4, 1, 4, 2, 4, 3, 4, 4, 4, 5, 4, 6, 4, 7, ...]. (End)

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A019433 Continued fraction for tan(1/10).

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A161011 Decimal expansion of tan(1/2).

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