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

Author

David W. Wilson

Keywords

• nonn
• cofr
• easy

Comments

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)

Examples

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

Links

• 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).

Crossrefs

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

Programs

Mathematica

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 *)

PARI

{ 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

PARI

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

Formula

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)