A384932 Decimal expansion of tan(1) + sec(1).

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

Offset

1,1

Comments

The continued fraction expansion of this constant - 1 is A133265.

More generally, for integer n >= 2, the simple continued fraction expansion of tan(1/n) + sec(1/n) is 1 + 1/(n-1 + 1/(2 + 1/(3*n-1 + 1/(2 + 1/(5*n-1 + 1/(2 + 1/(7*n-1 + 1/(2 + 1/(9*n-1 + ... ))))))))). - Peter Bala, Feb 14 2025

Links

Table of n, a(n) for n=1..87.

Peter Bala, The Simple Continued Fraction Expansion of tan(2/n) + sec(2/n)

Index entries for transcendental numbers.

Formula

Equals A049471 + A073448.

Equals 2 - Integral_{x = 0..1} sin(x)/(sin(x) - 1) dx.

Equals exp(Integral_{x = 0..1} sec(x) dx).

Equals exp(A248617). - Hugo Pfoertner, Jun 13 2025

From Peter Bala, Feb 14 2026: (Start)

Constant c = tan((Pi + 2)/4).

c = -i*(exp(i) + i)/(exp(i) - i), where i = sqrt(-1).

c = exp(arctanh(sin(1))).

c = (cos(1/2) + sin(1/2))/(cos(1/2) - sin(1/2)) = sqrt((1 + sin(1))/(1 - sin(1))).

(c - 1)/(c + 1) = tan(1/2).

c + 1/c = 2*sec(1).

Reciprocal 1/(tan(1) + sec(1)) = Integral_{x = 0..1} sin(x)/(sin(x) + 1) dx.

Related continued fraction expansions:

2*(tan(1) + sec(1)) = [6; {k, 4}, ...] and

(1/2)*(tan(1) + sec(1)) = [1; 1, 2, {2, 1, k, 1}, ...], both quasi-periodic with k >= 1. (End)

Example

3.408223442335827848481...

Mathematica

RealDigits[Tan[1] + Sec[1], 10, 100, 0][[1]]

Prog

(PARI) tan(1) + 1/cos(1) \\ Amiram Eldar, Jun 13 2025

Crossrefs

tan(1/k) + sec(1/k): A392146 (k = 2), A392147 (k = 3), A392148 (k = 4), A392149 (k = 5), A392150 (k = 6).

Cf. A049471, A073448, A100727, A133265, A248617, A346839, A392151, A392152, A392153.

Sequence in context: A011338 A214024 A215079 * A049251 A308642 A158674

Adjacent sequences: A384929 A384930 A384931 * A384933 A384934 A384935

Keyword

nonn,cons,easy

Author

Kritsada Moomuang, Jun 12 2025

Status

approved