1,2

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), May 24 2009: (Start)

The Taylor series for tan(x) appears to be identical to the quotient of the 'look-a-likes' of the numerator and denominator, i.e. A160469(n)/A156769(n).

(End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 75 (4.3.67).

G. W. Caunt, Infinitesimal Calculus, Oxford Univ. Press, 1914, p. 477.

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.

A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 74.

H. A. Rothe, in C. F. Hindenburg, editor, Sammlung Combinatorisch-Analytischer Abhandlungen, Vol. 2, Chap. XI. Fleischer, Leipzig, 1800, p. 329.

T. D. Noe, Table of n, a(n) for n=1..100

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 75 (4.3.67).

Eric Weisstein's World of Mathematics, Hyperbolic Tangent

Eric Weisstein's World of Mathematics, Tangent

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), May 24 2009: (Start)

a(n) = denom((-1)^(n-1)*2^(2*n)*(2^(2*n)-1)* bernoulli(2*n)/(2*n)!)

(End)

tan(x) = x + 2 x^3/3! + 16 x^5/5! + 272 x^7/7! + ... = x+1/3*x^3+2/15*x^5+17/315*x^7+62/2835*x^9+O(x^10).

The coefficients in the expansion of tan x are 0, 1, 0, 1/3, 0, 2/15, 0, 17/315, 0, 62/2835, 0, 1382/155925, 0, 21844/6081075, 0, 929569/638512875, 0, ... = A002430/A036279

tanh(x) = x-1/3*x^3+2/15*x^5-17/315*x^7+62/2835*x^9-1382/155925*x^11+...

The coefficients in the expansion of tanh x are 0, 1, 0, -1/3, 0, 2/15, 0, -17/315, 0, 62/2835, 0, -1382/155925, 0, 21844/6081075, 0, -929569/638512875, 0, 6404582/10854718875, 0, -443861162/1856156927625, ... = A002430/A036279

Cf. A002430, A000182.

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), May 24 2009: (Start)

Cf. A160469 and A156769

(End)

Sequence in context: A138896 A090627 A070234 this_sequence A156769 A029758 A103031

Adjacent sequences: A036276 A036277 A036278 this_sequence A036280 A036281 A036282

nonn,easy,frac

N. J. A. Sloane (njas(AT)research.att.com).

I deleted the comment by Stephen Crowley. His formula leads to incorrect values for higher values of this series Johannes W. Meijer (meijgia(AT)hotmail.com), Jan 19 2009