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A000281
Expansion of cos(x)/cos(2x).
(Formerly M3163 N1281)
17
1, 3, 57, 2763, 250737, 36581523, 7828053417, 2309644635483, 898621108880097, 445777636063460643, 274613643571568682777, 205676334188681975553003, 184053312545818735778213457, 193944394596325636374396208563
OFFSET
0,2
COMMENTS
a(n) is (2n)! times the coefficient of x^(2n) in the Taylor series for cos(x)/cos(2x).
REFERENCES
J. W. L. Glaisher, "On the coefficients in the expansions of cos x / cos 2x and sin x / cos 2x", Quart. J. Pure and Applied Math., 45 (1914), 187-222.
I. J. Schwatt, Intro. to Operations with Series, Chelsea, p. 278.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..215 (terms 0..50 from T. D. Noe)
Kwang-Wu Chen, An Interesting Lemma for Regular C-fractions, J. Integer Seqs., Vol. 6, 2003.
Matthieu Josuat-Vergès and Jang Soo Kim, Touchard-Riordan formulas, T-fractions, and Jacobi's triple product identity, arXiv:1101.5608 [math.CO] (2011).
D. Shanks, Generalized Euler and class numbers. Math. Comp. 21 (1967) 689-694, sequence c(2,n).
D. Shanks, Corrigenda to: "Generalized Euler and class numbers", Math. Comp. 22 (1968), 699.
D. Shanks, Generalized Euler and class numbers, Math. Comp. 21 (1967), 689-694; 22 (1968), 699. [Annotated scanned copy]
FORMULA
a(n) = Sum_{k=0..n} (-1)^k*binomial(2n, 2k)*A000364(n-k)*4^(n-k). - Philippe Deléham, Jan 26 2004
E.g.f.: Sum_{k>=0} a(k)x^(2k)/(2k)! = cos(x)/cos(2x).
a(n-1) is approximately 2^(4*n-3)*(2*n-1)!*sqrt(2)/((Pi^(2*n-1))*(2*n-1)). The approximation is quite good a(250) is of the order of 10^1181 and this formula is accurate to 238 digits. - Simon Plouffe, Jan 31 2007
G.f.: 1 / (1 - 1*3*x / (1 - 4*4*x / (1 - 5*7*x / (1 - 8*8*x / (1 - 9*11*x / ... ))))). - Michael Somos, May 12 2012
G.f.: 1/E(0) where E(k) = 1 - 3*x - 16*x*k*(2*k+1) - 16*x^2*(k+1)^2*(4*k+1)*(4*k+3)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
G.f.: T(0)/(1-3*x), where T(k) = 1 - 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2/( 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2 - (32*x*k^2+16*x*k+3*x-1 )*(32*x*k^2+80*x*k+51*x -1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
From Peter Bala, Mar 09 2015: (Start)
a(n) = (-1)^n*4^(2*n)*E(2*n,1/4), where E(n,x) denotes the n-th Euler polynomial.
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(4*k + 1)^2) = 1 + 3*x + 57*x^2 + 2763*x^3 + ....
We appear to have the asymptotic expansion Pi/(2*sqrt(2)) - Sum {k = 0..n - 1} (-1)^floor(k/2)/(2*k + 1) ~ 1/(2*n) - 3/(2*n)^3 + 57/(2*n)^5 - 2763/(2*n)^7 + .... See A093954.
Bisection of A001586. See also A188458 and A212435. Second row of A235605 (read as a square array).
The expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) appears to have integer coefficients. See A255883. (End)
From Peter Luschny, Mar 11 2015: (Start)
a(n) = ((-64)^n/((n+1/2)))*(B(2*n+1,7/8)-B(2*n+1,3/8)), B(n,x) Bernoulli polynomials.
a(n) = 2*(-16)^n*LerchPhi(-1, -2*n, 1/4).
a(n) = (-1)^n*Sum_{0..2*n} 2^k*C(2*n,k)*E(k), E(n) the Euler secant numbers A122045.
a(n) = (-4)^n*SKP(2*n,1/2) where SKP are the Swiss-Knife polynomials A153641.
a(n) = (-1)^n*2^(6*n+1)*(Zeta(-2*n,1/8) - Zeta(-2*n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. (End)
From Peter Bala, May 13 2017: (Start)
G.f.: 1/(1 + x - 4*x/(1 - 12*x/(1 + x - 40*x/(1 - 56*x/(1 + x - ... - 4*n(4*n - 3)*x/(1 - 4*n(4*n - 1)*x/(1 + x - ...
G.f.: 1/(1 + 9*x - 12*x/(1 - 4*x/(1 + 9*x - 56*x/(1 - 40*x/(1 + 9*x - ... - 4*n(4*n - 1)*x/(1 - 4*n(4*n - 3)*x/(1 + 9*x - .... (End)
From Peter Bala, Nov 08 2019: (Start)
a(n) = sqrt(2)*4^n*Integral_{x = 0..inf} x^(2*n)*cosh(Pi*x/2)/cosh(Pi*x) dx. Cf. A002437.
The L-series 1 + 1/3^(2*n+1) - 1/5^(2*n+1) - 1/7^(2*n+1) + + - - ... = sqrt(2)*(Pi/4)^(2*n+1)*a(n)/(2*n)! (see Shanks), which gives a(n) ~ (1/sqrt(2))*(2*n)!*(4/Pi)^(2*n+1). (End)
EXAMPLE
cos x / cos 2*x = 1 + 3*x^2/2 + 19*x^4/8 + 307*x^6/80 + ...
MAPLE
a := n -> (-1)^n*2^(6*n+1)*(Zeta(0, -2*n, 1/8)-Zeta(0, -2*n, 5/8)):
seq(a(n), n=0..13); # Peter Luschny, Mar 11 2015
MATHEMATICA
With[{nn=30}, Take[CoefficientList[Series[Cos[x]/Cos[2x], {x, 0, nn}], x] Range[0, nn]!, {1, -1, 2}]] (* Harvey P. Dale, Oct 06 2011 *)
PROG
(PARI) {a(n) = if( n<0, 0, n*=2; n! * polcoeff( cos(x + x * O(x^n)) / cos(2*x + x * O(x^n)), n))}; /* Michael Somos, Feb 09 2006 */
CROSSREFS
KEYWORD
nonn,easy,nice
STATUS
approved