A278080 -id:A278080 - cp's OEIS frontend

A003881 Decimal expansion of Pi/4.

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

Offset: 0

N. J. A. Sloane, Simon Plouffe

Also the ratio of the area of a circle to the circumscribed square. More generally, the ratio of the area of an ellipse to the circumscribed rectangle. Also the ratio of the volume of a cylinder to the circumscribed cube. - Omar E. Pol, Sep 25 2013
Also the surface area of a quarter-sphere of diameter 1. - Omar E. Pol, Oct 03 2013
Least positive solution to sin(x) = cos(x). - Franklin T. Adams-Watters, Jun 17 2014
Dirichlet L-series of the non-principal character modulo 4 (A101455) at 1. See e.g. Table 22 of arXiv:1008.2547. - R. J. Mathar, May 27 2016
This constant is also equal to the infinite sum of the arctangent functions with nested radicals consisting of square roots of two. Specifically, one of the Viete-like formulas for Pi is given by Pi/4 = Sum_{k = 2..oo} arctan(sqrt(2 - a_{k - 1})/a_k), where the nested radicals are defined by recurrence relations a_k = sqrt(2 + a_{k - 1}) and a_1 = sqrt(2) (see the article [Abrarov and Quine]). - Sanjar Abrarov, Jan 09 2017
Pi/4 is the area enclosed between circumcircle and incircle of a regular polygon of unit side. - Mohammed Yaseen, Nov 29 2023

			0.785398163397448309615660845819875721049292349843776455243736148...
N = 2, m = 6: Pi/4 = 4!*3^4 Sum_{k >= 0} (-1)^k/((2*k - 11)*(2*k - 5)*(2*k + 1)*(2*k + 7)*(2*k + 13)). - _Peter Bala_, Nov 15 2016

Jörg Arndt and Christoph Haenel, Pi: Algorithmen, Computer, Arithmetik, Springer 2000, p. 150.
Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 437.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 6.3 and 8.4, pp. 429 and 492.
Douglas R. Hofstadter, Gödel, Escher, Bach: An Eternal Golden Braid, Basic Books, p. 408.
J. Rivaud, Analyse, Séries, équations différentielles, Mathématiques supérieures et spéciales, Premier cycle universitaire, Vuibert, 1981, Exercice 3, p. 136.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 119.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Sanjar M. Abrarov and Brendan M. Quine, A Viète-like formula for pi based on infinite sum of the arctangent functions with nested radicals, figshare, 4509014, (2017).
Peter Bala, Arctanh(z) and the Legendre polynomials
Jonathan M. Borwein, Peter B. Borwein, and Karl Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Antonio Gracia Llorente, Shifting Constants Through Infinite Product Transformations, OSF Preprint, 2024.
Ronald K. Hoeflin, Titan Test.
Richard J. Mathar, Table of Dirichlet L-Series and Prime Zeta Modulo Functions for Small Moduli, arXiv:1008.2547 [math.NT], 2010-2015.
Literate Programs, Pi with Machin's formula (Haskell).
Michael Penn, A surprising appearance of pie!, YouTube video, 2020.
Michael Penn, Transforming normal identities into "crazy" ones, YouTube video, 2022.
Srinivasa Ramanujan, Question 353, J. Ind. Math. Soc.
Eric Weisstein's World of Mathematics, Prime Products.
Wikipedia, Leibniz formula for Pi.
Index entries for sequences from "Goedel, Escher, Bach".
Index entries for transcendental numbers.

Cf. A000796, A001586, A071904, A019669, A197723, A347152.
Cf. A000182, A000364, A024235, A278080, A278195.
Cf. A006752 (beta(2)=Catalan), A153071 (beta(3)), A175572 (beta(4)), A175571 (beta(5)), A175570 (beta(6)), A258814 (beta(7)), A258815 (beta(8)), A258816 (beta(9)).
Cf. A001622.

-- see link: Literate Programs
import Data.Char (digitToInt)
a003881_list len = map digitToInt $ show $ machin `div` (10 ^ 10) where
   machin = 4 * arccot 5 unity - arccot 239 unity
   unity = 10 ^ (len + 10)
   arccot x unity = arccot' x unity 0 (unity `div` x) 1 1 where
     arccot' x unity summa xpow n sign
    | term == 0 = summa
    | otherwise = arccot'
      x unity (summa + sign * term) (xpow `div` x ^ 2) (n + 2) (- sign)
    where term = xpow `div` n
-- Reinhard Zumkeller, Nov 20 2012

R:= RealField(100); Pi(R)/4; // G. C. Greubel, Mar 08 2018

Maple
```
evalf(Pi/4) ;
```

RealDigits[N[Pi/4,6! ]]  (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)
(* PROGRAM STARTS *)
(* Define the nested radicals a_k by recurrence *)
a[k_] := Nest[Sqrt[2 + #1] & , 0, k]
(* Example of Pi/4 approximation at K = 100 *)
Print["The actual value of Pi/4 is"]
N[Pi/4, 40]
Print["At K = 100 the approximated value of Pi/4 is"]
K := 100;  (* the truncating integer *)
N[Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 40] (* equation (8) *)
(* Error terms for Pi/4 approximations *)
Print["Error terms for Pi/4"]
k := 1; (* initial value of the index k *)
K := 10; (* initial value of the truncating integer K *)
sqn := {}; (* initiate the sequence *)
AppendTo[sqn, {"Truncating integer K ", " Error term in Pi/4"}];
While[K <= 30,
AppendTo[sqn, {K,
   N[Pi/4 - Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 1000] //
    N}]; K++]
Print[MatrixForm[sqn]]
(* Sanjar Abrarov, Jan 09 2017 *)

Pi/4 \\ Charles R Greathouse IV, Jul 07 2014

# Leibniz/Cohen/Villegas/Zagier/Arndt/Haenel
def FastLeibniz(n):
    b = 2^(2*n-1); c = b; s = 0
    for k in range(n-1,-1,-1):
        t = 2*k+1
        s = s + c/t if is_even(k) else s - c/t
        b *= (t*(k+1))/(2*(n-k)*(n+k))
        c += b
    return s/c
A003881 = RealField(3333)(FastLeibniz(1330))
print(A003881)  # Peter Luschny, Nov 20 2012

Equals Integral_{x=0..oo} sin(2x)/(2x) dx.
Equals lim_{n->oo} n*A001586(n-1)/A001586(n) (conjecture). - Mats Granvik, Feb 23 2011
Equals Integral_{x=0..1} 1/(1+x^2) dx. - Gary W. Adamson, Jun 22 2003
Equals Integral_{x=0..Pi/2} sin(x)^2 dx, or Integral_{x=0..Pi/2} cos(x)^2 dx. - Jean-François Alcover, Mar 26 2013
Equals (Sum_{x=0..oo} sin(x)*cos(x)/x) - 1/2. - Bruno Berselli, May 13 2013
Equals (-digamma(1/4) + digamma(3/4))/4. - Jean-François Alcover, May 31 2013
Equals Sum_{n>=0} (-1)^n/(2*n+1). - Geoffrey Critzer, Nov 03 2013
Equals Integral_{x=0..1} Product_{k>=1} (1-x^(8*k))^3 dx [cf. A258414]. - Vaclav Kotesovec, May 30 2015
Equals Product_{k in A071904} (if k mod 4 = 1 then (k-1)/(k+1)) else (if k mod 4 = 3 then (k+1)/(k-1)). - Dimitris Valianatos, Oct 05 2016
From Peter Bala, Nov 15 2016: (Start)
For N even: 2*(Pi/4 - Sum_{k = 1..N/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^(N/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. See Borwein et al., Theorem 1 (a).
For N odd: 2*(Pi/4 - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^((N-1)/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1.
For N = 0,1,2,... and m = 1,3,5,... there holds Pi/4 = (2*N)! * m^(2*N) * Sum_{k >= 0} ( (-1)^(N+k) * 1/Product_{j = -N..N} (2*k + 1 + 2*m*j) ); when N = 0 we get the Madhava-Gregory-Leibniz series for Pi/4.
For examples of asymptotic expansions for the tails of these series representations for Pi/4 see A024235 (case N = 1, m = 1), A278080 (case N = 2, m = 1) and A278195 (case N = 3, m = 1).
For N = 0,1,2,..., Pi/4 = 4^(N-1)*N!/(2*N)! * Sum_{k >= 0} 2^(k+1)*(k + N)!* (k + 2*N)!/(2*k + 2*N + 1)!, follows by applying Euler's series transformation to the above series representation for Pi/4 in the case m = 1. (End)
From Peter Bala, Nov 05 2019: (Start)
For k = 0,1,2,..., Pi/4 = k!*Sum_{n = -oo..oo} 1/((4*n+1)*(4*n+3)* ...*(4*n+2*k+1)), where Sum_{n = -oo..oo} f(n) is understood as lim_{j -> oo} Sum_{n = -j..j} f(n).
Equals Integral_{x = 0..oo} sin(x)^4/x^2 dx = Sum_{n >= 1} sin(n)^4/n^2, by the Abel-Plana formula.
Equals Integral_{x = 0..oo} sin(x)^3/x dx = Sum_{n >= 1} sin(n)^3/n, by the Abel-Plana formula. (End)
From Amiram Eldar, Aug 19 2020: (Start)
Equals arcsin(1/sqrt(2)).
Equals Product_{k>=1} (1 - 1/(2*k+1)^2).
Equals Integral_{x=0..oo} x/(x^4 + 1) dx.
Equals Integral_{x=0..oo} 1/(x^2 + 4) dx. (End)
With offset 1, equals 5 * Pi / 2. - Sean A. Irvine, Aug 19 2021
Equals (1/2)!^2 = Gamma(3/2)^2. - Gary W. Adamson, Aug 23 2021
Equals Integral_{x = 0..oo} exp(-x)*sin(x)/x dx (see Rivaud reference). - Bernard Schott, Jan 28 2022
From Amiram Eldar, Nov 06 2023: (Start)
Equals beta(1), where beta is the Dirichlet beta function.
Equals Product_{p prime >= 3} (1 - (-1)^((p-1)/2)/p)^(-1). (End)
Equals arctan( F(1)/F(4) ) + arctan( F(2)/F(3) ), where F(1), F(2), F(3), and F(4) are any four consecutive Fibonacci numbers. - Gary W. Adamson, Mar 03 2024
Pi/4 = Sum_{n >= 1} i/(n*P(n, i)*P(n-1, i)) = (1/2)*Sum_{n >= 1} (-1)^(n+1)*4^n/(n*A006139(n)*A006139(n-1)), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The n-th summand of the series is O( 1/(3 + 2*sqrt(3))^n ). - Peter Bala, Mar 16 2024
Equals arctan( phi^(-3) ) + arctan(phi^(-1) ). - Gary W. Adamson, Mar 27 2024
Equals Sum_{n>=1} eta(n)/2^n, where eta(n) is the Dirichlet eta function. - Antonio Graciá Llorente, Oct 04 2024
Equals Product_{k>=2} ((k + 1)^(k*(2*k + 1))*(k - 1)^(k*(2*k - 1)))/k^(4*k^2). - Antonio Graciá Llorente, Apr 12 2025
Equals Integral_{x=sqrt(2)..oo} dx/(x*sqrt(x^2 - 1)). - Kritsada Moomuang, May 29 2025

a(98) and a(99) corrected by Reinhard Zumkeller, Nov 20 2012

A019670 Decimal expansion of Pi/3.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane, Dec 11 1996

With an offset of zero, also the decimal expansion of Pi/30 ~ 0.104719... which is the average arithmetic area  of the 0-winding sectors enclosed by a closed Brownian planar path, of a given length t, according to Desbois, p. 1. - Jonathan Vos Post, Jan 23 2011
Polar angle (or apex angle) of the cone that subtends exactly one quarter of the full solid angle. See comments in A238238. - Stanislav Sykora, Jun 07 2014
60 degrees in radians. - M. F. Hasler, Jul 08 2016
Volume of a quarter sphere of radius 1. - Omar E. Pol, Aug 17 2019
Also smallest positive zero of Sum_{k>=1} cos(k*x)/k = -log(2*|sin(x/2)|). Proof of this identity: Sum_{k>=1} cos(k*x)/k = Re(Sum_{k>=1} exp(k*x*i)/k) = Re(-log(1-exp(x*i))) = -log(2*|sin(x/2)|), x != 2*m*Pi, where i = sqrt(-1). - Jianing Song, Nov 09 2019
The area of a circle circumscribing a unit-area regular dodecagon. - Amiram Eldar, Nov 05 2020

			Pi/3 = 1.04719755119659774615421446109316762806572313312503527365831486...
From _Peter Bala_, Nov 16 2016: (Start)
Case n = 1. Pi/3 = 18 * Sum_{k >= 0} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ).
Using the methods of Borwein et al. we can find the following asymptotic expansion for the tails of this series: for N divisible by 6 there holds Sum_{k >= N/6} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ) ~ 1/N^3 + 6/N^5 + 1671/N ^7 - 241604/N^9 + ..., where the sequence [1, 0, 6, 0, 1671, 0, -241604, 0, ...] is the sequence of coefficients in the expansion of ((1/18)*cosh(2*x)/cosh(3*x)) * sinh(3*x)^2 = x^2/2! + 6*x^4/4! + 1671*x^6/6! - 241604*x^8/8! + .... Cf. A024235, A278080 and A278195. (End)

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.3, p. 489.

Ivan Panchenko, Table of n, a(n) for n = 1..1000
Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Jean Desbois and Stéphane Ouvry, Algebraic and arithmetic area for m planar Brownian paths, arXiv:1101.4135 [math-ph], Jan 21, 2011.
Index entries for transcendental numbers.

Cf. A000796 (Pi), A019669 (Pi/2), A019692 (2*Pi), A122952 (3*Pi), A003881(Pi/4), A137914, A238238, A002388, A091925, A093954, A016910, A019694, A136017, A352324.

Integral_{x=0..oo} 1/(1+x^m) dx: A013661 (m=2), A248897 (m=3), A093954 (m=4), A352324 (m=5), this sequence (m=6), A352125 (m=8), A094888 (m=10).

RealDigits[N[Pi/3,6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)

Pi/3 \\ Charles R Greathouse IV, Sep 08 2011

N=150; default(realprecision, 3+N); digits(Pi*10^N\3) \\ M. F. Hasler, Jul 08 2016

A third of A000796, a sixth of A019692, the square root of A100044.
Sum_{k >= 0} (-1)^k/(6k+1) + (-1)^k/(6k+5). - Charles R Greathouse IV, Sep 08 2011
Product_{k >= 1}(1-(6k)^(-2))^(-1). - Fred Daniel Kline, May 30 2013
From Peter Bala, Feb 05 2015: (Start)
Pi/3 = Sum {k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k = 2F1(1/2,1/2;3/2;1/4). Similar series expansions hold for Pi^2 (A002388), Pi^3 (A091925) and Pi/(2*sqrt(2)) (A093954.)
The integer sequences A(n) := 4^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k ) both satisfy the second-order recurrence equation u(n) = (20*n^2 + 4*n + 1)*u(n-1) - 8*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/3 = 1 + 1/(24 - 8*3^3/(89 - 8*2*5^3/(193 - 8*3*7^3/(337 - ... - 8*(n - 1)*(2*n - 1)^3/((20*n^2 + 4*n + 1) - ... ))))). Cf. A002388 and A093954. (End)
Equals Sum_{k >= 1} arctan(sqrt(3)*L(2k)/L(4k)) where L=A000032. See also A005248 and A056854. - Michel Marcus, Mar 29 2016
Equals Product_{n >= 1} A016910(n) / A136017(n). - Fred Daniel Kline, Jun 09 2016
Equals Integral_{x=-oo..oo} sech(x)/3 dx. - Ilya Gutkovskiy, Jun 09 2016
From Peter Bala, Nov 16 2016: (Start)
Euler's series transformation applied to the series representation Pi/3 = Sum_{k >= 0} (-1)^k/(6*k + 1) + (-1)^k/(6*k + 5) given above by Greathouse produces the faster converging series Pi/3 = (1/2) * Sum_{n >= 0} 3^n*n!*( 1/(Product_{k = 0..n} (6*k + 1)) + 1/(Product_{k = 0..n} (6*k + 5)) ).
The series given above by Greathouse is the case n = 0 of the more general result Pi/3 = 9^n*(2*n)! * Sum_{k >= 0} (-1)^(k+n)*( 1/(Product_{j = -n..n} (6*k + 1 + 6*j)) + 1/(Product_{j = -n..n} (6*k + 5 + 6*j)) ) for n = 0,1,2,.... Cf. A003881. See the example section for notes on the case n = 1.(End)
Equals Product_{p>=5, p prime} p/sqrt(p^2-1). - Dimitris Valianatos, May 13 2017
Equals A019699/4 or A019693/2. - Omar E. Pol, Aug 17 2019
Equals Integral_{x >= 0} (sin(x)/x)^4 = 1/2 + Sum_{n >= 0} (sin(n)/n)^4, by the Abel-Plana formula. - Peter Bala, Nov 05 2019
Equals Integral_{x=0..oo} 1/(1 + x^6) dx. - Bernard Schott, Mar 12 2022
Pi/3 = -Sum_{n >= 1} i/(n*P(n, 1/sqrt(-3))*P(n-1, 1/sqrt(-3))), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The first twenty terms of the series gives the approximation Pi/3 = 1.04719755(06...) correct to 8 decimal places. - Peter Bala, Mar 16 2024
Equals Integral_{x >= 0} (2*x^2 + 1)/((x^2 + 1)*(4*x^2 + 1)) dx. - Peter Bala, Feb 12 2025

A024235 Expansion of e.g.f. tan(x)*sin(x)/2 (even powers only).

Original entry on oeis.org

0, 1, 2, 31, 692, 25261, 1351382, 99680491, 9695756072, 1202439837721, 185185594118762, 34674437196568951, 7757267081778543452, 2043536254646561946181, 626129820701814932734142, 220771946624511552276841411, 88759695789769644718332394832

Offset: 0

R. H. Hardin

From Peter Bala, Nov 10 2016: (Start)
This sequence gives the coefficients in an asymptotic expansion related to the constant Pi/8. Recall the Madhava-Gregory-Leibniz series Pi/4 = Sum_{k = 1..inf} (-1)^(k-1)/(2*k - 1). Borwein et al. gave an asymptotic expansion for the tails of this series: Pi/2 - 2*Sum_{k = 1..N/2} (-1)^(k-1)/(2*k - 1) ~ 1/N - 1/N^3 + 5/N^5 - 61/N^7 + ..., where N is an integer divisible by 4 and the sequence of unsigned coefficients [1, 1, 5, 61,...] is the sequence of Euler numbers A000364.
Similarly, we have the series representation Pi/8 = Sum_{k = 1..inf} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)): using the approach of Borwein et al. we can show the associated asymptotic expansion for the tails of the series is Pi/4 - 2*Sum_{k = 1..N/2} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)) ~ -1/N^3 + 2/N^5 - 31/N^7 + 692/N^9 - ..., where N is divisible by 4 and where the sequence of unsigned coefficients [1, 2, 31, 692,...] forms the present sequence. A numerical example is given below. Cf. A278080 and A278195. (End)

			tan(x)*sin(x)/2 = 1/2*x^2 + 1/12*x^4 + 31/720*x^6 + 173/10080*x^8 + ...
From _Peter Bala_, Nov 10 2016: (Start)
Asymptotic expansion at N = 100000.
The truncated series 2*Sum_{k = 1..N/2} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)) = 0.78539816339744(9)309615660(6)4581987(603) 104929(1657)84377... to 50 digits. The bracketed digits show where this decimal expansion differs from that of Pi/4. The numbers -1, 2, -31, 692 must be added to the bracketed numbers to give the correct decimal expansion to 50 digits: Pi/4 = 0.78539816339744(8)309615660(8)4581987(572)104929(2349)84377.... (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..50
J. M. Borwein, P. B. Borwein, K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Eric Weisstein's World of Mathematics, Euler Polynomial.

Cf. A009744, A000364, A004174, A019675, A278080, A278195.

A000364 := proc(n)
   abs(euler(2*n));
end proc:
seq(1/2*(A000364(n) - (-1)^n), n = 0..20); # Peter Bala, Nov 10 2016

With[{nn=30},Take[CoefficientList[Series[Tan[x]*Sin[x]/2,{x,0,nn}], x]Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Apr 27 2012 *)

G.f.: 1/2*(G(0) - 1/(1+x)) where G(k) = 1 - x*(2*k+1)^2/(1 - x*(2*k+2)^2/G(k+1) );  (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 09 2013
a(n) ~ (2*n)! * (2/Pi)^(2*n+1). - Vaclav Kotesovec, Jan 23 2015
From Peter Bala, Nov 10 2016: (Start)
a(n) = 1/2*(A000364(n) - (-1)^n).
a(n) = 1/8*(-4)^n*( -E(2*n,3/2) + 2*E(2*n,1/2) - E(2*n,-1/2) ), where E(n,x) is the Euler polynomial of order n.
G.f. 1/2!*sin^2(x)/cos(x) = x^2/2! + 2*x^4/4! + 31*x^6/6! + 692*x^8/8! + ....
O.g.f. for a signed version of the sequence: Sum_{n >= 0} ( 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n, k)/((1 - (2*k - 1)*x)*(1 - (2*k + 1)*x)*(1 - (2*k + 3)*x)) ) = 1 - 2*x^2 + 31*x^4 - 692*x^6 + .... (End)

Extended and signs tested Mar 15 1997.

More terms from Harvey P. Dale, Apr 27 2012

A278195 Expansion of e.g.f. (1/6!)*sin^6(x)/cos(x) (coefficients of even powers only).

Original entry on oeis.org

0, 0, 0, 1, -28, 882, -17116, 803803, 13713336, 3671012164, 506128123928, 96524822605365, 21542790273363260, 5676618945053498806, 1739246268204447115932, 613255488134158250903887, 246554708506039690689322544, 112115693433705109495581088008

Offset: 0

Peter Bala, Nov 15 2016

This sequence gives the coefficients in an asymptotic expansion of a series related to the constant Pi. It can be shown that (1/6!)*Pi/4 = Sum_{k >= 1} (-1)^k/((2*k - 7)*(2*k - 5)*(2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)*(2*k + 5)). Using Proposition 1 of Borwein et al. it can be shown that the following asymptotic expansion holds for the tails of this series: for N divisible by 4, 2*{ (1/6!)*Pi/4 - Sum_{k = 1..N/2} (-1)^k/((2*k - 7)*(2*k - 5)*(2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)*(2*k + 5)) } ~ -1/N^7 + (-28)/N^9 - 882/N^11 + (-17116)/N^13 - 803803/N^15 + .... An example is given below.

			Expansion of (1/6!)*sin^6(x)/cos(x) starts x^6/6! - 28*x^8/8! + 882*x^10/10! - 17116*x^12/12! + ....
Let N = 100000. The truncated series 2*Sum_{k = 1..N/2} (-1)^k/( (2*k - 7)*(2*k - 5)*(2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)*(2*k + 5) ) = 0.0021816615649929119711546134606107(7)5891803(617)860677(2450)20121.... The bracketed digits show where this decimal expansion differs from that of Pi/1440. The numbers -1, -28, -882 must be added to the bracketed numbers to give the correct decimal expansion: Pi/1440 = 0.0021816615649929119711546134606107(6)5891803(589)860677(1568)20121....

J. M. Borwein, P. B. Borwein, K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Eric Weisstein's World of Mathematics, Euler Polynomial.

Cf. A000364, A004174, A166984, A278079, A278080, A278194.

A000364 := n -> abs(euler(2*n)):
seq((1/6!)*(A000364(n) - (1/16)*((-25)^n - 7*(-9)^n + 22*(-1)^n) ), n = 0..20);

With[{nn=40},Take[CoefficientList[Series[Sin[x]^6/Cos[x] 1/6!,{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Sep 12 2019 *)

a(n) = [x^(2*n)/(2*n)!] ( 1/6!*sin^6(x)/cos(x) ).
a(n) = (1/6!)*( A000364(n) - 1/16*((-25)^n - 7*(-9)^n + 22*(-1)^n) ).
a(n) = (-1)^(n+1)/(2^6*6!) * 2^(2*n)*( E(2*n,7/2) - 6*E(2*n,5/2) + 15*E(2*n,3/2) - 20*E(2*n,1/2) + 15*E(2*n,-1/2) - 6*E(2*n,-3/2) + E(2*n,-5/2) ), where E(n,x) is the Euler polynomial of order n.

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A003881 Decimal expansion of Pi/4.

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A019670 Decimal expansion of Pi/3.

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A024235 Expansion of e.g.f. tan(x)*sin(x)/2 (even powers only).

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A278195 Expansion of e.g.f. (1/6!)*sin^6(x)/cos(x) (coefficients of even powers only).

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A278194 E.g.f. (1/5!)*sin^5(x)/cos(x) (coefficients of odd powers only).

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A278079 Expansion of e.g.f. (1/3!)*sin^3(x)/cos(x) (coefficients of odd powers only).

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