A181049 -id:A181049 - cp's OEIS frontend

A004767 a(n) = 4*n + 3.

3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, 103, 107, 111, 115, 119, 123, 127, 131, 135, 139, 143, 147, 151, 155, 159, 163, 167, 171, 175, 179, 183, 187, 191, 195, 199, 203, 207, 211, 215, 219, 223

Offset: 0

N. J. A. Sloane

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(12).
Binary expansion ends 11.
These the numbers for which zeta(2*x+1) needs just 2 terms to be evaluated. - Jorge Coveiro, Dec 16 2004 [This comment needs clarification]
a(n) is the smallest k such that for every r from 0 to 2n - 1 there exist j and i, k >= j > i > 2n - 1, such that j - i == r (mod (2n - 1)), with (k, (2n - 1)) = (j,(2n - 1)) = (i, (2n - 1)) = 1. - Amarnath Murthy, Sep 24 2003
Complement of A004773. - Reinhard Zumkeller, Aug 29 2005
Any (4n+3)-dimensional manifold endowed with a mixed 3-Sasakian structure is an Einstein space with Einstein constant lambda = 4n + 2 [Theorem 3, p. 10 of Ianus et al.]. - Jonathan Vos Post, Nov 24 2008
Solutions to the equation x^(2*x) = 3*x (mod 4*x). - Farideh Firoozbakht, May 02 2010
Subsequence of A022544. - Vincenzo Librandi, Nov 20 2010
First differences of A084849. - Reinhard Zumkeller, Apr 02 2011
Numbers n such that {1, 2, 3, ..., n} is a losing position in the game of Nim. - Franklin T. Adams-Watters, Jul 16 2011
Numbers n such that there are no primes p that satisfy the relationship p XOR n = p + n. - Brad Clardy, Jul 22 2012
The XOR of all numbers from 1 to a(n) is 0. - David W. Wilson, Apr 21 2013
A089911(4*a(n)) = 4. - Reinhard Zumkeller, Jul 05 2013
First differences of A014105. - Ivan N. Ianakiev, Sep 21 2013
All triangular numbers in the sequence are congruent to {3, 7} mod 8. - Ivan N. Ianakiev, Nov 12 2013
Apart from the initial term, length of minimal path on an n-dimensional cubic lattice (n > 1) of side length 2, until a self-avoiding walk gets stuck. Construct a path connecting all 2n points orthogonally adjacent from the center, ending at the center. Starting at any point adjacent to the center, there are 2 steps to reach each of the remaining 2n - 1 points, resulting in path length 4n - 2 with a final step connecting the center, for a total path length of 4n - 1, comprising 4n points. - Matthew Lehman, Dec 10 2013
a(n-1), n >= 1, appears as first column in the triangles A238476 and A239126 related to the Collatz problem. - Wolfdieter Lang, Mar 14 2014
For the Collatz Conjecture, we identify two types of odd numbers. This sequence contains all the ascenders: where (3*a(n) + 1) / 2 is odd and greater than a(n). See A016813 for the descenders. - Jaroslav Krizek, Jul 29 2016

			G.f. = 3 + 7*x + 11*x^2 + 15*x^3 + 19*x^4 + 23*x^5 + 27*x^6 + 31*x^7 + ...

Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 85.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999. See Theorem 8.1 on page 240.

Ivan Panchenko, Table of n, a(n) for n = 0..200
Guo-Niu Han, Enumeration of Standard Puzzles
Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]
Stere Ianus, Mihai Visinescu and Gabriel-Eduard Vilcu, Hidden symmetries and Killing tensors on curved spaces, arXiv:0811.3478 [math-ph], 2008. - _Jonathan Vos Post_, Nov 24 2008
Tanya Khovanova, Recursive Sequences
Juan F. Pulido, José L. Ramírez, and Andrés R. Vindas-Meléndez, Generating Trees and Fibonacci Polyominoes, arXiv:2411.17812 [math.CO], 2024. See p. 8.
William A. Stein, Dimensions of the spaces S_k(Gamma_0(N))
William A. Stein, The modular forms database
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A004773, A005408, A008545 (partial products), A008586, A014105, A016813, A016825, A017137, A017629, A022544, A084849, A181049, A238476, A239126.
Cf. A017101 and A004771 (bisection: 3 and 7 mod 8).
Cf. A016838 (square).

a004767 = (+ 3) . (* 4)
a004767_list = [3, 7 ..]  -- Reinhard Zumkeller, Oct 03 2012

[4*n+3: n in [0..50]]; // Wesley Ivan Hurt, Jul 09 2014

Maple
```
seq( 3+4*n, n=0..100 );
```

4 Range[50] - 1 (* Wesley Ivan Hurt, Jul 09 2014 *)

a(n)=4*n+3 \\ Charles R Greathouse IV, Jul 28 2015

Vec((3+x)/(1-x)^2 + O(x^200)) \\ Altug Alkan, Jan 15 2016

for n in range(0,50): print(4*n+3, end=', ') # Stefano Spezia, Dec 12 2018

[4*n+3 for n in range(50)] # G. C. Greubel, Dec 09 2018

(0 to 59).map(4 *  + 3) // _Alonso del Arte, Dec 12 2018

G.f.: (3+x)/(1-x)^2. - Paul Barry, Feb 27 2003
a(n) = 2*a(n-1) - a(n-2) for n > 1, a(0) = 3, a(1) = 7. - Philippe Deléham, Nov 03 2008
a(n) = A017137(n)/2. - Reinhard Zumkeller, Jul 13 2010
a(n) = 8*n - a(n-1) + 2 for n > 0, a(0) = 3. - Vincenzo Librandi, Nov 20 2010
a(n) = A005408(A005408(n)). - Reinhard Zumkeller, Jun 27 2011
a(n) = 3 + A008586(n). - Omar E. Pol, Jul 27 2012
a(n) = A014105(n+1) - A014105(n). - Michel Marcus, Sep 21 2013
a(n) = A016813(n) + 2. - Jean-Bernard François, Sep 27 2013
a(n) = 4*n - 1, with offset 1. - Wesley Ivan Hurt, Mar 12 2014
From Ilya Gutkovskiy, Jul 29 2016: (Start)
E.g.f.: (3 + 4*x)*exp(x).
Sum_{n >= 0} (-1)^n/a(n) = (Pi + 2*log(sqrt(2) - 1))/(4*sqrt(2)) = A181049. (End)

A093954 Decimal expansion of Pi/(2*sqrt(2)).

Original entry on oeis.org

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

Offset: 1

Eric W. Weisstein, Apr 19 2004

The value is the length Pi*sqrt(2)/4 of the diagonal in the square with side length Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) = A003881. The area of the circumcircle of this square is Pi*(Pi*sqrt(2)/8)^2 = Pi^3/32 = A153071. - Eric Desbiaux, Jan 18 2009
This is the value of the Dirichlet L-function of modulus m=8 at argument s=1 for the non-principal character (1,0,1,0,-1,0,-1,0). See arXiv:1008.2547. - R. J. Mathar, Mar 22 2011
Archimedes's-like scheme: set p(0) = sqrt(2), q(0) = 1; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (harmonic mean, i.e., 1/p(n+1) = (1/p(n) + 1/q(n))/2), q(n+1) = sqrt(p(n+1)*q(n)) (geometric mean, i.e., log(q(n+1)) = (log(p(n+1)) + log(q(n)))/2), for n >= 0. The error of p(n) and q(n) decreases by a factor of approximately 4 each iteration, i.e., approximately 2 bits are gained by each iteration. Set r(n) = (2*q(n) + p(n))/3, the error decreases by a factor of approximately 16 for each iteration, i.e., approximately 4 bits are gained by each iteration. For a similar scheme see also A244644. - A.H.M. Smeets, Jul 12 2018
The area of a circle circumscribing a unit-area regular octagon. - Amiram Eldar, Nov 05 2020

			1.11072073453959156175397...
From _Peter Bala_, Mar 03 2015: (Start)
Asymptotic expansion at n = 5000.
The truncated series Sum_{k = 0..5000 - 1} (-1)^floor(k/2)/(2*k + 1) = 1.110(6)207345(42)591561(18)3970(5238)1.... The bracketed digits show where this decimal expansion differs from that of Pi/(2*sqrt(2)). The numbers 1, -3, 57, -2763 must be added to the bracketed numbers to give the correct decimal expansion to 30 digits: Pi/(2*sqrt(2)) = 1.110(7)207345(39)591561(75)3970 (2475)1.... (End)
From _Peter Bala_, Nov 24 2016: (Start)
Case m = 1, n = 1:
Pi/(2*sqrt(2)) = 4*Sum_{k >= 0} (-1)^(1 + floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)).
We appear to have the following asymptotic expansion for the tails of this series: for N divisible by 4, Sum_{k >= N/2} (-1)^floor(k/2)/((2*k - 1)*(2*k + 1)*(2*k + 3)) ~ 1/N^3 - 14/N^5 + 691/N^7 - 62684/N^9 - ..., where the coefficient sequence [1, 0, -14, 0, 691, 0, -62684, ...] appears to come from the e.g.f. (1/2!)*cosh(x)/cosh(2*x)*sinh(x)^2 = x^2/2! - 14*x^4/4! + 691*x^6/6! - 62684*x^8/8! + .... Cf. A019670.
For example, take N = 10^5. The truncated series Sum_{k = 0..N/2 -1} (-1)^(1+floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)) = 0.27768018363489(8)89043849(11)61878(80026)6163(351171)58.... The bracketed digits show where this decimal expansion differs from that of (1/4)*Pi/(2*sqrt(2)). The numbers -1, 14, -691, 62684 must be added to the bracketed numbers to give the correct decimal expansion: (1/4)*Pi/(2*sqrt(2)) = 0.27768018363489(7) 89043849(25)61878(79335)6163(413855)58... (End)

J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, 1993, Exercice 5, p. 240.
George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press, 2006, p. 149.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.4.1, p. 20.
L. B. W. Jolley, Summation of Series, Dover, 1961, eq. 76, page 16.
Joel L. Schiff, The Laplace Transform: Theory and Applications, Springer-Verlag New York, Inc. (1999). See p. 149.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 53.

Harry J. Smith, Table of n, a(n) for n = 1..20000
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
R. J. Mathar, Table of Dirichlet L-series and prime zeta modulo functions for small moduli, arXiv:1008.2547 [math.NT], 2010-2015, table 7 and section 2.2, value of L(m=8,r=4,s=1).
Michael Penn, Newton's sum (2023), YouTube video.
Michael I. Shamos, A catalog of the real numbers, (2007). See p. 149.
Eric Weisstein's World of Mathematics, Bifoliate.
Index entries for transcendental numbers.

Cf. A161684 (continued fraction).
Cf. A002388, A019670.
Cf. A000281, A063448, A247719, A193887, A244976, A181048, A181049.
Cf. A003881, A251809, A188510.
Cf. A352324, A248897, A019669.

simplify( sum((cos((1/2)*k*Pi)+sin((1/2)*k*Pi))/(2*k+1), k = 0 .. infinity) );  # Peter Bala, Mar 09 2015

RealDigits[Pi/Sqrt@8, 10, 111][[1]] (* Michael De Vlieger, Sep 23 2016 and slightly modified by Robert G. Wilson v, Jul 23 2018 *)

default(realprecision, 20080); x=Pi*sqrt(2)/4; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b093954.txt", n, " ", d)); \\ Harry J. Smith, Jun 17 2009

Equals 1/A112628.
Equals Integral_{x=0..oo} 1/(x^4+1) dx. - Jean-François Alcover, Apr 29 2013
From Peter Bala, Feb 05 2015: (Start)
Pi/(2*sqrt(2)) = Sum_{k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k.
The integer sequences A(n) := 2^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k ) both satisfy the second order recurrence equation u(n) = (12*n^2 + 1)*u(n-1) - 4*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/(2*sqrt(2)) = 1 + 1/(12 - 4*3^3/(49 - 4*2*5^3/(109 - 4*3*7^3/(193 - ... - 4*(n - 1)*(2*n - 1)^3/((12*n^2 + 1) - ... ))))). Cf. A002388 and A019670. (End)
From Peter Bala, Mar 03 2015: (Start)
Pi/(2*sqrt(2)) = Sum_{k >= 0} (-1)^floor(k/2)/(2*k + 1) = limit (n -> infinity) Sum_{k = -n .. n - 1} (-1)^k/(4*k + 1). See Wells.
We conjecture 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 + ..., where n is a multiple of 4 and the sequence of unsigned coefficients [1, 3, 57, 2763, ...] is A000281. An example with n = 5000 is given below. (End)
From Peter Bala, Sep 21 2016: (Start)
c = 2 * Sum_{k >= 0} (-1)^k * (4*k + 2)/((4*k + 1)*(4*k + 3)) = A181048 + A181049. The asymptotic expansion conjectured above follows from the asymptotic expansions given in A181048 and A181049.
c = 1/2 * Integral_{x = 0..Pi/2} sqrt(tan(x)) dx. (End)
From Peter Bala, Nov 24 2016: (Start)
Let m be an odd integer and n a nonnegative integer. Then Pi/(2*sqrt(2)) = 2^n*m^(2*n)*(2*n)!*Sum_{k >= 0} (-1)^(n+floor(k/2)) * 1/Product_{j = -n..n} (2*k + 1 + 2*m*j). Cf. A003881.
In the particular case m = 1 the result has the equivalent form: for n a nonnegative integer, Pi/(2*sqrt(2)) = 2^n*(2*n)!*Sum_{k >= 0} (-1)^(n+k)*(8*k + 4)* 1/Product_{j = -n..n+1} (4*k + 2*j + 1). The case m = 1, n = 1 is considered in the Example section below.
Let m be an odd integer and n a nonnegative integer. Then Pi/(2*sqrt(2)) = 4^n*m^(2*n)*(2*n)!*Sum_{k >= 0} (-1)^(n+floor(k/2)) * 1/Product_{j = -n..n} (2*k + 1 + 4*m*j). (End)
Equals Integral_{x = 0..oo} cosh(x)/cosh(2*x) dx. - Peter Bala, Nov 01 2019
Equals Sum_{k>=1} A188510(k)/k = Sum_{k>=1} Kronecker(-8,k)/k = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - 1/13 - 1/15 + ... - Jianing Song, Nov 16 2019
From Amiram Eldar, Jul 16 2020: (Start)
Equals Product_{k>=1} (1 - (-1)^k/(2*k+1)).
Equals Integral_{x=0..oo} dx/(x^2 + 2).
Equals Integral_{x=0..Pi/2} dx/(sin(x)^2 + 1). (End)
Equals Integral_{x=0..oo} x^2/(x^4 + 1) dx (Arnaudiès). - Bernard Schott, May 19 2022
Equals Integral_{x = 0..1} 1/(2*x^2 + (1 - x)^2) dx. - Peter Bala, Jul 22 2022
Equals Integral_{x = 0..1} 1/(1 - x^4)^(1/4) dx. - Terry D. Grant, Mar 17 2023
Equals 1/Product_{p prime} (1 - Kronecker(-8,p)/p), where Kronecker(-8,p) = 0 if p = 2, 1 if p == 1 or 3 (mod 8) or -1 if p == 5 or 7 (mod 8). - Amiram Eldar, Dec 17 2023
Equals A068465*A068467. - R. J. Mathar, Jun 27 2024
From Stefano Spezia, Jun 05 2025: (Start)
Equals Sum_{k>=1} (-1)^(k+1)(1/(4*k - 3) + 1/(4*k - 1)).
Equals Product_{k=0..oo} (1 + (-1)^k/(2*k + 3)).
Equals Integral_{x=0..oo} 1/(2*x^2 + 1).
Equals Integral_{x=0..1} 1/((1 + x^2)*sqrt(1 - x^2)). (End)

A247719 Decimal expansion of Integral_{t=0..Pi/2} sqrt(tan(t)) dt.

Original entry on oeis.org

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

Offset: 1

Jean-François Alcover, Sep 23 2014

			2.22144146907918312350794049503034684930731...

G. C. Greubel, Table of n, a(n) for n = 1..10000
D. H. Bailey and J. M. Borwein, Highly Parallel, High-Precision Numerical Integration p. 7. (2005) Lawrence Berkeley National Laboratory.
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, pp. 235-236.

Cf. A063448, A093954, A193887, A244976, A181048, A181049, A186706.

SetDefaultRealField(RealField(100)); R:= RealField(); Pi(R)/Sqrt(2); // G. C. Greubel, Sep 07 2018

RealDigits[Pi/Sqrt[2], 10, 104] // First

default(realprecision, 100); Pi/sqrt(2) \\ G. C. Greubel, Sep 07 2018

Equals Pi/sqrt(2).
Equals A063448/2.
c = 2*( Sum_{k >= 0} (-1)^k/(4*k + 1) + Sum_{k >= 0} (-1)^k/(4*k + 3) ) = 2*(A181048 + A181049). - Peter Bala, Sep 21 2016
From Amiram Eldar, Aug 07 2020: (Start)
Equals Integral_{x=0..Pi} 1/(cos(x)^2 + 1) dx = Integral_{x=0..Pi} 1/(sin(x)^2 + 1) dx.
Equals Integral_{x=-oo..oo} 1/(x^4 + 1) dx.
Equals Integral_{x=-oo..oo} x^2/(x^4 + 1) dx.
Equals Integral_{x=0..oo} log(1 + 1/(2 * x^2)) dx. (End)
Equals Integral_{x=0..2*Pi} 1/(3 + sin(x)) dx; since for a>1: Integral_{x=0..2*Pi} 1/(a + sin(x)) dx = 2*Pi/sqrt(a^2-1). - Bernard Schott, Feb 19 2023
Equals 20/9 - 160*Sum_{n >= 1} 1/((64*n^2 - 1)*(64*n^2 - 4)*(64*n^2 - 9)). - Peter Bala, Nov 09 2023

A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+1).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 01 2010, Oct 06 2010

			0.86697298733991103757399516388287071365217536734524490433....
At N = 100000 the truncated series 2*Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) = 1.7339(3)5974(5)7982(5)075(25)79(846)27(404)7... to 32 digits. The bracketed numbers show where this decimal expansion differs from that of 2*A181048. The numbers 1, 1, -3, -11, 57, 361 must be added to the bracketed numbers to give the correct decimal expansion to 32 digits: 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) ) = 1.7339(4)5974(6)7982(2)075(14)79(903)27(765)7.... - _Peter Bala_, Sep 23 2016

Jolley, Summation of Series, Dover (1961) eq 82 page 16.
Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equation 21.17

Ivan Panchenko, Table of n, a(n) for n = 0..1000
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Eric W. Weisstein, Euler's Series Transformation.
Herbert S. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics & Theoretical Computer Science, Vol 3, No 4 (1999).

Cf. A001586, A003881, A024396, A091648, A093954, A113476, A181049, A181122, A188458, A193534, A262246.

RealDigits[(Pi Sqrt[2])/8 + (Sqrt[2] Log[1 + Sqrt[2]])/4, 10, 100][[1]] (* Alonso del Arte, Aug 11 2011 *)

(log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) \\ G. C. Greubel, Jul 05 2017

(asinh(1)+Pi/2)/sqrt(8) \\ Charles R Greathouse IV, Jul 06 2017

Equals (A093954 + A091648/sqrt(2))/2.
Integral_{x = 0..1} 1/(1+x^4) = Sum_{k >= 0} (-1)^k/(4*k+1) = (log(1+sqrt(2)) + Pi/2)/(2*sqrt(2)).
1 - 1/5 + 1/9 - 1/13 + 1/17 - ... = (Pi*sqrt(2))/8 + (sqrt(2)*log(1 + sqrt(2)))/4 = (Pi + 2*log(1 + sqrt(2)))/(4 sqrt(2)). The first two are the formulas as given in Spiegel et al., the third is how Mathematica rewrites the infinite sum. - Alonso del Arte, Aug 11 2011
Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) ) ~ 1/N + 1/N^2 - 3/N^3 - 11/N^4 + 57/N^5 + 361/N^6 - ..., where the sequence of coefficients [1, 1, -3, -11, 57, 361, ...] is A188458. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 1/4. An example is given below. Cf. A181049. - Peter Bala, Sep 23 2016
Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 1) = Sum_{n >= 0} 2^(n-1)*n!/A007696(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k + 1)). - Peter Bala, Dec 01 2021
From Peter Bala, Oct 23 2023: (Start)
The slowly converging series representation Sum_{n >= 0} (-1)^n/(4*n + 1) for the constant can be accelerated to give the following faster converging series:
1/2 + 2*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5));
7/10 + 8*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9));
71/90 + 48*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13));
971/1170 + 384*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13)*(4*n + 17)).
These results may be easily verified by taking the partial fraction expansions of the summands. The general result appears to be that for r >= 0, the constant equals
C(r) + (2^r)*r!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 5)*...*(4*n + 4*r + 1)), where C(r) is the rational number Sum_{k = 0..r-1} 2^(k-1)*k!/(1*5*9*...*(4*k + 1)). [added 19 Feb 2024: the general result can be proved by the WZ method as described in Wilf.]
In the limit as r -> oo we find that the constant equals Sum_{k >= 0} 2^(k-1)*k!/(Product_{i = 0..k} 4*i + 1) as noted above. (End)
From Peter Bala, Mar 03 2024: (Start)
Continued fraction: 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler.
Equals hypergeom([1/4, 1], [5/4], -1).
Gauss's continued fraction: 1/(1 + 1^2/(5 + 4^2/(9 + 5^2/(13 + 8^2/(17 + 9^2/(21 + 12^2/(25 + 13^2/(29 + 16^2/(33 + 17^2/(37 + ... )))))))))). (End)

A196525 Decimal expansion of log(1+sqrt(2))/sqrt(2).

Original entry on oeis.org

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

Offset: 0

R. J. Mathar, Oct 03 2011

			0.6232252401402305133940200802505680... = A091648/A002193.
From _Peter Bala_, Dec 01 2021: (Start)
With N = 10000, the truncated series Sum_{k = 0..N/4 - 1} (-1)^k/((4*k + 1)*(4*k+3)) = 0.6232252[3]014023[16]1339[3659]080... to 27 decimal places. The square bracketed numbers show where this decimal expansion differs from that of (1/sqrt(2))*log(1+sqrt(2)) = 0.6232252(4)014023(05) 1339(4020)080.... The numbers 1, -11, 361 must be added to the square bracketed numbers to give the correct decimal expansion to 27 decimal places. (End)

G. C. Greubel, Table of n, a(n) for n = 0..10000
R. J. Mathar, Table of Dirichlet L-series and Prime Zeta Modulo Functions for Small Moduli, arXiv:1008.2547 [math.NT], 2010-2015, Table section 2.2, L(m=8, r=2, s=1).
Paul J. Nahin, Inside interesting integrals, Undergrad. Lecture Notes in Physics, Springer (2020), (2.2.3)
Index entries for transcendental numbers.

Cf. A000464, A002193, A001539, A091648, A181048, A181049.

SetDefaultRealField(RealField(100)); Log(Sqrt(2)+1)/Sqrt(2); // G. C. Greubel, Oct 05 2018

RealDigits[Log[1+Sqrt[2]]/Sqrt[2],10,120][[1]] (* Harvey P. Dale, Dec 27 2011 *)
RealDigits[Sum[1/((2 n - 1) 2^n), {n, 1, Infinity}], 10, 120][[1]] (* Fred Daniel Kline, May 23 2019 *)

log(sqrt(2)+1)/sqrt(2) \\ Michel Marcus, Sep 27 2017

Equals Sum_{n>=1} A091337(n)/n = 1 - 1/3 - 1/5 + 1/7 + 1/9 - 1/11 - ...
Equals 2*Sum_{n>=1} (-1)^n/A001539(n). - Michel Marcus, Sep 27 2017
From Fred Daniel Kline, May 23 2019: (Start)
Equals arcsinh(1)/sqrt(2).
Equals Sum_{n>=1} 1/A118417(n-1) = Sum_{n>=1} 1/((2*n - 1)*2^n). (End)
From Peter Bala, Nov 01 2019: (Start)
Equals (1/sqrt(2))*arccoth(sqrt(2)).
Equals 1 - 8*Sum_{n >= 0} (-1)^(n+1)*n/(16*n^2 - 1).
Equals 1 - Integral_{x = 0..inf} exp(-2*x)*cosh(x)/cosh(2*x) dx.
Equals 2*Integral_{x = 0..inf} exp(x)*(exp(2*x) + 1)*(exp(4*x) - 1)/(exp(4*x) + 1)^2 dx - 1. (End)
From Amiram Eldar, Aug 16 2020: (Start)
Equals Sum_{k>=0} (-1)^k * (2*k)!!/(2*k+1)!!.
Equals Integral_{x=0..Pi/4} 1/(cos(x) + sin(x)) dx. (End)
From Peter Bala, Dec 01 2021: (Start)
Equals 2*Sum_{k >= 0} (-1)^k/((4*k + 1)*(4*k + 3)).
Let N be a positive integer divisible by 4. We have the asymptotic expansion (1/sqrt(2))*log(1 + sqrt(2)) - 2*Sum_{k = 0..N/4 - 1} (-1)^k/((4*k + 1)*(4*k + 3)) ~ 1/N^2 - 11/N^4 + 361/N^6 - 24611/N^8 + ..., where the sequence of unsigned coefficients [1, 11, 361, 24611, ...] is A000464. See A181048 and A181049. An example is given below. (End)
Equals 1/Product_{p prime} (1 - Kronecker(8,p)/p), where Kronecker(8,p) = 0 if p = 2, 1 if p == 1 or 7 (mod 8) or -1 if p == 3 or 5 (mod 8). - Amiram Eldar, Dec 17 2023
Equals integral_{x=0..Pi/2} sin^2(x)/(sin(x)+cos(x)) dx [Nahin]. - R. J. Mathar, May 16 2024

A181122 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+1).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 05 2010

			0.88831357265178863804075522702037934627811083077545817120597...

Gheorghe Coserea, Table of n, a(n) for n = 0..1000
H. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics and Theoretical Computer Science 3(4) (1999), 189-192.

Cf. A002162, A003881, A024396, A113476, A181048, A181049, A181122, A193534, A262246.

(int(1/(1+x^5),x=0..1));
evalf(LerchPhi(-1,1,1/5)/5) ; # R. J. Mathar, Oct 16 2011

(Sqrt[8 + 8/Sqrt[5]]*Pi + 2*Sqrt[5]*ArcCoth[3/Sqrt[5]] + Log[16])/20 // RealDigits[#, 10, 105]& // First (* Jean-François Alcover, Feb 13 2013 *)

default(realprecision, 106);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(5*n+1)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015

Sum_{k>=0} (-1)^k/(5k+1) = Integral_{x=0..1}dx/(1+x^5) = (1/10)*sqrt(10-2*sqrt(5))*arctan((3/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*sqrt(10-2*sqrt(5))*arctan(-(1/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*sqrt(10-2*sqrt(5))*sqrt(5)*arctan(-(1/4)*sqrt(10-2*sqrt(5)) + (1/4)*sqrt(10-2*sqrt(5))*sqrt(5)) + (1/20)*log(2)*sqrt(5) + (1/5)*log(2) - (1/20)*log(7-3*sqrt(5))*sqrt(5).
Equals Pi*sqrt(phi)/5^(5/4) + log(phi)/sqrt(5) + log(2)/5, where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Nov 01 2015
From Peter Bala, Feb 19 2024: (Start)
Equals (1/2)*Sum_{n >= 0} n!*(5/2)^n/(Product_{k = 0..n} 5*k + 1) = (1/2)*Sum_{n >= 0} n!*(5/2)^n/A008548(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(5*k + 1)).
Continued fraction: 1/(1 + 1^2/(5 + 6^2/(5 + 11^2/(5 + ... + (5*n + 1)^2/(5 + ... ))))).
The slowly converging series representation Sum_{n >= 0} (-1)^n/(5*n + 1) for the constant can be accelerated to give the following faster converging series:
1/2 + (5/2)*Sum_{n >= 0} (-1)^n/((5*n + 1)(5*n + 6)) and
17/24 + (25/2)*Sum_{n >= 0} (-1)^n/((5*n + 1)(5*n + 6)*(5*n + 11)).
These two series are the cases r = 1 and r = 2 of the general result: for r >= 0, the constant equals
C(r) + ((5/2)^r)*r!*Sum_{n >= 0} (-1)^n/((5*n + 1)*(5*n + 6)*...*(5*n + 5*r + 1)), where C(r) is the rational number (1/2)*Sum_{k = 0..r-1} (5/2)^k*k!/(1*6*11*...*(5*k + 1)). The general result can be proved by the WZ method as described in Wilf. (End)
From Peter Bala, Mar 03 2024: (Start)
Equals hypergeom([1/5, 1], [6/5], -1).
Gauss's continued fraction: 1/(1 + 1^2/(6 + 5^2/(11 + 6^2/(16 + 10^2/(21 + 11^2/(26 + 15^2/(31 + 16^2/(36 + 20^2/(41 + 21^2/(46 + ... )))))))))). (End)

A262246 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+2).

Original entry on oeis.org

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

Offset: 0

Gheorghe Coserea, Oct 06 2015

			0.4069016342...

Gheorghe Coserea, Table of n, a(n) for n = 0..2015
H. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics and Theoretical Computer Science 3(4) (1999), 189-192.

Cf. A003881, A024396, A113476, A181048, A181049, A181122, A193534.

N[(1/5)*((Sqrt[5]-1)*Log[2] + Sqrt[5]*Log[Sin[3*Pi/10]] + (Pi/2)*Sec[Pi/10]), 100] (* G. C. Greubel, Oct 07 2015 *) (* fixed by Vaclav Kotesovec, Dec 11 2017 *)

default(realprecision, 87);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(5*n+2)))), "3..-2"))

Sum_{n>=0} (-1)^n/(5n+2) = Integral_{x=0..1} x/(1+x^5)dx.
From G. C. Greubel, Oct 07 2015: (Start)
Sum_{n>=0} (-1)^n/(5n+2) = (1/5)*(sqrt(5)*log(phi) - log(2) + Pi*(5*phi^2)^(-1/4)), where 2*phi=1+sqrt(5).
Sum_{n>=0} (-1)^n/(5n+2) = (1/5)*(sqrt(5)*log(2*sin(3*Pi/10)) - log(2) + (Pi/2)*sec(Pi/10)).
(End)
Sum_{n>=0} (-1)^n/(5n+2) = (Psi(1/5) - Psi(7/10))/10 , see A200135 and A354643. - Robert Israel, Oct 08 2015
From Peter Bala, Feb 19 2024: (Start)
Equals (1/2)*Sum_{n >= 0} n!*(5/2)^n/(Product_{k = 0..n} 5*k + 2) = (1/2)*Sum_{n >= 0} n!*(5/2)^n/A047055(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(5*k + 2)).
Continued fraction: 1/(2 + 2^2/(5 + 7^2/(5 + 12^2/(5 + ... + (5*n + 2)^2/(5 + ... ))))).
The slowly converging series representation Sum_{n >= 0} (-1)^n/(5*n + 2) for the constant can be accelerated to give the following faster converging series
1/4 + (5/2)*Sum_{n >= 0} (-1)^n/((5*n + 2)*(5*n + 7)) and
19/56 + (5^2/2)*Sum_{n >= 0} (-1)^n/((5*n + 2)*(5*n + 7)*(5*n + 12)).
These two series are the cases r = 1 and r = 2 of the general result:
for r >= 0, the constant equals C(r) + ((5/2)^r)*r!*Sum_{n >= 0} (-1)^n/((5*n + 2)*(5*n + 7)*...*(5*n + 5*r + 2)), where C(r) is the rational number (1/2)*Sum_{k = 0..r-1} (5/2)^k*k!/(2*7*12*...*(5*k + 2)). The general result can be proved by the WZ method as described in Wilf. (End)
From Peter Bala, Mar 03 2024: (Start)
Equals (1/2)*hypergeom([2/5, 1], [7/5], -1).
Gauss's continued fraction: 1/(2 + 2^2/(7 + 5^2/(12 + 7^2/(17 + 10^2/(22 + 12^2/(27 + 15^2/(32 + 17^2/(37 + 20^2/(42 + 22^2/(47 + ... )))))))))). (End)

A237841 Decimal expansion of Ramanujan's AGM Continued Fraction R(2) = R_1(2,2).

Original entry on oeis.org

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

Offset: 0

Jean-François Alcover, Feb 14 2014

Other closed form evaluations of R(p/q):
R(1/4) = Pi/2 - 4/3,
R(1/3) = 1 - log(2),
R(1/2) = 2 - Pi/2,
R(2/3) = 4 - Pi/sqrt(2),
R(1) = log(2),
R(3/2) = Pi + sqrt(3)*log(2 - sqrt(3)),
R(3) = Pi/sqrt(3) - log(2).

			0.97499098879872209671990033452921084400592...

D. H. Bailey, J. M. Borwein, V. Kapoor and E. Weisstein, Ten Problems in Experimental Mathematics, page 13.
Jonathan M. Borwein, Ramanujan's Arithmetic-Geometric Mean Continued Fractions and Dynamics

Cf. A002162: R(1) = log(2); A180434: R(1/2) = 2-Pi/2.

RealDigits[Sqrt[2]*(Pi/2 - Log[1 + Sqrt[2]]), 10, 100] // First

(psi(7/8)-psi(3/8))/2 \\ Charles R Greathouse IV, Mar 03 2016

Equivalent formulas:
sqrt(2)*(Pi/2 - log(1 + sqrt(2))),
(Pi - 2*arccoth(sqrt(2)))/sqrt(2),
Integral_{x >= 0} sech(Pi*x/4)/(1 + x^2) dx,
2*Integral_{x = 0..1} sqrt(x)/(1 + x^2) dx,
Integral_{x >= 0} exp(-x/2)*sech(x) dx,
4*Sum_{k >= 1} (-1)^(k+1)/(4*k - 1),
1/2*(-psi(3/8) + psi(7/8)), where psi is the digamma function,
4/3 * 2F1(3/4, 1, 7/4, -1), where 2F1 is the hypergeometric function,
(H(-1/8) - H(-5/8))/2, where H(n) is the n-th harmonic number.
General formula:
The Borwein's closed form formula for R(n) with n integer simplifies to:
R(n) = Pi/2*sec(Pi/(2n)) - 2*sum( cos((k*(n+1)*Pi)/(2*n))*log(2*sin((k*Pi)/(4*n))), {k, 1, 2n-1, 2} ).
Equals 4*A181049. - Peter Bala, Apr 02 2024

cp's OEIS Frontend

A004767 a(n) = 4*n + 3.

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A093954 Decimal expansion of Pi/(2*sqrt(2)).

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A247719 Decimal expansion of Integral_{t=0..Pi/2} sqrt(tan(t)) dt.

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A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4*k+1).

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A196525 Decimal expansion of log(1+sqrt(2))/sqrt(2).

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A181122 Decimal expansion of Sum_{k>=0} (-1)^k/(5k+1).

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A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+1).