A181048 -id:A181048 - cp's OEIS frontend

A016813 a(n) = 4*n + 1.

1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 85, 89, 93, 97, 101, 105, 109, 113, 117, 121, 125, 129, 133, 137, 141, 145, 149, 153, 157, 161, 165, 169, 173, 177, 181, 185, 189, 193, 197, 201, 205, 209, 213, 217, 221, 225, 229, 233, 237

Offset: 0

N. J. A. Sloane

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0( 23 ).
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 64 ).
Numbers k such that k and (k+1) have the same binary digital sum. - Benoit Cloitre, Jun 05 2002
Numbers k such that (1 + sqrt(k))/2 is an algebraic integer. - Alonso del Arte, Jun 04 2012
Numbers k such that 2 is the only prime p that satisfies the relationship p XOR k = p + k. - Brad Clardy, Jul 22 2012
This may also be interpreted as the array T(n,k) = A001844(n+k) + A008586(k) read by antidiagonals:
    1,   9,  21,  37,  57,  81, ...
    5,  17,  33,  53,  77, 105, ...
   13,  29,  49,  73, 101, 133, ...
   25,  45,  69,  97, 129, 165, ...
   41,  65,  93, 125, 161, 201, ...
   61,  89, 121, 157, 197, 241, ...
   ...
- R. J. Mathar, Jul 10 2013
With leading term 2 instead of 1, 1/a(n) is the largest tolerance of form 1/k, where k is a positive integer, so that the nearest integer to (n - 1/k)^2 and to (n + 1/k)^2 is n^2. In other words, if interval arithmetic is used to square [n - 1/k, n + 1/k], every value in the resulting interval of length 4n/k rounds to n^2 if and only if k >= a(n). - Rick L. Shepherd, Jan 20 2014
Odd numbers for which the number of prime factors congruent to 3 (mod 4) is even. - Daniel Forgues, Sep 20 2014
For the Collatz conjecture, we identify two types of odd numbers. This sequence contains all the descenders: where (3*a(n) + 1) / 2 is even and requires additional divisions by 2. See A004767 for the ascenders. - Fred Daniel Kline, Nov 29 2014 [corrected by Jaroslav Krizek, Jul 29 2016]
a(n-1), n >= 1, is also the complex dimension of the manifold M(S), the set of all conjugacy classes of irreducible representations of the fundamental group pi_1(X,x_0) of rank 2, where S = {a_1, ..., a_{n}, a_{n+1} = oo}, a subset of P^1 = C U {oo}, X = X(S) = P^1 \ S, and x_0 a base point in X. See the Iwasaki et al. reference, Proposition 2.1.4. p. 150. - Wolfdieter Lang, Apr 22 2016
For n > 3, also the number of (not necessarily maximal) cliques in the n-sunlet graph. - Eric W. Weisstein, Nov 29 2017
For integers k with absolute value in A047202, also exponents of the powers of k having the same unit digit of k in base 10. - Stefano Spezia, Feb 23 2021
Starting with a(1) = 5, numbers ending with 01 in base 2. - John Keith, May 09 2022

			From _Leo Tavares_, Jul 02 2021: (Start)
Illustration of initial terms:
                                        o
                        o               o
            o           o               o
    o     o o o     o o o o o     o o o o o o o
            o           o               o
                        o               o
                                        o
(End)

K. Iwasaki, H. Kimura, S. Shimomura and M. Yoshida, From Gauss to Painlevé, Vieweg, 1991. p. 150.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Colin Defant and Noah Kravitz, Loops and Regions in Hitomezashi Patterns, arXiv:2201.03461 [math.CO], 2022. Theorem 1.3.
Gennady Eremin, Partitioning the set of natural numbers into Mersenne trees and into arithmetic progressions; Natural Matrix and Linnik's constant, arXiv:2405.16143 [math.CO], 2024. See pp. 5, 8, 14.
Gennady Eremin, Infinite matrix of odd natural numbers. A bit about Sophie Germain prime numbers, arXiv:2501.17090 [math.GM], 2025. See pp. 3, 11.
L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 16.
Tanya Khovanova, Recursive Sequences
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original German edition of "Theory and Application of Infinite Series")
Konrad Knopp, Theory and Application of Infinite Series, Dover, p. 269.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
William A. Stein's The Modular Forms Database, PARI-readable dimension tables for Gamma_0(N)
Eric Weisstein's World of Mathematics, Clique
Eric Weisstein's World of Mathematics, Hilbert Number
Eric Weisstein's World of Mathematics, Sunlet Graph
Wikipedia, Interval arithmetic
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Subsequence of A042963 and of A079523.
a(n) = A093561(n+1, 1), (4, 1)-Pascal column.
Cf. A016921, A017281, A017533, A047202, A158057, A161705, A161709, A161714, A128470. - Reinhard Zumkeller, Jun 17 2009
Cf. A004772 (complement).
Cf. A017557.

List([0..70],n->4*n+1); # Muniru A Asiru, Aug 08 2018

a016813 = (+ 1) . (* 4)
a016813_list = [1, 5 ..]  -- Reinhard Zumkeller, Feb 14 2012

Magma
```
[n: n in [1..250 by 4]];
```

seq(4*k+1, k=0..100); # Wesley Ivan Hurt, Sep 28 2013

Range[1, 237, 4] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)
Table[4 n + 1, {n, 0, 20}] (* Eric W. Weisstein, Nov 29 2017 *)
4 Range[0, 20] + 1 (* Eric W. Weisstein, Nov 29 2017 *)
LinearRecurrence[{2, -1}, {5, 9}, {0, 20}] (* Eric W. Weisstein, Nov 29 2017 *)
CoefficientList[Series[(1 + 3 x)/(-1 + x)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Nov 29 2017 *)

a(n)=4*n+1 \\ Charles R Greathouse IV, Mar 22 2013

x='x+O('x^100); Vec((1+3*x)/(1-x)^2) \\ Altug Alkan, Oct 22 2015

(0 to 59).map(4 *  + 1) // _Alonso del Arte, Aug 08 2018

a(n) = A005408(2*n).
Sum_{n>=0} (-1)^n/a(n) = (1/(4*sqrt(2)))*(Pi+2*log(sqrt(2)+1)) = A181048 [Jolley]. - Benoit Cloitre, Apr 05 2002 [corrected by Amiram Eldar, Jul 30 2023]
G.f.: (1+3*x)/(1-x)^2. - Paul Barry, Feb 27 2003 [corrected for offset 0 by Wolfdieter Lang, Oct 03 2014]
(1 + 5*x + 9*x^2 + 13*x^3 + ...) = (1 + 2*x + 3*x^2 + ...) / (1 - 3*x + 9*x^2 - 27*x^3 + ...). - Gary W. Adamson, Jul 03 2003
a(n) = A001969(n) + A000069(n). - Philippe Deléham, Feb 04 2004
a(n) = A004766(n-1). - R. J. Mathar, Oct 26 2008
a(n) = 2*a(n-1) - a(n-2); a(0)=1, a(1)=5. a(n) = 4 + a(n-1). - Philippe Deléham, Nov 03 2008
A056753(a(n)) = 3. - Reinhard Zumkeller, Aug 23 2009
A179821(a(n)) = a(A179821(n)). - Reinhard Zumkeller, Jul 31 2010
a(n) = 8*n - 2 - a(n-1) for n > 0, a(0) = 1. - Vincenzo Librandi, Nov 20 2010
The identity (4*n+1)^2 - (4*n^2+2*n)*(2)^2 = 1 can be written as a(n)^2 - A002943(n)*2^2 = 1. - Vincenzo Librandi, Mar 11 2009 - Nov 25 2012
A089911(6*a(n)) = 8. - Reinhard Zumkeller, Jul 05 2013
a(n) = A004767(n) - 2. - Jean-Bernard François, Sep 27 2013
a(n) = A058281(3n+1). - Eli Jaffe, Jun 07 2016
From Ilya Gutkovskiy, Jul 29 2016: (Start)
E.g.f.: (1 + 4*x)*exp(x).
a(n) = Sum_{k = 0..n} A123932(k).
a(A005098(k)) = x^2 + y^2.
Inverse binomial transform of A014480. (End)
Dirichlet g.f.: 4*Zeta(-1 + s) + Zeta(s). - Stefano Spezia, Nov 02 2018

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)

A091648 Decimal expansion of arccosh(sqrt(2)), the inflection point of sech(x).

Original entry on oeis.org

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

Offset: 0

Eric W. Weisstein, Jan 24 2004

Asymptotic growth constant in the exponent for the number of spanning trees on the 2 X infinity strip on the square lattice. - R. J. Mathar, May 14 2006
Arccosh(sqrt(2)) = (1/2)*log((sqrt(2)+1)/(sqrt(2)-1)) = log(tan(3*Pi/8)) = int(1/cos(x),x=0..Pi/4). Therefore, in Gerardus Mercator's (conformal) map this is the value of the ordinate y/R (R radius of the spherical earth) for latitude phi = 45 degrees north, or Pi/4. See, e.g., the Eli Maor reference, eqs. (5) and (6). This is the latitude of, e.g., the Mission Point Lighthouse, Michigan, U.S.A. - Wolfdieter Lang, Mar 05 2013
Decimal expansion of the arclength on the hyperbola y^2 - x^2 = 1 from (0,0) to (1,sqrt(2)). - Clark Kimberling, Jul 04 2020

			0.8813735870195430252326093249797923090281603282616...

L. B. W. Jolley, Summation of Series, Dover (1961), Eq. (85) page 16-17.
E. Maor, Trigonometric Delights, Princeton University Press, NJ, 1998, chapter 13, A Mapmaker's Paradise, pp. 163-180.
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 30, equation 30:10:4 at page 283.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
E. D. Krupnikov, K. S. Kölbig, Some special cases of the generalized hypergeometric function (q+1)Fq, J. Comp. Appl. Math. 78 (1997) 79-95.
D. H. Lehmer, Interesting Series Involving the Central Binomial Coefficient, Am. Math. Monthly 92 (1985) 449.
Michael I. Shamos, A catalog of the real numbers, (2007). See pp. 614-615.
R. Shrock and F. Y. Wu, Spanning trees on graphs and lattices in d dimensions, J Phys A: Math Gen 33 (2000) 3881-3902.
Eric Weisstein's World of Mathematics, Hyperbolic Secant.
Eric Weisstein's World of Mathematics, Universal Parabolic Constant.
Index entries for transcendental numbers.

Cf. A014176, A103710, A103711, A103712, A181048.

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

fpprec : 100$ ev(bfloat(log(1 + sqrt(2)))); /* Martin Ettl, Oct 17 2012 */

PARI
```
asinh(1) \\ Michel Marcus, Oct 19 2014
```

Equals log(1 + sqrt(2)). - Jonathan Sondow, Mar 15 2005
Equals (1/2)*log(3+2*sqrt(2)) = A244920/2. - R. J. Mathar, May 14 2006
Equals Sum_{n>=1, n odd} binomial(2*n,n)/(n*4^n) [see Lehmer link]. - R. J. Mathar, Mar 04 2009
Equals arcsinh(1), since arcsinh(x) = log(x+sqrt(x^2+1)). - Stanislav Sykora, Nov 01 2013
Equals asin(i)/i. - L. Edson Jeffery, Oct 19 2014
Equals (Pi/4) * 3F2(1/4, 1/2, 3/4; 1, 3/2; 1). - Jean-François Alcover, Apr 23 2015
Equals arctanh(sqrt(2)/2). - Amiram Eldar, Apr 22 2022
Equals lim_{n->oo} Sum_{k=1..n} 1/sqrt(n^2+k^2). - Amiram Eldar, May 19 2022
Equals Sum_{n >= 1} 1/(n*P(n, sqrt(2))*P(n-1, sqrt(2))), where P(n, x) denotes the n-th Legendre polynomial. The first twenty terms of the series gives the approximate value 0.88137358701954(24...), correct to 14 decimal places. - Peter Bala, Mar 16 2024
Equals 2F1(1/2,1/2;3/2;-1) [Krupnikov]. - R. J. Mathar, May 13 2024
Equals Integral_{x=0..1} (x^sqrt(2) - 1)/log(x) dx. - Kritsada Moomuang, Jun 06 2025

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

A181049 Decimal expansion of (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+3).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 01 2010, Oct 05 2010

Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*((Pi/2 - log(1 + sqrt(2)))/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3)) ~ 1/N - 1/N^2 - 3/N^3 + 11/N^4 + 57/N^5 - ..., where the sequence of coefficients [1, -1, -3, 11, 57, ...] is A212435. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 3/4. An example is given below. Cf. A181048. - Peter Bala, Sep 23 2016

			0.2437477471996805241799750836323027110...
From _Peter Bala_, Sep 23 2016: (Start)
At N = 100000 the truncated series 2*Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3) = 0.4874(8)5494(4)9936(4)048(24)99(444)67(625)6... to 32 digits. The bracketed numbers show where this decimal expansion differs from that of 2*A181049. 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*( Pi/2 - log(1+sqrt(2)))/(2*sqrt(2) ) = 0.4874(9)5494(3)9936(1)048(35)99(501)67(264)6.... (End)

Gheorghe Coserea, Table of n, a(n) for n = 0..2015
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.

Cf. A113476, A001586, A093954, A181048, A212435, A247719.

C := ComplexField(); [(Pi(C)/2 - Log(1+Sqrt(2)))/(2*Sqrt(2))]; // G. C. Greubel, Nov 28 2017

Mathematica

First@ RealDigits[N[(Pi/2 - Log[1 + Sqrt@ 2])/(2 Sqrt@ 2), 105]] (* Michael De Vlieger, Oct 07 2015 *)

PARI

default(realprecision, 106); eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+3)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015

PARI

(Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) \\ G. C. Greubel, Nov 28 2017

Formula

Equals Integral_{x=0..1} (x^2 dx)/(1+x^4).

Equals (1/2) * Integral_{x = 0..Pi/4} sqrt(tan(x)) dx. Cf. A247719. - Peter Bala, Sep 23 2016

Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 3) = Sum_{n >= 0} 2^(n-1)*n!/A008545(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k+3)). - Peter Bala, Dec 01 2021

From Peter Bala, Mar 03 2024: (Start)

Continued fraction: 1/(3 + 3^2/(4 + 7^2/(4 + 11^2/(4 + 15^2/(4 + ... ))))) due to Euler.

Equals (1/3)*hypergeom([3/4, 1], [7/4], -1).

Gauss's continued fraction: 1/(3 + 3^2/(7 + 4^2/(11 + 7^2/(15 + 8^2/(19 + 11^2/(23 + 12^2/(27 + 15^2/(31 + 16^2/(35 + 19^2/(39 + ... )))))))))). (End)

Equals Integral_{x=1..oo, y=1..oo} 1/(x^4 + y^4) dx. - Vaclav Kotesovec, Jun 13 2024

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

A193887 Decimal expansion of Pi * sqrt(2)/8.

Original entry on oeis.org

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

Offset: 0

Alonso del Arte, Aug 07 2011

This number arises as an addend in one way of giving the closed form of sum(k>=0, (-1)^k/(4*k + 1) ), for example, in Spiegel et al. (2009).

			0.55536036726979578088...

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

G. C. Greubel, Table of n, a(n) for n = 0..10000
Piotr Garbaczewski and Vladimir A. Stephanovich, Semigroup modeling of confined Levy flights, arXiv:1106.1530 [cond-mat.stat-mech], 2011, p. 8, equation 40.
Piotr Garbaczewski and Vladimir A. Stephanovich, Dynamics of Confined Lévy Flights in Terms of (Lévy) Semigroups, Acta Phys. Pol. B 43, 977-999 (2012). See p. 992.
Index entries for transcendental numbers.

Cf. A181048, A063448, A247719, A093954, A244976.

R:= RealField(); Pi(R)*Sqrt(2)/8; // G. C. Greubel, Feb 02 2018

RealDigits[(Pi Sqrt[2])/8, 10, 100][[1]]

Pi*sqrt(2)/8 \\ G. C. Greubel, Feb 02 2018

Equals Pi/(4*sqrt(2)).
Equals Sum_{k >= 0} (-1)^k * (4*k + 2)/((4*k + 1)*(4*k + 3)). - Peter Bala, Sep 21 2016
From Amiram Eldar, Aug 15 2020: (Start)
Equals Integral_{x=0..oo} 1/(x^2 + 8) dx.
Equals Integral_{x=0..oo} 1/(8*x^2 + 1) dx.
Equals Integral_{x=0..oo} 1/(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7) dx. (End)

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)

A024383 a(n) = s(1)s(2)...s(n)(1/s(1) - 1/s(2) + ... + c/s(n)), where c = (-1)^(n+1) and s(k) = 4*k - 3 for k = 1, 2, 3, ....

Original entry on oeis.org

1, 4, 41, 488, 8881, 176556, 4622745, 128838480, 4403082465, 157917434580, 6659489632905, 292097166060600, 14653855170875025, 759940716395000700, 44202442040567948025, 2645857155729629066400, 175060715455871850866625

Offset: 1

Clark Kimberling

Cf. A024199, A024217, A024396, A007696, A181048.

a := proc(n) option remember; if n = 0 then 1 elif n = 1 then 4 else 4*a(n-1) + (4*n - 3)^2*a(n-2) end if; end:
seq(a(n), n = 0..20);

Table[Product[4*k - 3, {k, 1, n}] * Sum[(-1)^(k+1)/(4*k - 3), {k, 1, n}], {n, 1, 20}] (* Vaclav Kotesovec, Jan 02 2020 *)

a(n) = prod(k=1, n, 4*k-3)*sum(k=1, n, (-1)^(k+1)/(4*k-3)); \\ Michel Marcus, Jul 06 2019

a(n) ~ (Pi^(3/2) + 2*sqrt(Pi)*log(1 + sqrt(2))) * 2^(2*n - 2) * n^(n - 1/4) / (Gamma(1/4) * exp(n)). - Vaclav Kotesovec, Jan 02 2020
From Peter Bala, Mar 21 2024: (Start)
a(n) = Product_{k = 0..n} (4*k + 1) * Sum_{k = 0..n} (-1)^k/(4*k + 1).
a(n) = 4*a(n-1) + (4*n - 3)^2*a(n-2) with a(0) = 1 and a(1) = 4.
b(n) := Product_{k = 0..n} (4*k + 1) = A007696(n+1) satisfies the same 3-term recurrence with b(0) = 1 and b(1) = 5, leading to the continued fraction expansion for the constant A181048 = Sum_{k >= 0} (-1)^k/(4*k + 1) = 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler. (End)

More terms from Sean A. Irvine, Jul 06 2019

cp's OEIS Frontend

A016813 a(n) = 4*n + 1.

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

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A091648 Decimal expansion of arccosh(sqrt(2)), the inflection point of sech(x).

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

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A181049 Decimal expansion of (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+3).

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

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A024383 a(n) = s(1)s(2)...s(n)(1/s(1) - 1/s(2) + ... + c/s(n)), where c = (-1)^(n+1) and s(k) = 4*k - 3 for k = 1, 2, 3, ....