Search: a181049 -id:a181049
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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
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0,1
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COMMENTS
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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
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
Numbers n such that there are no primes p that satisfy the relationship p XOR n = p + n. - Brad Clardy, Jul 22 2012
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
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
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LINKS
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FORMULA
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a(n) = 2*a(n-1) - a(n-2) for n > 1, a(0) = 3, a(1) = 7. - Philippe Deléham, Nov 03 2008
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)
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EXAMPLE
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G.f. = 3 + 7*x + 11*x^2 + 15*x^3 + 19*x^4 + 23*x^5 + 27*x^6 + 31*x^7 + ...
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MAPLE
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seq( 3+4*n, n=0..100 );
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MATHEMATICA
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PROG
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(Haskell)
a004767 = (+ 3) . (* 4)
(PARI) Vec((3+x)/(1-x)^2 + O(x^200)) \\ Altug Alkan, Jan 15 2016
(Python) for n in range(0, 50): print(4*n+3, end=', ') # Stefano Spezia, Dec 12 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A093954
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Decimal expansion of Pi/(2*sqrt(2)).
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+10
32
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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
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1,5
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COMMENTS
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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
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REFERENCES
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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.
L. B. W. Jolley, Summation of Series, Dover (1961), eq 76 page 16.
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LINKS
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Eric Weisstein's World of Mathematics, Bifoliate.
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FORMULA
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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)
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).
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)
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)
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
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
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EXAMPLE
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1.11072073453959156175397...
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)
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)
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MAPLE
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simplify( sum((cos((1/2)*k*Pi)+sin((1/2)*k*Pi))/(2*k+1), k = 0 .. infinity) ); # Peter Bala, Mar 09 2015
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MATHEMATICA
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PROG
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(PARI) 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
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CROSSREFS
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STATUS
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approved
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A181048
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Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4*k+1).
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+10
11
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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
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REFERENCES
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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
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LINKS
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FORMULA
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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
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)
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)
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EXAMPLE
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0.86697298733991103757399516388287071365217536734524490433....
At N = 100000 the truncated series 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
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MATHEMATICA
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RealDigits[(Pi Sqrt[2])/8 + (Sqrt[2] Log[1 + Sqrt[2]])/4, 10, 100][[1]] (* Alonso del Arte, Aug 11 2011 *)
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PROG
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(PARI) (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) \\ G. C. Greubel, Jul 05 2017
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CROSSREFS
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Cf. A001586, A003881, A024396, A091648, A093954, A113476, A181049, A181122, A188458, A193534, A262246.
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approved
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A247719
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Decimal expansion of Integral_{t=0..Pi/2} sqrt(tan(t)) dt.
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+10
11
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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
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LINKS
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FORMULA
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Equals Pi/sqrt(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
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
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EXAMPLE
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2.22144146907918312350794049503034684930731...
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MATHEMATICA
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RealDigits[Pi/Sqrt[2], 10, 104] // First
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PROG
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(PARI) default(realprecision, 100); Pi/sqrt(2) \\ G. C. Greubel, Sep 07 2018
(Magma) SetDefaultRealField(RealField(100)); R:= RealField(); Pi(R)/Sqrt(2); // G. C. Greubel, Sep 07 2018
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CROSSREFS
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approved
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A196525
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Decimal expansion of log(1+sqrt(2))/sqrt(2).
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+10
8
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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
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FORMULA
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Equals Sum_{n>=1} A091337(n)/n = 1 - 1/3 - 1/5 + 1/7 + 1/9 - 1/11 - ...
Equals arcsinh(1)/sqrt(2).
Equals Sum_{n>=1} 1/A118417(n-1) = Sum_{n>=1} 1/((2*n - 1)*2^n). (End)
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)
Equals Sum_{k>=0} (-1)^k * (2*k)!!/(2*k+1)!!.
Equals Integral_{x=0..Pi/4} 1/(cos(x) + sin(x)) dx. (End)
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
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EXAMPLE
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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)
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MATHEMATICA
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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 *)
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PROG
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(Magma) SetDefaultRealField(RealField(100)); Log(Sqrt(2)+1)/Sqrt(2); // G. C. Greubel, Oct 05 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A181122
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Decimal expansion of Sum_{k>=0} (-1)^k/(5k+1).
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+10
3
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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
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OFFSET
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0,1
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LINKS
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FORMULA
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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
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)
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)
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EXAMPLE
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0.88831357265178863804075522702037934627811083077545817120597...
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MAPLE
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(int(1/(1+x^5), x=0..1));
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MATHEMATICA
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(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 *)
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PROG
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(PARI)
default(realprecision, 106);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(5*n+1)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A262246
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Decimal expansion of Sum_{k>=0} (-1)^k/(5k+2).
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+10
3
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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
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OFFSET
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0,1
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LINKS
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FORMULA
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Sum_{n>=0} (-1)^n/(5n+2) = Integral_{x=0..1} x/(1+x^5)dx.
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)
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)
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)
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EXAMPLE
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0.4069016342...
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MATHEMATICA
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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 *)
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PROG
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(PARI)
default(realprecision, 87);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(5*n+2)))), "3..-2"))
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A237841
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Decimal expansion of Ramanujan's AGM Continued Fraction R(2) = R_1(2,2).
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+10
0
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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
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OFFSET
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0,1
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COMMENTS
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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).
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LINKS
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FORMULA
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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} ).
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EXAMPLE
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0.97499098879872209671990033452921084400592...
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MATHEMATICA
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RealDigits[Sqrt[2]*(Pi/2 - Log[1 + Sqrt[2]]), 10, 100] // First
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PROG
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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