A019670 -id:A019670 - cp's OEIS frontend

A007310 Numbers congruent to 1 or 5 mod 6.

1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 163, 167, 169, 173, 175

Offset: 1

C. Christofferson (Magpie56(AT)aol.com)

Numbers n such that phi(4n) = phi(3n). - Benoit Cloitre, Aug 06 2003
Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5-rough numbers. (Edited by M. F. Hasler, Nov 01 2014: merged with comments from Zak Seidov, Apr 26 2007 and Michael B. Porter, Oct 09 2009)
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 38 ).
Numbers k such that k mod 2 = 1 and (k+1) mod 3 <> 1. - Klaus Brockhaus, Jun 15 2004
Also numbers n such that the sum of the squares of the first n integers is divisible by n, or A000330(n) = n*(n+1)*(2*n+1)/6 is divisible by n. - Alexander Adamchuk, Jan 04 2007
Numbers n such that the sum of squares of n consecutive integers is divisible by n, because A000330(m+n) - A000330(m) = n*(n+1)*(2*n+1)/6 + n*(m^2+n*m+m) is divisible by n independent of m. - Kaupo Palo, Dec 10 2016
A126759(a(n)) = n + 1. - Reinhard Zumkeller, Jun 16 2008
Terms of this sequence (starting from the second term) are equal to the result of the expression sqrt(4!*(k+1) + 1) - but only when this expression yields integral values (that is when the parameter k takes values, which are terms of A144065). - Alexander R. Povolotsky, Sep 09 2008
For n > 1: a(n) is prime if and only if A075743(n-2) = 1; a(2*n-1) = A016969(n-1), a(2*n) = A016921(n-1). - Reinhard Zumkeller, Oct 02 2008
A156543 is a subsequence. - Reinhard Zumkeller, Feb 10 2009
Numbers n such that ChebyshevT(x, x/2) is not an integer (is integer/2). - Artur Jasinski, Feb 13 2010
If 12*k + 1 is a perfect square (k = 0, 2, 4, 10, 14, 24, 30, 44, ... = A152749) then the square root of 12*k + 1 = a(n). - Gary Detlefs, Feb 22 2010
A089128(a(n)) = 1. Complement of A047229(n+1) for n >= 1. See A164576 for corresponding values A175485(a(n)). - Jaroslav Krizek, May 28 2010
Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (with h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 6). Also a(n)^2 - 1 == 0 (mod 12). - Bruno Berselli, Nov 05 2010 - Nov 17 2010
Numbers n such that ( Sum_{k = 1..n} k^14 ) mod n = 0. (Conjectured) - Gary Detlefs, Dec 27 2011
From Peter Bala, May 02 2018: (Start)
The above conjecture is true. Apply Ireland and Rosen, Proposition 15.2.2. with m = 14 to obtain the congruence 6*( Sum_{k = 1..n} k^14 )/n = 7 (mod n), true for all n >= 1.  Suppose n is coprime to 6, then 6 is a unit in Z/nZ, and it follows from the congruence that ( Sum_{k = 1..n} k^14 )/n is an integer. On the other hand, if either 2 divides n or 3 divides n then the congruence shows that ( Sum_{k = 1..n} k^14 )/n cannot be integral. (End)
A126759(a(n)) = n and A126759(m) < n for m < a(n). - Reinhard Zumkeller, May 23 2013
(a(n-1)^2 - 1)/24 = A001318(n), the generalized pentagonal numbers. - Richard R. Forberg, May 30 2013
Numbers k for which A001580(k) is divisible by 3. - Bruno Berselli, Jun 18 2014
Numbers n such that sigma(n) + sigma(2n) = sigma(3n). - Jahangeer Kholdi and Farideh Firoozbakht, Aug 15 2014
a(n) are values of k such that Sum_{m = 1..k-1} m*(k-m)/k is an integer. Sums for those k are given by A062717. Also see Detlefs formula below based on A062717. - Richard R. Forberg, Feb 16 2015
a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)/6 is integral, and also such that the sequence c(n) = n*(n + m)*(n + 2*m)*(n + 3*m)/24 is integral. Cf. A007775. - Peter Bala, Nov 13 2015
Along with 2, these are the numbers k such that the k-th Fibonacci number is coprime to every Lucas number. - Clark Kimberling, Jun 21 2016
This sequence is the Engel expansion of 1F2(1; 5/6, 7/6; 1/36) + 1F2(1; 7/6, 11/6; 1/36)/5. - Benedict W. J. Irwin, Dec 16 2016
The sequence a(n), n >= 4 is generated by the successor of the pair of polygonal numbers {P_s(4) + 1, P_(2*s - 1)(3) + 1}, s >= 3. - Ralf Steiner, May 25 2018
The asymptotic density of this sequence is 1/3. - Amiram Eldar, Oct 18 2020
Also, the only vertices in the odd Collatz tree A088975 that are branch values to other odd nodes t == 1 (mod 2) of A005408. - Heinz Ebert, Apr 14 2021
From Flávio V. Fernandes, Aug 01 2021: (Start)
For any two terms j and k, the product j*k is also a term (the same property as p^n and smooth numbers).
From a(2) to a(phi(A033845(n))), or a((A033845(n))/3), the terms are the totatives of the A033845(n) itself. (End)
Also orders n for which cyclic and semicyclic diagonal Latin squares exist (see A123565 and A342990). - Eduard I. Vatutin, Jul 11 2023
If k is in the sequence, then k*2^m + 3 is also in the sequence, for all m > 0. - Jules Beauchamp, Aug 29 2024

			G.f. = x + 5*x^2 + 7*x^3 + 11*x^4 + 13*x^5 + 17*x^6 + 19*x^7 + 23*x^8 + ...

K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1980.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Andreas Enge, William Hart, and Fredrik Johansson, Short addition sequences for theta functions, arXiv:1608.06810 [math.NT], 2016-2018.
B. W. J. Irwin, Constants Whose Engel Expansions are the k-rough Numbers.
L. B. W. Jolley, Summation of Series, Dover, 1961
Cedric A. B. Smith, Prime factors and recurring duodecimals, Math. Gaz. 59 (408) (1975) 106-109.
William A. Stein's The Modular Forms Database, PARI-readable dimension tables for Gamma_0(N).
Eric Weisstein's World of Mathematics, Rough Number.
Eric Weisstein's World of Mathematics, Pi Formulas. [_Jaume Oliver Lafont_, Oct 23 2009]
Index entries for sequences related to smooth numbers [_Michael B. Porter_, Oct 09 2009]
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

A005408 \ A016945. Union of A016921 and A016969; union of A038509 and A140475. Essentially the same as A038179. Complement of A047229. Subsequence of A186422.
Cf. A000330, A001580, A002194, A019670, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885-A175887.
For k-rough numbers with other values of k, see A000027, A005408, A007775, A008364-A008366, A166061, A166063.
Cf. A126760 (a left inverse).
Row 3 of A260717 (without the initial 1).
Cf. A105397 (first differences).

Filtered([1..150],n->n mod 6=1 or n mod 6=5); # Muniru A Asiru, Dec 19 2018

a007310 n = a007310_list !! (n-1)
a007310_list = 1 : 5 : map (+ 6) a007310_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..250] | n mod 6 in [1, 5]]; // Vincenzo Librandi, Feb 12 2016

seq(seq(6*i+j,j=[1,5]),i=0..100); # Robert Israel, Sep 08 2014

Select[Range[200], MemberQ[{1, 5}, Mod[#, 6]] &] (* Harvey P. Dale, Aug 27 2013 *)
a[n_] := (6 n + (-1)^n - 3)/2; a[rem156, 60] (* Robert G. Wilson v, May 26 2014 from a suggestion by N. J. A. Sloane *)
Flatten[Table[6n + {1, 5}, {n, 0, 24}]] (* Alonso del Arte, Feb 06 2016 *)
Table[2*Floor[3*n/2] - 1, {n, 1000}] (* Mikk Heidemaa, Feb 11 2016 *)

isA007310(n) = gcd(n,6)==1 \\ Michael B. Porter, Oct 09 2009

A007310(n)=n\2*6-(-1)^n \\ M. F. Hasler, Oct 31 2014

\\ given an element from the sequence, find the next term in the sequence.
nxt(n) = n + 9/2 - (n%6)/2 \\ David A. Corneth, Nov 01 2016

def A007310(n): return (n+(n>>1)<<1)-1 # Chai Wah Wu, Oct 10 2023

[i for i in range(150) if gcd(6,i) == 1] # Zerinvary Lajos, Apr 21 2009

a(n) = (6*n + (-1)^n - 3)/2. - Antonio Esposito, Jan 18 2002
a(n) = a(n-1) + a(n-2) - a(n-3), n >= 4. - Roger L. Bagula
a(n) = 3*n - 1 - (n mod 2). - Zak Seidov, Jan 18 2006
a(1) = 1 then alternatively add 4 and 2. a(1) = 1, a(n) = a(n-1) + 3 + (-1)^n. - Zak Seidov, Mar 25 2006
1 + 1/5^2 + 1/7^2 + 1/11^2 + ... = Pi^2/9 [Jolley]. - Gary W. Adamson, Dec 20 2006
For n >= 3 a(n) = a(n-2) + 6. - Zak Seidov, Apr 18 2007
From R. J. Mathar, May 23 2008: (Start)
Expand (x+x^5)/(1-x^6) = x + x^5 + x^7 + x^11 + x^13 + ...
O.g.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^2). (End)
a(n) = 6*floor(n/2) - 1 + 2*(n mod 2). - Reinhard Zumkeller, Oct 02 2008
1 + 1/5 - 1/7 - 1/11 + + - - ... = Pi/3 = A019670 [Jolley eq (315)]. - Jaume Oliver Lafont, Oct 23 2009
a(n) = ( 6*A062717(n)+1 )^(1/2). - Gary Detlefs, Feb 22 2010
a(n) = 6*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), with n > 1. - Bruno Berselli, Nov 05 2010
a(n) = 6*n - a(n-1) - 6 for n>1, a(1) = 1. - Vincenzo Librandi, Nov 18 2010
Sum_{n >= 1} (-1)^(n+1)/a(n) = A093766 [Jolley eq (84)]. - R. J. Mathar, Mar 24 2011
a(n) = 6*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
a(n) = 3*n + ((n+1) mod 2) - 2. - Gary Detlefs, Jan 08 2012
a(n) = 2*n + 1 + 2*floor((n-2)/2) = 2*n - 1 + 2*floor(n/2), leading to the o.g.f. given by R. J. Mathar above. - Wolfdieter Lang, Jan 20 2012
1 - 1/5 + 1/7 - 1/11 + - ... = Pi*sqrt(3)/6 = A093766 (L. Euler). - Philippe Deléham, Mar 09 2013
1 - 1/5^3 + 1/7^3 - 1/11^3 + - ... = Pi^3*sqrt(3)/54 (L. Euler). - Philippe Deléham, Mar 09 2013
gcd(a(n), 6) = 1. - Reinhard Zumkeller, Nov 14 2013
a(n) = sqrt(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5)/sqrt(2). - Alexander R. Povolotsky, May 16 2014
a(n) = 3*n + 6/(9*n mod 6 - 6). - Mikk Heidemaa, Feb 05 2016
From Mikk Heidemaa, Feb 11 2016: (Start)
a(n) = 2*floor(3*n/2) - 1.
a(n) = A047238(n+1) - 1. (suggested by Michel Marcus) (End)
E.g.f.: (2 + (6*x - 3)*exp(x) + exp(-x))/2. - Ilya Gutkovskiy, Jun 18 2016
From Bruno Berselli, Apr 27 2017: (Start)
a(k*n) = k*a(n) + (4*k + (-1)^k - 3)/2 for k>0 and odd n, a(k*n) = k*a(n) + k - 1 for even n. Some special cases:
k=2: a(2*n) = 2*a(n) + 3 for odd n, a(2*n) = 2*a(n) + 1 for even n;
k=3: a(3*n) = 3*a(n) + 4 for odd n, a(3*n) = 3*a(n) + 2 for even n;
k=4: a(4*n) = 4*a(n) + 7 for odd n, a(4*n) = 4*a(n) + 3 for even n;
k=5: a(5*n) = 5*a(n) + 8 for odd n, a(5*n) = 5*a(n) + 4 for even n, etc. (End)
From Antti Karttunen, May 20 2017: (Start)
a(A273669(n)) = 5*a(n) = A084967(n).
a((5*n)-3) = A255413(n).
A126760(a(n)) = n. (End)
a(2*m) = 6*m - 1, m >= 1; a(2*m + 1) = 6*m + 1, m >= 0. - Ralf Steiner, May 17 2018
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(3) (A002194).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi/3 (A019670). (End)

A002388 Decimal expansion of Pi^2.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

Also equals the volume of revolution of the sine or cosine curve for one full period, Integral_{x=0..2*Pi} sin(x)^2 dx. - Robert G. Wilson v, Dec 15 2005

Equals Sum_{n>0} 20/A026424(n)^2 where A026424 are the integers such that the number of prime divisors (counted with multiplicity) is odd. - Michel Lagneau, Oct 23 2015

			9.869604401089358618834490999876151135313699407240790626413349376220044...

W. E. Mansell, Tables of Natural and Common Logarithms. Royal Society Mathematical Tables, Vol. 8, Cambridge Univ. Press, 1964, p. XVIII.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 76.

Harry J. Smith, Table of n, a(n) for n = 1..20000
Mohammad K. Azarian, Al-Risala al-Muhitiyya: A Summary (The Treatise on the Circumference), Missouri Journal of Mathematical Sciences, Vol. 22, No. 2, 2010, pp. 64-85.
D. H. Bailey and J. M. Borwein, Experimental Mathematics: Examples, Methods and Implications, Notices AMS, 52 (No. 5 2005), 502-514.
David H. Bailey, Jonathan M. Borwein, Andrew Mattingly, and Glenn Wightwick, The Computation of Previously Inaccessible Digits of (Pi)^2 and Catalan's Constant, Notices AMS, 60 (No. 7 2013), 844-854.
N. D. Elkies, Why is (Pi)^2 so close to 10?
Melissa Larson, Verifying and discovering BBP-type formulas, 2008.
Simon Plouffe, Pi^2 to 10000 digits.
Simon Plouffe, Plouffe's Inverter, Pi^2 to 10000 digits.
Wikipedia, Bailey-Borwein-Plouffe formula.
Herbert S. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics and Theoretical Computer Science, Vol 3, No 4 (1999).
Index entries for sequences related to the number Pi.
Index entries for transcendental numbers.

Cf. A102753, A058284, A019670, A091925, A093954, A093602, A304656.

R:= RealField(100); Pi(R)^2; // G. C. Greubel, Mar 08 2018

Digits:=100: evalf(Pi^2); # Wesley Ivan Hurt, Jul 13 2014

RealDigits[Pi^2, 10, 111][[1]] (* Robert G. Wilson v, Dec 15 2005 *)

default(realprecision, 20080); x=Pi^2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b002388.txt", n, " ", d)); \\ Harry J. Smith, May 31 2009

# Use some guard digits when computing.
# BBP formula (9 / 8) P(2, 64, 6, (16, -24, -8, -6,  1, 0)).
from decimal import Decimal as dec, getcontext
def BBPpi2(n: int) -> dec:
    getcontext().prec = n
    s = dec(0); f = dec(1); g = dec(64)
    for k in range(int(n * 0.5536546824812272) + 1):
        sixk = dec(6 * k)
        s += f * ( dec(16) / (sixk + 1) ** 2 - dec(24) / (sixk + 2) ** 2
                 - dec(8)  / (sixk + 3) ** 2 - dec(6)  / (sixk + 4) ** 2
                 + dec(1)  / (sixk + 5) ** 2 )
        f /= g
    return (s * dec(9)) / dec(8)
print(BBPpi2(200))  # Peter Luschny, Nov 03 2023

Pi^2 = 11/2 + 16 * Sum_{k>=2} (1+k-k^3)/(1-k^2)^3. - Alexander R. Povolotsky, May 04 2009
Pi^2 = 3*(Sum_{n>=1} ((2*n+1)^2/Sum_{k=1..n} k^3)/4 - 1). - Alexander R. Povolotsky, Jan 14 2011
Pi^2 = (3/2)*(Sum_{n>=1} ((7*n^2+2*n-2)/(2*n^2-1)/(n+1)^5) - zeta(3) - 3*zeta(5) + 22 - 7*polygamma(0,1-1/sqrt(2)) + 5*sqrt(2)*polygamma(0,1-1/sqrt(2)) - 7*polygamma(0,1+1/sqrt(2)) - 5*sqrt(2)*polygamma(0,1+1/sqrt(2)) - 14*EulerGamma). - Alexander R. Povolotsky, Aug 13 2011
Also equals 32*Integral_{x=0..1} arctan(x)/(1+x^2) dx. - Jean-François Alcover, Mar 25 2013
From Peter Bala, Feb 05 2015: (Start)
Pi^2 = 20 * Integral_{x = 0 .. log(phi)} x*coth(x) dx, where phi = (1/2)*(1 + sqrt(5)) is the golden ratio.
Pi^2 = 10 * Sum_{k >= 0} binomial(2*k,k)*(1/(2*k + 1)^2)*(-1/16)^k. Similar series expansions hold for Pi/3 (see A019670) and (7*/216)*Pi^3 (see A091925).
The integer sequences A(n) := 2^n*(2*n + 1)!^2/n! and B(n) := A(n)*( Sum_{k = 0..n} binomial(2*k,k)*1/(2*k + 1)^2*(-1/16)^k ) both satisfy the second order recurrence equation u(n) = (24*n^3 + 44*n^2 + 2*n + 1)*u(n-1) + 8*(n - 1)*(2*n - 1)^5*u(n-2). From this observation we can obtain the continued fraction expansion Pi^2/10 = 1 - 1/(72 + 8*3^5/(373 + 8*2*5^5/(1051 + ... + 8*(n - 1)*(2*n - 1)^5/((24*n^3 + 44*n^2 + 2*n + 1) + ... )))). Cf. A093954. (End)
Pi^2 = A304656 * A093602 = (gamma(0, 1/6) - gamma(0, 5/6))*(gamma(0, 2/6) - gamma(0, 4/6)), where gamma(n,x) are the generalized Stieltjes constants. This formula can also be expressed by the polygamma function. - Peter Luschny, May 16 2018
Equals 8 + Sum_{k>=1} 1/(k^2 - 1/4)^2 = -8 + Sum_{k>=0} 1/(k^2 - 1/4)^2. - Amiram Eldar, Aug 21 2020
From Peter Bala, Dec 10 2021: (Start)
Pi^2 = (2^6)*Sum_{n >= 1} n^2/(4*n^2 - 1)^2 = (2^11)*Sum_{n >= 1} n^2/ ((4*n^2 - 1)^2*(4*n^2 - 3^2)^2) = ((2^19)*(3^2)/7) * Sum_{n >= 1} n^2/((4*n^2 - 1)^2*(4*n^2 - 3^2)^2*(4*n^2 - 5^2)^2).
More generally, it appears that for k >= 0 we have Pi^2 = (2*k+1)*2^(4*k+6) * (2*k)!^4/(4*k)! * Sum_{n >= 1} n^2/((4*n^2 - 1)^2*...*(4*n^2 - (2*k+1)^2)^2).
It also appears that for k >= 0 we have Pi^2 = (-1)^k * 2^(6*k+8)*(2*k+1)^3/(6*k+1) * ((2*k)!^6 * (3*k)!)/(k!^3 * (6*k)!) * Sum_{n >= 1} n^2/((4*n^2 - 1)^3*...*(4*n^2 - (2*k+1)^2)^3). (End)
From Peter Bala, Oct 27 2023: (Start)
Pi^2 = 10 - Sum_{n >= 1} 1/(n*(n + 1))^3.
Pi^2 = 6217/630 + (648/35)*Sum_{n >= 1} 1/(n*(n + 1)*(n + 2)*(n + 3))^3.
The general result (verified using the WZ method - see Wilf) is : for n >= 0,
Pi^2 = A(n) + (-1)^(n+1) * B(n)*Sum_{k >= 1} 1/(k*(k + 1)*...*(k + 2*n + 1))^3, where A(n) = 10 - Sum_{i = 1..n} (-1)^(i+1) * (56*i^2 + 24*i + 3)*(2*i)!^3*(3*i)!/(2*i^2*(2*i + 1)*(6*i + 1)!*i!^3) and B(n) = (2*n + 1)!^6 * (3*n)! / ( (2*n + 1)*(6*n + 1)!*n!^3 ).
Letting n -> oo gives the fast converging alternating series
Pi^2 = 10 - Sum_{i >= 1} (-1)^(i+1) * (56*i^2 + 24*i + 3)*(2*i)!^3 * (3*i)!/(2*i^2*(2*i + 1)*(6*i + 1)!*i!^3). The i-th summand of the series is asymptotic to (14/3) * 1/(i^2 * 27^i) so taking 70 terms of the series gives a value for Pi^2 accurate to more than 100 decimal places.
The series representation Pi^2 = 3*Sum_{k >= 1} (2*k)/k^3 can be accelerated to give the faster converging series
Pi^2 = 99/10 - (8/5)*Sum_{k >= 1} (2*k + 2)/(k*(k + 1)*(k + 2))^3 and
Pi^2 = 54715/5544 + (41472/385)*Sum_{k >= 1} (2*k + 4)/(k*(k + 1)*(k + 2)*(k + 3)*(k + 4))^3.
The general result is: for n >= 1, Pi^2 = C(n) + (-1)^n * D(n)*Sum_{k >= 1} (2*k + 2*n)/(k*(k + 1)*...*(k + 2*n))^3, where C(n) = A(n) - 10*(-1)^n*(3*n)!*(2*n)!^3/((2*n + 1)*n!^3*(6*n + 1)!) and D(n) = (2*n)!^6 * (3*n)! / ( 2*n*(6*n - 1)!*n!^3 ). (End)
Equals 9 + 3*Sum_{n>=1} 1/((n^2*(n+1)^2)). - Davide Rotondo, May 29 2025

More terms from Robert G. Wilson v, Dec 15 2005

A197476 Decimal expansion of least x>0 having cos(x) = cos(2*x)^2.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Oct 15 2011

The Mathematica program includes a graph.  Guide for least x>0 satisfying cos(b*x) = cos(c*x)^2, for selected b and c:
b.....c......x
....2.......A197476
....3.......A197477
....4.......A197478
....1.......A000796, Pi
....3.......A197479
....4.......A197480
....1.......A019669, Pi/2
....2.......A197482
....4.......A197483
....1.......A168229, arctan(sqrt(7))
....2.......A019669, Pi/2
....3.......A019679, Pi/12
....6.......A197485
....8.......A197486
....2.......A003881, Pi/4
....3.......A019670, Pi/3, tangency point
....4.......A197488
....8.......A197489
....4*Pi....A197334
....3*Pi....A197335
....2*Pi....A197490
....3*Pi/2..A197491
....Pi......A197492
....Pi/2....A197493
....Pi/3....A197494
....Pi/4....A197495
....2*Pi/3..A197506
....3*Pi....A197507
....3*Pi/2..A197508
....2*Pi....A197509
....Pi......A197510
....Pi/2....A197511
....Pi/3....A197512
....Pi/4....A197513
....Pi/6....A197514
Pi....1.......A197515
Pi....2.......A197516
Pi....1/2.....A197517
2*Pi..1.......A197518
2*Pi..2.......A197519
2*Pi..3.......A197520
Pi/2..Pi/3....A197521
Pi/2..Pi/6....3
Pi/3..1.......A197582
Pi/3..2.......A197583
Pi/3..1/3.....A197584
See A197133 for a guide for least x>0 satisfying sin(b*x) = sin(c*x)^2 for selected b and c.

			1.137743932905455557789449860055008349584...

Index entries for transcendental numbers.

Cf. A197133.

b = 1; c = 2; f[x_] := Cos[x]
t = x /. FindRoot[f[b*x] == f[c*x]^2, {x, 1.1, 1.3}, WorkingPrecision -> 200]
RealDigits[t] (* A197476 *)
Plot[{f[b*x], f[c*x]^2}, {x, 0, 2}]
(* or *)
RealDigits[ ArcCos[ ((19 - 3*Sqrt[33])^(1/3) + (19 + 3*Sqrt[33])^(1/3) - 2)/6], 10, 99] // First (* Jean-François Alcover, Feb 19 2013 *)

Edited by Georg Fischer, Jul 28 2021

A091925 Decimal expansion of Pi^3.

Original entry on oeis.org

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

Offset: 2

Mohammad K. Azarian, Mar 16 2004

Surface area of the 6-dimensional unit sphere. - Stanislav Sykora, Nov 08 2013
Surface area of a sphere of diameter Pi equals the volume of the circumscribed cube. - Omar E. Pol, Dec 25 2013
Area of a circle of radius Pi. - Omar E. Pol, Jan 31 2016

			31.00627668029982017547631506710139520222528856588510769414453810380639...

Harry J. Smith, Table of n, a(n) for n = 2..20000
G. L. Honaker, Jr. and Chris Caldwell, Prime Curios! 31
Jean-Christophe Pain, Series representations for Pi^3 involving the golden ratio, arXiv:2206.15281 [math.NT], 2022.
Khodabakhsh Hessami Pilehrood, Tatiana Hessami Pilehrood, Series acceleration formulas for beta values, Discrete Mathematics & Theoretical Computer Science 12:2 (2010), pp. 223-236.
Stanislav Sykora, Surface Integrals over n-Dimensional Spheres.
Index entries for transcendental numbers

Cf. A000796, A002388, A058285 (continued fraction), A019670, A093954, A092731, A092735.

R:= RealField(100); (Pi(R))^3; // G. C. Greubel, Mar 09 2018

First@ RealDigits@ N[Pi^3, 120] (* Michael De Vlieger, Jan 31 2016 *)

default(realprecision, 20080); x=Pi^3/10; for (n=2, 20000, d=floor(x); x=(x-d)*10; write("b091925.txt", n, " ", d)); \\ Harry J. Smith, Jun 22 2009

Sum_{k >= 0} binomial(2*k,k)/((2*k + 1)^3*16^k) = 7*Pi^3/216. (Kh. Hessami Pilehrood and T. Hessami Pilehrood).
From Peter Bala, Feb 05 2015: (Start)
The integer sequences A(n) := 2^n*(2*n + 1)!^3/n!^2 and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)^3*(1/16)^k ) both satisfy the second order recurrence equation u(n) = (160*n^4 + 128*n^3 + 144*n^2 + 2)*u(n-1) - 32*(n - 1)*(2*n - 1)^7*u(n-2). From this observation we can obtain the continued fraction expansion 7/216*Pi^3 = 1 + 2/(432 - 32*3^7/(4162 - 32*2*5^7/(17714 - ... - 32*(n - 1)*(2*n - 1)^7/((160*n^4 + 128*n^3 + 144*n^2 + 2) - ... )))). Cf. A002388, A019670 and A093954. (End)
From Peter Bala, Oct 31 2019: (Start)
Pi^3 = (1/7) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/6)^3 + 1/(n + 5/6)^3 ).
Pi^3 = (1/31) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/10)^3 - 1/(n + 3/10)^3 - 1/(n + 7/10)^3 + 1/(n + 9/10)^3 ). Cf. A019692, A092731 and A092735. (End)
Equals Integral_{x=-oo..oo} x^2/(exp(x/2) + exp(-x/2)) dx. - Amiram Eldar, May 21 2021

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)

A197739 Decimal expansion of least x>0 having sin(2x)=3*sin(6x).

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Oct 18 2011

This constant is the least x>0 for which the function f(x)=(sin(x))^2+(cos(3x))^2 has its maximal value.  Least positive solutions of the equations f(x)=m/2, f(x)=m/3, f(x)=1, and f(x)=1/2 are given by sequences shown in the guide below.
In general, suppose that b and c are distinct positive real numbers.  Let f(x)=(sin(bx))^2+cos((cx))^2.  The extrema of f are the solutions of b*sin(2bx)=c*sin(2cx).
In the following guide, constants x given by the sequences (or explicit number) listed for each b,c are, in this order:
(1) least x>0 such that f(x)=(its maximum, m)
(2) m, the maximum of f
(3) least x>0 having f(x)=m/2
(4) least x>0 having f(x)=m/3
(5) least x>0 having f(x)=1
(6) least x>0 having f(x)=1/2
...
(b,c)=(1,2):  A195700, x=25/16, A197589, A197591,
  A019670, A197592
(b,c)=(1,3):  A197739, A197588, A197590, A197755,
  A003881, A197488
(b,c)=(1,4):  A197758, A197759, A197760, A197761,
  A019692 (x=pi/5), A003881
(b,c)=(1,pi): A197821, A197822, A197823, A197824,
  A197726, A197826
(b,c)=(1,2*pi): A197827, A197828, A197829, A197830,
  A197700, A197832
(b,c)=(1,3*pi): A197833, A197834, A197835, A197836,
  A197837, A197838

			0.47765830906225463908192855125787887712170734750500...

Index entries for transcendental numbers.

Cf. A197739, A197588.

b = 1; c = 3;
f[x_] := Cos[b*x]^2; g[x_] := Sin[c*x]^2; s[x_] := f[x] + g[x];
r = x /. FindRoot[b*Sin[2 b*x] == c*Sin[2 c*x], {x, .47, .48}, WorkingPrecision -> 110]
RealDigits[r]  (* A197739 *)
m = s[r]
RealDigits[m]  (* A197588 *)
Plot[{b*Sin[2 b*x], c*Sin[2 c*x]}, {x, 0, Pi}]
d = m/2; t = x /. FindRoot[s[x] == d, {x, 0.7, 0.8}, WorkingPrecision -> 110]
RealDigits[t]  (* A197590 *)
Plot[{s[x], d}, {x, 0, Pi}, AxesOrigin -> {0, 0}]
d = m/3; t = x /. FindRoot[s[x] == d, {x, 0.8, 0.9}, WorkingPrecision -> 110]
RealDigits[t]  (* A197755 *)
Plot[{s[x], d}, {x, 0, Pi}, AxesOrigin -> {0, 0}]
d = 1; t = x /. FindRoot[s[x] == d, {x, 0.7, 0.8}, WorkingPrecision -> 110]
RealDigits[t]  (* A003881 *)
Plot[{s[x], d}, {x, 0, Pi}, AxesOrigin -> {0, 0}]
d = 1/2; t = x /. FindRoot[s[x] == d, {x, .9, .93}, WorkingPrecision -> 110]
RealDigits[t]  (* A197488 *)
Plot[{s[x], d}, {x, 0, Pi}, AxesOrigin -> {0, 0}]
RealDigits[ ArcTan[ Sqrt[ 2-Sqrt[3] ] ], 10, 99] // First (* Jean-François Alcover, Feb 27 2013 *)

A019675 Decimal expansion of Pi/8.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Equals Integral_{x>=0} sin(4*x)/(4*x) dx. - Jean-François Alcover, Feb 28 2013

Consider 4 circles inscribed in a square. Inscribe a square in each circle. And finally, inscribe 4 circles inside each four small squares. Totally we get 16 small circles. Pi/8 is the ratio of the area of the 16 small circles to the area of initial square. See the link. - Kirill Ustyantsev, Apr 30 2020

			Pi/8 = 0.392699081698724154807830422909937860524646174921888227621868... - _Vladimir Joseph Stephan Orlovsky_, Dec 02 2009

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.4, p. 492.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Kirill Ustyantsev, Geometric sense of Pi/8.
Index entries for transcendental numbers.

Cf. A195909, A195913, A195697. - Mohammad K. Azarian, Oct 11 2011
Cf. A000796, A003881, A019670, A024235, A244978, A334422 (Pi/128).
Cf. A001539, A061550.

pi:=Pi(RealField(110)); Reverse(Intseq(Floor(10^100*(pi)/8))); // Vincenzo Librandi, Oct 07 2015

RealDigits[N[Pi/8,6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)

default(realprecision, 1002);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+2)))), "3..-2"))  \\ Gheorghe Coserea, Oct 06 2015

From Peter Bala, Nov 15 2016: (Start)
Pi/8 = Sum_{k >= 1} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)).
More generally, for n >= 0 we have 1/(2*n)! * Pi/4 = Sum_{k >= 1} (-1)^(k+n-1) * 1/Product_{j = -n..n} (2*k + 2*j - 1): when n = 0 we get the Madhava-Gregory-Leibniz series for Pi/4.
For N divisible by 4, we have the asymptotic expansion Pi/8 - Sum_{k = 1..N/2} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)) ~ -1/2*(1/N^3 - 2/N^5 + 31/N^7 - 692/N^9 + ...), where the sequence of unsigned coefficients [1, 2, 31, 692, ...] equals A024235. (End)
Equals Integral_{x = 0..1} x*sqrt(1 - x^4) dx. - Peter Bala, Oct 27 2019
Equals Integral_{x = 0..oo} sin(x)^6/x^4 dx = Sum_{n >= 1} sin(n)^6/n^4, by the Abel-Plana formula. - Peter Bala, Nov 04 2019
From Amiram Eldar, Jul 12 2020: (Start)
Equals arctan(sqrt(2) - 1).
Equals Sum_{k>=0} (-1)^k/(4*k+2).
Equals Sum_{k>=0} 1/((4*k+1)*(4*k+3)) = Sum_{k>=0} 1/A001539(k).
Equals Integral_{x=0..oo} dx/(x^2 + 16).
Equals Integral_{x=0..oo} dx/(x^4 + 4) = Integral_{x=0..oo} x/(x^4 + 4) dx.
Equals Integral_{x=0..oo} x/(x^4 + 1)^2 dx = Integral_{x=0..1} x/(x^4 + 1) dx.
Equals Integral_{x=0..1} x * arcsin(x) dx. (End)
From Kritsada Moomuang, Jun 18 2025: (Start)
Equals Integral_{x=0..oo} (x*log(x + 1))/((x^2 + 1)^2) dx.
Equals Integral_{x=0..oo} (x^3 - 3*x + 3*arctan(x))/(3*x^5) dx. (End)

A093825 Decimal expansion of Pi/(3*sqrt(2)).

Original entry on oeis.org

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

Offset: 0

Eric W. Weisstein, Apr 16 2004

Density of densest packing of equal spheres in three dimensions (achieved for example by the fcc lattice).

Atomic packing factor (APF) of the face-centered-cubic (fcc) and the hexagonal-close-packed (hcp) crystal lattices filled with spheres of the same diameter. - Stanislav Sykora, Sep 29 2014

			0.74048048969306104116931349834344894976910361489594837...

J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer, 3rd. ed., 1998. See p. 15, line n = 3.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.7, p. 506.
Clifford A. Pickover, The Math Book: From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics (2009), at p. 126.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 29.

Harry J. Smith, Table of n, a(n) for n = 0..20000
James Grime and Brady Haran, The Best Way to Pack Spheres, Numberphile video (2018).
J. H. Conway and N. J. A. Sloane, What are all the best sphere packings in low dimensions?, Discr. Comp. Geom., 13 (1995), 383-403.
Thomas C. Hales, Dense Sphere Packings, Cambridge University Press, 2012.
G. Nebe and N. J. A. Sloane, Home page for fcc lattice.
N. J. A. Sloane and Andrey Zabolotskiy, Table of maximal density of a packing of equal spheres in n-dimensional Euclidean space (some values are only conjectural).
Eric Weisstein's World of Mathematics, Cubic Close Packing.
Eric Weisstein's World of Mathematics, Ellipsoid Packing.
Eric Weisstein's World of Mathematics, Sphere Packing.
Wikipedia, Atomic packing factor.
Index entries for sequences related to f.c.c. lattice.
Index entries for transcendental numbers.

Cf. A093824.
Cf. APF's of other crystal lattices: A019673 (simple cubic), A247446 (diamond cubic).
Cf. A161686 (continued fraction).
Related constants: A020769, A020789, A093766, A222066, A222067, A222068, A222069, A222070, A222071, A222072, A260646.

RealDigits[Pi/(3 Sqrt[2]), 10, 120][[1]] (* Harvey P. Dale, Feb 03 2012 *)

default(realprecision, 20080); x=10*Pi*sqrt(2)/6; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b093825.txt", n, " ", d)); \\ Harry J. Smith, Jun 18 2009

Pi/sqrt(18) \\ Charles R Greathouse IV, May 11 2017

Equals A019670*A010503. - R. J. Mathar, Feb 05 2009

Equals Integral_{x >= 0} (4*x^2 + 1)/((2*x^2 + 1)*(8*x^2 + 1)) dx. - Peter Bala, Feb 12 2025

Entry revised by N. J. A. Sloane, Feb 10 2013

cp's OEIS Frontend

A387195 Decimal expansion of J_0(Pi/3), Bessel Function of the first kind of index 0 at A019670.

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A071005 Binary expansion of Pi/3 (A019670).

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A007310 Numbers congruent to 1 or 5 mod 6.

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A002388 Decimal expansion of Pi^2.

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A197476 Decimal expansion of least x>0 having cos(x) = cos(2*x)^2.

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A091925 Decimal expansion of Pi^3.

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