A304656 -id:A304656 - cp's OEIS frontend

A002388 Decimal expansion of Pi^2.

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

A002391 Decimal expansion of natural logarithm of 3.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

			1.098612288668109691395245236922525704647490557822749451734694333637494...

Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 221.
W. E. Mansell, Tables of Natural and Common Logarithms. Royal Society Mathematical Tables, Vol. 8, Cambridge Univ. Press, 1964, p. 2.
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).

Harry J. Smith, Table of n, a(n) for n = 1..20000
D. H. Bailey, Compendium to BBP formulas. [Broken link]
Peter Bala, New series for old functions.
G. Huvent, Formules BBP en base 3. - _Jaume Oliver Lafont_, Oct 12 2009
Melissa Larson, Verifying and discovering BBP-type formulas, 2008.
Simon Plouffe, Plouffe's Inverter, The natural logarithm of 3 to 10000 digits.
Simon Plouffe, log(3), natural logarithm of 3 to 2000 places.
S. Ramanujan, Notebook entry.
Horace S. Uhler, Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17, Proc. Nat. Acad. Sci. U. S. A. 26, (1940). 205-212.
Eric Weisstein's World of Mathematics, BBP-Type Formula.
Wikipedia, Bailey-Borwein-Plouffe formula.
Index entries for transcendental numbers

Cf. A058962, A154920, A002162, A016731 (continued fraction), A073000, A105531, A254619.

RealDigits[Log[3],10,120][[1]]  (* Harvey P. Dale, Apr 23 2011 *)

log(3) \\ Charles R Greathouse IV, Jan 24 2012

# Use some guard digits when computing.
# BBP formula P(1, 4, 2, (1, 0)).
from decimal import Decimal as dec, getcontext
def BBPlog3(n: int) -> dec:
    getcontext().prec = n
    s = dec(0); f = dec(1); g = dec(4)
    for k in range(2 * n):
        s += f / dec(2 * k + 1)
        f /= g
    return s
print(BBPlog3(200))  # Peter Luschny, Nov 03 2023

log(3) = Sum_{n>=1} (9*n-4)/((3*n-2)*(3*n-1)*3*n). [Jolley, Summation of Series, Dover (1961) eq 74]
log(3) = (1/4)*(1 + Sum_{m>=0} (1/9)^(k+1)*(27/(2*k+1) + 4/(2*k+2) + 1/(2*k+3))) (a BBP-type formula). - Alexander R. Povolotsky, Dec 01 2008
log(3) = 4/5 + (1/5)*Sum_{n>=0} (1/4)^n*(1/(2*n+1) + 1/(2*n+3)). - Alexander R. Povolotsky, Dec 18 2008
log(3) = Sum_{k>=0} (1/9)^(k+1)*(9/(2k+1) + 1/(2k+2)). - Jaume Oliver Lafont, Dec 22 2008
Sum_{i>=1} 1/(9^i*i) + Sum_{i>=0} 1/(9^i*(i+1/2)) = 2*log(3) (Huvent 2001). - Jaume Oliver Lafont, Oct 12 2009
Conjecture: log(3) = Sum_{k>=1} A191907(3,k)/k. - Mats Granvik, Jun 19 2011
log(3) = lim_{n->oo} Sum_{k=3^n..3^(n+1)-1} 1/k. Also see A002162. By analogy to the integral of 1/x, log(m) = lim_{n->oo} Sum_{k=m^n..m^(n+1)-1} 1/k, for any value of m > 1. - Richard R. Forberg, Aug 16 2014
From Peter Bala, Feb 04 2015: (Start)
log(3) = Sum {k >= 0} 1/((2*k + 1)*4^k).
Define a pair of integer sequences A(n) = 4^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*4^k). Both sequences satisfy the same second-order recurrence equation u(n) = (20*n + 6)*u(n-1) - 16*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(3) = 1 + 2/(24 - 16*3^2/(46 - 16*5^2/(66 - ... - 16*(2*n - 1)^2/((20*n + 6) - ... )))). Cf. A002162, A073000 and A105531 for similar expansions.
log(3) = 2 * Sum_{k >= 1} (-1)^(k+1)*(4/3)^k/(k*binomial(2*k,k)).
log(3) = (1/4) * Sum_{k >= 1} (-1)^(k+1) (55*k - 23)*(8/9)^k/( 2*k*(2*k - 1)*binomial(3*k,k) ).
log(3) = (1/4) * Sum_{k >= 1} (7*k + 1)*(8/3)^k/( 2*k*(2*k - 1)*binomial(3*k,k) ). (End)
log(3) = -lim_{n->oo} (n+1)th derivative of zeta(n) / n-th derivative of zeta(n). By n = 1000 there is convergence to 25 digits. A related expression: lim_{n->oo} n-th derivative of zeta(n-1) / n-th derivative of zeta(n) = 3. Also see A002581. - Richard R. Forberg, Feb 24 2015
From Peter Bala, Nov 02 2019: (Start)
log(3) = 2*Integral_{x = 0..1} (1 - x^2)/(1 + x^2 + x^4) dx = 2*( 1 - (2/3) + 1/5 + 1/7 - (2/9) + 1/11 + 1/13 - (2/15) + ... ).
log(3) = 16*Sum_{n >= 0} 1/( (6*n + 1)*(6*n + 3)*(6*n + 5) ).
log(3) = 4/5 + 64*Sum_{n >= 0} (18*n + 1)/((6*n - 5)*(6*n - 3)*(6*n - 1)*(6*n + 1)*(6*n + 7)). (End)
From Amiram Eldar, Jul 05 2020: (Start)
Equals 2*arctanh(1/2).
Equals Sum_{k>=1} (2/3)^k/k.
Equals Integral_{x=0..Pi} sin(x)dx/(2 + cos(x)). (End)
log(3) = Integral_{x = 0..1} (x^2 - 1)/log(x) dx. - Peter Bala, Nov 14 2020
From Peter Bala, Oct 28 2023: (Start)
The series representation log(3) = 16*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*(6*n + 5)) given above appears to be the case k = 0 of the following infinite family of series representations for log(3):
log(3) = c(k) + (-1)^k*d(k)*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*...*(6*n + 12*k + 5)), where c(k) is a rational approximation to log(3) and d(k) = 2^(6*k+3)/27^k * (6*k + 2)!.
The first few values of c(k) for k >= 0 are [0, 2996/2673, 89195548/81236115, 23239436137364/21153065697225, 3345533089100222564/3045237239236561677, ...]. Cf A304656. (End)
log(3) = 1 + 2*Sum_{k>=1} 1/((3*k)^3 - 3*k) [Ramanujan]. - Stefano Spezia, Jul 01 2024

Editing and more terms from Charles R Greathouse IV, Apr 20 2010

A113660 Expansion of phi(x)^3 / phi(x^3) where phi() is a Ramanujan theta function.

Original entry on oeis.org

1, 6, 12, 6, -6, 0, 12, 12, 12, 6, 0, 0, -6, 12, 24, 0, -6, 0, 12, 12, 0, 12, 0, 0, 12, 6, 24, 6, -12, 0, 0, 12, 12, 0, 0, 0, -6, 12, 24, 12, 0, 0, 24, 12, 0, 0, 0, 0, -6, 18, 12, 0, -12, 0, 12, 0, 24, 12, 0, 0, 0, 12, 24, 12, -6, 0, 0, 12, 0, 0, 0, 0, 12, 12

Offset: 0

Michael Somos, Nov 03 2005

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Cubic AGM theta functions: a(q) (see A004016), b(q) (A005928), c(q) (A005882).

			G.f. = 1 + 6*q + 12*q^2 + 6*q^3 - 6*q^4 + 12*q^6 + 12*q^7 + 12*q^8 + 6*q^9 + ...

Bruce C. Berndt, Ramanujan's Notebooks Part III, Springer-Verlag, 1991, see p. 227, Entry 4(iv).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael Somos, Introduction to Ramanujan theta functions, 2019.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.

Cf. A113661, A113973, A304656.

Cf. A000122, A000700, A004016, A005882, A005928, A010054, A121373.

A := Basis( ModularForms( Gamma1(12), 1), 74); A[1] + 6*A[2] + 12*A[3] + 6*A[4] - 6*A[5]; /* Michael Somos, May 20 2015 */

a[ n_] := SeriesCoefficient[ EllipticTheta[ 3, 0, q]^3 / EllipticTheta[ 3, 0, q^3], {q, 0, n}]; (* Michael Somos, May 20 2015 *)

{a(n) = my(x); if( n<1, n==0, x = valuation(n, 2); if( n%2, 2, (1 - 3*(-1)^x))*3 * sumdiv(n/2^x, d, kronecker(-3, d)))};

{a(n) = my(A, p, e); if( n<1, n==0, A = factor(n); 6*prod(k=1,matsize(A)[1], [p, e] = A[k,]; if( p==2, (1 - 3*(-1)^e) / 2, p==3, 1, p%6==1, e+1, !(e%2))))};

{a(n) = if( n<1, n==0, 6 * direuler(p=2, n, if( p==2, 2 - (1 - 2*X) / (1 - X^2), 1 / ((1 - X) * (1 - kronecker(-3, p) * X))))[n])};

{a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A)^15 * eta(x^3 + A)^2 * eta(x^12 + A)^2 / (eta(x + A)^6 * eta(x^4 + A)^6 * eta(x^6 + A)^5), n))};

Expansion of a(q) + 2*q(q^2) - 2*a(q^4) = b(-q)^2 / b(q^2) = (b(q) - 2*b(q^4))^2 / b(q^2) = (c(q) + 2*c(q^4))^2 / (3 * c(q^2)) in powers of q where a(), b(), c() are cubic AGM functions. - Michael Somos, May 20 2015
Expansion of (eta(q^2)^15 * eta(q^3)^2 * eta(q^12)^2) / (eta(q)^6 * eta(q^4)^6 * eta(q^6)^5) in powers of q.
a(n) = 6*b(n) where b(n) is multiplicative with a(0) = 1, b(2^e) = (1 - 3(-1)^e) / 2 if e>0, b(3^e) = 1, b(p^e) = e+1 if p == 1 (mod 6), b(p^e) = (1 + (-1)^e) / 2 if p == 5 (mod 6).
Euler transform of period 12 sequence [ 6, -9, 4, -3, 6, -6, 6, -3, 4, -9, 6, -2, ...].
Moebius transform is period 12 sequence [ 6, 6, 0, -18, -6, 0, 6, 18, 0, -6, -6, 0, ...].
G.f. is a period 1 Fourier series which satisfies f(-1 / (12 t)) = 108^(1/2) (t/i) g(t) where q = exp(2 Pi i t) and g() is the g.f. for A113973. - Michael Somos, May 20 2015
G.f.: 1 + 6 * ( Sum_{k>0} x^k / (1 + x^k + x^(2*k)) + 2*x^(4*k - 2) / (1 + x^(4*k - 2) + x^(8*k - 4)) ).
a(n) = 6 * A113661(n), if n>0.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi*sqrt(3) = 5.441398... (A304656). - Amiram Eldar, Dec 25 2023

Corrected by Charles R Greathouse IV, Sep 02 2009

A227354 Expansion of 2 * a(q) - a(q^2) in powers of q where a() is a cubic AGM theta function.

Original entry on oeis.org

1, 12, -6, 12, 12, 0, -6, 24, -6, 12, 0, 0, 12, 24, -12, 0, 12, 0, -6, 24, 0, 24, 0, 0, -6, 12, -12, 12, 24, 0, 0, 24, -6, 0, 0, 0, 12, 24, -12, 24, 0, 0, -12, 24, 0, 0, 0, 0, 12, 36, -6, 0, 24, 0, -6, 0, -12, 24, 0, 0, 0, 24, -12, 24, 12, 0, 0, 24, 0, 0, 0

Offset: 0

Michael Somos, Jul 08 2013

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Cubic AGM theta functions: a(q) (see A004016), b(q) (A005928), c(q) (A005882).

			G.f. = 1 + 12*q - 6*q^2 + 12*q^3 + 12*q^4 - 6*q^6 + 24*q^7 - 6*q^8 + 12*q^9 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael Somos, Introduction to Ramanujan theta functions, 2019.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.

Cf. A033762, A112604, A112605, A122859, A304656.

Cf. A000122, A000700, A004016, A005882, A005928, A010054, A121373.

a[ n_] := SeriesCoefficient[ EllipticTheta[ 4, 0, q^2]^3 / EllipticTheta[ 4, 0, q^6] + 3 EllipticTheta[ 2, 0, q] EllipticTheta[ 2, 0, q^3], {q, 0, n}];
a[ n_] := If[ n < 1, Boole[n == 0], 6 Sum[ JacobiSymbol[ d, 3] (Mod[ n/d, 2] + 1), {d, Divisors@n}]]; (* Michael Somos, Jan 09 2015 *)

{a(n) = if( n<1, n==0, 12 * sumdiv( n, d, kronecker( d, 3)) - 6 * sumdiv( 2*n, d, kronecker( d, 3)))};

{a(n) = local(A, p, e); if( n<1, n==0, A = factor(n); 12 * prod( k=1, matsize(A)[1], if( p=A[k, 1], e=A[k, 2]; if( p==2, (1 + 3*(-1)^e) / 4, if( p == 3, 1, if( p%6 == 1, e+1, (1 + (-1)^e) / 2 ))))))};

Expansion of (4 * b(q^4)^2 - 2 * b(q) * b(q^4) - b(q)^2) / b(q^2) in powers of q where b() is a cubic AGM theta function.
Expansion of phi(-q^2)^3 / phi(-q^6) + 12 * q * psi(q^2) * psi(q^6) in powers of q where phi(), psi() are Ramanujan theta functions. - Michael Somos, Jan 09 2015
Expansion of theta_4(q^2)^3 / theta_4(q^6) + 3 * theta_2(q) * theta_2(q^3) in powers of q.
Moebius transform is period 6 sequence [ 12, -18, 0, 18, -12, 0, ...].
a(n) = 12 * b(n) where b(n) is multiplicative with b(2^e) = (1 + 3*(-1)^e) / 4, b(3^e) = 1, b(p^e) = e+1 if p == 1 (mod 6), b(p^e) = (1 + (-1)^e) / 2 if p == 5 (mod 6).
a(n) = A122859(8*n). a(2*n) = A122859(n). a(2*n + 1) = 12 * A033762(n). a(4*n) = a(n). a(4*n + 1) = 12 * A112604(n). a(4*n + 2) = -6 * A033762(n). a(4*n + 3) = 12 * A112605(n).
G.f.: 1 + 6 * Sum_{k>0} ((k mod 2) + 1) * x^k / (1 + x^k + x^(2*k)). - Michael Somos, Jan 09 2015
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi*sqrt(3) = 5.441398... (A304656). - Amiram Eldar, Nov 23 2023

A386732 Decimal expansion of Integral_{x>=2} 1/(x^12-1) dx.

Original entry on oeis.org

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

Offset: 0

Jason Bard, Jul 31 2025

			0.000044394388389732931619793708861045902941185047688518...

Jason Bard, Table of n, a(n) for n = 0..1003
Michael I. Shamos, A catalog of the real numbers, (2007). See p. 21.

Cf. A016644, A105199, A123868, A304656.

Join[{0, 0, 0, 0}, RealDigits[1/72 (-4 (3 + Sqrt[3]) Pi + 3 (4 ArcTan[2] + 2 Sqrt[3] ArcTan[5/Sqrt[3]] + 2 ArcTan[4 - Sqrt[3]] + 2 ArcTan[4 + Sqrt[3]] + Log[21] - Sqrt[3] Log[5 - 2 Sqrt[3]] + Sqrt[3] Log[5 + 2 Sqrt[3]])), 10, 100][[1]]]
(* or *)
Join[{0, 0, 0, 0}, RealDigits[Integrate[1/(x^12 - 1), {x, 2, Infinity}], 10, 100][[1]]]
(* or *)
Join[{0, 0, 0, 0}, RealDigits[1/22528*Hypergeometric2F1[11/12, 1, 23/12, 1/4096], 10, 100][[1]]]

Equals (1/22528) * hypergeometric(11/12, 1; 23/12; 1/4096).

Equals (-6*Pi - 4*sqrt(3)*Pi + 12*arctan(2) - 3*arctan(12/5) + 6*sqrt(3) * arctan(5/sqrt(3)) + 6*sqrt(3) * arctanh((2*sqrt(3))/5) + log(9261))/72.

A145431 Decimal expansion of Sum_{n>=1} n^3/binomial(2n,n).

Original entry on oeis.org

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

Offset: 1

R. J. Mathar, Feb 08 2009

			4.990384604362124...

Alexander Apelblat, Tables of Integrals and Series, Harri Deutsch, 1996, 4.1.37.

Cf. A000984, A304656.

Maple
```
10/3+74/243*Pi*3^(1/2);
```

RealDigits[2*(405+37*Pi*Sqrt[3])/243, 10, 120][[1]] (* Amiram Eldar, Aug 23 2024 *)

Equals (2/243)*(405+37*Pi*sqrt(3)).

A347055 Decimal expansion of Pi * (sqrt(3) - sqrt(2)).

Original entry on oeis.org

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

Offset: 0

Sean A. Irvine, Aug 14 2021

			0.9985151545442873047663537828657734982375770549025315...

Michael I. Shamos, Shamos's catalog of the real numbers (2011).
Index entries for transcendental numbers.

Cf. A063448, A304656.

RealDigits[Pi * (Sqrt[3] - Sqrt[2]), 10, 120][[1]] (* Amiram Eldar, Jun 20 2023 *)

Pi*(sqrt(3)-sqrt(2)) \\ Charles R Greathouse IV, Oct 02 2022

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

A379664 Decimal expansion of hypergeom([1/2, 1/2], [1], -2).

Original entry on oeis.org

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

Offset: 0

Stefano Spezia, Dec 29 2024

			0.74574918731632960996248206535345110430267519798...

Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 17, page 143.

Cf. A000796, A002194, A304656.

RealDigits[Hypergeometric2F1[1/2,1/2,1,-2],10,100][[1]] (* or *)
RealDigits[Hypergeometric2F1[1/2,1/2,1,2/3]/Sqrt[3],10,100][[1]] (* or *)
RealDigits[2EllipticK[2/3]/(Pi Sqrt[3]),10,100][[1]]

hypergeom([1/2,1/2],1,2/3)/sqrt(3) \\ Hugo Pfoertner, Dec 29 2024

hypergeom([1,1]/2,1,-2) \\ Charles R Greathouse IV, Feb 05 2025

2*ellK(sqrt(2/3))/Pi/sqrt(3) \\ Charles R Greathouse IV, Feb 05 2025

Equals hypergeom([1/2, 1/2], [1], 2/3)/sqrt(3).

Equals 2*EllipticK(2/3)/(Pi*sqrt(3)).

cp's OEIS Frontend

A002388 Decimal expansion of Pi^2.

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A002391 Decimal expansion of natural logarithm of 3.

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A113660 Expansion of phi(x)^3 / phi(x^3) where phi() is a Ramanujan theta function.

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A227354 Expansion of 2 * a(q) - a(q^2) in powers of q where a() is a cubic AGM theta function.

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A386732 Decimal expansion of Integral_{x>=2} 1/(x^12-1) dx.

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A145431 Decimal expansion of Sum_{n>=1} n^3/binomial(2n,n).

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