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A002391
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Decimal expansion of natural logarithm of 3.
(Formerly M4595 N1960)
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46
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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
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OFFSET
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1,3
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REFERENCES
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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).
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LINKS
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FORMULA
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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
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
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
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
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)
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
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)
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EXAMPLE
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1.098612288668109691395245236922525704647490557822749451734694333637494...
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MATHEMATICA
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PROG
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(Python) # 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
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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