A120755 -id:A120755 - cp's OEIS frontend

A016730 Continued fraction for log(2).

0, 1, 2, 3, 1, 6, 3, 1, 1, 2, 1, 1, 1, 1, 3, 10, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 3, 1, 13, 7, 4, 1, 1, 1, 7, 2, 4, 1, 1, 2, 5, 14, 1, 10, 1, 4, 2, 18, 3, 1, 4, 1, 6, 2, 7, 3, 3, 1, 13, 3, 1, 4, 4, 1, 3, 1, 1, 1, 1, 2, 17, 3, 1, 2, 32, 1, 1, 1

Offset: 0

N. J. A. Sloane

Continued fraction for 1/log(2) is the same but without the initial zero.

			log(2) = 0.6931471805599453094... = 0 + 1/(1 + 1/(2 + 1/(3 + 1/(1 + ...)))). - _Harry J. Smith_, Apr 21 2009

Harry J. Smith, Table of n, a(n) for n = 0..19999
G. Xiao, Contfrac
Eric Weisstein's World of Mathematics, Natural Logarithm of 2
Index entries for continued fractions for constants

Cf. A120754, A120755, A002162 (decimal expansion).

ContinuedFraction(Log(2)); // G. C. Greubel, Sep 15 2018

ContinuedFraction[Log[2], 80] (* Alonso del Arte, Oct 03 2017 *)

{ allocatemem(932245000); default(realprecision, 21000); x=contfrac(log(2)); for (n=1, 20000, write("b016730.txt", n-1, " ", x[n])); } \\ Harry J. Smith, Apr 21 2009

Offset changed by Andrew Howroyd, Jul 10 2024

A129935 Numbers n such that ceiling( 2/(2^(1/n)-1) ) is not equal to floor( 2n/log(2) ).

Original entry on oeis.org

777451915729368, 140894092055857794, 1526223088619171207, 3052446177238342414, 54545811706258836911039145, 624965662836733496131286135873807507, 1667672249427111806462471627630318921648499, 36465374036664559522628534720215805439659141

Offset: 1

Richard Stanley, Apr 30 2007 (who sent a(1))

nonn

If n belongs to this sequence and m = ceiling(2/(2^(1/n)-1)), then 0 < m/(2n) - 1/log(2) < (log(2)/3) * (1/(2n)^2) implying that m/(2n) is a convergent of 1/log(2) (note that m and 2n are not necessarily coprime). - Max Alekseyev, Jun 06 2007
From David Applegate, Jun 07 2007: (Start)
"Some background to Max Alekseyev's comments: The key point is that the Laurent series for 2/(2^(1/n)-1) about n=infinity is 2/log(2)*n - 1 + (1/6)*log(2)/n + O(1/n^3).
"Also, since 2/log(2) is irrational, 2n/log(2) is never integral, so floor(2n/log(2)) = ceiling(2n/log(2)-1).
"So the question becomes: when is 2n/log(2)-1 so close to an integer that 2/(2^(1/n)-1) is on the other side of the integer? That is why the continued fraction expansion of 2/log(2) is relevant." (End)
The appropriate generalization of ceiling(2/(2^(1/n)-1)) = ? floor(2n/log(2)) is floor(a/(b^(1/n)-1)+a/2) = ceiling(an/log(b)). When a=2, the a/2 can be hidden in floor() + 1 = ceiling(). - David Applegate, Jun 08 2007 [edited Jun 11 2007]

S. W. Golomb and A. W. Hales, "Hypercube Tic-Tac-Toe", in "More Games of No Chance", ed. R. J. Nowakowski, MSRI Publications 42, Cambridge University Press, 2002, pp. 167-182. Here it is stated that the first counterexample is at n=6847196937, an error due to faulty multiprecision arithmetic. The correct value was found by J. Buhler in 2004 and is reported in S. Golomb, "Martin Gardner and Tictacktoe," in Demaine, Demaine, and Rodgers, eds., A Lifetime of Puzzles, A K Peters, 2008, pp. 293-301.
Dean Hickerson, Email to Jon Perry and N. J. A. Sloane, Dec 16 2002. Gives first three terms: 777451915729368, 140894092055857794, 1526223088619171207, as well as five later terms. - N. J. A. Sloane, Apr 30 2014

Max Alekseyev and Robert Gerbicz, Table of n, a(n) for n = 1..100
K. O'Bryant, The sequence of fractional parts of roots, arXiv preprint arXiv:1410.2927 [math.NT], 2014-2015.
Max Alekseyev and others, Integer Parts [in Russian]
Art of Problem Solving, Logarithmic identity
S. W. Golomb and A. W. Hales, Hypercube Tic-Tac-Toe, More Games of No Chance, MSRI Publications, Vol. 42, 2002.
N. J. A. Sloane, Two Sequences that Agree for a Long Time (Vugraph from a talk about the OEIS)

Cf. A078608 for the sequence ceiling( 2/(2^(1/n)-1) ).

Cf. A016730, A120754, A120755.

(* Mma 9.0.1 code from Bill Gosper, Mar 15 2013. He comments: "This reproduces the hundred values in the b-file, and probably works up to around half a billion digits. When Mathematica gets fixed, change 999999999 to infinity." *)
$MaxExtraPrecision = 999999999; For[{lo = {0, 1}, hi = {1, 0}, nu = {0, 0}, n = 0}, nu[[2]] < 10^386, nu = lo + hi; For[{k = nu[[2]]}, Floor[k*2/Log[2]] != Ceiling[2/(2^(1/k) - 1)], k += nu[[2]], Print[{++n, k}]];
  If[nu[[1]]*Log[2] > 2*nu[[2]], hi = nu, lo = nu]]

prec=1500;default(realprecision,prec);c=contfrac(log(2)/2);default(realprecision,prec*2+50); i=0;for(n=2,#c-1, cand=contfracpnqn(vecextract(c,2^n-1))[1,1];forstep(m=cand,c[n+1]*cand,cand, if(ceil(2/(2^(1/m)-1)) != floor(2*m/log(2)), i++;print(i" "m), break))) /* Phil Carmody, Mar 20 2013 */

More terms from Max Alekseyev, Jun 06 2007

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 08 2007

A120754 Records in continued fraction expansion of 1/log(2) (cf. A016730).

Original entry on oeis.org

1, 2, 3, 6, 10, 13, 14, 18, 32, 52, 113, 485, 3377, 4714, 963664, 10467647, 14779710, 15407967, 70919074, 73672410, 363144903, 409121736, 628298429, 2803904265, 4054561652, 53155160769

Offset: 1

N. J. A. Sloane, Jun 08 2007

a(n) are also the incrementally largest terms in the continued fraction of log(2) = [a_0; a_1, a_2, ...] = [0; 1, 2, 3, 1, 6, 3, 1, 1, 2] (excluding the a_0 term). - Eric W. Weisstein, Aug 20 2013

Eric Weisstein's World of Mathematics, Natural Logarithm of 2 Continued Fraction

Cf. A016730 (continued fraction of log(2)).
Cf. A120755 (positions of records in the continued fraction of 1/log(2) and log(2)).
Cf. A129935.

with(numtheory); read transforms; t1:= cfrac(1/log(2),2000,'quotients'); RECORDS(t1);

a(14)-a(22) from Eric W. Weisstein, Aug 20 2013

a(23)-a(26) from Eric W. Weisstein, Aug 21 2013

cp's OEIS Frontend

A016730 Continued fraction for log(2).

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A129935 Numbers n such that ceiling( 2/(2^(1/n)-1) ) is not equal to floor( 2n/log(2) ).

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A120754 Records in continued fraction expansion of 1/log(2) (cf. A016730).

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