Search: a120754 -id:a120754
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A016730
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Continued fraction for log(2).
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+10
10
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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
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OFFSET
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0,3
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COMMENTS
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Continued fraction for 1/log(2) is the same but without the initial zero.
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LINKS
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EXAMPLE
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log(2) = 0.6931471805599453094... = 0 + 1/(1 + 1/(2 + 1/(3 + 1/(1 + ...)))). - Harry J. Smith, Apr 21 2009
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MATHEMATICA
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PROG
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(PARI) { 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
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CROSSREFS
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KEYWORD
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nonn,cofr
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A129935
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Numbers n such that ceiling( 2/(2^(1/n)-1) ) is not equal to floor( 2n/log(2) ).
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+10
5
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777451915729368, 140894092055857794, 1526223088619171207, 3052446177238342414, 54545811706258836911039145, 624965662836733496131286135873807507, 1667672249427111806462471627630318921648499, 36465374036664559522628534720215805439659141
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OFFSET
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1,1
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COMMENTS
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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
"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]
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REFERENCES
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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
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LINKS
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S. W. Golomb and A. W. Hales, Hypercube Tic-Tac-Toe, More Games of No Chance, MSRI Publications, Vol. 42, 2002.
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MATHEMATICA
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(* 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]]
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PROG
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(PARI) 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 */
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CROSSREFS
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Cf. A078608 for the sequence ceiling( 2/(2^(1/n)-1) ).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A120755
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Where records occur in continued fraction expansion of 1/log(2) (cf. A016730).
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+10
3
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1, 2, 3, 5, 15, 28, 41, 47, 74, 109, 123, 166, 501, 6528, 9168, 465506, 3456790, 28568688, 47066208, 146052963, 201331652, 415612810, 1047079803, 1464289355, 2294768489, 2565310827
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OFFSET
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1,2
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COMMENTS
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a(n) are also the positions of incrementally highest 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 term a_0) - Eric W. Weisstein, Aug 20 2013
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LINKS
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CROSSREFS
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Cf. A016730 (continued fraction of log(2)).
Cf. A120754 (records in the continued fraction expansion of 1/log(2) and log(2).
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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