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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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4
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777451915729368, 140894092055857794, 1526223088619171207, 3052446177238342414, 54545811706258836911039145, 624965662836733496131286135873807507, 1667672249427111806462471627630318921648499
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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/ln(2) < ln(2)/3 * 1/(2n)^2 implying that m/(2n) is a convergent of 1/ln(2) (note that m and 2n are not necessary coprime). - Max Alekseyev, Jun 06 2007
Comment 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/ln(2)*n-1+(1/6)*ln(2)/n+O(1/n^3).
"Also, since 2/log(2) is irrational, 2n/log(2) is never integral, so floor(2n/log(2)) = ceil(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)
Comment from David Applegate, Jun 08 2007, edited Jun 11 2007: The appropriate generalization of ceil(2/(2^(1/n)-1)) =? floor(2n/log(2)) is floor(a/(b^(1/n)-1)+a/2) = ceil(an/log(b)). When a=2, the a/2 can be hidden in floor() + 1 = ceil().
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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 The correct value was found by J. Buhler in 2004 and is reported in S. Golomb, "Martin Gardner and Tictacktoe" (unpublished).
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LINKS
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Max Alekseyev and Robert Gerbicz, Table of n, a(n) for n = 1..100
Authors?, Discussion in Russian
Authors?, Discussion in English
N. J. A. Sloane, Two Sequences that Agree for a Long Time (Vugraph from a talk about the OEIS)
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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) ).
Cf. A016730, A120754, A120755.
Sequence in context: A172585 A088867 A159042 * A173405 A104835 A128446
Adjacent sequences: A129932 A129933 A129934 * A129936 A129937 A129938
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KEYWORD
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nonn
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AUTHOR
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Richard Stanley, Apr 30 2007 (who sent a(1)).
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EXTENSIONS
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More terms from Max Alekseyev, Jun 06 2007
Edited by N. J. A. Sloane at the suggestion of Andrew Plewe, Jun 08 2007
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STATUS
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approved
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