cp's OEIS Frontend

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

%I A129935 #69 Jul 09 2025 04:27:01
%S A129935 777451915729368,140894092055857794,1526223088619171207,
%T A129935 3052446177238342414,54545811706258836911039145,
%U A129935 624965662836733496131286135873807507,1667672249427111806462471627630318921648499,36465374036664559522628534720215805439659141
%N A129935 Numbers n such that ceiling( 2/(2^(1/n)-1) ) is not equal to floor( 2n/log(2) ).
%C A129935 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
%C A129935 From _David Applegate_, Jun 07 2007: (Start)
%C A129935 "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).
%C A129935 "Also, since 2/log(2) is irrational, 2n/log(2) is never integral, so floor(2n/log(2)) = ceiling(2n/log(2)-1).
%C A129935 "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)
%C A129935 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]
%D A129935 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.
%D A129935 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
%H A129935 Max Alekseyev and Robert Gerbicz, <a href="/A129935/b129935.txt">Table of n, a(n) for n = 1..100</a>
%H A129935 K. O'Bryant, <a href="http://arxiv.org/abs/1410.2927">The sequence of fractional parts of roots</a>, arXiv preprint arXiv:1410.2927 [math.NT], 2014-2015.
%H A129935 Max Alekseyev and others, <a href="http://dxdy.ru/topic6838.html">Integer Parts</a> [in Russian]
%H A129935 Art of Problem Solving, <a href="https://artofproblemsolving.com/community/c6h140737">Logarithmic identity</a>
%H A129935 S. W. Golomb and A. W. Hales, <a href="http://library.msri.org/books/Book42/files/golomb.pdf">Hypercube Tic-Tac-Toe</a>, More Games of No Chance, MSRI Publications, Vol. 42, 2002.
%H A129935 N. J. A. Sloane, <a href="/A078608/a078608.pdf">Two Sequences that Agree for a Long Time</a> (Vugraph from a talk about the OEIS)
%t A129935 (* 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." *)
%t A129935 $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}]];
%t A129935   If[nu[[1]]*Log[2] > 2*nu[[2]], hi = nu, lo = nu]]
%o A129935 (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 */
%Y A129935 Cf. A078608 for the sequence ceiling( 2/(2^(1/n)-1) ).
%Y A129935 Cf. A016730, A120754, A120755.
%K A129935 nonn
%O A129935 1,1
%A A129935 _Richard Stanley_, Apr 30 2007 (who sent a(1))
%E A129935 More terms from _Max Alekseyev_, Jun 06 2007
%E A129935 Edited by _N. J. A. Sloane_ at the suggestion of _Andrew S. Plewe_, Jun 08 2007