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

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

Offset: 1

Author

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

Keywords

• nonn

Comments

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]

References

• 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

Links

• 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)

Crossrefs

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

Cf. A016730, A120754, A120755.

Programs

Mathematica

(* 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]]

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 */

Extensions

More terms from Max Alekseyev, Jun 06 2007

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