

A129935


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


4



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

1,1


COMMENTS

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)*n1+(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().


REFERENCES

S. W. Golomb and A. W. Hales, "Hypercube TicTacToe", in "More Games of No Chance", ed. R. J. Nowakowski, MSRI Publications 42, Cambridge University Press, 2002, pp. 167182. 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. 293301.
Dean Hickerson, Email to Jon Perry and N. J. A. Sloane, Dec 16 2002. Gives first three terms of A129932: 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
Max Alekseyev and others, Integer Parts [in Russian]
Art of Problem Solving, Logarithmic Identity
S. W. Golomb and A. W. Hales, Hypercube TicTacToe
N. J. A. Sloane, Two Sequences that Agree for a Long Time (Vugraph from a talk about the OEIS)


MATHEMATICA

Mma 9.0.1 code from Bill Gosper, Mar 15 2013. He comments: "This reproduces the hundred values in the bfile, 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]]


PROG

(PARI) prec=1500; default(realprecision, prec); c=contfrac(log(2)/2); default(realprecision, prec*2+50); i=0; for(n=2, #c1, cand=contfracpnqn(vecextract(c, 2^n1))[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 */


CROSSREFS

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


KEYWORD

nonn


AUTHOR

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


EXTENSIONS

More terms from Max Alekseyev, Jun 06 2007
Edited by N. J. A. Sloane at the suggestion of Andrew Plewe, Jun 08 2007


STATUS

approved



