cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A361502 Index of n-th prime in A359804.

Original entry on oeis.org

2, 3, 4, 8, 13, 42, 347, 3466, 49012, 528231, 717126, 63056215, 1375559400, 7038527851
Offset: 1

Author

N. J. A. Sloane, Mar 18 2023, based on a comment made by Michael De Vlieger in A359804 in which he gave the values of a(1) to a(12)

Keywords

Comments

Theorem: Every prime appears in A359804. For proof see A359804.
It appears that the primes in A359804 appear in order.

Crossrefs

Programs

  • Mathematica
    nn = 2^20; c[] = False; q[] = 1;
     i = 1; j = 2; c[1] = c[2] = True; u = 3;
     {2}~Join~Reap[Monitor[Do[
          (k = q[#]; While[c[k #], k++]; k *= #;
             While[c[# q[#]], q[#]++]) &[(p = 2;
            While[Divisible[i j, p], p = NextPrime[p]]; p)];
          If[PrimeQ[k], Sow[n]; Print[n]];
          Set[{c[k], i, j}, {True, j, k}];
    If[k == u, While[c[u], u++]], {n, 3, nn}], n]][[-1, -1]] (* Michael De Vlieger, Mar 19 2023 *)

Extensions

a(13)-a(14) from Rémy Sigrist, Mar 19 2023

A292766 Numbers n whose trajectory under iteration of the map k -> (sigma(k)+phi(k))/2 consists only of integers and is unbounded, excluding numbers n whose trajectory merges with the trajectory of a smaller number.

Original entry on oeis.org

270, 440, 496, 702, 737, 813, 828, 897, 905, 1027, 1066, 1099, 1240, 1241, 1260, 1331, 1353, 1368, 1371, 1422, 1507, 1537, 1754, 1760, 1834, 1848, 2002, 2016, 2282
Offset: 1

Author

N. J. A. Sloane, Sep 27 2017, based on emails from Sean A. Irvine, Sep 14 2017, who computed a(1)-a(9), and Hans Havermann, same date, who computed a(10)-a(29). Hugo Pfoertner also computed many of these terms

Keywords

Comments

These are the "seeds" in A291790, that is, every number which blows up under iteration of the map k -> (sigma(k)+phi(k))/2 belongs to one of these trajectories. AT PRESENT ALL TERMS ARE CONJECTURAL.
The trajectories of these numbers are pairwise disjoint for the first 400 steps.
This is unsatisfactory because it is possible that, at some later step, these trajectories may merge, reach a prime (a fixed point), or reach a fraction (and die). However, this seems unlikely on probabilistic grounds - see the remarks of Andrew R. Booker in A292108.
Normally such a sequence would not be included in the OEIS, but exceptions have been made for this and A291790 because a number of people have worked on them, and also in the hope that this will encourage resolution of some of the open questions.
Needs a b-file.

Crossrefs

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

Original entry on oeis.org

777451915729368, 140894092055857794, 1526223088619171207, 3052446177238342414, 54545811706258836911039145, 624965662836733496131286135873807507, 1667672249427111806462471627630318921648499, 36465374036664559522628534720215805439659141
Offset: 1

Author

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

Keywords

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

Crossrefs

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

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