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%I M2155 #141 Feb 08 2024 07:10:53
%S 1,2,33,242,11605,28374,171893,1043710445721,2197379769820,
%T 2642166652554075
%N Start of first run of n consecutive integers with same number of divisors.
%C The entry 40311 given by Guy and by Wells is incorrect. - _Jud McCranie_, Jan 20 2002
%C a(10) <= 2642166652554075, a(11) <= 17707503256664346, a(12) <= 9827470582657267545. - _David Wasserman_, Feb 22 2008
%C a(10) > 10^13. - _Giovanni Resta_, Jul 13 2015
%C a(12) <= 3842083249515874843. - _Hugo van der Sanden_, Sep 20 2022
%C a(13) <= 34169215324203592637988571. - _Hugo van der Sanden_, Apr 13 2022
%C a(14) <= 9721439902882994590514319997146. - _Hugo van der Sanden_, Jun 14 2022
%C a(15) <= 80215613469168729088982885848674841. - _Natalia Makarova_, Sep 18 2022
%C a(16) <= 37981337212463143311694743672867136611416. - _Vladimir Letsko_, Mar 17 2017
%C a(17) <= 768369049267672356024049141254832375543516. - _Vladimir Letsko_, Sep 12 2017
%C a(18) <= 488900003598703704335810037459507226590256411. - _Vladimir Letsko_, Jun 03 2022
%C a(19) <= 5908388043825578351730345292813071711296723319324. - _Vladimir Letsko_, Apr 09 2022
%C a(20) <= 17668887847524548413038893976018715843277693308027547. _Vladimir Letsko_, May 30 2022
%C Spătaru proves that the longest such run up to N is at most exp(C*sqrt(log N log log N)) for some constant C, hence a(n) >> exp(exp(W((log^2 n)/C))) which is approximately exp(log^2 n/(2 log log n)). - _Charles R Greathouse IV_, Feb 06 2023
%D J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 33, pp 12, Ellipses, Paris 2008.
%D R. K. Guy, Unsolved Problems in Number Theory, section B18.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, NY, 1986, pages 147 and 176.
%H Pentti Haukkanen, <a href="https://www.researchgate.net/publication/282603270_SOME_COMPUTATIONAL_RESULTS_CONCERNING_THE_DIVISOR_FUNCTIONS_dn_AND_SIGMAn">Some computational results concerning the divisor functions d(n) and sigma(n)</a>, The Mathematics Student, Vol. 62 Nos. 1-4 (1993) pp. 166-168. See p. 167.
%H Vladimir A. Letsko, <a href="http://arxiv.org/abs/1510.07081">Some new results on consecutive equidivisible integers</a>, arXiv:1510.07081 [math.NT], 2015.
%H Vladimir A. Letsko and Vasilii Dziubenko <a href="http://grani.vspu.ru/files/publics/1464684416.pdf">On consecutive equidivisible integers</a> (in Russian), Boundaries of knowledge, 2 (45) 2016.
%H Carlos Rivera, <a href="http://www.primepuzzles.net/problems/prob_020.htm">Problem 20: k consecutive numbers with the same number of divisors</a>, The Prime Puzzles and Problems Connection.
%H Carlos Rivera, <a href="http://www.primepuzzles.net/problems/prob_061.htm">Problem 61: problem 20 revisited</a>, The Prime Puzzles and Problems Connection.
%H Vlad-Titus Spătaru, <a href="https://arxiv.org/abs/2301.04464">Runs of consecutive integers having the same number of divisors</a>, arXiv preprint (2023). arXiv:2301.04464 [math.NT]
%e 33 has four divisors (1, 3, 11, and 33), 34 has four divisors (1, 2, 17, and 34), 35 has four divisors (1, 5, 7, and 35). These are the first three consecutive numbers with the same number of divisors, so a(3)=33.
%t tau = DivisorSigma[0, #]&;
%t A006558[q_, w_] := Module[{a, k, j, ok, n}, For[j = 0, j <= w, j++, For[n = 1, n <= q, n++, ok = 1; a = tau[n]; For[k = 1, k <= j, k++, If[a != tau[n + k], ok = 0; Break[]]]; If [ok == 1, Print[n]; Break[]]]]];
%t A006558[2*10^5, 7] (* _Jean-François Alcover_, Dec 10 2017 *)
%o (PARI) isok(n, k)=nb = numdiv(k); for (j=k+1, k+n-1, if (numdiv(j) != nb, return(0));); 1;
%o a(n) = {k=1; while (!isok(n, k), k++); k;} \\ _Michel Marcus_, Feb 17 2016
%Y Cf. A000005, A005237, A005238, A006601, A049051, A019273, A039665.
%Y Cf. A034173, A115158, A119479.
%K nonn,more,hard
%O 1,2
%A _N. J. A. Sloane_, _Robert G. Wilson v_
%E a(8) from _Jud McCranie_, Jan 20 2002
%E a(9) conjectured by _David Wasserman_, Jan 08 2006
%E a(9) confirmed by _Jud McCranie_, Jan 14 2006
%E a(10) by _Jud McCranie_, Nov 27 2018
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