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From Mertens's conjecture (2): floor(sqrt(n)) - Mertens's function A002321(n).
2

%I #18 May 21 2023 14:50:16

%S 0,1,2,3,4,3,4,4,5,4,5,5,6,5,4,5,6,6,7,7,6,5,6,6,7,6,6,6,7,8,9,9,8,7,

%T 6,7,8,7,6,6,7,8,9,9,9,8,9,9,10,10,9,9,10,10,9,9,8,7,8,8,9,8,8,9,8,9,

%U 10,10,9,10,11,11,12,11,11,11,10,11,12,12,13,12,13

%N From Mertens's conjecture (2): floor(sqrt(n)) - Mertens's function A002321(n).

%C Mertens conjectured that |A002321(n)| < sqrt(n) for all n > 1. This is now known to be false.

%D D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section VI.2.

%H Paolo Xausa, <a href="/A059572/b059572.txt">Table of n, a(n) for n = 1..10000</a>

%H A. M. Odlyzko and H. J. J. te Riele, <a href="http://www.dtc.umn.edu/~odlyzko/doc/zeta.html">Disproof of the Mertens conjecture</a>, J. reine angew. Math., 357 (1985), pp. 138-160.

%t Table[Floor[Sqrt[n]] - Plus @@ MoebiusMu[Range[n]], {n, 1, 80}] (* _Carl Najafi_, Aug 17 2011 *)

%Y Cf. A002321, A059571.

%K nonn

%O 1,3

%A _N. J. A. Sloane_, Feb 16 2001