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%I #28 Dec 26 2018 03:34:17
%S 3,36,361,3503,34065,332712,3252966
%N Number of primitive (squarefree) congruent numbers (A006991) <= 10^n.
%C Conjecture: the limit of a(n)/10^n tends to 3/Pi^2 (A104141). This is based on the assumption, conditional on the Birch Swinnerton-Dyer conjecture, that all squarefree integers congruent to {5, 6, 7} mod 8 (A273929) are a subset of primitive congruent numbers (A006991) and have a natural density of 3/Pi^2. However, squarefree integers congruent to {1, 2, 3} mod 8 are conjecturally sparsely congruent numbers with a natural density of 0. It has been proved without the BSD conjecture that the natural density of congruent numbers is at least 55.9% the natural density of squarefree numbers congruent to {5, 6, 7} mod 8 (see A. Smith link).
%C The Mathematica program below is a slow implementation of the Tunnell criteria for determining congruent numbers. It will give counts for up to 10^5 in realistic time. Counts for 10^6 and 10^7 have been derived from tables generated by _Giovanni Resta_ (see link).
%H Keith Conrad, <a href="http://www.math.uconn.edu/~kconrad/articles/congruentnumber.pdf">The Congruent Number Problem</a>, The Harvard College Mathematics Review, (2008).
%H Giovanni Resta, <a href="http://www.numbersaplenty.com/data/primitive-congruent-mod123.txt">Table of primitive congruent numbers {1, 2, 3} mod 8</a>
%H Alexander Smith, <a href="http://arxiv.org/abs/1603.08479">The congruent numbers have positive natural density</a>, arXiv:1603.08479 [math.NT], 2016.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Congruent_number">Congruent number</a>
%H Shou-Wu Zhang, <a href="http://www.asiapacific-mathnews.com/03/0302/0012_0015.pdf">The Congruent Numbers and Heegner Points</a>, Asian Pacific Mathematics Newsletter, Vol 3(2) (2013).
%t CongruentQ[n_] := Module[{x, y, z, ok=False}, (Which[!SquareFreeQ[n], Null[], MemberQ[{5,6,7}, Mod[n, 8]], ok=True, OddQ@n&&Length@Solve[x^2 + 2 y^2 + 8 z^2 == n, {x, y, z}, Integers]==2Length@Solve[x^2+2y^2+32z^2==n, {x, y, z}, Integers], ok=True, EvenQ@n&&Length@Solve[x^2+4y^2+8z^2==n/2, {x, y, z}, Integers]==2Length@Solve[x^2+4y^2+32z^2==n/2, {x, y, z}, Integers], ok=True]; ok)]; Table[Length@Select[Range[10^n], CongruentQ], {n, 1, 5}]
%Y Cf. A006991, A062695, A071172, A104141, A273929, A274043, A274264.
%K nonn,more
%O 1,1
%A _Frank M Jackson_, Jun 20 2016
%E a(7) corrected by _Frank M Jackson_, Jul 25 2016
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