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.
%I A290349 #18 Aug 03 2017 04:57:21
%S A290349 5,3,2,5,1,1,1,3,5,2,2,2,1,1,1,5,2,3,2,1,1,1,1,1,5,2,2,1,1,1,1,3,5,1,
%T A290349 2,5,1,1,1,2,1,2,2,2,1,1,1,2,5,3,2,1,1,1,1,1,5,2,2,1,1,1,1,5,1,3,2,2,
%U A290349 1,1,1,3,3,2,2,2,1,1,1,1,5,2,2,1,1,1,1,1
%N A290349 Least multiplier of n such that n*a(n) becomes a congruent number A003273.
%C A290349 Conditional on the Birch and Swinnerton-Dyer conjecture, it can be shown that the only members of this sequence are the Fibonacci numbers {1,2,3,5}. The underlying pattern of three consecutive 1's per octet shows that numbers congruent to {5,6,7} mod 8 are congruent numbers. Also if n is a square then a(n)=5. This is because all congruent numbers can be obtained by multiplying a primitive congruent number A006991 by a positive square number A000290 and 5 is the least congruent number.
%H A290349 Frank M Jackson, <a href="/A290349/b290349.txt">Table of n, a(n) for n = 1..100000</a>
%H A290349 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 A290349 Giovanni Resta, <a href="http://www.numbersaplenty.com/set/congruent_number/">Congruent numbers</a> Primitive congruent numbers up to 10^7.
%H A290349 Frank M Jackson, <a href="/A290349/a290349_1.txt">Mma program for generating b-file</a>
%e A290349 a(10)=2 as 10*2=20 and 10*3=30 are congruent numbers but 2 is the least multiplier.
%t A290349 Sfcore[n_] := Module[{m, fac=Select[FactorInteger[n], OddQ[#[[2]]] &]}, If[!SquareFreeQ[n], Times@@Table[fac[[m]][[1]], {m, Length[fac]}], n]]; 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+2y^2+8z^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)]; lst = {}; Do[AppendTo[lst, (Min[Select[n {1, 2, 3, 5}, CongruentQ[Sfcore[#]] &]])/n], {n, 1, 200}]; lst
%Y A290349 Cf. A003273, A006991, A273929.
%K A290349 nonn
%O A290349 1,1
%A A290349 _Frank M Jackson_, Jul 28 2017