A290349 Least multiplier of n such that n*a(n) becomes a congruent number A003273.

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, 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, 1, 1, 1, 3, 3, 2, 2, 2, 1, 1, 1, 1, 5, 2, 2, 1, 1, 1, 1, 1

Offset: 1

Author

Frank M Jackson, Jul 28 2017

Keywords

• nonn

Comments

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.

Examples

			a(10)=2 as 10*2=20 and 10*3=30 are congruent numbers but 2 is the least multiplier.
		

Links

• Frank M Jackson, Table of n, a(n) for n = 1..100000
• Keith Conrad, The Congruent Number Problem, The Harvard College Mathematics Review, 2008.
• Giovanni Resta, Congruent numbers Primitive congruent numbers up to 10^7.
• Frank M Jackson, Mma program for generating b-file

Crossrefs

Cf. A003273, A006991, A273929.

Programs

Mathematica

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