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A290349
Least multiplier of n such that n*a(n) becomes a congruent number A003273.
1
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
1,1
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.
LINKS
Keith Conrad, The Congruent Number Problem, The Harvard College Mathematics Review, 2008.
Giovanni Resta, Congruent numbers Primitive congruent numbers up to 10^7.
EXAMPLE
a(10)=2 as 10*2=20 and 10*3=30 are congruent numbers but 2 is the least multiplier.
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Frank M Jackson, Jul 28 2017
STATUS
approved