A301640 Largest integer k such that n^2 - 3*2^k can be written as x^2 + 2*y^2 with x and y integers, or -1 if no such k exists.
-1, 0, 1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 6, 7, 7, 7, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 8, 9, 9, 9, 9, 8, 9, 9, 7, 9, 7, 9, 9, 8, 9, 10, 10, 10, 10, 10, 10, 10, 9, 10, 10, 10, 9, 10, 10, 10
Offset: 1
Keywords
Examples
a(2) = 0 since 2^2 - 3*2^0 = 1^2 + 2*0^2. a(3) = 1 since 3^2 - 3*2^1 = 2^2 + 2*1^2. a(5) = 3 since 5^2 - 3*2^3 = 1^2 + 2*0^2. a(6434567) = 10 since 6434567^2 - 3*2^10 = 5921293^2 + 2*1780722^2.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
- Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
- Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.
Crossrefs
Programs
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Maple
f:= proc(n) local k,t; for k from floor(log[2](n^2/3)) by -1 to 0 do if g(n^2 - 3*2^k) then return k fi od; -1 end proc: map(f, [$1..100]); # Robert Israel, Mar 26 2018
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Mathematica
f[n_]:=f[n]=FactorInteger[n]; g[n_]:=g[n]=Sum[Boole[(Mod[Part[Part[f[n],i],1],8]==5||Mod[Part[Part[f[n],i],1],8]==7)&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0; QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]); tab={};Do[Do[If[QQ[n^2-3*2^(Floor[Log[2,n^2/3]]-k)],tab=Append[tab,Floor[Log[2,n^2/3]]-k];Goto[aa]],{k,0,Log[2,n^2/3]}];tab=Append[tab,-1];Label[aa],{n,1,70}];Print[tab]
Comments