OFFSET
1,2
COMMENTS
Conjecture: a(n) > 0 for all n > 0. In other words, for any prime p there are nonnegative integers x, y and z such that x^2 + 2*y^2 + 3*2^z = p^2.
As mentioned in A301471, for the composite number m = 5884015571 = 7*17*49445509 there are no nonnegative integers x,y,z such that x^2 + 2*y^2 + 3*2^z = m^2.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..6000
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.
EXAMPLE
a(1) = 1 with prime(1)^2 = 4 = 1^2 + 2*0^2 + 3*2^0.
a(2) = 2 with prime(2)^2 = 9 = 2^2 + 2*1^2 + 3*2^0 = 1^2 + 2*1^2 + 3*2^1.
MATHEMATICA
p[n_]:=p[n]=Prime[n];
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
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[r=0; Do[If[QQ[p[n]^2-3*2^k], Do[If[SQ[p[n]^2-3*2^k-2x^2], r=r+1], {x, 0, Sqrt[(p[n]^2-3*2^k)/2]}]], {k, 0, Log[2, p[n]^2/3]}]; tab=Append[tab, r], {n, 1, 70}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 15 2018
STATUS
approved