OFFSET
1,4
COMMENTS
Conjecture: a(n) > 0 for all n > 1. Moreover, any integer n > 1 can be written as x*(3*x+2) + y*(3*y+2) + 3^z + 3^w, where x is an integer and y,z,w are nonnegative integers.
a(n) > 0 for all n = 2..10^8. Note that those x*(3*x-2) with x integral are called generalized octagonal numbers (A001082).
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.
EXAMPLE
a(2) = 1 with 2 = 0*(3*0-2) + 0*(3*0-2) + 3^0 + 3^0.
a(3) = 1 with 3 = 0*(3*0-2) + 1*(3*1-2) + 3^0 + 3^0.
a(4) = 2 with 4 = 1*(3*1-2) + 1*(3*1-2) + 3^0 + 3^0 = 0*(3*0-2) + 0*(3*0-2) + 3^0 + 3^1.
a(5) = 1 with 5 = 0*(3*0-2) + 1*(3*1-2) + 3^0 + 3^1.
a(9) = 1 with 9 = (-1)*(3*(-1)-2) + 0*(3*0-2) + 3^0 + 3^1.
MATHEMATICA
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], 4]==3&&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[3(n-3^j-3^k)+2], Do[If[SQ[3(n-3^j-3^k-x(3x-2))+1], r=r+1], {x, -Floor[(Sqrt[3(n-3^j-3^k)/2+1]-1)/3], (Sqrt[3(n-3^j-3^k)/2+1]+1)/3}]],
{j, 0, Log[3, n/2]}, {k, j, Log[3, n-3^j]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 23 2018
STATUS
proposed