OFFSET
0,2
COMMENTS
Conjecture: For each n = 0,1,2,... we can write 12*n + 5 as 2*x^2 + 5*y^2 + 9*z^2 + x*y*z with x,y,z nonnegative integers.
This has been verified for all n = 0..10^6.
It seems that a(n) = 1 only for n = 0, 2, 5, 12, 65, 86, 155, 338, 21030.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
EXAMPLE
a(0) = 1 with 12*0 + 5 = 5 = 2*0^2 + 5*1^2 + 9*0^2 + 0*1*0.
a(2) = 1 with 12*2 + 5 = 29 = 2*0^2 + 5*2^2 + 9*1^2 + 0*2*1.
a(5) = 1 with 12*5 + 5 = 65 = 2*3^2 + 5*1^2 + 9*2^2 + 3*1*2.
a(12) = 1 with 12*12 + 5 = 149 = 2*0^2 + 5*1^2 + 9*4^2 + 0*1*4.
a(65) = 1 with 12*65 + 5 = 785 = 2*1^2 + 5*9^2 + 9*6^2 + 1*9*6.
a(86) = 1 with 12*86 + 5 = 1037 = 2*6^2 + 5*1^2 + 9*10^2 + 6*1*10.
a(155) = 1 with 12*155 + 5 = 1865 = 2*2^2 + 5*6^2 + 9*13^2 + 2*6*13.
a(338) = 1 with 12*338 + 5 = 4061 = 2*20^2 + 5*6^2 + 9*13^2 + 20*6*13.
a(21030) = 1 with 12*21030 + 5 = 252365 = 2*32^2 + 5*126^2 + 9*39^2 + 32*126*39.
MATHEMATICA
SQ[n_]:=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[8(12n+5-5y^2-9z^2)+y^2*z^2]&&Mod[Sqrt[8(12n+5-5y^2-9z^2)+y^2*z^2]-y*z, 4]==0, r=r+1], {y, 0, Sqrt[(12n+5)/5]}, {z, 0, Sqrt[(12n+5-5y^2)/9]}]; tab=Append[tab, r], {n, 0, 100}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 13 2022
STATUS
approved