OFFSET
0,9
COMMENTS
Conjecture: a(n) > 0 for all n = 0,1,2,....
In the a-file, we list the tuples (m,r,a,b,c) with 30 >= m > max{2,r} >= 0, 100 >= a >= b >= c > 0, gcd(a,b,c) = 1, and the form a*x^2+b*y^2+c*z^2 irregular, such that all the numbers m*n+r (n = 0,1,2,...) should be representable by a*x^2+b*y^2+c*z^2 with x,y,z integers.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Tomáš Hejda and Vítezslav Kala, Ternary quadratic forms representing arithmetic progressions, arXiv:1906.02538 [math.NT], 2019.
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), 1367-1396.
Zhi-Wei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017.
Hai-Liang Wu and Zhi-Wei Sun, Some universal quadratic sums over the integers, arXiv:1707.06223 [math.NT], 2017.
Hai-Liang Wu and Zhi-Wei Sun, Arithmetic progressions represented by diagonal ternary quadratic forms, arXiv:1811.05855 [math.NT], 2018.
EXAMPLE
a(9) = 1 since 6*9 + 1 = 1^2 + 3*0^2 + 54*1^2.
a(34) = 1 since 6*34 + 1 = 2^2 + 3*7^2 + 54*1^2.
a(125) = 1 since 6*125 + 1 = 26^2 + 3*5^2 + 54*0^2.
a(130) = 1 since 6*130 + 1 = 22^2 + 3*9^2 + 54*1^2.
a(133) = 1 since 6*133 + 1 = 11^2 + 3*8^2 + 54*3^2.
a(203) = 1 since 6*203 + 1 = 25^2 + 3*6^2 + 54*3^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
table={}; Do[r=0; Do[If[SQ[6n+1-3y^2-54z^2], r=r+1], {y, 0, Sqrt[(6n+1)/3]}, {z, 0, Sqrt[(6n+1-3y^2)/54]}]; table=Append[table, r], {n, 0, 70}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 02 2017
STATUS
approved