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%I #30 Jul 21 2023 06:52:31
%S 1,1,1,1,1,1,1,1,2,1,2,2,2,2,1,3,2,3,1,1,2,2,3,1,2,2,2,2,2,2,2,2,2,1,
%T 1,2,4,4,3,2,2,4,2,3,3,3,3,3,2,2,4,3,4,1,3,2,3,4,3,3,3,3,2,3,3,2,4,3,
%U 2,3,2
%N Number of ways to write 6*n+1 as x^2 + 3*y^2 + 54*z^2 with x,y,z nonnegative integers.
%C Conjecture: a(n) > 0 for all n = 0,1,2,....
%C 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.
%H Zhi-Wei Sun, <a href="/A286885/b286885.txt">Table of n, a(n) for n = 0..10000</a>
%H Tomáš Hejda and Vítezslav Kala, <a href="https://arxiv.org/abs/1906.02538">Ternary quadratic forms representing arithmetic progressions</a>, arXiv:1906.02538 [math.NT], 2019.
%H Zhi-Wei Sun, <a href="/A286885/a286885_1.txt">Tuples (m,r,a,b,c) with 30 >= m > max{2,r} >= 0 and 100 >= a >= b >= c > 0, for which 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. </a>
%H Zhi-Wei Sun, <a href="https://doi.org/10.1007/s11425-015-4994-4">On universal sums of polygonal numbers</a>, Sci. China Math. 58 (2015), 1367-1396.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1502.03056">On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2</a>, arXiv:1502.03056 [math.NT], 2015-2017.
%H Hai-Liang Wu and Zhi-Wei Sun, <a href="http://arxiv.org/abs/1707.06223">Some universal quadratic sums over the integers</a>, arXiv:1707.06223 [math.NT], 2017.
%H Hai-Liang Wu and Zhi-Wei Sun, <a href="https://arxiv.org/abs/1811.05855">Arithmetic progressions represented by diagonal ternary quadratic forms</a>, arXiv:1811.05855 [math.NT], 2018.
%e a(9) = 1 since 6*9 + 1 = 1^2 + 3*0^2 + 54*1^2.
%e a(34) = 1 since 6*34 + 1 = 2^2 + 3*7^2 + 54*1^2.
%e a(125) = 1 since 6*125 + 1 = 26^2 + 3*5^2 + 54*0^2.
%e a(130) = 1 since 6*130 + 1 = 22^2 + 3*9^2 + 54*1^2.
%e a(133) = 1 since 6*133 + 1 = 11^2 + 3*8^2 + 54*3^2.
%e a(203) = 1 since 6*203 + 1 = 25^2 + 3*6^2 + 54*3^2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t 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}]
%Y Cf. A000290, A286944, A287616, A290342.
%K nonn
%O 0,9
%A _Zhi-Wei Sun_, Aug 02 2017
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