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%I #7 Feb 03 2015 11:52:01
%S 1,1,1,1,1,1,3,2,1,2,1,3,2,1,1,2,3,1,3,1,3,5,2,1,3,3,2,3,2,3,3,3,1,2,
%T 4,1,5,1,2,5,2,3,5,4,1,4,4,3,4,4,2,5,2,1,4,5,5,3,1,1,7,5,1,3,4,2,5,3,
%U 2,6,5,3,4,4,5,5,4,4,5,3,1,8,2,4,7,3,4,3,5,3,6,3,3,3,6,4,5,5,2,9,2
%N Number of ways to write n as x^2 + y*(y+1) + z*(4*z+1) with x,y,z nonnegative integers.
%C Conjecture: (i) a(n) > 0 for all n, and a(n) > 1 for all n > 338.
%C (ii) If f(x) is one of the polynomials 3x^2+x, (7x^2+3x)/2, (9x^2-x)/2, (11x^2-7x)/2, (15x^2-7x)/2, (15x^2-11x)/2, then any nonnegative integer n can be written as x^2 + y*(y+1) + f(z) with x,y,z nonnegative integers.
%C We have proved that for each n = 0,1,... there are integers x,y,z such that n = x^2 + y*(y+1) + z*(4z+1).
%C It is known that {x^2+y*(y+1): x,y=0,1,...} = {x*(x+1)/2+y*(y+1)/2: x,y=0,1,...}.
%H Zhi-Wei Sun, <a href="/A254629/b254629.txt">Table of n, a(n) for n = 0..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/0905.0635">On universal sums of polygonal numbers</a>, arXiv:0905.0635 [math.NT], 2009-2015.
%e a(103) = 1 since 103 = 8^2 + 0*1 + 3*(4*3+1).
%e a(122) = 1 since 122 = 9^2 + 1*2 +3*(4*3+1).
%e a(143) = 1 since 143 = 6^2 + 1*2 + 5*(4*5+1).
%e a(167) = 1 since 167 = 3^2 + 9*10 + 4*(4*4+1).
%e a(248) = 1 since 248 = 5^2 + 4*5 + 7*(4*7+1).
%e a(338) = 1 since 338 = 5^2 + 10*11 + 7*(4*7+1).
%t SQ[n_]:=IntegerQ[Sqrt[n]]
%t Do[r=0;Do[If[SQ[n-y(y+1)-z*(4z+1)],r=r+1],{y,0,(Sqrt[4n+1]-1)/2},{z,0,(Sqrt[16(n-y(y+1))+1]-1)/8}];
%t Print[n," ",r];Continue,{n,0,100}]
%Y Cf. A000217, A000290, A007742, A254573, A254574, A254623.
%K nonn
%O 0,7
%A _Zhi-Wei Sun_, Feb 03 2015
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