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A254573 Number of ways to write n = x*(x+1) + y*(3*y+1)/2 + z*(3*z-1)/2 with x,y,z nonnegative integers.

%I A254573 #14 Jul 23 2025 14:50:25
%S A254573 1,1,2,2,1,2,1,4,2,3,1,1,3,3,5,2,2,2,3,3,4,3,4,1,4,2,4,5,4,3,2,4,5,4,
%T A254573 2,4,2,6,3,5,3,3,6,5,5,3,3,6,2,6,5,3,4,3,6,2,4,9,6,4,4,5,5,5,7,3,2,3,
%U A254573 8,4,6
%N A254573 Number of ways to write n = x*(x+1) + y*(3*y+1)/2 + z*(3*z-1)/2 with x,y,z nonnegative integers.
%C A254573 Conjecture: a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 1, 4, 6, 10, 11, 23.
%C A254573 This has been verified for all n = 0..10^7. We have proved that every nonnegative integer can be written as x*(x+1) + y*(3*y+1)/2 + z*(3*z-1)/2 with x,y,z integers.
%H A254573 Zhi-Wei Sun, <a href="/A254573/b254573.txt">Table of n, a(n) for n = 0..10000</a>
%H A254573 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 A254573 a(10) = 1 since 10 = 1*2 + 2*(3*2+1)/2 + 1*(3*1-1)/2.
%e A254573 a(11) = 1 since 11 = 2*3 + 0*(3*0+1)/2 + 2*(3*2-1)/2.
%e A254573 a(23) = 1 since 23 = 4*5 + 1*(3*1+1)/2 + 1*(3*1-1)/2.
%e A254573 a(34) = 2 since 34 = 3*4 + 0*(3*0+1)/2 + 4*(3*4-1)/2 = 4*5 + 1*(3*1+1)/2 + 3*(3*3-1)/2.
%t A254573 sQ[n_]:=IntegerQ[Sqrt[4n+1]]
%t A254573 Do[r=0;Do[If[sQ[n-y(3y+1)/2-z(3z-1)/2],r=r+1],{y,0,(Sqrt[24n+1]-1)/6},{z,0,(Sqrt[24(n-y(3y+1)/2)+1]+1)/6}];
%t A254573 Print[n," ",r];Continue,{n,0,70}]
%Y A254573 Cf. A000326, A002378, A005449, A254574.
%K A254573 nonn
%O A254573 0,3
%A A254573 _Zhi-Wei Sun_, Feb 01 2015