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%I A306227 #12 Jan 30 2019 09:09:06
%S A306227 1,1,2,2,2,2,2,2,2,2,3,2,2,1,2,3,4,4,2,3,3,4,5,3,3,4,4,3,4,3,4,5,4,2,
%T A306227 2,3,5,7,4,3,3,2,3,4,4,4,5,5,2,3,4,4,5,2,4,4,4,4,4,3,3,5,3,2,4,5,6,5,
%U A306227 2,2,4,5,4,4,2,3,4,4,2,3,6,7,8,4,5,4,3,5,5,3,4,7,7,6,6,4,7,6,4,5
%N A306227 Number of ways to write n as w + x^4 + y*(y+1)/2 + z*(z+1)/2, where w is 0 or 1, and x, y, z are nonnegative integers with x >= w and y < z.
%C A306227 Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer can be written as the sum of two fourth powers one of which is 0 or 1, and two distinct triangular numbers.
%C A306227 We have verified a(n) > 0 for all n = 1..10^6. The conjecture implies that the set S = {x^4 + y*(y+1)/2: x,y = 0,1,2,...} is an additive basis of order two (i.e., the sumset S + S coincides with {0,1,2,...}).
%C A306227 See also A306225 for a similar conjecture.
%H A306227 Zhi-Wei Sun, <a href="/A306227/b306227.txt">Table of n, a(n) for n = 1..10000</a>
%e A306227 a(1) = 1 with 1 = 0 + 0^4 + 0*1/2 + 1*2/2.
%e A306227 a(2) = 1 with 2 = 0 + 1^4 + 0*1/2 + 1*2/2.
%e A306227 a(14) = 1 with 14 = 0 + 1^4 + 2*3/2 + 4*5/2.
%e A306227 a(3774) = 1 with 3774 = 1 + 5^4 + 52*53/2 + 59*60/2.
%e A306227 a(7035) = 1 with 7035 = 0 + 3^4 + 48*49/2 + 107*108/2.
%t A306227 TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
%t A306227 tab={};Do[r=0;Do[If[TQ[n-x-y^4-z(z+1)/2],r=r+1],{x,0,Min[1,(n-1)/2]},{y,x,(n-1-x)^(1/4)},{z,0,(Sqrt[4(n-1-x-y^4)+1]-1)/2}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y A306227 Cf. A000217, A000583, A262813, A262941, A262954, A262956, A262959, A270566, A306225.
%K A306227 nonn
%O A306227 1,3
%A A306227 _Zhi-Wei Sun_, Jan 30 2019

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