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%I #16 Aug 03 2023 08:22:05
%S 1,1,1,1,2,1,1,1,2,2,2,4,1,2,2,2,2,1,4,1,2,2,1,2,1,4,3,3,2,2,5,3,3,2,
%T 3,3,3,4,2,3,5,2,2,1,3,3,5,2,1,3,2,4,3,6,1,3,5,2,1,3,6,2,2,3,3,3,6,4,
%U 4,2
%N Number of ordered ways to write n as x^3 + y^2 + z*(3z+1), where x, y and z are integers with x positive and y nonnegative.
%C We guess that a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 4, 6, 7, 8, 13, 18, 20, 23, 25, 44, 49, 55, 59, 121, 238.
%C Based on our computation, we propose the following general conjecture (which extends the conjectures in A262813 and A270469).
%C Conjecture: Let T(x) = x*(x+1)/2 and pen(x) = x*(3x+1)/2. Every natural number can be written as P(x,y,z) with x a nonnegative integer and y and z integers, where P(x,y,z) is any of the following cubic polynomials: x^3+T(y)+z^2, a*x^3+T(y)+pen(z) (a = 1,2,3,4), x^3+T(y)+z*(5z+1)/2, x^3+T(y)+z*(3z+r) (r = 1,2), x^3+T(y)+z*(7z+3)/2, x^3+T(y)+z*(9z+j)/2 (j = 5,7), x^3+T(y)+z*(5z+r) (r = 2,3), x^3+T(y)+2z*(3z+r) (r = 1,2), x^3+T(y)+z*(6z+5), x^3+T(y)+z*(13z+j)/2 (j = 3,7,9), x^3+T(y)+z*(7z+k) (k = 2,6), a*x^3+y^2+pen(z) (a = 1,2,3,4), x^3+y^2+z*(5z+3)/2, x^3+y^2+2*pen(z), x^3+2*T(y)+pen(z), x^3+2*T(y)+z(5z+j)/2 (j = 1,3), a*x^3+2*T(y)+z*(3z+2) (a = 1,2,3), x^3+2*T(y)+z*(7z+3)/2, x^3+4*T(y)+pen(z), x^3+2y^2+pen(z), x^3+pen(y)+c*pen(z) (c = 1,2,3,4), x^3+b*pen(y)+z*(5z+j)/2 (b = 1,2; j = 1,3), x^3+pen(y)+z*(7z+k)/2 (k = 1,3,5), x^3+pen(y)+z*(4z+j) (j = 1,3), x^3+pen(y)+z*(9z+5)/2, a*x^3+pen(y)+z*(9z+r)/2 (a = 1,2; r = 1,7), x^3+pen(y)+z*(5z+r) (r = 1,2,3,4), a*x^3+pen(y)+z*(11z+9)/2 (a = 1,2), x^3+pen(y)+2z*(3z+2),x^3+pen(y)+z*(13z+11)/2, x^3+pen(y)+z*(7z+k) (k = 4,5,6), x^3+pen(y)+3z*(5z+3)/2, x^3+pen(y)+z*(15z+11)/2, x^3+pen(y)+z*(8z+7), x^3+pen(y)+z*(11z+7), x^3+2*pen(y)+z*(7z+j)/2 (j = 1,5), x^3+2*pen(y)+3*pen(z), x^3+2*pen(y)+z*(4z+1), x^3+2*pen(y)+z*(7z+2), x^3+y*(5y+j)/2+z*(7z+k)/2 (j = 1,3; k = 3,5), x^3+y*(5y+3)/2+z*(9z+7)/2, x^3+y*(3y+2)+z*(4z+1), x^3+y*(3y+2)+z*(5z+1)/2, x^3+y*(7y+3)/2+z*(7z+5)/2, 2x^3+T(y)+z*(5z+3)/2, 2x^3+T(y)+z*(3z+r) (r = 1,2), 2x^3+T(y)+z*(5z+4), 2x^3+2*T(y)+z*(5z+3)/2, 2x^3+3*T(y)+pen(z), 2x^3+y^2+2*pen(z), 2x^3+pen(y)+pen(z), a*x^3+pen(y)+3*pen(z) (a = 2,3,4), a*x^3+pen(y)+z*(7z+5)/2 (a = 2,3,4), 2x^3+pen(y)+z*(5z+k) (k = 1,3), 2x^3+y*(5y+3)/2+z*(7z+5)/2, 2x^3+2*pen(y)+z*(3z+2), 2x^3+2*pen(y)+z*(7z+5)/2, 2x^3+y*(3y+2)+z*(4z+3), 3x^3+pen(y)+z*(7z+3)/2, 4x^3+y^2+z*(5z+1)/2, 4x^3+pen(y)+z*(4z+3).
%C The listed ternary polynomials in the conjecture should exhaust all those P(x,y,z) = a*x^3+y*(s*y+t)/2+z*(u*z+v)/2 with a,s,u > 0, 0 <= t <= s, 0 <= v <= u, s == t (mod 2), u == v (mod 2), and (s-2t)*(u-2v) nonzero, such that any natural number can be written as P(x,y,z) with x a nonnegative integer and y and z integers. Note that those numbers y*(2y+1) with y integral are just triangular numbers.
%C Conjecture verified for all polynomials up to 10^11. - _Mauro Fiorentini_, Aug 03 2023
%C See also A271106 for another general conjecture on universal sums.
%H Zhi-Wei Sun, <a href="/A271325/b271325.txt">Table of n, a(n) for n = 1..10000</a>
%H Z.-W. Sun, <a href="http://dx.doi.org/10.4064/aa127-2-1">Mixed sums of squares and triangular numbers</a>, Acta Arith. 127(2007), 103-113.
%H Z.-W. Sun, <a href="http://math.scichina.com:8081/sciAe/EN/abstract/abstract517007.shtml">On universal sums of polygonal numbers</a>, Sci. China Math. 58(2015), 1367-1396.
%H Z.-W. Sun, <a href="http://arxiv.org/abs/1502.03056">On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z)</a>, preprint, arXiv:1502.03056 [math.NT], 2015.
%e a(13) = 1 since 13 = 2^3 + 1^2 + 1*(3*1+1).
%e a(18) = 1 since 18 = 2^3 + 0^2 + (-2)*(3*(-2)+1).
%e a(20) = 1 since 20 = 1^3 + 3^2 + (-2)*(3*(-2)+1).
%e a(23) = 1 since 23 = 2^3 + 1^2 + 2*(3*2+1).
%e a(25) = 1 since 25 = 1^3 + 0^2 + (-3)*(3*(-3)+1).
%e a(44) = 1 since 44 = 2^3 + 6^2 + 0*(3*0+1).
%e a(49) = 1 since 49 = 1^3 + 2^2 + (-4)*(3*(-4)+1).
%e a(55) = 1 since 55 = 3^3+ 2^2 + (-3)*(3*(-3)+1).
%e a(59) = 1 since 59 = 2^3 + 7^2 + (-1)*(3*(-1)+1).
%e a(121) = 1 since 121 = 3^3 + 8^2 + 3*(3*3+1).
%e a(238) = 1 since 238 = 4^3 + 12^2 + 3*(3*3+1).
%t pQ[n_]:=pQ[n]=IntegerQ[Sqrt[12n+1]]
%t Do[r=0;Do[If[pQ[n-x^3-y^2],r=r+1],{x,1,n^(1/3)},{y,0,Sqrt[n-x^3]}];Print[n," ",r];Label[aa];Continue,{n,1,70}]
%Y Cf. A000217, A000290, A000326, A000578, A001318, A160325, A160326, A262813, A262815, A262816, A270488, A270469, A270516, A270533, A270559, A270566, A271106.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Apr 04 2016
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