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A270921
Number of ordered ways to write n as x*(3x+2) + y*(5y+1)/2 - z^4, where x and y are integers, and z is a nonnegative integer with z^4 <= n.
2
1, 2, 3, 3, 2, 1, 1, 3, 3, 3, 4, 3, 1, 1, 1, 2, 5, 3, 3, 3, 3, 4, 3, 4, 6, 3, 6, 4, 3, 4, 2, 3, 3, 2, 2, 3, 2, 2, 4, 3, 3, 5, 9, 6, 3, 4, 2, 2, 2, 6, 3, 3, 2, 2, 3, 2, 2, 4, 5, 4, 5, 3, 2, 2, 6, 7, 4, 4, 2, 2, 4, 3, 3, 3, 2, 3, 4, 4, 4, 2, 5
OFFSET
0,2
COMMENTS
Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 0, 5, 6, 12, 13, 14, 112, 193, 194, 200, 242, 333, 345, 376, 492, 528, 550, 551, 613, 797, 1178, 1195, 1222, 1663, 3380, 3635, 6508, 8755, 9132, 12434, 20087.
Compare this conjecture with the conjecture in A270566.
LINKS
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Zhi-Wei Sun, On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z), preprint, arXiv:1502.03056 [math.NT], 2015.
EXAMPLE
a(5) = 1 since 5 = 1*(3*1+2) + 0*(5*0+1)/2 - 0^4.
a(6) = 1 since 6 = 1*(3*1+2) + (-1)*(5*(-1)+1)/2 - 1^4.
a(13) = 1 since 13 = 1*(3*1+2) + (-2)*(5*(-2)+1)/2 - 1^4.
a(376) = 1 since 376 = 0*(3*0+2) + (-16)*(5*(-16)+1)/2 - 4^4.
a(9132) = 1 since 9132 = (-13)*(3*(-13)+2) + 59*(5*59+1)/2 - 3^4.
a(12434) = 1 since 12434 = (-21)*(3*(-21)+2) + 78*(5*78+1)/2 - 8^4.
a(20087) = 1 since 20087 = 19*(3*19+2) + 87*(5*87+1)/2 - 0^4.
5, 6, 12, 13, 14, 112, 193, 194, 200, 242, 333, 345, 376, 492, 528, 550, 551, 613, 797, 1178, 1195, 1222, 1663, 3380, 3635, 6508, 8755, 9132, 12434, 20087
MATHEMATICA
pQ[n_]:=pQ[n]=IntegerQ[Sqrt[40n+1]]&&(Mod[Sqrt[40n+1], 10]==1||Mod[Sqrt[40n+1], 10]==9)
Do[r=0; Do[If[pQ[n+x^4-y(3y+2)], r=r+1], {x, 0, n^(1/4)}, {y, -Floor[(Sqrt[3(n+x^4)+1]+1)/3], (Sqrt[3(n+x^4)+1]-1)/3}]; Print[n, " ", r]; Continue, {n, 0, 80}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 25 2016
STATUS
approved