OFFSET
0,3
COMMENTS
Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 5, 7, 19, 35, 54, 62, 63, 197, 285, 339, 479, 505, 917. Moreover, any integer n > 2 can be written as x^2 + y*(y+1) + z*(z^2+1), where x is a positive integer, and y and z are nonnegative integers.
We also guess that each n = 0,1,2,... can be expressed as x*(x+1)/2 + P(y,z) with x, y and z nonnegative integers, where P(y,z) is any of the polynomials y(y+1) + z^2*(z+1), y^2 + z*(z^2+2), y^2 + z*(z^2+7), y^2 + z*(z^2+z+2), y^2 + z*(z^2+2z+3), y^2 + z*(2z^2+z+1).
It is known that every n = 0,1,2,... can be written as x^2 + y*(y+1) + z*(z+1), where x, y and z are nonnegative integers.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.
EXAMPLE
a(35) = 1 since 35 = 5^2 + 0*1 + 2*(2^2+1).
a(54) = 1 since 54 = 2^2 + 4*5 + 3*(3^2+1).
a(62) = 1 since 62 = 2^2 + 7*8 + 1*(1^2+1).
a(63) = 1 since 63 = 7^2 + 3*4 + 1*(1^2+1).
a(197) = 1 since 197 = 5^2 + 6*7 + 5*(5^2+1).
a(285) = 1 since 285 = 15^2 + 5*6 + 3*(3^2+1).
a(339) = 1 since 339 = 17^2 + 4*5 + 3*(3^2+1).
a(479) = 1 since 479 = 7^2 + 20*21 + 2*(2^2+1).
a(505) = 1 since 505 = 13^2 + 17*18 + 3*(3^2+1).
a(917) = 1 since 917 = 15^2 + 18*19 + 7*(7^2+1).
MATHEMATICA
SQ[x_]:=SQ[x]=IntegerQ[Sqrt[x]]
Do[r=0; Do[If[SQ[n-y(y+1)-z(z^2+1)], r=r+1], {y, 0, (Sqrt[4n+1]-1)/2}, {z, 0, (n-y(y+1))^(1/3)}]; Print[n, " ", r]; Continue, {n, 0, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 17 2016
STATUS
approved