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A270559
Number of ordered ways to write n as x^4 + x^3 + y^2 + z*(z+1)/2, where x, y and z are integers with x nonzero, y nonnegative and z positive.
19
1, 1, 2, 2, 2, 2, 3, 1, 3, 4, 2, 5, 2, 3, 4, 2, 3, 4, 5, 1, 4, 3, 3, 4, 3, 4, 5, 5, 3, 6, 5, 3, 3, 6, 2, 4, 6, 3, 9, 4, 2, 3, 4, 3, 7, 6, 3, 6, 2, 4, 2, 6, 5, 7, 6, 4, 5, 3, 6, 4, 11, 1, 5, 9, 3, 6, 5, 3, 8, 8
OFFSET
1,3
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0. In other words, for each n = 1,2,3,... there are integers x and y such that n-(x^4+x^3+y^2) is a positive triangular number.
(ii) a(n) = 1 only for n = 1, 2, 8, 20, 62, 97, 296, 1493, 4283, 4346, 5433.
In contrast, the author conjectured in A262813 that any positive integer can be expressed as the sum of a nonnegative cube, a square and a positive triangular number.
LINKS
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
EXAMPLE
a(1) = 1 since 1 = (-1)^4 + (-1)^3 + 0^2 + 1*2/2.
a(2) = 1 since 2 = (-1)^4 + (-1)^3 + 1^2 + 1*2/2.
a(8) = 1 since 8 = 1^4 + 1^3 + 0^2 + 3*4/2.
a(20) = 1 since 20 = (-2)^4 + (-2)^3 + 3^2 + 2*3/2.
a(62) = 1 since 62 = (-2)^4 + (-2)^3 + 3^2 + 9*10/2.
a(97) = 1 since 97 = 1^4 + 1^3 + 2^2 + 13*14/2.
a(296) = 1 since 296 = (-4)^4 + (-4)^3 + 7^2 + 10*11/2.
a(1493) = 1 since 1493 = (-2)^4 + (-2)^3 + 0^2 + 54*55/2.
a(4283) = 1 since 4283 = (-6)^4 + (-6)^3 + 50^2 + 37*38/2.
a(4346) = 1 since 4346 = (-3)^4 + (-3)^3 + 49^2 + 61*62/2.
a(5433) = 1 since 5433 = (-8)^4 + (-8)^3 + 14^2 + 57*58/2.
MATHEMATICA
TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[x!=0&&TQ[n-y^2-x^4-x^3], r=r+1], {y, 0, Sqrt[n]}, {x, -1-Floor[(n-y^2)^(1/4)], (n-y^2)^(1/4)}]; Print[n, " ", r]; Continue, {n, 1, 10000}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 18 2016
STATUS
approved