login
A270616
Number of ordered ways to write n as the sum of a positive square, the square of a triangular number, and a generalized pentagonal number (A001318).
3
1, 2, 2, 2, 2, 3, 2, 1, 3, 4, 4, 3, 2, 3, 3, 4, 6, 4, 3, 3, 2, 3, 3, 3, 6, 4, 5, 4, 1, 4, 4, 5, 2, 1, 3, 5, 6, 5, 6, 5, 5, 5, 2, 5, 6, 3, 5, 3, 5, 6, 6, 10, 4, 2, 3, 4, 5, 4, 5, 7, 6, 5, 4, 4, 6, 6, 7, 2, 3, 3, 6, 6, 5, 6, 5, 6, 5, 3, 4, 8
OFFSET
1,2
COMMENTS
Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 29, 34, 5949, 10913.
See also A270566 and A270594 for more similar conjectures.
By the author's work in Sci. China Math. 58(2015), any natural number can be written as the sum of a triangular number, a square and a generalized pentagonal number.
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(1) = 1 since 1 = 1^2 + (0*1/2)^2 + 0*(3*0+1)/2.
a(8) = 1 since 8 = 1^2 + (0*1/2)^2 + 2*(3*2+1)/2.
a(29) = 1 since 29 = 4^2 + (1*2/2)^2 + 3*(3*3-1)/2.
a(34) = 1 since 34 = 5^2 + (2*3/2)^2 + 0*(3*0+1)/2.
a(5949) = 1 since 5949 = 47^2 + (10*11/2)^2 + 22*(3*22-1)/2.
a(10913) = 1 since 10913 = 23^2 + (2*3/2)^2 +83*(3*83+1)/2.
MATHEMATICA
pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
Do[r=0; Do[If[pQ[n-x^2-(y(y+1)/2)^2], r=r+1], {x, 1, Sqrt[n]}, {y, 0, (Sqrt[8*Sqrt[n-x^2]+1]-1)/2}]; Print[n, " ", r]; Continue, {n, 1, 80}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 20 2016
STATUS
approved