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).

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, Table of n, a(n) for n = 1..10000

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}]

Crossrefs

Cf. A000217, A000290, A001318, A262813, A262815, A262816, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559, A270566, A270594.

Sequence in context: A270559 A231727 A368876 * A304523 A368541 A020649

Adjacent sequences: A270613 A270614 A270615 * A270617 A270618 A270619

Keyword

nonn

Author

Zhi-Wei Sun, Mar 20 2016

Status

approved