A230121 Number of ways to write n = x + y + z (0 < x <= y <= z) such that x*(x+1)/2 + y*(y+1)/2 + z*(z+1)/2 is a triangular number.

0, 0, 1, 0, 0, 1, 0, 1, 2, 1, 1, 0, 2, 1, 2, 1, 2, 3, 2, 2, 6, 1, 3, 5, 1, 2, 3, 5, 2, 1, 3, 3, 3, 4, 3, 8, 2, 5, 11, 2, 5, 8, 4, 6, 4, 9, 4, 6, 5, 4, 6, 3, 8, 8, 5, 8, 10, 7, 7, 11, 8, 6, 7, 8, 5, 9, 7, 6, 8, 7, 7, 8, 13, 9, 11, 10, 7, 22, 9, 10, 13, 3, 6, 10, 8, 17, 12, 7, 9, 10, 16, 6, 18, 18, 10, 15, 9, 12, 20, 5

Offset: 1

Author

Zhi-Wei Sun, Oct 10 2013

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 except for n = 1, 2, 4, 5, 7, 12. Moreover, for each n = 20, 21, ... there are three distinct positive integers x, y and z with x + y + z = n such that x*(x+1)/2 + y*(y+1)/2 + z*(z+1)/2 is a triangular number.

(ii) A positive integer n cannot be written as x + y + z (x, y, z > 0) with x^2 + y^2 + z^2 a square if and only if n has the form 2^r*3^s or the form 2^r*7, where r and s are nonnegative integers.

(iii) Any integer n > 14 can be written as a + b + c + d, where a, b, c, d are positive integers with a^2 + b^2 + c^2 + d^2 a square. If n > 20 is not among 22, 28, 30, 38, 44, 60, then we may require additionally that a, b, c, d are pairwise distinct.

(iv) For each integer n > 50 not equal to 71, there are positive integers a, b, c, d with a + b + c + d = n such that both a^2 + b^2 and c^2 + d^2 are squares.

Part (ii) and the first assertion in part (iii) were confirmed by Chao Huang and Zhi-Wei Sun in 2021. - Zhi-Wei Sun, May 09 2021

Examples

			a(16) = 1 since 16 = 3 + 6 + 7 and 3*4/2 + 6*7/2 + 7*8/2 = 55 = 10*11/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..3000
• Chao Huang and Zhi-Wei Sun, On partitions of integers with restrictions involving squares, arXiv:2105.03416 [math.NT], 2021.
• Zhi-Wei Sun, Diophantine problems involving triangular numbers and squares, a message to Number Theory List, Oct. 11, 2013.

Crossrefs

Cf. A000217, A000290.

Cf. A008443, A053604, A053603, A052343, A052344, A008441.

Programs

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
T[n_]:=n(n+1)/2
a[n_]:=Sum[If[SQ[8(T[i]+T[j]+T[n-i-j])+1],1,0],{i,1,n/3},{j,i,(n-i)/2}]
Table[a[n],{n,1,100}]

PARI

a(n)=my(t=(n+1)*n/2,s);sum(x=1,n\3,s=t-n--*x;sum(y=x,n\2,is_A000217(s-(n-y)*y))) \\ - M. F. Hasler, Oct 11 2013