A264025 Number of ways to write n as x^2 + y*(2*y+1) + z*(z+1)/2 where x, y and z are nonnegative integers with z or z+1 prime.

1, 1, 1, 2, 2, 2, 3, 1, 1, 5, 2, 2, 4, 3, 4, 2, 4, 2, 4, 4, 2, 7, 1, 4, 6, 4, 3, 5, 6, 1, 8, 5, 2, 3, 4, 4, 5, 5, 3, 9, 3, 5, 5, 1, 3, 6, 7, 1, 5, 4, 4, 5, 4, 2, 6, 6, 3, 8, 4, 5, 4, 7, 2, 5, 8, 4, 11, 2, 4, 7, 4, 2, 7, 9, 3, 5, 7, 4, 4, 10, 5, 8, 4, 4, 11, 4, 7, 8, 4, 5, 9, 11, 3, 8, 9, 2, 7, 2, 4, 8

Offset: 1

Author

Zhi-Wei Sun, Nov 01 2015

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 8, 9, 23, 30, 44, 48, 198, 219, 1344.

(ii) Any positive integer n not equal to 8 can be written as x*(2*x+1) + y*(y+1)/2 + z*(z+1)/2, where x, y and z are nonnegative integers with z or z+1 prime.

(iii) Any integer n > 1 can be written as x^2 + y*(y+1) + z*(z+1) (or 2*x^2 + y*(y+1)/2 + z*(z+1)), where x, y and z are nonnegative integers with z or z+1 prime.

(iv) Each integer n > 2 can be written as x^2 + y*(y+1)/2 + 3*z*(z+1)/2, where x, y and z are nonnegative integers with z or z+1 prime.

(v) Every n = 1,2,3,... can be written as 2*x^2 + y*(y+1)/2 + z*(z+1)/2, where x, y and z are nonnegative integers with z or z+1 prime. Also, any integer n > 4 can be written as 2*x^2 + y*(y+1) + z*(z+1)/2, where x, y and z are nonnegative integers with z or z+1 prime.

Note that the integers n*(2*n+1) = 2n*(2n+1)/2 (n = 0,1,2,...) are second hexagonal numbers.

See also A262785, A263998 and A264010 for similar conjectures.

Examples

			a(1) = 1 since 1 = 0^2 + 0*(2*0+1) + 1*2/2 with 2 prime.
a(2) = 1 since 2 = 1^2 + 0*(2*0+1) + 1*2/2 with 2 prime.
a(3) = 1 since 3 = 0^2 + 0*(2*0+1) + 2*3/2 with 2 prime.
a(8) = 1 since 8 = 2^2 + 1*(2*1+1) + 1*2/2 with 2 prime.
a(9) = 1 since 9 = 0^2 + 1*(2*1+1) + 3*4/2 with 3 prime.
a(23) = 1 since 23 = 1^2 + 3*(2*3+1) + 1*2/2 with 2 prime.
a(30) = 1 since 30 = 3^2 + 0*(2*0+1) + 6*7/2 with 7 prime.
a(44) = 1 since 44 = 4^2 + 0*(2*0+1) + 7*8/2 with 7 prime.
a(48) = 1 since 48 = 3^2 + 4*(2*4+1) + 2*3/2 with 2 prime.
a(198) = 1 since 198 = 3^2 + 4*(2*4+1) + 17*18/2 with 17 prime.
a(219) = 1 since 219 = 6^2 + 7*(2*7+1) + 12*13/2 with 13 prime.
a(1344) = 1 since 1344 = 21^2 + 0*(2*0+1) + 42*43/2 with 43 prime.
		

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 ax^2+by^2+f(z), aT_x+bT_y+f(z) and aT_x+by^2+f(z), arXiv:1502.03056 [math.NT], 2015.

Crossrefs

Cf. A000040, A000217, A000290, A014105, A262785, A263998, A264010.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[(PrimeQ[z]||PrimeQ[z+1])==False,Goto[aa]];Do[If[SQ[n-z(z+1)/2-y(2y+1)],r=r+1],{y,0,(Sqrt[8(n-z(z+1)/2)+1]-1)/4}];Label[aa];Continue,{z,1,(Sqrt[8n+1]-1)/2}];Print[n, " ", r];Continue, {n,1,100}]