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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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.

Original entry on oeis.org

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

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Author

Zhi-Wei Sun, Nov 01 2015

Keywords

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.
		

Crossrefs

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}]
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