cp's OEIS Frontend

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.

A262785 Number of ordered ways to write n as x^2 + y^2 + p*(p+d)/2, where 0 <= x <= y, d is 1 or -1, and p is prime.

Original entry on oeis.org

1, 1, 3, 2, 3, 2, 3, 3, 1, 3, 5, 3, 2, 3, 3, 4, 2, 2, 6, 4, 4, 2, 6, 2, 2, 5, 2, 7, 4, 4, 4, 5, 3, 1, 7, 2, 5, 4, 4, 5, 4, 3, 3, 5, 1, 6, 4, 3, 1, 3, 5, 3, 8, 2, 7, 6, 3, 2, 4, 5, 3, 4, 2, 6, 5, 4, 5, 9, 2, 3, 9, 1, 5, 5, 7, 4, 3, 5, 5, 7, 3, 5, 7, 5, 3, 8, 4, 7, 4, 2, 9, 7, 6, 2, 9, 6, 1, 3, 3, 9
Offset: 1

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Author

Zhi-Wei Sun, Oct 01 2015

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 9, 34, 45, 49, 72, 97, 241, 337, 538.
(ii) Any integer n > 9 can be written as x^2 + y^2 + z*(z+1), where x,y,z are nonnegative integers with z-1 or z+1 prime.
In 2015, the author refined a result of Euler by proving that any positive integer can be written as the sum of two squares and a positive triangular number.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 2*(2-1)/2 with 2 prime.
a(2) = 1 since 2 = 0^2 + 1^2 + 2*(2-1)/2 with 2 prime.
a(3) = 3 since 3 = 0^2 + 0^2 + 2*(2+1)/2 = 0^2 + 0^2 + 3*(3-1)/2 = 1^2 + 1^2 + 2*(2-1)/2 with 2 and 3 both prime.
a(9) = 1 since 9 = 2^2 + 2^2 + 2*(2-1)/2 with 2 prime.
a(34) = 1 since 34 = 2^2 + 3^2 + 7*(7-1)/2 with 7 prime.
a(45) = 1 since 45 = 1^2 + 4^2 + 7*(7+1)/2 with 7 prime.
a(49) = 1 since 49 = 3^2 + 5^2 + 5*(5+1)/2 with 5 prime.
a(72) = 1 since 72 = 1^2 + 4^2 + 11*(11-1)/2 with 11 prime.
a(97) = 1 since 97 = 1^2 + 9^2 + 5(5+1)/2 with 5 prime.
a(241) = 1 since 241 = 1^2 + 15^2 + 5*(5+1)/2 with 5 prime.
a(337) = 1 since 337 = 5^2 + 6^2 + 23*(23+1)/2 with 23 prime.
a(538) = 1 since 538 = 3^2 + 8^2 + 31*(31-1)/2 with 31 prime.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
    f[d_,n_]:=Prime[n](Prime[n]+(-1)^d)/2
    Do[r=0;Do[If[SQ[n-f[d,k]-x^2],r=r+1],{d,0,1},{k,1,PrimePi[(Sqrt[8n+1]-(-1)^d)/2]},{x,0,Sqrt[(n-f[d,k])/2]}];Print[n," ",r];Continue,{n,1,100}]