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.
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
Keywords
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.
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.
Programs
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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}]
Comments