OFFSET
1,3
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.
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.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 01 2015
STATUS
approved