OFFSET
1,7
COMMENTS
Conjectures: (i) a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3, 4, 5, 6, 10, 11, 15, 20, 29, 1125.
(ii) Any integer n > 2 can be written as x*(x+1) + y*(y+1)/2 + z*(z+1)/2, where x, y and z are nonnegative integers such that x or x+1 is prime, and y or y+1 is prime.
(iii) Any integer n > 7 can be written as x*(x+1) + y*(y+1)/2 + 3*z*(z+1)/2, where x, y and z are nonnegative integers such that y or y+1 is prime, and z or z+1 is prime.
It is known that any natural number can be written as x^2 + y*(y+1)+ z*(z+1)/2 (or x*(x+1) + y*(y+1)/2 + z*(z+1)/2, or x*(x+1) + y*(y+1)/2 + 3*z(z+1)/2) with x, y and z nonnegative integers.
See also A264025 for similar conjectures.
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.
EXAMPLE
a(5) = 1 since 5 = 0^2 + 1*2 + 2*3/2 with 2 prime.
a(6) = 1 since 6 = 1^2 + 1*2 + 2*3/2 with 2 prime.
a(10) = 1 since 10 = 1^2 + 2*3 + 2*3/2 with 2 prime.
a(11) = 1 since 11 = 2^2 + 2*3 + 1*2/2 with 2 prime.
a(15) = 1 since 15 = 0^2 + 3*4 + 2*3/2 with 3 prime.
a(20) = 1 since 20 = 2^2 + 2*3 + 4*5/2 with 2 and 5 both prime.
a(29) = 1 since 29 = 4^2 + 3*4 + 1*2/2 with 3 and 2 both prime.
a(1125) = 1 since 1125 = 33^2 + 5*6 + 3*4/2 with 5 and 3 both prime.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[(PrimeQ[y]||PrimeQ[y+1])==False, Goto[aa]]; Do[If[(PrimeQ[z]||PrimeQ[z+1])&&SQ[n-y(y+1)-z(z+1)/2], r=r+1], {z, 1, (Sqrt[8(n-y(y+1))+1]-1)/2}]; Label[aa]; Continue, {y, 1, (Sqrt[4n+1]-1)/2}]; Print[n, " ", r]; Continue, {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 31 2015
STATUS
approved