A216503 a(n) = number of positive integers k such that n = x^2 + k*y^2 has a solution with x>0, y>0.
0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 5, 3, 3, 4, 6, 5, 5, 5, 5, 5, 4, 6, 7, 5, 5, 6, 8, 5, 6, 7, 7, 7, 5, 9, 10, 6, 6, 9, 11, 6, 8, 9, 10, 7, 6, 10, 11, 7, 7, 11, 11, 8, 7, 10, 11, 9, 8, 9, 14, 7, 9, 10, 11, 8, 10, 13, 11, 9, 8, 15, 17, 9, 9, 14, 11, 8, 11, 12, 14
Offset: 1
Keywords
Examples
a(17) = 6 because 17 = 4^2 + 1*1^2. 17 = 3^2 + 2*2^2. 17 = 1^2 + 4*2^2. 17 = 3^2 + 8*1^2. 17 = 2^2 + 13*1^2. 17 = 1^2 + 16*1^2. Therefore there are 6 different values of k for which 17 can be written in the form a^2 + k*b^2.
Links
- T. D. Noe, Table of n, a(n) for n = 1..1000
Programs
-
Mathematica
Table[cnt = 0; Do[b = 1; found = False; While[q = n - k*b^2; ! found && q > 0, If[IntegerQ[Sqrt[q]], cnt++; found = True]; b++], {k, n}]; cnt, {n, 100}] (* T. D. Noe, Sep 11 2012 *) -
PARI
for(n=1, 100, sol=0; for(k=1, n-1, for(x=1, n, if(issquare(n-k*x*x)&&n-k*x*x>0, sol++; break))); print1(sol", ")) /* V. Raman, Oct 16 2012 */