A067274 Number of ordered integer pairs (b,c), with -n<=b<=n, -n<=c<=n, such that both roots of x^2+bx+c=0 are integers.

1, 4, 10, 16, 25, 31, 41, 47, 57, 66, 76, 82, 96, 102, 112, 122, 135, 141, 155, 161, 175, 185, 195, 201, 219, 228, 238, 248, 262, 268, 286, 292, 306, 316, 326, 336, 357, 363, 373, 383, 401, 407, 425, 431, 445, 459, 469, 475, 497, 506, 520, 530, 544, 550, 568

Offset: 0

Author

John W. Layman, Feb 21 2002

Keywords

• nonn

Comments

Conjecture: The difference a(n)-a(n-1) is 6 if and only if n is a prime number. This has been checked up to about n=300 and may be easy to prove.

Preceding conjecture is a corollary of Jovovic's formula below. - Eric M. Schmidt, Aug 19 2012

Links

• Eric M. Schmidt, Table of n, a(n) for n = 0..10000
• Eric Weisstein's World of Mathematics, Quadratic Equation

Crossrefs

Cf. A010052.

Programs

Mathematica

a[n_] := If[n >= 1, 2 Sum[Length[Divisors[k]], {k, n}] + Floor[Sqrt[n]] + 2 n - 1]; Join[{1}, Table[a[n], {n, 50}]] (* Lorenz H. Menke, Jr., Apr 13 2016 *)

Sage

def A067274(max) :
    res = [1]
    term = 4
    for ii in range(1, max+1) :
        res += [term]
        term += 2 * (number_of_divisors(ii+1) + 1) + is_square(ii+1)
    return res
# Eric M. Schmidt, Aug 19 2012

Formula

a(n) = a(n-1)+2*(tau(n)+1)+A010052(n), n>1, a(1) = 4. - Vladeta Jovovic, Mar 05 2002, edited by Eric M. Schmidt, Aug 19 2012