A239064 Positive integers whose square can be expressed as k*x^2+(k+1)*x+(k+2) with k and x positive integers.

3, 5, 6, 9, 10, 12, 15, 16, 18, 19, 21, 23, 24, 26, 27, 30, 33, 36, 37, 39, 40, 41, 42, 44, 45, 47, 48, 51, 52, 54, 57, 58, 60, 61, 63, 64, 65, 66, 67, 68, 69, 72, 75, 78, 79, 81, 82, 83, 84, 86, 87, 89, 90, 92, 93, 96, 99, 100, 102, 103, 105, 107, 108, 110, 111, 114, 117, 120, 121, 123, 124, 126, 128, 129, 130, 131, 132, 134, 135, 138, 141, 142, 144, 145, 147, 149, 150, 152, 153, 156, 159, 162, 163, 165, 166, 168, 170, 171, 173, 174, 176, 177, 180, 183, 184, 186, 187, 189, 191, 192, 194, 195, 196, 198, 201, 204

Offset: 1

Author

Carmine Suriano, Mar 28 2014

Keywords

• nonn

Comments

Some can be expressed in more than one way. E.g., a(46) = 81 for x = 1, k = 26 and for x = 2 and k = 11.

x=1 for k=3*m^2-1 that is A080663, with a = 3*m.

From Wolfdieter Lang, Apr 26 2014: (Start)

The solution of this problem can be found as follows. Consider k*x^2 + (k+1)*x + (k+2) - a^2 = 0. Solve for x (positive) as function of k: x = (-(k+1) + sqrt((2*a)^2*k - c(k)))/(2*k), where c(k) = 3*k(k+2) - 1 is the sequence [8, 23, 44, 71, 104, 143, 188, 239, 296, 359,...]. A necessary condition for a solution is (2*a)^2*k - c(k) = b^2, or k*X^2 - Y^2 = c(k), with X = 2*a and Y = b. This is a binary indefinite quadratic form with discriminant D = (2^2)*k > 0. If k is a square then there is no solution with even X because (K*X)^2 - Y^2 = c(K^2) has only the solution with K=1, X=3, Y=1. For k not a square there are either no solutions for given k (e.g., k = 5, 6, 10, ...) or countable infinitely many ones of the representation k*X^2 - Y^2 = c(k). From the solutions one has to pick first the ones with even X (a = X/2) and then to test whether 2*k divides Y - (k+1). The solution for x is then x = (Y - (k+1))/(2*k).

For representations of integers by indefinite binary quadratic forms see, e.g., the D. A. Buell reference, and also a W. Lang link (with further references) with an on-line program.

(End)

Examples

			a(10)=19 from 19^2=361=51*2^2+(51+1)*2+(51+2).
From _Wolfdieter Lang_, Apr 21 2014: (Start)
k = 2: There are two proper solutions of 2*X^2 - Y^2 = 23, namely [4, 3], [6, 7]. Both generate infinitely many new solutions, all with even X, namely [4, 3], [18, 25], [104, 147], [606, 857],[3532, 4995], [20586, 29113], ... and [6, 7], [32, 45], [186, 263], [1084, 1533],[6318, 8935], [36824, 52077], ... . Only every other solution has 2*k = 4 dividing Y - (2+1) = Y-3, giving the positive solutions for (a=X/2, b=Y; x), starting with the second proper solution and then alternating between the two sets of solutions (3, 7; 1), (52, 147; 36), (93, 263; 65), (1766, 4995; 1248), (3159, 8935; 2233), ... . Thus the positive x solutions for k = 2 are 1, 36, 65, 1248, 2233, ..., with a = 3, 52, 93, 1766, 3159, ... .
k = 3: the positive solutions for x are 2, 5, 38, 79, 538, 1109, ..., with a = 5, 10, 67, 138, 933, 1922, ..., coming from the even X solutions of 3*X^2 - Y^2 = 44, [4, 2], [10, 16], [36, 62], [134, 232], [500, 866], [1866, 3232], ... and [6, 8], [20, 34], [74, 128], [276, 478], [1030, 1784], [3844, 6658], ... . Then 2*k = 6 has to divide Y - 4, leaving every other of these solutions with (a = X/2, b=Y; x) given by (5, 16; 2), (10, 34; 5), (67, 232; 38), (138, 478; 79), (933, 3232; 538), (1922, 6658; 1109), ... .
k = 5: there are no solutions of 5*X^2 - Y2 = 104.
k = 6: there are no solutions of 6*X^2 - Y2 = 143.
(End)
		

References

• D. A. Buell, Binary Quadratic Forms. Springer-Verlag, NY, 1989.

Links

• Wolfdieter Lang, Binary Quadratic Forms (indefinite case).

Programs

BASIC

for k=1 to z
  for x=1 to z
    a=k*x*x+(k+1)*x+(k+2)
    if sqr(a)-int(sqr(a))=0 then
      j=j+1
      a_n(j)=sqr(a)
    endif
   next x
next k

Formula

Increasingly ordered set {a from the positive integers: a^2 = k*x^2 + (k+1)*x + (k+2), with k and x positive integers}. - Wolfdieter Lang, Apr 21 2014

Extensions

Edited, name specified, comment changed, reference and link added by Wolfdieter Lang, Apr 21 2014