

A248194


Positive integers n such that the equation x^2  n*y^2 = n*(n+1)/2 has integer solutions.


2



1, 3, 8, 9, 17, 19, 24, 25, 33, 49, 51, 57, 67, 72, 73, 81, 88, 89, 96, 97, 99, 121, 129, 136, 147, 152, 163, 169, 177, 179, 193, 201, 211, 225, 233, 241, 243, 249, 264, 288, 289, 297, 313, 337, 339, 352, 361, 369, 387, 393, 408, 409, 441, 449, 451, 456, 457
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OFFSET

1,2


COMMENTS

All odd squares are in this sequence. Proof: Set n = (2k + 1)^2, then we have x^2  (2k + 1)^2 * y^2 = (2k + 1)^2 * (2k^2 + 2k + 1). Rearranging gives x^2 = (2k + 1)^2 * (y^2 + 2k^2 + 2k + 1). As 2k^2 + 2k + 1 is odd, a careful selection of y makes the RHS square. So [(2k+1) * (k(k + 1) + 1), k(k + 1)]. E.g., if k=2, then (5*7)^2  25*6^2 = 1225  900 = 325 = 25*26/2.  Jon Perry, Nov 07 2014
No even squares are in the sequence. Proof: Rearrange the equation to read x^2 = n(n + 1 + 2y^2)/2, with n = 4k^2. n + 1 + 2y^2 is always odd and so the RHS contains an odd exponent of 2, and therefore cannot be square.  Jon Perry, Nov 15 2014
From Jon Perry, Nov 15 2014: (Start)
Odd squares + 8 are always in this sequence. Proof: Let m = 4k^2 + 4k + 9 and let n = (m+1)/2 = 2k^2 + 2k + 5.
Rearranging the equation x^2  m*y^2 = m(m + 1)/2, we get x^2 = m(m + 1 + 2y^2)/2, and so x^2 = m(n + y^2) = (4k^2 + 4k + 9)(2k^2 + 2k + 5 + y^2).
We aim to find a y such that the last bracket on the RHS is z^2 * (4k^2 + 4k + 9), so that x equals z*m. We claim that if we let Y = ((n3)/2)^2*m  n, then Y is a square, and letting Y = y^2, we have y^2 + n = Y + n = z^2 * m as required, with z = (n3)/2 = k^2 + k + 1.
To prove that Y is a square, Y = [(n^2  6n + 9)*(2n  1)  4n]/4 = [2n^3  13n^2 + 20n  9]/4 = [(n1)^2*(2n9)]/4, and with n as it is, 2n  9 = 4k^2 + 4k + 1 = (2k + 1)^2, and so we arrive at Y = [(n1)^2*(2k+1)^2]/4 = [(n1)(2k+1)/2]^2 = [(k^2 + k + 2)(2k + 1)]^2, a square as required, with y = (k^2 + k + 2)(2k + 1). Also GCD(n3,2n1)=1 as required.
This gives a solution as [(k^2 + k + 1)*(4k^2 + 4k + 9), (k^2 + k + 2)*(2k + 1)]. E.g., if k=4, n=45 and a solution is [21*89, 22*9] = [1869, 198]. To validate, 1869^2  89*198^2 = 3493161  3489156 = 4005 = 89*45.
(End)


LINKS

Lars Blomberg, Table of n, a(n) for n = 1..10000


EXAMPLE

3 is in the sequence because x^2  3*y^2 = 6 has integer solutions, including (x, y) = (3, 1) and (9, 5).


CROSSREFS

Cf. A134419, A245226.
Sequence in context: A152411 A080517 A264898 * A295289 A212849 A191487
Adjacent sequences: A248191 A248192 A248193 * A248195 A248196 A248197


KEYWORD

nonn


AUTHOR

Colin Barker, Oct 03 2014


EXTENSIONS

More terms from Lars Blomberg, Nov 02 2014
"Positive" added to definition by N. J. A. Sloane, Nov 02 2014


STATUS

approved



