

A259288


Odd numbers of the form (m*k)^2/(m^2k^2) for distinct integers m and k.


1



147, 225, 405, 1323, 2025, 3645, 3675, 4225, 5625, 7203, 7623, 10125, 11025, 11907, 14415, 17457, 17787, 18225, 18513, 19845, 24375, 24843, 27225, 30625, 32805, 33075, 38025, 42483, 49005, 50625, 53067, 61347, 64827, 65025, 68445, 68607, 77763, 81225, 91125, 91875, 98397, 99225, 105625, 107163, 117045, 119025
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OFFSET

1,1


COMMENTS

The first term ending in a 9 seems to be 1225449, and the first term ending in a 1 is 136161.
For 1 <= m <= 10^4 and 1 <= k <= m, there are 9217 numbers of the form (m*k)^2/(m^2k^2). Of these numbers, only 679 are odd.
If a(n) is not a square, then m = 9*k or m = 7*k. If a(n) is a square, m does not appear to be a multiple of k.
Let a(n) be a square generated by m_1 and k_1. If a(n1) is generated by m_2 and k_2, then k_1 = k_2 and m_1 < m_2.
The reciprocals of these numbers can be represented as the difference in the reciprocals of two squares (i.e., there exists two distinct integers m and k satisfying 1/a(n) = 1/m^2  1/k^2).


LINKS

Table of n, a(n) for n=1..46.


EXAMPLE

(84*12)^2/(84^212^2) = 84^2/48 = 147. So 147 is a member of this sequence. (Note that k=12 and m=84 and so m=7*k.)


PROG

(PARI) v=[]; for(m=1, 7500, for(n=1, m1, if(type(s=(m*n)^2/(m^2n^2))=="t_INT"&&(s%2), v=concat(v, s)))); vecsort(v, , 8)


CROSSREFS

Cf. A259263, A111200.
Sequence in context: A195239 A122064 A320511 * A325608 A255103 A184542
Adjacent sequences: A259285 A259286 A259287 * A259289 A259290 A259291


KEYWORD

nonn


AUTHOR

Derek Orr, Jun 23 2015


STATUS

approved



