# [seqfan] (no subject)

Ron Hardin rhhardin at att.net
Mon May 13 18:03:51 CEST 2013

```Question at end

/tmp/dhi
T(n,k)=Number of pairs of orthogonal (-x,y) vectors of length k*(x+y), where x/y
is the n-th rational number <= 1, ordered first by y and then x, e.g. 1/1, 1/2,
1/3, 2/3, 1/4, 3/4 .

Table starts
...4......48..........640.............8960...............129024
...9.....390........20160..........1106424.............62606544
..20....3192.......652784........134432480..........28244153600
..60...48720.....41801760......36951200000.......33792065269760
..45...26010.....21594300......15593341800.......11432293516320
.420.8168160.106322711040.1496349265582080.22046445329510891520

row 1 (two column vectors)
All solutions for k=1
.-1.-1...-1..1...-1..1....1..1
.-1..1....1..1...-1.-1...-1..1

row 2 (two column vectors)
All solutions for k=1
.-1.-1...-1..2...-1..2...-1..2....2..2...-1..2...-1.-1....2..2...-1..2
.-1..2...-1.-1....2.-1....2..2...-1..2....2..2...-1.-1...-1..2...-1..2
.-1.-1...-1.-1....2..2....2.-1....2.-1...-1..2...-1..2...-1..2....2..2

row 3 (two column vectors)
Some solutions for k=1
.-1..3...-1..3....3..3...-1..3...-1..3...-1..3...-1..3...-1.-1....3..3...-1..3
..3..3...-1..3...-1..3....3..3...-1..3....3.-1...-1..3...-1.-1...-1..3....3.-1
.-1..3....3..3...-1..3....3.-1....3.-1....3..3...-1..3...-1.-1....3.-1....3..3
..3.-1....3.-1....3.-1...-1..3....3..3...-1..3....3..3...-1..3....3.-1....3.-1

row 4 (two column vectors)
Some solutions for k=1
.-2..3...-2..3...-2..3....3..3...-2.-2...-2..3...-2.-2...-2.-2...-2..3...-2..3
.-2..3....3.-2....3..3...-2..3...-2..3...-2..3...-2..3...-2..3....3.-2....3.-2
..3..3....3..3...-2..3...-2..3...-2.-2...-2..3...-2.-2...-2.-2....3..3...-2..3
.-2..3....3..3....3..3....3..3...-2..3....3..3...-2.-2....3.-2....3..3....3..3
..3..3....3.-2...-2..3....3.-2...-2.-2....3..3....3.-2...-2.-2...-2..3....3..3

row 5 (two column vectors)
Some solutions for k=1
.-1..4...-1..4....4..4...-1..4...-1..4...-1.-1...-1..4....4..4...-1..4...-1..4
..4..4...-1..4...-1..4....4.-1....4..4...-1.-1....4.-1...-1..4....4.-1....4..4
..4.-1...-1..4...-1..4....4.-1...-1..4...-1.-1....4..4...-1..4....4..4...-1..4
..4.-1....4..4...-1..4...-1..4...-1..4...-1..4...-1..4....4.-1....4.-1...-1..4
.-1..4...-1..4...-1..4....4..4...-1..4...-1.-1....4.-1....4.-1...-1..4....4.-1

row 6 (two column vectors)
Some solutions for k=1
.-3..4...-3..4...-3..4...-3..4....4..4...-3..4....4..4...-3.-3...-3.-3....4..4
.-3..4....4..4...-3.-3....4..4...-3..4....4.-3...-3..4...-3.-3...-3.-3....4..4
.-3..4....4.-3...-3.-3....4..4....4..4....4..4....4..4...-3..4...-3..4...-3..4
..4..4....4..4...-3..4...-3..4...-3..4....4.-3...-3..4...-3..4...-3..4....4.-3
..4.-3....4.-3...-3.-3....4.-3....4.-3...-3..4....4..4....4.-3...-3.-3...-3..4
..4..4...-3..4...-3.-3....4..4....4..4....4..4....4.-3...-3.-3...-3.-3....4..4
..4..4....4..4...-3..4...-3..4....4.-3....4..4...-3..4...-3.-3...-3..4....4.-3

row 7 (two column vectors)
Some solutions for k=1
.-1..5...-1..5...-1..5...-1..5...-1..5...-1..5...-1..5...-1..5...-1..5...-1..5
.-1..5...-1..5...-1.-1...-1..5....5.-1...-1..5...-1..5....5..5....5.-1....5.-1
.-1..5...-1..5...-1.-1....5.-1....5.-1...-1..5....5..5...-1..5...-1..5....5..5
..5..5...-1..5...-1.-1....5..5...-1..5....5.-1...-1..5...-1..5...-1..5...-1..5
.-1..5....5..5...-1.-1...-1..5...-1..5....5.-1...-1..5....5.-1....5..5....5.-1
.-1..5....5.-1...-1.-1...-1..5....5..5....5..5....5.-1...-1..5....5.-1....5.-1

Entries appear to be zero only when the vector lengths of (-x,y) vectors is a
multiple of x+y.

Is this easy to prove?  Does a formula for column 1 fall out of it?

Column 1 starts
4 9 20 60 45 420 102 378 1120 3024 231 22176 520 2376 8160 23760 63360 164736
1161 21780 185328 1235520 2570

Row 1 is http://oeis.org/A098402

rhhardin at mindspring.com
rhhardin at att.net (either)

```