OFFSET
1,2
COMMENTS
The length of row n is 2*A273165(n).
The number of divisors 1 and -1 (mod 4) in each row are identical, namely A273165(n). See the Jan 05 2004 Jovovic comment on A078703. For prime numbers 3 (mod 4) this is obvious. For the proof see a comment on A091236 with the Grosswald reference.
From Paul Curtz, Jul 31 2016: (Start)
For each row n of length 2*r(n) one has:
T(n, m)*T(n, 2*r(n)-m+1) = T(n, 2*r(n)),for m=1, 2, ... , r(n).
From the second comment it follows that the row sums are congruent to 0 modulo 4. (End)
FORMULA
T(n, k) gives the k-th divisor of A091236(n) in increasing order.
EXAMPLE
The irregular array T(n, k) begins:
n\k 1 2 3 4 5 6 7 8 ...
1: 1 3 5 15
2: 1 3 9 27
3: 1 5 7 35
4: 1 3 13 39
5: 1 3 17 51
6: 1 5 11 55
7: 1 3 7 9 21 63
8: 1 3 5 15 25 75
9: 1 3 29 87
10: 1 7 13 91
11: 1 5 19 95
12: 1 3 9 11 33 99
13: 1 3 37 111
14: 1 5 23 115
15: 1 7 17 119
16: 1 3 41 123
17: 1 3 5 9 15 27 45 135
18: 1 11 13 143
19: 1 3 7 21 49 147
20: 1 5 31 155
...
The irregular array modulo 4 gives (-1 for 3 (mod 4)):
n\k 1 2 3 4 5 6 7 8 ...
1: 1 -1 1 -1
2: 1 -1 1 -1
3: 1 1 -1 -1
4: 1 -1 1 -1
5: 1 -1 1 -1
6: 1 1 -1 -1
7: 1 -1 1 -1 1 -1
8: 1 -1 1 -1 1 -1
9: 1 -1 1 -1
10: 1 -1 1 -1
11: 1 1 -1 -1
12: 1 -1 1 -1 1 -1
13: 1 -1 1 -1
14: 1 1 -1 -1
15: 1 -1 1 -1
16: 1 -1 1 -1
17: 1 -1 1 1 -1 -1 1 -1
18: 1 -1 1 -1
19: 1 -1 -1 1 1 -1
20: 1 1 -1 -1
...
MATHEMATICA
Divisors@ Select[Range@ 155, CompositeQ@ # && Mod[#, 4] == 3 &] // Flatten (* Michael De Vlieger, Aug 01 2016 *)
CROSSREFS
KEYWORD
nonn,easy,tabf
AUTHOR
Wolfdieter Lang, Jul 29 2016
STATUS
approved