cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-1 of 1 results.

A273164 Irregular array read by rows: T(n, k) giving in row n the divisors of nonprime numbers that are 3 (mod 4).

Original entry on oeis.org

1, 3, 5, 15, 1, 3, 9, 27, 1, 5, 7, 35, 1, 3, 13, 39, 1, 3, 17, 51, 1, 5, 11, 55, 1, 3, 7, 9, 21, 63, 1, 3, 5, 15, 25, 75, 1, 3, 29, 87, 1, 7, 13, 91, 1, 5, 19, 95, 1, 3, 9, 11, 33, 99, 1, 3, 37, 111, 1, 5, 23, 115, 1, 7, 17, 119, 1, 3, 41, 123, 1, 3, 5, 9, 15, 27, 45, 135, 1, 11, 13, 143, 1, 3, 7, 21, 49, 147, 1, 5, 31, 155
Offset: 1

Views

Author

Wolfdieter Lang, Jul 29 2016

Keywords

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)

Examples

			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
...
		

Crossrefs

Programs

  • Mathematica
    Divisors@ Select[Range@ 155, CompositeQ@ # && Mod[#, 4] == 3 &] // Flatten (* Michael De Vlieger, Aug 01 2016 *)

Formula

T(n, k) gives the k-th divisor of A091236(n) in increasing order.
Showing 1-1 of 1 results.