OFFSET
1,4
COMMENTS
Conjecture: the positive terms in row n are the odd divisors of n.
Note that the elements appear with an unusual ordering, for example; row 45 is 1, 45, 3, 0, 5, 15, 0, 0, 9.
The positive terms give A261697.
The number of positive terms in row n is A001227(n).
The sum of row n is A000593(n).
The connection with the symmetric representation of sigma is as follows: A237048 --> A235791 --> A237591 --> A237593.
Proof of the conjecture: let n = 2^m*s*t with s and t odd. The property stated in A237048 verifies the conjecture with odd divisor k <= A003056(n) of n in position k and odd divisor t > A003056(n) in position 2^(m+1)*s. Therefore reading in row n the nonzero odd positions from left to right and then the nonzero even positions from right to left gives all odd divisors of n in increasing order. - Hartmut F. W. Hoft, Oct 25 2015
From Peter Munn, Jul 30 2017: (Start)
Each odd divisor d of n corresponds to n written as a sum of consecutive integers (n/d - (d-1)/2) .. (n/d + (d-1)/2). After canceling any corresponding negative and positive terms and deleting any zero term, the lower bound becomes abs(n/d - d/2) + 1/2, leaving k terms where k = n/d + d/2 - abs(n/d - d/2). It can be shown T(n,k) = d.
This sequence thereby defines a one to one relationship between odd divisors of n and partitions of n into k consecutive parts.
The relationship is expressed below using 4 sequences (with matching row lengths), starting with this one:
A261699(n,k) = d, the odd divisor.
A211343(n,k) = abs(n/d - d/2) + 1/2, smallest part.
A285914(n,k) = k, number of parts.
A286013(n,k) = n/d + (d-1)/2, largest part.
If no partition of n into k consecutive parts exists, the corresponding sequence terms are 0.
(End)
FORMULA
From Hartmut F. W. Hoft, Oct 25 2015: (Start)
T(n, k) = 2n/k, if A237048(n, k) = 1 and k even,
and in accordance with the definition:
T(n, k) = k, if A237048(n, k) = 1 and k odd,
T(n, k) = 0 otherwise; for k <= A003056(n).
(End)
For m >= 1, d >= 1 and odd, T(m*d, m + d/2 - abs(m - d/2)) = d. - Peter Munn, Jul 24 2017
EXAMPLE
Triangle begins:
1;
1;
1, 3;
1, 0;
1, 5;
1, 0, 3;
1, 7, 0;
1, 0, 0;
1, 9, 3;
1, 0, 0, 5;
1, 11, 0, 0;
1, 0, 3, 0;
1, 13, 0, 0;
1, 0, 0, 7;
1, 15, 3, 0, 5;
1, 0, 0, 0, 0;
1, 17, 0, 0, 0;
1, 0, 3, 9, 0;
1, 19, 0, 0, 0;
1, 0, 0, 0, 5;
1, 21, 3, 0, 0, 7;
1, 0, 0, 11, 0, 0;
1, 23, 0, 0, 0, 0;
1, 0, 3, 0, 0, 0;
1, 25, 0, 0, 5, 0;
1, 0, 0, 13, 0, 0;
1, 27, 3, 0, 0, 9;
1, 0, 0, 0, 0, 0, 7;
...
From Omar E. Pol, Dec 19 2016: (Start)
Illustration of initial terms in a right triangle whose structure is the same as the structure of A237591:
Row _
1 _|1|
2 _|1 _|
3 _|1 |3|
4 _|1 _|0|
5 _|1 |5 _|
6 _|1 _|0|3|
7 _|1 |7 |0|
8 _|1 _|0 _|0|
9 _|1 |9 |3 _|
10 _|1 _|0 |0|5|
11 _|1 |11 _|0|0|
12 _|1 _|0 |3 |0|
13 _|1 |13 |0 _|0|
14 _|1 _|0 _|0|7 _|
15 _|1 |15 |3 |0|5|
16 _|1 _|0 |0 |0|0|
17 _|1 |17 _|0 _|0|0|
18 _|1 _|0 |3 |9 |0|
19 _|1 |19 |0 |0 _|0|
20 _|1 _|0 _|0 |0|5 _|
21 _|1 |21 |3 _|0|0|7|
22 _|1 _|0 |0 |11 |0|0|
23 _|1 |23 _|0 |0 |0|0|
24 _|1 _|0 |3 |0 _|0|0|
25 _|1 |25 |0 _|0|5 |0|
26 _|1 _|0 _|0 |13 |0 _|0|
27 _|1 |27 |3 |0 |0|9 _|
28 |1 |0 |0 |0 |0|0|7|
... (End)
MATHEMATICA
T[n_, k_?OddQ] /; n == k (k + 1)/2 := k; T[n_, k_?OddQ] /; Mod[n - k (k + 1)/2, k] == 0 := k; T[n_, k_?EvenQ] /; n == k (k + 1)/2 := k + 1; T[n_, k_?EvenQ] /; Mod[n - k (k + 1)/2, k] == 0 := T[n - k, k] + 2; T[_, _] = 0; Table[T[n, k], {n, 1, 26}, {k, 1, Floor[(Sqrt[1 + 8 n] - 1)/2]}] // Flatten (* Jean-François Alcover, Sep 21 2015 *)
(* alternate definition using function a237048 *)
T[n_, k_] := If[a237048[n, k] == 1, If[OddQ[k], k, 2n/k], 0] (* Hartmut F. W. Hoft, Oct 25 2015 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Sep 20 2015
STATUS
approved