A383147 -id:A383147 - cp's OEIS frontend

A239657 Number of odd divisors m of n such that there is a divisor d of n with d < m < 2*d.

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 3, 0, 0, 0, 0, 1, 2, 0, 0, 0, 1, 0, 3, 0, 0, 3, 0, 0, 1, 0, 0, 0, 0, 0, 3, 0, 1, 0, 0, 0, 3, 0, 0, 1, 0, 0, 3, 0, 0, 0, 1, 0, 2, 0, 0, 2, 0, 1, 2, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 5, 1, 0, 0, 0, 0, 1, 0, 0, 1, 2, 0, 2, 0, 1, 4, 0, 0, 3, 0, 1, 0, 1, 0, 2, 0, 0, 1, 0, 0, 3, 0, 0, 0, 0, 0, 5, 0, 0

Offset: 1

Omar E. Pol, Mar 23 2014

nonn

The original name was: Number of odd divisors of n minus the number of parts in the symmetric representation of sigma(n).
Observation: at least the indices of the first 42 positive elements coincide with A005279: 6, 12, 15, 18, 20, 24..., checked (by hand) up to n = 2^7.
The observation is true for the indices of all positive elements. Hence the indices of the zeros give A174905. - Omar E. Pol, Jan 06 2017
a(n) is the number of subparts minus the number of parts in the symmetric representation of sigma(n). For the definition of "subpart" see A279387. - Omar E. Pol, Sep 26 2018
a(n) is the number of subparts of the symmetric representation of sigma(n) that are not in the first layer. - Omar E. Pol, Jan 26 2025

			Illustration of the symmetric representation of sigma(15) = 24 in the third quadrant:
.      _
.     | |
.     | |
.     | |
.     | |
.     | |
.     | |
.     | |
.     |_|_ _ _
.    8      | |_ _
.           |_    |
.             |_  |_
.            8  |_ _|
.                   |
.                   |_ _ _ _ _ _ _ _
.                   |_ _ _ _ _ _ _ _|
.                 8
.
For n = 15 the divisors of 15 are 1, 3, 5, 15, so the number of odd divisors of 15 is equal to 4. On the other hand the parts of the symmetric representation of sigma(15) are [8, 8, 8], there are three parts, so a(15) = 4 - 3 = 1.
From _Omar E. Pol_, Sep 26 2018: (Start)
Also the number of odd divisors of 15 equals the number of partitions of 15 into consecutive parts and equals the number of subparts in the symmetric representation of sigma(15). Then we have that the number of subparts minus the number of parts is  4 - 3 = 1, so a(15) = 1.
.      _
.     | |
.     | |
.     | |
.     | |
.     | |
.     | |
.     | |
.     |_|_ _ _
.    8      | |_ _
.           |_ _  |
.          7  |_| |_
.            1  |_ _|
.                   |
.                   |_ _ _ _ _ _ _ _
.                   |_ _ _ _ _ _ _ _|
.                 8
.
The above diagram shows the symmetric representation of sigma(15) with its four subparts: [8, 7, 1, 8]. (End)
From _Omar E. Pol_, Mar 30 2025: (Start)
The above diagram also shows that in the first layer there are three parts (having sizes [8, 7, 8]). Also there is another part that is not in the first layer, so a(15) = 1.
On the other hand for n = 15 there is only one odd divisor m of 15 such that  d < m < 2*d and d divides 15. That odd divisor is 5 as shown below, so a(15) = 1.
   d  <  m  <  2*d
--------------------
   1            2
   3     5      6
   5           10
  15           30
.
For n = 18 there are two odd divisors m of 18 such that  d < m < 2*d and d divides 18. Those odd divisors are 3 and 9 as shown below, so a(18) = 2.
   d  <  m  <  2*d
--------------------
   1            2
   2     3      4
   3            6
   6     9     12
   9           18
  18           36
.
(End)

Antti Karttunen, Table of n, a(n) for n = 1..5000 (computed from the b-file of A237271 provided by Michel Marcus)

Cf. A000203, A001227, A005279, A174905, A196020, A236104, A235791, A237048, A237270, A237271, A237591, A237593, A245092, A262626, A279387, A279391, A262626, A286000, A286001, A296508, A347186, A383147.

a[n_] := Module[{d = Partition[Divisors[n], 2, 1]}, Count[d, ?(OddQ[#[[2]]] && #[[2]] < 2*#[[1]] &)]]; Array[a, 100] (* _Amiram Eldar, Apr 01 2025 *)

a(n) = A001227(n) - A237271(n).

New Name from Omar E. Pol, Jan 26 2025

A383209 Irregular triangle read by rows in which row n lists the odd divisors m of n such that there is a divisor d of n with d < m < 2*d, or 0 if such odd divisors do not exist.

Original entry on oeis.org

0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 3, 0, 0, 5, 0, 0, 3, 9, 0, 5, 0, 0, 0, 3, 0, 0, 0, 7, 0, 3, 5, 15, 0, 0, 0, 0, 7, 3, 9, 0, 0, 0, 5, 0, 3, 7, 21, 0, 0, 5, 9, 15, 0, 0, 3, 0, 0, 0, 0, 0, 3, 9, 27, 0, 7, 0, 0, 0, 3, 5, 15, 0, 0, 9, 0, 0, 3, 11, 33, 0, 0, 0, 7, 0, 3, 9, 0, 0, 5, 25, 0, 11, 3, 39

Offset: 1

Omar E. Pol, Apr 19 2025

			For n = 1..17 every row of the triangle has only one term.
For n = 18..30 the triangle is as shown below:
  3, 9;
  0;
  5;
  0;
  0;
  0;
  3;
  0;
  0;
  0;
  7;
  0;
  3, 5, 15;
  ...
For n = 30 there are three odd divisors m of 30 such that there is a divisor d of 30 with d < m < 2*d. Those odd divisors are 3, 5 and 15 as shown below:
   d  <  m  <  2*d
--------------------
   1            2
   2     3      4
   3     5      6
   5           10
   6           12
  10    15     20
  15           30
  30           60
.
So the 30th row of the triangle is [3, 5, 15].
.
For n = 78 there are two odd divisors m of 78 such that there is a divisor d of 78 with d < m < 2*d. Those odd divisors are 3 and 39 as shown below:
   d  <  m  <  2*d
--------------------
   1            2
   2     3      4
   3            6
   6           12
  13           26
  26    39     52
  39           78
  78          156
.
Note that 13 is an odd divisor of 78 but 13 does not qualify.
So the 78th row of the triangle is [3, 39].

Also zeros and odd terms of A379461.
Row sums give A383147.
The number of positive terms in row n is A239657(n).
Cf. A027750, A237593, A379288, A379374.

row[n_] := Module[{d = Partition[Divisors[n], 2, 1], r}, r = Select[d, OddQ[#[[2]]] && #[[2]] < 2*#[[1]] &][[;; , 2]]; If[r == {}, {0}, r]]; Array[row, 80] // Flatten (* Amiram Eldar, Apr 19 2025 *)

cp's OEIS Frontend

A239657 Number of odd divisors m of n such that there is a divisor d of n with d < m < 2*d.

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