VOOZH about

URL: https://oeis.org/A239657

⇱ A239657 - OEIS

login
A239657
Number of odd divisors m of n such that there is a divisor d of n with d < m < 2*d.
23
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
1,18
COMMENTS
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
Conjecture: number of partitions of n into consecutive parts minus the number of 2-dense sublists of divisors of n. - Omar E. Pol, Nov 22 2025
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..5000 (computed from the b-file of A237271 provided by Michel Marcus)
FORMULA
a(n) = A001227(n) - A237271(n).
EXAMPLE
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)
MATHEMATICA
a[n_] := Module[{d = Partition[Divisors[n], 2, 1]}, Count[d, _?(OddQ[#[[2]]] && #[[2]] < 2*#[[1]] &)]]; Array[a, 100] (* Amiram Eldar, Apr 01 2025 *)
KEYWORD
nonn
AUTHOR
Omar E. Pol, Mar 23 2014
EXTENSIONS
New Name from Omar E. Pol, Jan 26 2025
STATUS
approved