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A239657
Number of odd divisors of n minus the number of parts in the symmetric representation of sigma(n).
18
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
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.
For more information about the symmetric representation of sigma see A237270, A237591.
The first comment 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
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:
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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.
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The above diagram shows the symmetric representation of sigma(15) with its four subparts: [8, 7, 1, 8]. (End)
KEYWORD
nonn
AUTHOR
Omar E. Pol, Mar 23 2014
STATUS
approved