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A365410
Parity of A000070.
1
1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1
OFFSET
0
COMMENTS
Parity of the number of 1's in all partitions of (n + 1).
Hence parity of the number of 1's in all divisors of the terms in the (n + 1)st row of the triangle A176206.
FORMULA
a(n) = A000035(A000070(n)).
EXAMPLE
For n = 5 the partitions of 6 and the divisors of terms of the 6th row of A176206 show an example of the correspondence divisor/part in an arrangement as shown below:
.
6
3, 3
4, 2
2, 2, 2
5, 1
3, 2, 1
4, 1, 1
2, 2, 1, 1
3, 1, 1, 1
2, 1, 1, 1, 1
1, 1, 1, 1, 1, 1
------------------
1 2 3 6
1 5
1 2 4
1 2 4
1 3
1 3
1 3
1 2
1 2
1 2
1 2
1 2
1
1
1
1
1
1
1
.
In the above arrangement appear the partitions of 6 and below the divisors of [6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1] the 6th row of A176206.
The parts of all partitions of 6 are also the divisors of the terms of the 6th row of A176206 hence the number of 1's in all partitions of 6 equals the number of 1's in all divisors of the terms in the 6th row of A176206. In each case there are nineteen 1's. The parity of 19 is 1 so a(5) = 1.
MATHEMATICA
Mod[Accumulate@ Array[PartitionsP, 120, 0], 2] (* Michael De Vlieger, Oct 01 2023 *)
KEYWORD
nonn
AUTHOR
Omar E. Pol, Sep 25 2023
STATUS
approved