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%I #28 Oct 08 2023 09:35:15
%S 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,
%T 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,
%U 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
%N Parity of A000070.
%C Parity of the number of 1's in all partitions of (n + 1).
%C Hence parity of the number of 1's in all divisors of the terms in the (n + 1)st row of the triangle A176206.
%F a(n) = A000035(A000070(n)).
%e 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:
%e .
%e 6
%e 3, 3
%e 4, 2
%e 2, 2, 2
%e 5, 1
%e 3, 2, 1
%e 4, 1, 1
%e 2, 2, 1, 1
%e 3, 1, 1, 1
%e 2, 1, 1, 1, 1
%e 1, 1, 1, 1, 1, 1
%e ------------------
%e 1 2 3 6
%e 1 5
%e 1 2 4
%e 1 2 4
%e 1 3
%e 1 3
%e 1 3
%e 1 2
%e 1 2
%e 1 2
%e 1 2
%e 1 2
%e 1
%e 1
%e 1
%e 1
%e 1
%e 1
%e 1
%e .
%e 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.
%e 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.
%t Mod[Accumulate@ Array[PartitionsP, 120, 0], 2] (* _Michael De Vlieger_, Oct 01 2023 *)
%Y Cf. A000035, A000041, A000070, A040051, A176206, A336811, A338156.
%K nonn
%O 0
%A _Omar E. Pol_, Sep 25 2023
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