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A309511
Number of odd parts in the partitions of n into 3 parts.
1
0, 0, 0, 3, 2, 4, 4, 8, 8, 13, 12, 18, 18, 24, 24, 33, 32, 40, 40, 50, 50, 61, 60, 72, 72, 84, 84, 99, 98, 112, 112, 128, 128, 145, 144, 162, 162, 180, 180, 201, 200, 220, 220, 242, 242, 265, 264, 288, 288, 312, 312, 339, 338, 364, 364, 392, 392, 421, 420
OFFSET
0,4
FORMULA
a(n) = Sum_{j=1..floor(n/3)} Sum_{i=j..floor((n-j)/2)} ((i mod 2) + (j mod 2) + ((n-i-j) mod 2)).
Conjectures from Colin Barker, Aug 06 2019: (Start)
G.f.: x^3*(3 - x + 2*x^2 + x^4 + x^5) / ((1 - x)^3*(1 + x)^2*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-4) - a(n-5) + a(n-6) - a(n-7) - a(n-10) + a(n-11) for n>10.
(End)
EXAMPLE
Figure 1: The partitions of n into 3 parts for n = 3, 4, ...
1+1+8
1+1+7 1+2+7
1+2+6 1+3+6
1+1+6 1+3+5 1+4+5
1+1+5 1+2+5 1+4+4 2+2+6
1+1+4 1+2+4 1+3+4 2+2+5 2+3+5
1+1+3 1+2+3 1+3+3 2+2+4 2+3+4 2+4+4
1+1+1 1+1+2 1+2+2 2+2+2 2+2+3 2+3+3 3+3+3 3+3+4 ...
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n | 3 4 5 6 7 8 9 10 ...
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a(n) | 3 2 4 4 8 8 13 12 ...
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MATHEMATICA
Table[Sum[Sum[Mod[i, 2] + Mod[j, 2] + Mod[n - i - j, 2], {i, j, Floor[(n - j)/2]}], {j, Floor[n/3]}], {n, 0, 80}]
Table[Count[Flatten[IntegerPartitions[n, {3}]], _?OddQ], {n, 0, 60}] (* Harvey P. Dale, Jan 16 2022 *)
CROSSREFS
Sequence in context: A371180 A147604 A095401 * A195472 A370806 A240538
KEYWORD
nonn
AUTHOR
Wesley Ivan Hurt, Aug 05 2019
STATUS
approved