A308034 Number of partitions of n into 3 parts with at least 1 part that is a nondivisor of n.
0, 0, 0, 0, 2, 1, 4, 4, 6, 8, 10, 9, 14, 16, 18, 20, 24, 25, 30, 32, 36, 40, 44, 45, 52, 56, 60, 64, 70, 73, 80, 84, 90, 96, 102, 105, 114, 120, 126, 132, 140, 145, 154, 160, 168, 176, 184, 189, 200, 208, 216, 224, 234, 241, 252, 260, 270, 280, 290, 297, 310
Offset: 1
Examples
7 = 2 + (1 + 4) = 2 + (2 + 3) = 3 + (1 + 3) = 5 + (1 + 1); the first integer corresponds to one part that is a nondivisor of 7. So a(7) = 4. - _Bernard Schott_, May 12 2019
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) | 0 0 2 1 4 4 6 8 ...
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- _Wesley Ivan Hurt_, Sep 07 2019
Crossrefs
Cf. A284825.
Programs
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Mathematica
Table[Sum[Sum[1 - (1 - Ceiling[n/i] + Floor[n/i])*(1 - Ceiling[n/k] + Floor[n/k])*(1 - Ceiling[n/(n - i - k)] + Floor[n/(n - i - k)]), {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}]
Formula
a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} 1 - chi(n/i) * chi(n/k) * chi(n/(n-i-k)), where chi(n) = 1 - ceiling(n) + floor(n).
Conjectures from Colin Barker, May 11 2019: (Start)
G.f.: x^5*(2 - 3*x + 6*x^2 - 6*x^3 + 6*x^4 - 3*x^5) / ((1 - x)^3*(1 + x)*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)).
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4) + a(n-6) - 2*a(n-7) + 2*a(n-8) - 2*a(n-9) + a(n-10) for n>10.
(End)