OFFSET
1,5
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)
EXAMPLE
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
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}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, May 10 2019
STATUS
approved