OFFSET
0,6
LINKS
Jinyuan Wang, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1,0,0,2,-2,-2,2,0,0,-1,1,1,-1).
FORMULA
a(n) = Sum_{j=1..floor(n/3)} Sum_{i=j..floor((n-j)/2)} j * (j mod 2).
From Colin Barker, Aug 22 2019: (Start)
G.f.: x^3*(1 + x^2)*(1 - x^2 + x^4) / ((1 - x)^4*(1 + x)^3*(1 - x + x^2)^2*(1 + x + x^2)^2).
a(n) = a(n-1) + a(n-2) - a(n-3) + 2*a(n-6) - 2*a(n-7) - 2*a(n-8) + 2*a(n-9) - a(n-12) + a(n-13) + a(n-14) - a(n-15) for n > 14.
(End)
a(n) = (-4*s^3+(2*t-7)*s^2+(4*t-1)*s+2*t+2)/2, where s = floor((n-3)/6) and t = floor((n-3)/2). - Wesley Ivan Hurt, Oct 27 2021
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) | 1 1 2 2 3 3 7 7 ...
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MATHEMATICA
Table[Sum[Sum[j*Mod[j, 2], {i, j, Floor[(n - j)/2]}], {j, Floor[n/3]}], {n, 0, 80}]
LinearRecurrence[{1, 1, -1, 0, 0, 2, -2, -2, 2, 0, 0, -1, 1, 1, -1}, {0, 0, 0, 1, 1, 2, 2, 3, 3, 7, 7, 11, 11, 15, 15}, 20] (* Wesley Ivan Hurt, Aug 29 2019 *)
PROG
(PARI) a(n) = sum(j = 1, floor(n/3), sum(i = j, floor((n-j)/2), j * (j%2))); \\ Jinyuan Wang, Aug 29 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Aug 12 2019
STATUS
approved