OFFSET
1,3
COMMENTS
Sum of the areas of the distinct rectangles with even length and integer width such that L + W = n, W <= L. For example, a(8) = 28; the rectangles are 2 X 6 and 4 X 4, so 2*6 + 4*4 = 28.
Sum of the ordinates from the ordered pairs (n-k,n*k-k^2) corresponding to integer points along the right side of the parabola b_k = n*k-k^2 where n-k is an even integer such that 0 < k <= floor(n/2).
LINKS
FORMULA
a(n) = Sum_{i=1..floor(n/2)} i * (n-i) * ((n-i+1) mod 2).
Conjectures from Colin Barker, Nov 20 2017: (Start)
G.f.: 2*x^3*(1 + x + 2*x^3 + 2*x^4 + 2*x^5 - x^6 + x^7) / ((1 - x)^4*(1 + x)^3*(1 + x^2)^3).
a(n) = a(n-1) + 3*a(n-4) - 3*a(n-5) - 3*a(n-8) + 3*a(n-9) + a(n-12) - a(n-13) for n>13.
(End)
Conjectures verified by Robert Israel, Dec 05 2017.
a(n) = (1/384)*((2-2*(-1)^n)*(1+(-1)^n-6*(-1)^((2*n-3-(-1)^n)/4))-16*n*(1+3*(-1)^n)+12*n^2*(1+(-1)^n-2*(-1)^((3+(-1)^n)/4)*(-1)^(3*n/2))+16*n^3). - Wesley Ivan Hurt, Dec 04 2017
EXAMPLE
a(8) = 28; the partitions of 8 into two parts are (7,1), (6,2), (5,3) and (4,4). There are two partitions with the larger part even: (6,2) and (4,4), so 6*2 + 4*4 = 28.
MAPLE
# Alternative:
F[0]:= (1/24)*n^3+(1/8)*n^2-(1/6)*n:
F[1]:= -1/16+(1/24)*n^3-(1/16)*n^2+(1/12)*n:
F[2]:= (1/24)*n^3-(1/6)*n:
F[3]:= 1/16+(1/24)*n^3+(1/16)*n^2+(1/12)*n:
seq(F[n mod 4], n=1..100); # Robert Israel, Dec 05 2017
MATHEMATICA
Table[Sum[i (n - i) Mod[n - i + 1, 2], {i, Floor[n/2]}], {n, 80}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Nov 19 2017
STATUS
approved