OFFSET
1,4
COMMENTS
Sum of the areas of the distinct rectangles with integer length and even 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 (k,n*k-k^2) corresponding to integer points along the left side of the parabola b_k = n*k-k^2 where 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) * ((i+1) mod 2).
Conjectures from Colin Barker, Nov 20 2017: (Start)
G.f.: 2*x^4*(2 + x + x^2 + x^3 + 3*x^4) / ((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) = (1/96)*(3*n^2*(1+(-1)^n)-16*n+4*n^3+3*I^n*(1+I)*(-1)^(-(-1)^n/4)*((1+2*n^2)*(-1)^n-1)/sqrt(2)) where I=sqrt(-1). - Wesley Ivan Hurt, Dec 02 2017
EXAMPLE
a(10) = 40; the partitions of 10 into two parts are (9,1), (8,2), (7,3), (6,4), (5,5). Two of these partitions have their smaller part even, namely (8,2) and (6,4). So, a(10) = 8*2 + 6*4 = 40.
MAPLE
# Alternative:
for j from 0 to 3 do
F[j]:= expand(simplify(eval(sum(2*i*(4*k+j-2*i), i=1..k)), {k=(n-j)/4}))
od:
seq(F[n mod 4], n=1..30); # Robert Israel, Dec 05 2017
MATHEMATICA
Table[Sum[i (n - i) Mod[i + 1, 2], {i, Floor[n/2]}], {n, 80}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Nov 19 2017
STATUS
approved