OFFSET
1,3
COMMENTS
Sum of the areas of the distinct rectangles with integer length and odd width such that L + W = n, W <= L. For example, a(6) = 14; the rectangles are 1 X 5 and 3 X 3, so 5 + 9 = 14.
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 odd integer such that 0 < k <= floor(n/2).
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,3,-3,0,0,-3,3,0,0,1,-1).
FORMULA
a(n) = Sum_{i=1..floor(n/2)} i * (n - i) * (i mod 2).
Conjectures from Colin Barker, Nov 20 2017: (Start)
G.f.: x^2*(1 + x + x^2 + x^3 + 7*x^4 + x^5 + x^6 + x^7 + 2*x^8) / ((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)/4+(-1)^n/4))+32*n+12*n^2*(1+(-1)^n+2*(-1)^((2*n+3)/4+(-1)^n/4))+16*n^3). - Wesley Ivan Hurt, Dec 02 2017
E.g.f.: (3*(x - 1)*x*cos(x) + x*(2*x^2 + 9*x + 6)*cosh(x) + 3*(x^2 + x - 1)*sin(x) + x*(2*x^2 + 6*x + 9)*sinh(x))/48. - Stefano Spezia, Nov 13 2021
EXAMPLE
a(10) = 55; the partitions of 10 into two parts are (9,1), (8,2), (7,3), (6,4), (5,5). Three of these partitions have odd numbers as their smaller parts, namely 1,3,5. Then the sum of the products of the smaller and larger parts of these partitions is 9*1 + 7*3 + 5*5 = 55.
MAPLE
# Alternate:
for j from 0 to 3 do
F[j]:= expand(simplify(eval(sum((2*i-1)*(4*k+j-2*i+1), i=1..k+floor(j/2))), {k=(n-j)/4}))
od:
seq(F[n mod 4], n=1..100); # Robert Israel, Dec 05 2017
MATHEMATICA
Table[Sum[i (n - i) Mod[i, 2], {i, Floor[n/2]}], {n, 80}]
Table[Total[Times@@@Select[IntegerPartitions[n, {2}], OddQ[#[[2]]]&]], {n, 60}] (* Harvey P. Dale, Sep 15 2022 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Nov 19 2017
STATUS
approved