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Sum of the products of the smaller and larger parts of the partitions of n into two distinct parts with the larger part odd.
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%I #40 Apr 16 2018 19:01:02

%S 0,0,0,3,6,5,10,22,34,30,46,73,100,91,124,172,220,204,260,335,410,385,

%T 470,578,686,650,770,917,1064,1015,1176,1368,1560,1496,1704,1947,2190,

%U 2109,2370,2670,2970,2870,3190,3553,3916,3795,4180,4612,5044,4900,5356

%N Sum of the products of the smaller and larger parts of the partitions of n into two distinct parts with the larger part odd.

%C Sum of the areas of the distinct rectangles with odd length and integer width such that L + W = n, W < L. For example, a(10) = 30; the rectangles are 1 X 9 and 3 X 7 (5 X 5 is not included since we have W < L), so 1*9 + 3*7 = 30.

%C 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 odd integer such that 0 < k < floor(n/2).

%C Sum of the areas of the trapezoids with bases n and n-2i and height i for odd values of n-i where i is in 0 <= i <= floor((n-1)/2). For a(n) the area formula for a trapezoid becomes (n+n-2i)*i/2 = (2n-2i)*i/2 = i*(n-i). For n=10, n-i is odd when i=1,3 so a(10) = 1*(10-1) + 3*(10-3) = 30. - _Wesley Ivan Hurt_, Mar 21 2018

%C Sum of the areas of the symmetric L-shaped polygons with long side n/2 and width i such that n-i is odd for i in 0 <= i <= floor((n-1)/2). The area of each polygon is given by i^2+2i(n/2-i) = i^2+ni-2i^2 = i(n-i). For n=8, 8-i is odd for i=1,3 so 1(8-1) + 3(8-3) = 7 + 15 = 22. - _Wesley Ivan Hurt_, Mar 26 2018

%H Vincenzo Librandi, <a href="/A295320/b295320.txt">Table of n, a(n) for n = 1..5000</a>

%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>

%F a(n) = Sum_{i=1..floor((n-1)/2)} i * (n-i) * ((n-i) mod 2).

%F Conjectures from _Colin Barker_, Nov 20 2017: (Start)

%F G.f.: x^4*(3+3*x-x^2+5*x^3+3*x^4+3*x^5-x^6+x^7) / ((1-x)^4*(1+x)^3*(1+x^2)^3).

%F 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.

%F (End)

%F a(n) = (1/384)*(-1)^(-(-1)^n/4)*((2-2*(-1)^n)*(-(-1)^((4*n+2-(-1)^n)/4)+6*(-1)^((6*n+1)/4)-(-1)^((2-(-1)^n)/4))+4*n*(12*(-1)^((-1)^n/4)*(-1)^n-6*n*(-1)^((2*n-1-2*(-1)^n)/4)+(-1)^((-1)^n/4)*(-4-3*n*(1+(-1)^n)+4*n^2))). - _Wesley Ivan Hurt_, Dec 03 2017

%e For n=8, the partitions into two distinct parts are 7 + 1, 6 + 2, and 5 + 3. Of these, 7 + 1 and 5 + 3 have the smaller part odd, so a(8) = 7*1 + 5*3 = 22. - _Michael B. Porter_, Dec 05 2017

%p A295320:=n->add(i*(n-i)*((n-i) mod 2), i=1..floor((n-1)/2)): seq(A295320(n), n=1..100);

%t Table[Sum[i (n - i) Mod[n - i, 2], {i, Floor[(n - 1)/2]}], {n, 80}]

%o (PARI) a(n) = sum(i=1, (n-1)\2, i*(n-i)*((n-i) % 2)); \\ _Michel Marcus_, Mar 26 2018

%Y Cf. A295321.

%K nonn,easy

%O 1,4

%A _Wesley Ivan Hurt_, Nov 19 2017