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%I #35 Sep 15 2022 13:18:30
%S 0,1,2,3,4,14,18,22,26,55,64,73,82,140,156,172,188,285,310,335,360,
%T 506,542,578,614,819,868,917,966,1240,1304,1368,1432,1785,1866,1947,
%U 2028,2470,2570,2670,2770,3311,3432,3553,3674,4324,4468,4612,4756,5525,5694
%N Sum of the products of the smaller and larger parts of the partitions of n into two parts with the smaller part odd.
%C 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.
%C 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).
%H Robert Israel, <a href="/A295286/b295286.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>
%H <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,3,-3,0,0,-3,3,0,0,1,-1).
%F a(n) = Sum_{i=1..floor(n/2)} i * (n - i) * (i mod 2).
%F Conjectures from _Colin Barker_, Nov 20 2017: (Start)
%F 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).
%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 Conjectures verified by _Robert Israel_, Dec 05 2017.
%F 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
%F 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
%e 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.
%p A295286:=n->add(i*(n-i)*(i mod 2), i=1..floor(n/2)): seq(A295286(n), n=1..100);
%p # Alternate:
%p for j from 0 to 3 do
%p 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}))
%p od:
%p seq(F[n mod 4],n=1..100); # _Robert Israel_, Dec 05 2017
%t Table[Sum[i (n - i) Mod[i, 2], {i, Floor[n/2]}], {n, 80}]
%t Table[Total[Times@@@Select[IntegerPartitions[n,{2}],OddQ[#[[2]]]&]],{n,60}] (* _Harvey P. Dale_, Sep 15 2022 *)
%Y Cf. A295287.
%K nonn,easy
%O 1,3
%A _Wesley Ivan Hurt_, Nov 19 2017
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