A295286 Sum of the products of the smaller and larger parts of the partitions of n into two parts with the smaller part odd.

0, 1, 2, 3, 4, 14, 18, 22, 26, 55, 64, 73, 82, 140, 156, 172, 188, 285, 310, 335, 360, 506, 542, 578, 614, 819, 868, 917, 966, 1240, 1304, 1368, 1432, 1785, 1866, 1947, 2028, 2470, 2570, 2670, 2770, 3311, 3432, 3553, 3674, 4324, 4468, 4612, 4756, 5525, 5694

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 sequences related to partitions

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

A295286:=n->add(i*(n-i)*(i mod 2), i=1..floor(n/2)): seq(A295286(n), n=1..100);
# 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

Cf. A295287.

Sequence in context: A140128 A167906 A100998 * A127283 A047193 A333622

Adjacent sequences: A295283 A295284 A295285 * A295287 A295288 A295289

Keyword

nonn,easy

Author

Wesley Ivan Hurt, Nov 19 2017

Status

approved