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A296159
Sum of the smaller parts in the partitions of n into two distinct parts with the larger part odd.
0
0, 0, 0, 1, 2, 1, 2, 4, 6, 4, 6, 9, 12, 9, 12, 16, 20, 16, 20, 25, 30, 25, 30, 36, 42, 36, 42, 49, 56, 49, 56, 64, 72, 64, 72, 81, 90, 81, 90, 100, 110, 100, 110, 121, 132, 121, 132, 144, 156, 144, 156, 169, 182, 169, 182, 196, 210, 196, 210, 225, 240, 225
OFFSET
1,5
COMMENTS
a(n+1) is the sum of the smaller parts in the partitions of n into two parts with the larger part even. For example, the partitions of 8 into two parts are 7 + 1, 6 + 2, 5 + 3 and 4 + 4. Two of these partitions have their larger part even, so the sum of the smaller parts gives a(8+1) = a(9) = 6.
FORMULA
a(n) = Sum_{i=1..floor((n-1)/2)} i * ((n-i) mod 2).
Conjectures from Colin Barker, Dec 06 2017: (Start)
G.f.: x^4*(1 + x - x^2 + x^3) / ((1 - x)^3*(1 + x)^2*(1 + x^2)^2).
a(n) = a(n-1) + 2*a(n-4) - 2*a(n-5) - a(n-8) + a(n-9) for n>9.
(End)
a(n) = floor(n/4)*(floor(n/4)+1)*(n mod 2)+floor(n/4)^2*((n+1) mod 2). - Wesley Ivan Hurt, Dec 06 2017
EXAMPLE
a(12) = 9; the partitions of 12 into two distinct parts are 11 + 1, 10 + 2, 9 + 3, 8 + 4 and 7 + 5. Three of these partitions have their larger part odd, and the sum of the smaller parts of these partitions gives 1 + 3 + 5 = 9.
MATHEMATICA
Table[Sum[i Mod[n - i, 2], {i, Floor[(n - 1)/2]}], {n, 80}]
CROSSREFS
Sequence in context: A368058 A324225 A214739 * A283334 A301413 A305056
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Dec 06 2017
STATUS
approved