|
|
A330915
|
|
Sum of the "middle" side lengths (b such that a <= b <= c) of all Heronian triangles with perimeter A051518(n).
|
|
7
|
|
|
4, 5, 5, 8, 12, 23, 45, 15, 29, 13, 48, 30, 77, 24, 69, 117, 25, 25, 46, 119, 20, 26, 110, 246, 26, 167, 172, 205, 169, 79, 468, 33, 229, 38, 222, 167, 429, 41, 429, 101, 270, 560, 416, 100, 153, 276, 390, 717, 50, 615, 61, 61, 60, 404, 634, 214, 130, 130, 1033, 975, 382
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
LINKS
|
|
|
FORMULA
|
a(n) = Sum_{k=1..floor(c(n)/3)} Sum_{i=k..floor((c(n)-k)/2)} sign(floor((i+k)/(c(n)-i-k+1))) * chi(sqrt((c(n)/2)*(c(n)/2-i)*(c(n)/2-k)*(c(n)/2-(c(n)-i-k))) * i, where chi(n) = 1 - ceiling(n) + floor(n) and c(n) = A051518(n). - Wesley Ivan Hurt, May 12 2020
|
|
EXAMPLE
|
a(1) = 4; there is one Heronian triangle with perimeter A051518(1) = 12, which is [3,4,5] and its "middle" side length is 4.
a(6) = 23; there are two Heronian triangles with perimeter A051518(6) = 32, [4,13,15] and [10,10,12]. The sum is 13 + 10 = 23.
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|