|
|
A128549
|
|
Difference between triangular number and next perfect square.
|
|
2
|
|
|
3, 1, 3, 6, 1, 4, 8, 13, 4, 9, 15, 3, 9, 16, 1, 8, 16, 25, 6, 15, 25, 3, 13, 24, 36, 10, 22, 35, 6, 19, 33, 1, 15, 30, 46, 10, 26, 43, 4, 21, 39, 58, 15, 34, 54, 8, 28, 49, 71, 21, 43, 66, 13, 36, 60, 4, 28, 53, 79, 19, 45, 72, 9, 36, 64, 93, 26, 55, 85, 15, 45, 76, 3, 34, 66, 99, 22
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
If a(n)=1 then such n gives the sequence A006451 (triangular numbers whose distance to the nearest bigger perfect square is 1). [From Ctibor O. Zizka, Oct 07 2009]
|
|
LINKS
|
|
|
FORMULA
|
a(n) = (floor(sqrt(n(n+1)/2))+1)^2-n(n+1)/2.
|
|
EXAMPLE
|
a(1)=2^2-1(1+1)/2=3, a(2)=2^2-2(2+1)/2=1, a(3)=3^2-3(3+1)/2=3, a(3)=4^2-4(4+1)/2=6.
|
|
MAPLE
|
f:= n -> (floor(sqrt(n*(n+1)/2))+1)^2-n*(n+1)/2:
|
|
MATHEMATICA
|
Table[(Floor[Sqrt[n(n+1)/2]]+1)^2-n(n+1)/2, {n, 100}]
(Floor[Sqrt[#]]+1)^2-#&/@Accumulate[Range[100]] (* Harvey P. Dale, Oct 15 2014 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|