A128549 Difference between triangular number and next perfect square.

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

Offset: 1

Author

Zak Seidov, May 08 2007

Keywords

• nonn

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]

Examples

			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.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000217, A000290, A006451.

Programs

Maple

f:= n -> (floor(sqrt(n*(n+1)/2))+1)^2-n*(n+1)/2:
map(f, [$1..100]); # Robert Israel, Jan 21 2020

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 *)

Python

from math import isqrt
def A128549(n): return (isqrt(m:=n*(n+1)>>1)+1)**2-m # Chai Wah Wu, Jun 01 2024

Formula

a(n) = (floor(sqrt(n(n+1)/2))+1)^2-n(n+1)/2.