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

1,1

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

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

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:
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 *)

Prog

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

Crossrefs

Cf. A000217, A000290, A006451.

Sequence in context: A181843 A131111 A336574 * A055885 A181425 A174505

Adjacent sequences: A128546 A128547 A128548 * A128550 A128551 A128552

Keyword

nonn

Author

Zak Seidov, May 08 2007

Status

approved