A squareless sequence

Number Theory Level 3

A new sequence is obtained from positive integers ${1,2,3\dots}$ by deleting all the perfect squares .Find the 2016th term of this sequence.

Bonus question : The answer to this problem resembles the occurrence of a celestial body. Find it and its significance.

The answer is 2061.

2 solutions

Nihar Mahajan
Apr 11, 2015

We see that

$1936<2016<2025 \\ 44^2<2016<45^2$

Since the 44 numbers which are perfect squares are deleted , we get $2016^{th}$ term as $2016+44=2060$

But , in the meanwhile, $45^2$ $(2025)$ also gets counted. So it must also be deleted. Hence finally $2016^{th}$ term is :

$2060+1=\huge\boxed{\color{#3D99F6}{2061}}$

Answer to bonus question: This is the year when Halley's comet will be seen from earth. Ahh! I would be $61$ years old then!

I like your solution better, but here's python code:

from math import *
c=1
array=[]
while True:
    if len(array)==2016:
        print array[-1]
        break
    elif not sqrt(c).is_integer():
        array.append(c)
    c+=1

David Holcer - 6 years, 2 months ago

$correct !$

Vaibhav Prasad - 6 years, 2 months ago

$\large \ddot\smile \ddot\smile \ddot\smile$

Nihar Mahajan - 6 years, 2 months ago

I got 2060, but I forgot about $45^2$ . My Bad, Anyways, what's about that celestial occurrence??

Vaibhav Kandwal - 6 years, 2 months ago

Wait, Are you talking about Halley's Comet??

Vaibhav Kandwal - 6 years, 2 months ago

Yeah , that's right! I have given the statement regarding it at the end of the solution.

Nihar Mahajan - 6 years, 2 months ago

LOL I done same in hard way

Yash Patel - 6 years, 1 month ago

Aravind Vishnu
Apr 23, 2015

$\text{The number of perfect squares from}\, 1\, \text{to}\, 2016 \,\text{is}\, 44\\ \text{So the}\, 2016^{\text{th}} \, \text{term is} \, 2016+44+1\,(\text{since}\,2025\, \text{is a perfect square.})=2061$

A squareless sequence

The answer is 2061.

2 solutions

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