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How do I make a function that generates nth natural number that isn't a perfect square?

Convergent subsequencesHow to find a “better description” (e.g. recurrence relation) for this sequence?Determine the value for which a sequence is an arithmetic progression.Convergence of a sequence in a Hilbert space2-cycles and limitsFind a recursive definition for the sequencesMonotonicity of $fracnsqrt[n](n!)$$2021^textst$ term of a SequenceGetting a specific element of a non-recursive sequenceHow to calculate the general formula for nth term of this recursion?

1

1

$begingroup$

So I want to make a function such that for every n that you input it generates nth natural number that isn't a perfect square, like 2, 3, 5,...? I tried recurrance relation and I can't seem to find the proper relation between the members of sequence. Then I tried making a function but I don't know what to use actually... Any help?

sequences-and-series roots

share|cite|improve this question

asked 3 hours ago

Adnan CAdnan C

335

$endgroup$

add a comment | 

1

1

$begingroup$

So I want to make a function such that for every n that you input it generates nth natural number that isn't a perfect square, like 2, 3, 5,...? I tried recurrance relation and I can't seem to find the proper relation between the members of sequence. Then I tried making a function but I don't know what to use actually... Any help?

sequences-and-series roots

share|cite|improve this question

asked 3 hours ago

Adnan CAdnan C

335

$endgroup$

add a comment | 

$begingroup$

So I want to make a function such that for every n that you input it generates nth natural number that isn't a perfect square, like 2, 3, 5,...? I tried recurrance relation and I can't seem to find the proper relation between the members of sequence. Then I tried making a function but I don't know what to use actually... Any help?

sequences-and-series roots

share|cite|improve this question

asked 3 hours ago

Adnan CAdnan C

335

$endgroup$

share|cite|improve this question

asked 3 hours ago

Adnan CAdnan C

335

share|cite|improve this question

asked 3 hours ago

Adnan CAdnan C

335

asked 3 hours ago

Adnan CAdnan C

335

asked 3 hours ago

Adnan CAdnan C

335

2 Answers
2

active

oldest

votes

6

$begingroup$

OEIS to the rescue.

It gives the formula
$$
n+leftlfloorfrac12+sqrt nrightrfloor.
$$
where $lfloorcdotrfloor$ is the floor function.

share|cite|improve this answer

answered 3 hours ago

ArthurArthur

124k7122211

$endgroup$

add a comment | 

0

$begingroup$

Let $Bbb N = 1,2,3,dots$.

Let $pi_1$ and $pi_2$ be the two coordinate projection mappings on $Bbb N times Bbb N$.

We will define a function $f$ using recursion.

Define $f(1) = (2,2)$.

For $n ge 1$ define

$$
f(n+1) = leftbeginarraylr
left ( , pi_1(f(n)) + 1 ,pi_2(f(n)), right ) & textwhen pi_1(f(n)) + 1 lt [pi_2(f(n))]^2 \
left ( , pi_1(f(n)) + 2 ,pi_2(f(n))+1 , right ) & textelse
endarrayright
$$

The function $pi_1 circ f$ has the desired properties.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

CopyPasteItCopyPasteIt

4,4471828

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add a comment | 

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6

$begingroup$

OEIS to the rescue.

It gives the formula
$$
n+leftlfloorfrac12+sqrt nrightrfloor.
$$
where $lfloorcdotrfloor$ is the floor function.

share|cite|improve this answer

answered 3 hours ago

ArthurArthur

124k7122211

$endgroup$

add a comment | 

6

$begingroup$

OEIS to the rescue.

It gives the formula
$$
n+leftlfloorfrac12+sqrt nrightrfloor.
$$
where $lfloorcdotrfloor$ is the floor function.

share|cite|improve this answer

answered 3 hours ago

ArthurArthur

124k7122211

$endgroup$

add a comment | 

$begingroup$

OEIS to the rescue.

It gives the formula
$$
n+leftlfloorfrac12+sqrt nrightrfloor.
$$
where $lfloorcdotrfloor$ is the floor function.

share|cite|improve this answer

answered 3 hours ago

ArthurArthur

124k7122211

$endgroup$

share|cite|improve this answer

answered 3 hours ago

ArthurArthur

124k7122211

answered 3 hours ago

ArthurArthur

124k7122211

answered 3 hours ago

ArthurArthur

124k7122211

0

$begingroup$

Let $Bbb N = 1,2,3,dots$.

Let $pi_1$ and $pi_2$ be the two coordinate projection mappings on $Bbb N times Bbb N$.

We will define a function $f$ using recursion.

Define $f(1) = (2,2)$.

For $n ge 1$ define

$$
f(n+1) = leftbeginarraylr
left ( , pi_1(f(n)) + 1 ,pi_2(f(n)), right ) & textwhen pi_1(f(n)) + 1 lt [pi_2(f(n))]^2 \
left ( , pi_1(f(n)) + 2 ,pi_2(f(n))+1 , right ) & textelse
endarrayright
$$

The function $pi_1 circ f$ has the desired properties.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

CopyPasteItCopyPasteIt

4,4471828

$endgroup$

add a comment | 

0

$begingroup$

Let $Bbb N = 1,2,3,dots$.

Let $pi_1$ and $pi_2$ be the two coordinate projection mappings on $Bbb N times Bbb N$.

We will define a function $f$ using recursion.

Define $f(1) = (2,2)$.

For $n ge 1$ define

$$
f(n+1) = leftbeginarraylr
left ( , pi_1(f(n)) + 1 ,pi_2(f(n)), right ) & textwhen pi_1(f(n)) + 1 lt [pi_2(f(n))]^2 \
left ( , pi_1(f(n)) + 2 ,pi_2(f(n))+1 , right ) & textelse
endarrayright
$$

The function $pi_1 circ f$ has the desired properties.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

CopyPasteItCopyPasteIt

4,4471828

$endgroup$

add a comment | 

$begingroup$

Let $Bbb N = 1,2,3,dots$.

Let $pi_1$ and $pi_2$ be the two coordinate projection mappings on $Bbb N times Bbb N$.

We will define a function $f$ using recursion.

Define $f(1) = (2,2)$.

For $n ge 1$ define

$$
f(n+1) = leftbeginarraylr
left ( , pi_1(f(n)) + 1 ,pi_2(f(n)), right ) & textwhen pi_1(f(n)) + 1 lt [pi_2(f(n))]^2 \
left ( , pi_1(f(n)) + 2 ,pi_2(f(n))+1 , right ) & textelse
endarrayright
$$

The function $pi_1 circ f$ has the desired properties.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

CopyPasteItCopyPasteIt

4,4471828

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

CopyPasteItCopyPasteIt

4,4471828

answered 1 hour ago

CopyPasteItCopyPasteIt

4,4471828

answered 1 hour ago

CopyPasteItCopyPasteIt

4,4471828