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Given a Fibonacci number , find the next Fibonacci number

how to find nth term in a fibonacci series or sum of a series of fibonacci numbersdetermine the number of terms in a fibonacci sequence that are divisible by $3$How do I choose first terms of a Fibonacci sequence?Is fibonacci sequence a member of more broad family of sequences?First Fibonacci Number with Given RemainderNumber of ways to write $n$ as sum of odd or even number of Fibonacci numbersModification of the Fibonacci Sequences

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6

4

$begingroup$

The Fibonacci sequence is $0, 1, 1, 2, 3, 5, 8, 13, 21, 34,ldots$, where each term after the first two is the sum of the two previous terms.

Can we find the next Fibonacci number if we are given any Fibonacci number?

For example, if $n = 8$ then it's answer should be $13$ because $13$ is the next Fibonacci number after $8$.

sequences-and-series combinatorics discrete-mathematics fibonacci-numbers

share|cite|improve this question

edited 1 hour ago

Matthew Daly

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asked yesterday

sr123sr123

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$endgroup$

add a comment
 | 

6

4

$begingroup$

The Fibonacci sequence is $0, 1, 1, 2, 3, 5, 8, 13, 21, 34,ldots$, where each term after the first two is the sum of the two previous terms.

Can we find the next Fibonacci number if we are given any Fibonacci number?

For example, if $n = 8$ then it's answer should be $13$ because $13$ is the next Fibonacci number after $8$.

sequences-and-series combinatorics discrete-mathematics fibonacci-numbers

share|cite|improve this question

edited 1 hour ago

Matthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

asked yesterday

sr123sr123

9666 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

The Fibonacci sequence is $0, 1, 1, 2, 3, 5, 8, 13, 21, 34,ldots$, where each term after the first two is the sum of the two previous terms.

Can we find the next Fibonacci number if we are given any Fibonacci number?

For example, if $n = 8$ then it's answer should be $13$ because $13$ is the next Fibonacci number after $8$.

sequences-and-series combinatorics discrete-mathematics fibonacci-numbers

share|cite|improve this question

edited 1 hour ago

Matthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

asked yesterday

sr123sr123

9666 bronze badges

$endgroup$

share|cite|improve this question

edited 1 hour ago

Matthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

asked yesterday

sr123sr123

9666 bronze badges

share|cite|improve this question

edited 1 hour ago

Matthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

asked yesterday

sr123sr123

9666 bronze badges

edited 1 hour ago

Matthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

edited 1 hour ago

Matthew Daly

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asked yesterday

sr123sr123

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asked yesterday

sr123sr123

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3 Answers
3

active

oldest

votes

20

$begingroup$

The ratio of any two consecutive entries in the Fibonacci sequence rapidly approaches $varphi=frac{1+sqrt5}2$. So if you multiply your number by $frac{1+sqrt5}2$ and round to the nearest integer, you will get the next term unless you're at the very beginning of the sequence.

share|cite|improve this answer

edited 1 hour ago

answered yesterday

Matthew DalyMatthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  float ans = ((1+sqrt(5))/2.0)*n; return ceil(ans); This is not giving correct result.
  $endgroup$
  – sr123
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  17
  

  
  
  
  
  
  

  
  $begingroup$
  @sr123 He did say "round to the nearest integer". That's not what ceil does, so of course you don't get the correct result. From checking the first few values, actually rounding will get you correct results unless you're starting from 0 or from 1.
  $endgroup$
  – HTNW
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes float ans = ((1+sqrt(5))/2.0)*n; return round(ans); is working !!
  $endgroup$
  – sr123
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Trying to do this in floating point is a bad idea; Fibonacci numbers quickly grow beyond the ability of floating point to accurately represent, so your computation will suffer from rounding error.
  $endgroup$
  – user2357112
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  This is a math site, not a computer site, and we work with abstract real number arithmetic here, not floating point. However, anyone actually trying to compute a Fibonacci number by multiplying another Fibonacci number by $(1+sqrt{5})/2$ is probably going to use a computer, and representation issues will be important.
  $endgroup$
  – user2357112
  5 hours ago
  

  
  
  
  

 | 
show 2 more comments

15

$begingroup$

$ninmathbb{N}$ is a Fibonacci number iff $5n^2-4$ or $5n^2+4$ is a square. In the former case $n=F_{2k+1}$ while in the latter case $n=F_{2k}$. Assuming $ngeq 2$, in the former case $F_{2k+2}=lfloor varphi n rfloor$ and in the latter case $F_{2k+1}=lceil varphi nrceil $, with $varphi=frac{1+sqrt{5}}{2}$.

Example: if $n=8$ we have that $5cdot 8^2+4=18^2$, hence $n$ is a Fibonacci number with even index and the next Fibonacci number is $lceil 8varphi rceil =13$.

share|cite|improve this answer

answered yesterday

Jack D'AurizioJack D'Aurizio

302k3333 gold badges297297 silver badges695695 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  If you've gone to the trouble to find the square root $m$ of $5n^2 - 4$ or $5n^2 + 4$, you can avoid floating-point arithmetic by computing the next Fibonacci number as $(n + m)/2$.
  $endgroup$
  – Mark Dickinson
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkDickinson How does that work?
  $endgroup$
  – Varad Mahashabde
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Good question, but it looks like @TonyK has saved me from answering ...
  $endgroup$
  – Mark Dickinson
  2 hours ago
  

  
  
  
  

add a comment
 | 

8

$begingroup$

Given a Fibonacci number $n$, let $m$ be the next Fibonacci number. The Fibonacci sequence has the property that for any three consecutive elements $r,s,t$, we have $rt-s^2=pm 1$ (proof is by induction, which you might like to try $-$ the choice of signs alternates). And we know that the previous Fibonacci number is $m-n$. So we have
$$m(m-n)=n^2pm 1$$
This is a quadratic equation in $m$, with solutions
$m=frac12(npmsqrt{5n^2pm 4})$. We know that $m>n$, so $m$ must equal $frac12(n+sqrt{5n^2pm 4})$. And we can choose between $+4$ and $-4$ because only one of $sqrt{5n^2+4}$ and $sqrt{5n^2-4}$ can be an integer (with the single exception of $n=1$).

So the answer is whichever one of $frac12(n+sqrt{5n^2+4})$ and $frac12(n+sqrt{5n^2-4})$ is an integer.

Note that the single exception $n=1$ occurs twice in the Fibonacci sequence, so there are indeed two possible answers.

share|cite|improve this answer

edited 1 hour ago

answered 3 hours ago

TonyKTonyK

47.4k33 gold badges6262 silver badges140140 bronze badges

$endgroup$

add a comment
 | 

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20

$begingroup$

The ratio of any two consecutive entries in the Fibonacci sequence rapidly approaches $varphi=frac{1+sqrt5}2$. So if you multiply your number by $frac{1+sqrt5}2$ and round to the nearest integer, you will get the next term unless you're at the very beginning of the sequence.

share|cite|improve this answer

edited 1 hour ago

answered yesterday

Matthew DalyMatthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  float ans = ((1+sqrt(5))/2.0)*n; return ceil(ans); This is not giving correct result.
  $endgroup$
  – sr123
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  17
  

  
  
  
  
  
  

  
  $begingroup$
  @sr123 He did say "round to the nearest integer". That's not what ceil does, so of course you don't get the correct result. From checking the first few values, actually rounding will get you correct results unless you're starting from 0 or from 1.
  $endgroup$
  – HTNW
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes float ans = ((1+sqrt(5))/2.0)*n; return round(ans); is working !!
  $endgroup$
  – sr123
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Trying to do this in floating point is a bad idea; Fibonacci numbers quickly grow beyond the ability of floating point to accurately represent, so your computation will suffer from rounding error.
  $endgroup$
  – user2357112
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  This is a math site, not a computer site, and we work with abstract real number arithmetic here, not floating point. However, anyone actually trying to compute a Fibonacci number by multiplying another Fibonacci number by $(1+sqrt{5})/2$ is probably going to use a computer, and representation issues will be important.
  $endgroup$
  – user2357112
  5 hours ago
  

  
  
  
  

 | 
show 2 more comments

20

$begingroup$

The ratio of any two consecutive entries in the Fibonacci sequence rapidly approaches $varphi=frac{1+sqrt5}2$. So if you multiply your number by $frac{1+sqrt5}2$ and round to the nearest integer, you will get the next term unless you're at the very beginning of the sequence.

share|cite|improve this answer

edited 1 hour ago

answered yesterday

Matthew DalyMatthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  float ans = ((1+sqrt(5))/2.0)*n; return ceil(ans); This is not giving correct result.
  $endgroup$
  – sr123
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  17
  

  
  
  
  
  
  

  
  $begingroup$
  @sr123 He did say "round to the nearest integer". That's not what ceil does, so of course you don't get the correct result. From checking the first few values, actually rounding will get you correct results unless you're starting from 0 or from 1.
  $endgroup$
  – HTNW
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes float ans = ((1+sqrt(5))/2.0)*n; return round(ans); is working !!
  $endgroup$
  – sr123
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Trying to do this in floating point is a bad idea; Fibonacci numbers quickly grow beyond the ability of floating point to accurately represent, so your computation will suffer from rounding error.
  $endgroup$
  – user2357112
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  This is a math site, not a computer site, and we work with abstract real number arithmetic here, not floating point. However, anyone actually trying to compute a Fibonacci number by multiplying another Fibonacci number by $(1+sqrt{5})/2$ is probably going to use a computer, and representation issues will be important.
  $endgroup$
  – user2357112
  5 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

The ratio of any two consecutive entries in the Fibonacci sequence rapidly approaches $varphi=frac{1+sqrt5}2$. So if you multiply your number by $frac{1+sqrt5}2$ and round to the nearest integer, you will get the next term unless you're at the very beginning of the sequence.

share|cite|improve this answer

edited 1 hour ago

answered yesterday

Matthew DalyMatthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered yesterday

Matthew DalyMatthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

answered yesterday

Matthew DalyMatthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

answered yesterday

Matthew DalyMatthew Daly

7,03011 gold badge1010 silver badges2929 bronze badges

15

$begingroup$

$ninmathbb{N}$ is a Fibonacci number iff $5n^2-4$ or $5n^2+4$ is a square. In the former case $n=F_{2k+1}$ while in the latter case $n=F_{2k}$. Assuming $ngeq 2$, in the former case $F_{2k+2}=lfloor varphi n rfloor$ and in the latter case $F_{2k+1}=lceil varphi nrceil $, with $varphi=frac{1+sqrt{5}}{2}$.

Example: if $n=8$ we have that $5cdot 8^2+4=18^2$, hence $n$ is a Fibonacci number with even index and the next Fibonacci number is $lceil 8varphi rceil =13$.

share|cite|improve this answer

answered yesterday

Jack D'AurizioJack D'Aurizio

302k3333 gold badges297297 silver badges695695 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  If you've gone to the trouble to find the square root $m$ of $5n^2 - 4$ or $5n^2 + 4$, you can avoid floating-point arithmetic by computing the next Fibonacci number as $(n + m)/2$.
  $endgroup$
  – Mark Dickinson
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkDickinson How does that work?
  $endgroup$
  – Varad Mahashabde
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Good question, but it looks like @TonyK has saved me from answering ...
  $endgroup$
  – Mark Dickinson
  2 hours ago
  

  
  
  
  

add a comment
 | 

15

$begingroup$

$ninmathbb{N}$ is a Fibonacci number iff $5n^2-4$ or $5n^2+4$ is a square. In the former case $n=F_{2k+1}$ while in the latter case $n=F_{2k}$. Assuming $ngeq 2$, in the former case $F_{2k+2}=lfloor varphi n rfloor$ and in the latter case $F_{2k+1}=lceil varphi nrceil $, with $varphi=frac{1+sqrt{5}}{2}$.

Example: if $n=8$ we have that $5cdot 8^2+4=18^2$, hence $n$ is a Fibonacci number with even index and the next Fibonacci number is $lceil 8varphi rceil =13$.

share|cite|improve this answer

answered yesterday

Jack D'AurizioJack D'Aurizio

302k3333 gold badges297297 silver badges695695 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  If you've gone to the trouble to find the square root $m$ of $5n^2 - 4$ or $5n^2 + 4$, you can avoid floating-point arithmetic by computing the next Fibonacci number as $(n + m)/2$.
  $endgroup$
  – Mark Dickinson
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkDickinson How does that work?
  $endgroup$
  – Varad Mahashabde
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Good question, but it looks like @TonyK has saved me from answering ...
  $endgroup$
  – Mark Dickinson
  2 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

$ninmathbb{N}$ is a Fibonacci number iff $5n^2-4$ or $5n^2+4$ is a square. In the former case $n=F_{2k+1}$ while in the latter case $n=F_{2k}$. Assuming $ngeq 2$, in the former case $F_{2k+2}=lfloor varphi n rfloor$ and in the latter case $F_{2k+1}=lceil varphi nrceil $, with $varphi=frac{1+sqrt{5}}{2}$.

Example: if $n=8$ we have that $5cdot 8^2+4=18^2$, hence $n$ is a Fibonacci number with even index and the next Fibonacci number is $lceil 8varphi rceil =13$.

share|cite|improve this answer

answered yesterday

Jack D'AurizioJack D'Aurizio

302k3333 gold badges297297 silver badges695695 bronze badges

$endgroup$

share|cite|improve this answer

answered yesterday

Jack D'AurizioJack D'Aurizio

302k3333 gold badges297297 silver badges695695 bronze badges

answered yesterday

Jack D'AurizioJack D'Aurizio

302k3333 gold badges297297 silver badges695695 bronze badges

answered yesterday

Jack D'AurizioJack D'Aurizio

302k3333 gold badges297297 silver badges695695 bronze badges

8

$begingroup$

Given a Fibonacci number $n$, let $m$ be the next Fibonacci number. The Fibonacci sequence has the property that for any three consecutive elements $r,s,t$, we have $rt-s^2=pm 1$ (proof is by induction, which you might like to try $-$ the choice of signs alternates). And we know that the previous Fibonacci number is $m-n$. So we have
$$m(m-n)=n^2pm 1$$
This is a quadratic equation in $m$, with solutions
$m=frac12(npmsqrt{5n^2pm 4})$. We know that $m>n$, so $m$ must equal $frac12(n+sqrt{5n^2pm 4})$. And we can choose between $+4$ and $-4$ because only one of $sqrt{5n^2+4}$ and $sqrt{5n^2-4}$ can be an integer (with the single exception of $n=1$).

So the answer is whichever one of $frac12(n+sqrt{5n^2+4})$ and $frac12(n+sqrt{5n^2-4})$ is an integer.

Note that the single exception $n=1$ occurs twice in the Fibonacci sequence, so there are indeed two possible answers.

share|cite|improve this answer

edited 1 hour ago

answered 3 hours ago

TonyKTonyK

47.4k33 gold badges6262 silver badges140140 bronze badges

$endgroup$

add a comment
 | 

8

$begingroup$

Given a Fibonacci number $n$, let $m$ be the next Fibonacci number. The Fibonacci sequence has the property that for any three consecutive elements $r,s,t$, we have $rt-s^2=pm 1$ (proof is by induction, which you might like to try $-$ the choice of signs alternates). And we know that the previous Fibonacci number is $m-n$. So we have
$$m(m-n)=n^2pm 1$$
This is a quadratic equation in $m$, with solutions
$m=frac12(npmsqrt{5n^2pm 4})$. We know that $m>n$, so $m$ must equal $frac12(n+sqrt{5n^2pm 4})$. And we can choose between $+4$ and $-4$ because only one of $sqrt{5n^2+4}$ and $sqrt{5n^2-4}$ can be an integer (with the single exception of $n=1$).

So the answer is whichever one of $frac12(n+sqrt{5n^2+4})$ and $frac12(n+sqrt{5n^2-4})$ is an integer.

Note that the single exception $n=1$ occurs twice in the Fibonacci sequence, so there are indeed two possible answers.

share|cite|improve this answer

edited 1 hour ago

answered 3 hours ago

TonyKTonyK

47.4k33 gold badges6262 silver badges140140 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Given a Fibonacci number $n$, let $m$ be the next Fibonacci number. The Fibonacci sequence has the property that for any three consecutive elements $r,s,t$, we have $rt-s^2=pm 1$ (proof is by induction, which you might like to try $-$ the choice of signs alternates). And we know that the previous Fibonacci number is $m-n$. So we have
$$m(m-n)=n^2pm 1$$
This is a quadratic equation in $m$, with solutions
$m=frac12(npmsqrt{5n^2pm 4})$. We know that $m>n$, so $m$ must equal $frac12(n+sqrt{5n^2pm 4})$. And we can choose between $+4$ and $-4$ because only one of $sqrt{5n^2+4}$ and $sqrt{5n^2-4}$ can be an integer (with the single exception of $n=1$).

So the answer is whichever one of $frac12(n+sqrt{5n^2+4})$ and $frac12(n+sqrt{5n^2-4})$ is an integer.

Note that the single exception $n=1$ occurs twice in the Fibonacci sequence, so there are indeed two possible answers.

share|cite|improve this answer

edited 1 hour ago

answered 3 hours ago

TonyKTonyK

47.4k33 gold badges6262 silver badges140140 bronze badges

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 3 hours ago

TonyKTonyK

47.4k33 gold badges6262 silver badges140140 bronze badges

answered 3 hours ago

TonyKTonyK

47.4k33 gold badges6262 silver badges140140 bronze badges

answered 3 hours ago

TonyKTonyK

47.4k33 gold badges6262 silver badges140140 bronze badges