How many sequnces can be made with 5 digits so that the difference between any two consecutive digits is $1$?












5












$begingroup$


Using the digits $0$, $1$, $2$, $3$, and $4$, how many ten digit sequences can be written so
that the difference between any two consecutive digits is $1$?



I was wondering if my solution is right.



Let $a(n)$ be the number of n digit sequences that end with $0$ or $4$ so
that the difference between any two consecutive digits is $1$.



$b(n)$ be the number of n digit sequences that end with $1$ or $3$ so
that the difference between any two consecutive digits is $1$.



$c(n)$ be the number of n digit sequences that end with $2$ so
that the difference between any two consecutive digits is $1$.



$x(n)$ be the number of n digit sequences so
that the difference between any two consecutive digits is $1$.



$x(n) = a(n) + b(n) + c(n)$



$a(n) = b(n-1)$



$b(n) = a(n-1) + 2c(n-1)$



$c(n) = b(n-1)$



By substituting $a(n-1)$ and $c(n-1)$ in the formula for $b(n)$ we get $b(n) = 3b(n-2)$



We know that $b(1) = 2, b(2) = 4$.



The characteristic equation for this recurssion is $x^2-3 = 0$ with have the roots $3^{1/2}$ and $-3^{1/2}$, so $b(n) = A{(3^{1/2})}^{n} + B{(-3^{1/2})}^{n}$ where $A = {(3^{1/2}+2)}/{3}$ and $B = {(2-3^{1/2})}/{3}$. I think this is an integer.



We get $x(n) = 2b(n-1) + 3b(n-2)$ and by substituting we get something.






combinatorics proof-verification







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$endgroup$












  • $begingroup$
    You get $b(n)=3b(n-2)$
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 7:26










  • $begingroup$
    thanks, i edited it
    $endgroup$
    – Lazar Ionut Radu
    Dec 18 '18 at 8:06










  • $begingroup$
    Looks correct to me
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 8:14
















5












$begingroup$


Using the digits $0$, $1$, $2$, $3$, and $4$, how many ten digit sequences can be written so
that the difference between any two consecutive digits is $1$?



I was wondering if my solution is right.



Let $a(n)$ be the number of n digit sequences that end with $0$ or $4$ so
that the difference between any two consecutive digits is $1$.



$b(n)$ be the number of n digit sequences that end with $1$ or $3$ so
that the difference between any two consecutive digits is $1$.



$c(n)$ be the number of n digit sequences that end with $2$ so
that the difference between any two consecutive digits is $1$.



$x(n)$ be the number of n digit sequences so
that the difference between any two consecutive digits is $1$.



$x(n) = a(n) + b(n) + c(n)$



$a(n) = b(n-1)$



$b(n) = a(n-1) + 2c(n-1)$



$c(n) = b(n-1)$



By substituting $a(n-1)$ and $c(n-1)$ in the formula for $b(n)$ we get $b(n) = 3b(n-2)$



We know that $b(1) = 2, b(2) = 4$.



The characteristic equation for this recurssion is $x^2-3 = 0$ with have the roots $3^{1/2}$ and $-3^{1/2}$, so $b(n) = A{(3^{1/2})}^{n} + B{(-3^{1/2})}^{n}$ where $A = {(3^{1/2}+2)}/{3}$ and $B = {(2-3^{1/2})}/{3}$. I think this is an integer.



We get $x(n) = 2b(n-1) + 3b(n-2)$ and by substituting we get something.






combinatorics proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    You get $b(n)=3b(n-2)$
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 7:26










  • $begingroup$
    thanks, i edited it
    $endgroup$
    – Lazar Ionut Radu
    Dec 18 '18 at 8:06










  • $begingroup$
    Looks correct to me
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 8:14














5












5








5


2



$begingroup$


Using the digits $0$, $1$, $2$, $3$, and $4$, how many ten digit sequences can be written so
that the difference between any two consecutive digits is $1$?



I was wondering if my solution is right.



Let $a(n)$ be the number of n digit sequences that end with $0$ or $4$ so
that the difference between any two consecutive digits is $1$.



$b(n)$ be the number of n digit sequences that end with $1$ or $3$ so
that the difference between any two consecutive digits is $1$.



$c(n)$ be the number of n digit sequences that end with $2$ so
that the difference between any two consecutive digits is $1$.



$x(n)$ be the number of n digit sequences so
that the difference between any two consecutive digits is $1$.



$x(n) = a(n) + b(n) + c(n)$



$a(n) = b(n-1)$



$b(n) = a(n-1) + 2c(n-1)$



$c(n) = b(n-1)$



By substituting $a(n-1)$ and $c(n-1)$ in the formula for $b(n)$ we get $b(n) = 3b(n-2)$



We know that $b(1) = 2, b(2) = 4$.



The characteristic equation for this recurssion is $x^2-3 = 0$ with have the roots $3^{1/2}$ and $-3^{1/2}$, so $b(n) = A{(3^{1/2})}^{n} + B{(-3^{1/2})}^{n}$ where $A = {(3^{1/2}+2)}/{3}$ and $B = {(2-3^{1/2})}/{3}$. I think this is an integer.



We get $x(n) = 2b(n-1) + 3b(n-2)$ and by substituting we get something.






combinatorics proof-verification







share|cite|improve this question











$endgroup$




Using the digits $0$, $1$, $2$, $3$, and $4$, how many ten digit sequences can be written so
that the difference between any two consecutive digits is $1$?



I was wondering if my solution is right.



Let $a(n)$ be the number of n digit sequences that end with $0$ or $4$ so
that the difference between any two consecutive digits is $1$.



$b(n)$ be the number of n digit sequences that end with $1$ or $3$ so
that the difference between any two consecutive digits is $1$.



$c(n)$ be the number of n digit sequences that end with $2$ so
that the difference between any two consecutive digits is $1$.



$x(n)$ be the number of n digit sequences so
that the difference between any two consecutive digits is $1$.



$x(n) = a(n) + b(n) + c(n)$



$a(n) = b(n-1)$



$b(n) = a(n-1) + 2c(n-1)$



$c(n) = b(n-1)$



By substituting $a(n-1)$ and $c(n-1)$ in the formula for $b(n)$ we get $b(n) = 3b(n-2)$



We know that $b(1) = 2, b(2) = 4$.



The characteristic equation for this recurssion is $x^2-3 = 0$ with have the roots $3^{1/2}$ and $-3^{1/2}$, so $b(n) = A{(3^{1/2})}^{n} + B{(-3^{1/2})}^{n}$ where $A = {(3^{1/2}+2)}/{3}$ and $B = {(2-3^{1/2})}/{3}$. I think this is an integer.



We get $x(n) = 2b(n-1) + 3b(n-2)$ and by substituting we get something.






combinatorics proof-verification




combinatorics proof-verification






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share|cite|improve this question




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edited Dec 20 '18 at 3:39









Alex Ravsky

42.2k32383




42.2k32383










asked Dec 18 '18 at 7:12









Lazar Ionut RaduLazar Ionut Radu

866




866












  • $begingroup$
    You get $b(n)=3b(n-2)$
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 7:26










  • $begingroup$
    thanks, i edited it
    $endgroup$
    – Lazar Ionut Radu
    Dec 18 '18 at 8:06










  • $begingroup$
    Looks correct to me
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 8:14


















  • $begingroup$
    You get $b(n)=3b(n-2)$
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 7:26










  • $begingroup$
    thanks, i edited it
    $endgroup$
    – Lazar Ionut Radu
    Dec 18 '18 at 8:06










  • $begingroup$
    Looks correct to me
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 8:14
















$begingroup$
You get $b(n)=3b(n-2)$
$endgroup$
– Shubham Johri
Dec 18 '18 at 7:26




$begingroup$
You get $b(n)=3b(n-2)$
$endgroup$
– Shubham Johri
Dec 18 '18 at 7:26












$begingroup$
thanks, i edited it
$endgroup$
– Lazar Ionut Radu
Dec 18 '18 at 8:06




$begingroup$
thanks, i edited it
$endgroup$
– Lazar Ionut Radu
Dec 18 '18 at 8:06












$begingroup$
Looks correct to me
$endgroup$
– Shubham Johri
Dec 18 '18 at 8:14




$begingroup$
Looks correct to me
$endgroup$
– Shubham Johri
Dec 18 '18 at 8:14










2 Answers
2






active

oldest

votes


















0












$begingroup$

Your approach looks good to me but you have to check the final result somehow. You can do that with 15 lines of code:



#include <iostream>

using namespace std;



int count(int startDigit, int length) {

    if(startDigit < 0 || startDigit > 4) {

        return 0;

    }

    if(length == 1)

        return 1;

    return count(startDigit - 1, length - 1) + count(startDigit + 1, length - 1);

}



int main() {

    cout << count(1, 10) + count(2, 10) + count(3, 10) + count(4, 10);

}


...and the result is 567.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$


    here is your answer. Your approach was absolutely correct.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Your approach looks good to me but you have to check the final result somehow. You can do that with 15 lines of code:



      #include <iostream>
      
using namespace std;
      

      
int count(int startDigit, int length) {
      
    if(startDigit < 0 || startDigit > 4) {
      
        return 0;
      
    }
      
    if(length == 1)
      
        return 1;
      
    return count(startDigit - 1, length - 1) + count(startDigit + 1, length - 1);
      
}
      

      
int main() {
      
    cout << count(1, 10) + count(2, 10) + count(3, 10) + count(4, 10);
      
}


      ...and the result is 567.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Your approach looks good to me but you have to check the final result somehow. You can do that with 15 lines of code:



        #include <iostream>
        
using namespace std;
        

        
int count(int startDigit, int length) {
        
    if(startDigit < 0 || startDigit > 4) {
        
        return 0;
        
    }
        
    if(length == 1)
        
        return 1;
        
    return count(startDigit - 1, length - 1) + count(startDigit + 1, length - 1);
        
}
        

        
int main() {
        
    cout << count(1, 10) + count(2, 10) + count(3, 10) + count(4, 10);
        
}


        ...and the result is 567.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Your approach looks good to me but you have to check the final result somehow. You can do that with 15 lines of code:



          #include <iostream>
          
using namespace std;
          

          
int count(int startDigit, int length) {
          
    if(startDigit < 0 || startDigit > 4) {
          
        return 0;
          
    }
          
    if(length == 1)
          
        return 1;
          
    return count(startDigit - 1, length - 1) + count(startDigit + 1, length - 1);
          
}
          

          
int main() {
          
    cout << count(1, 10) + count(2, 10) + count(3, 10) + count(4, 10);
          
}


          ...and the result is 567.






          share|cite|improve this answer









          $endgroup$



          Your approach looks good to me but you have to check the final result somehow. You can do that with 15 lines of code:



          #include <iostream>
          
using namespace std;
          

          
int count(int startDigit, int length) {
          
    if(startDigit < 0 || startDigit > 4) {
          
        return 0;
          
    }
          
    if(length == 1)
          
        return 1;
          
    return count(startDigit - 1, length - 1) + count(startDigit + 1, length - 1);
          
}
          

          
int main() {
          
    cout << count(1, 10) + count(2, 10) + count(3, 10) + count(4, 10);
          
}


          ...and the result is 567.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 16:35









          OldboyOldboy

          8,4801936




          8,4801936























              0












              $begingroup$


              here is your answer. Your approach was absolutely correct.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$


                here is your answer. Your approach was absolutely correct.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$


                  here is your answer. Your approach was absolutely correct.






                  share|cite|improve this answer









                  $endgroup$




                  here is your answer. Your approach was absolutely correct.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 1 at 4:10









                  Sarah janeSarah jane

                  1797




                  1797






























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