Words of length $n$ containing an odd number of zeros












1












$begingroup$



Let $a_n$ be the number of words of length $n$ with letters from the alphabet ${0, 1, 2, 3}$
containing an odd number of zeros.




I have already verified that this is given by the recurrence relation:
$$a_{n+1} = 2 a_n + 4^n.$$ where $a_0=0$.



I then used the method of a generating function to solve this and arrive at:
$$a_n= 2^{2n-1} - 2^{n-1}.$$
Which seems to work for the first couple of values, also Wolfram Alpha agrees with my deduction.



This problem can also be solved alternatively, using a new sequence $B_n$, I find this method more difficult but wish to learn it as well. It's more of a combinatorics method, probably using a clever counting argument.




Another way to find the same result is as follows:



1) Consider the set $B_n$ of
words of length $n$ with at least one $0$ or $1$.



How many of those are there?
I am not sure how to express this in terms of $a_n$



2) In
this set there are exactly as many words with an even number of zeros as there are with
an odd number of zeros. Just change the first occuring $0$ or $1$ with its dierence
with $1$ $dots$ (not sure what is meant here, the wording is a bit vague)



I am not sure what the author is getting at, can anybody see where the argument is going and give me some pointers?






combinatorics recurrence-relations







share|cite|improve this question









$endgroup$

















    1












    $begingroup$



    Let $a_n$ be the number of words of length $n$ with letters from the alphabet ${0, 1, 2, 3}$
    containing an odd number of zeros.




    I have already verified that this is given by the recurrence relation:
    $$a_{n+1} = 2 a_n + 4^n.$$ where $a_0=0$.



    I then used the method of a generating function to solve this and arrive at:
    $$a_n= 2^{2n-1} - 2^{n-1}.$$
    Which seems to work for the first couple of values, also Wolfram Alpha agrees with my deduction.



    This problem can also be solved alternatively, using a new sequence $B_n$, I find this method more difficult but wish to learn it as well. It's more of a combinatorics method, probably using a clever counting argument.




    Another way to find the same result is as follows:



    1) Consider the set $B_n$ of
    words of length $n$ with at least one $0$ or $1$.



    How many of those are there?
    I am not sure how to express this in terms of $a_n$



    2) In
    this set there are exactly as many words with an even number of zeros as there are with
    an odd number of zeros. Just change the first occuring $0$ or $1$ with its dierence
    with $1$ $dots$ (not sure what is meant here, the wording is a bit vague)



    I am not sure what the author is getting at, can anybody see where the argument is going and give me some pointers?






    combinatorics recurrence-relations







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$



      Let $a_n$ be the number of words of length $n$ with letters from the alphabet ${0, 1, 2, 3}$
      containing an odd number of zeros.




      I have already verified that this is given by the recurrence relation:
      $$a_{n+1} = 2 a_n + 4^n.$$ where $a_0=0$.



      I then used the method of a generating function to solve this and arrive at:
      $$a_n= 2^{2n-1} - 2^{n-1}.$$
      Which seems to work for the first couple of values, also Wolfram Alpha agrees with my deduction.



      This problem can also be solved alternatively, using a new sequence $B_n$, I find this method more difficult but wish to learn it as well. It's more of a combinatorics method, probably using a clever counting argument.




      Another way to find the same result is as follows:



      1) Consider the set $B_n$ of
      words of length $n$ with at least one $0$ or $1$.



      How many of those are there?
      I am not sure how to express this in terms of $a_n$



      2) In
      this set there are exactly as many words with an even number of zeros as there are with
      an odd number of zeros. Just change the first occuring $0$ or $1$ with its dierence
      with $1$ $dots$ (not sure what is meant here, the wording is a bit vague)



      I am not sure what the author is getting at, can anybody see where the argument is going and give me some pointers?






      combinatorics recurrence-relations







      share|cite|improve this question









      $endgroup$





      Let $a_n$ be the number of words of length $n$ with letters from the alphabet ${0, 1, 2, 3}$
      containing an odd number of zeros.




      I have already verified that this is given by the recurrence relation:
      $$a_{n+1} = 2 a_n + 4^n.$$ where $a_0=0$.



      I then used the method of a generating function to solve this and arrive at:
      $$a_n= 2^{2n-1} - 2^{n-1}.$$
      Which seems to work for the first couple of values, also Wolfram Alpha agrees with my deduction.



      This problem can also be solved alternatively, using a new sequence $B_n$, I find this method more difficult but wish to learn it as well. It's more of a combinatorics method, probably using a clever counting argument.




      Another way to find the same result is as follows:



      1) Consider the set $B_n$ of
      words of length $n$ with at least one $0$ or $1$.



      How many of those are there?
      I am not sure how to express this in terms of $a_n$



      2) In
      this set there are exactly as many words with an even number of zeros as there are with
      an odd number of zeros. Just change the first occuring $0$ or $1$ with its dierence
      with $1$ $dots$ (not sure what is meant here, the wording is a bit vague)



      I am not sure what the author is getting at, can anybody see where the argument is going and give me some pointers?






      combinatorics recurrence-relations




      combinatorics recurrence-relations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 22:46









      Wesley StrikWesley Strik

      2,113423




      2,113423






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          For (1), we have $|B_n| = 4^n - 2^n$ since there are $4^n$ total strings over the alphabet, and $2^n$ of them have no zeros and no ones.



          To expand on (2), let $O_n subset B_n$ be the subset of strings with an odd number of zeros and $E_n subset B_n$ be the subset with an even number of zeros. Note that the function $f : O_n to E_n$ which on input $x$ replaces the first $0$ or $1$ in $x$ with $1$ or $0$, respectively, is a bijection. Thus $a_n = |O_n| = |E_n|$, and further, $O_n$ and $E_n$ partition $B_n$, so each has size $frac{1}{2}(4^n - 2^n)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That is indeed an elegant way to derive the same result, the bijection is very satisfying, thank you.
            $endgroup$
            – Wesley Strik
            Dec 14 '18 at 6:44











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038688%2fwords-of-length-n-containing-an-odd-number-of-zeros%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          For (1), we have $|B_n| = 4^n - 2^n$ since there are $4^n$ total strings over the alphabet, and $2^n$ of them have no zeros and no ones.



          To expand on (2), let $O_n subset B_n$ be the subset of strings with an odd number of zeros and $E_n subset B_n$ be the subset with an even number of zeros. Note that the function $f : O_n to E_n$ which on input $x$ replaces the first $0$ or $1$ in $x$ with $1$ or $0$, respectively, is a bijection. Thus $a_n = |O_n| = |E_n|$, and further, $O_n$ and $E_n$ partition $B_n$, so each has size $frac{1}{2}(4^n - 2^n)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That is indeed an elegant way to derive the same result, the bijection is very satisfying, thank you.
            $endgroup$
            – Wesley Strik
            Dec 14 '18 at 6:44
















          3












          $begingroup$

          For (1), we have $|B_n| = 4^n - 2^n$ since there are $4^n$ total strings over the alphabet, and $2^n$ of them have no zeros and no ones.



          To expand on (2), let $O_n subset B_n$ be the subset of strings with an odd number of zeros and $E_n subset B_n$ be the subset with an even number of zeros. Note that the function $f : O_n to E_n$ which on input $x$ replaces the first $0$ or $1$ in $x$ with $1$ or $0$, respectively, is a bijection. Thus $a_n = |O_n| = |E_n|$, and further, $O_n$ and $E_n$ partition $B_n$, so each has size $frac{1}{2}(4^n - 2^n)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That is indeed an elegant way to derive the same result, the bijection is very satisfying, thank you.
            $endgroup$
            – Wesley Strik
            Dec 14 '18 at 6:44














          3












          3








          3





          $begingroup$

          For (1), we have $|B_n| = 4^n - 2^n$ since there are $4^n$ total strings over the alphabet, and $2^n$ of them have no zeros and no ones.



          To expand on (2), let $O_n subset B_n$ be the subset of strings with an odd number of zeros and $E_n subset B_n$ be the subset with an even number of zeros. Note that the function $f : O_n to E_n$ which on input $x$ replaces the first $0$ or $1$ in $x$ with $1$ or $0$, respectively, is a bijection. Thus $a_n = |O_n| = |E_n|$, and further, $O_n$ and $E_n$ partition $B_n$, so each has size $frac{1}{2}(4^n - 2^n)$.






          share|cite|improve this answer











          $endgroup$



          For (1), we have $|B_n| = 4^n - 2^n$ since there are $4^n$ total strings over the alphabet, and $2^n$ of them have no zeros and no ones.



          To expand on (2), let $O_n subset B_n$ be the subset of strings with an odd number of zeros and $E_n subset B_n$ be the subset with an even number of zeros. Note that the function $f : O_n to E_n$ which on input $x$ replaces the first $0$ or $1$ in $x$ with $1$ or $0$, respectively, is a bijection. Thus $a_n = |O_n| = |E_n|$, and further, $O_n$ and $E_n$ partition $B_n$, so each has size $frac{1}{2}(4^n - 2^n)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 1 at 21:00

























          answered Dec 13 '18 at 23:07









          Zach LangleyZach Langley

          9731019




          9731019












          • $begingroup$
            That is indeed an elegant way to derive the same result, the bijection is very satisfying, thank you.
            $endgroup$
            – Wesley Strik
            Dec 14 '18 at 6:44


















          • $begingroup$
            That is indeed an elegant way to derive the same result, the bijection is very satisfying, thank you.
            $endgroup$
            – Wesley Strik
            Dec 14 '18 at 6:44
















          $begingroup$
          That is indeed an elegant way to derive the same result, the bijection is very satisfying, thank you.
          $endgroup$
          – Wesley Strik
          Dec 14 '18 at 6:44




          $begingroup$
          That is indeed an elegant way to derive the same result, the bijection is very satisfying, thank you.
          $endgroup$
          – Wesley Strik
          Dec 14 '18 at 6:44


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038688%2fwords-of-length-n-containing-an-odd-number-of-zeros%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Plaza Victoria

          Image
          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
          Read more

          Puebla de Zaragoza

          Image
          Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
          Read more

          Musa

          Image
          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
          Read more