Proof that Sylvester numbers, when reduced modulo 864 , form an arithmetic progression...











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The following observation has been made:



Numbers in Sylvester's sequence,when reduced $modulo 864$, form an arithmetic progression, namely $$7,43,79,115,151,187,223,259,295,331,.....$$



This has been checked for the first ten members of the sequence:



$$7≡7(mod864)$$
$$43≡43(mod864)$$
$$1807≡79(mod864)$$
$$3263443≡115(mod864)$$
$$10650056950807≡151(mod864)$$
$$113423713055421844361000443≡187(mod864)$$
$$12864938683278671740537145998360961546653259485195807≡223(mod864)$$



I have been unable to check other numbers in this sequence, due to the rapid growth of the sequence, the numbers become too large to handle.
However, we can use congruence relations, congruence arithmetic and arithmetic of residue classes to prove that Sylvester numbers ,when reduced $modulo 864$, form an arithmetic progression.
Consider the following:
One may define the sequence by the recurrence relation:



$$si=si−1(si−1−1)+1$$



Sylvester's sequence can also be defined by the formula:



$$sn=1+∏n−1i=0si$$



$$7≡7 (mod864)$$
$$7x6+1=43≡43 (mod864)$$
$$43x42+1=1807≡79 (mod 864)$$
$$79x78+1=6163≡115 (mod 864)$$
$$115x114+1=13111≡151 (mod 864)$$
$$151x150+1=22651≡187 (mod 864)$$
$$187x186+1=34783≡223 (mod 864)$$
$$223x222+1=49507≡259 (mod 864)$$
$$259x258+1=66823≡295 (mod 864)$$
$$295x294+1=86731≡331 (mod 864)$$
$$331x330+1=109231≡367 (mod 864)$$
$$367x366+1=134323≡403 (mod 864)$$
$$403x402+1=162007≡439 (mod 864)$$
$$439x438+1=192283≡475 (mod 864)$$
$$475x474+1=225151≡511 (mod 864)$$
$$511x510+1=260611≡547 (mod 864)$$
$$547x546+1=298663≡583 (mod 864)$$
$$583x582+1=339307≡619 (mod 864)$$
$$619x618+1=382543≡655 (mod 864)$$
$$655x654+1=428371≡691 (mod 864)$$
$$691x690+1=476791≡727 (mod 864)$$
$$727x726+1=527803≡763 (mod 864)$$
$$763x762+1=581407≡799 (mod 864)$$
$$799x798+1=637603≡835 (mod 864)$$
$$835x834+1=696391≡7 (mod 864)$$
$$7x6+1 =43≡43 (mod 864)$$
$$43x42+1 =1807≡79 (mod 864)$$
$$79x78+1=6163≡115 (mod 864)$$
etc.



Notice that after $24$ cycles , we get back to where we started. Hence Sylvester numbers , reduced $modulo 864$ form an arithmetic progression of $24$ terms which will then repeat until infinity. Therefore Sylvester sequence , reduced $mod 864$, forms an arithmetic progression of $24$ terms, which will repeat until infinity.
QED










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  • Please see math.meta.stackexchange.com/questions/5020 Incidentally, have you an OEIS number for this sequence?
    – Lord Shark the Unknown
    Nov 16 at 21:00












  • It's easy to compute the Sylvester numbers modulo any small number, just use the recursive definition...$a_{n+1}=a_n^2-a_n+1$. $pmod {864}$ the sequence goes ${2,3,7,43,79,115,151,187,223,259,295,331,367,403,439,475,511,547,583,619,655,cdots }$ and so on.
    – lulu
    Nov 16 at 21:09












  • Sylvester's sequence at OEIS: oeis.org/A000058
    – nickgard
    Nov 16 at 21:15










  • I expanded my answer to show how it arises simply via the Binomial Theorem.
    – Bill Dubuque
    Nov 18 at 0:09










  • I don't see a question here.
    – Gerry Myerson
    Nov 18 at 0:11















up vote
1
down vote

favorite












The following observation has been made:



Numbers in Sylvester's sequence,when reduced $modulo 864$, form an arithmetic progression, namely $$7,43,79,115,151,187,223,259,295,331,.....$$



This has been checked for the first ten members of the sequence:



$$7≡7(mod864)$$
$$43≡43(mod864)$$
$$1807≡79(mod864)$$
$$3263443≡115(mod864)$$
$$10650056950807≡151(mod864)$$
$$113423713055421844361000443≡187(mod864)$$
$$12864938683278671740537145998360961546653259485195807≡223(mod864)$$



I have been unable to check other numbers in this sequence, due to the rapid growth of the sequence, the numbers become too large to handle.
However, we can use congruence relations, congruence arithmetic and arithmetic of residue classes to prove that Sylvester numbers ,when reduced $modulo 864$, form an arithmetic progression.
Consider the following:
One may define the sequence by the recurrence relation:



$$si=si−1(si−1−1)+1$$



Sylvester's sequence can also be defined by the formula:



$$sn=1+∏n−1i=0si$$



$$7≡7 (mod864)$$
$$7x6+1=43≡43 (mod864)$$
$$43x42+1=1807≡79 (mod 864)$$
$$79x78+1=6163≡115 (mod 864)$$
$$115x114+1=13111≡151 (mod 864)$$
$$151x150+1=22651≡187 (mod 864)$$
$$187x186+1=34783≡223 (mod 864)$$
$$223x222+1=49507≡259 (mod 864)$$
$$259x258+1=66823≡295 (mod 864)$$
$$295x294+1=86731≡331 (mod 864)$$
$$331x330+1=109231≡367 (mod 864)$$
$$367x366+1=134323≡403 (mod 864)$$
$$403x402+1=162007≡439 (mod 864)$$
$$439x438+1=192283≡475 (mod 864)$$
$$475x474+1=225151≡511 (mod 864)$$
$$511x510+1=260611≡547 (mod 864)$$
$$547x546+1=298663≡583 (mod 864)$$
$$583x582+1=339307≡619 (mod 864)$$
$$619x618+1=382543≡655 (mod 864)$$
$$655x654+1=428371≡691 (mod 864)$$
$$691x690+1=476791≡727 (mod 864)$$
$$727x726+1=527803≡763 (mod 864)$$
$$763x762+1=581407≡799 (mod 864)$$
$$799x798+1=637603≡835 (mod 864)$$
$$835x834+1=696391≡7 (mod 864)$$
$$7x6+1 =43≡43 (mod 864)$$
$$43x42+1 =1807≡79 (mod 864)$$
$$79x78+1=6163≡115 (mod 864)$$
etc.



Notice that after $24$ cycles , we get back to where we started. Hence Sylvester numbers , reduced $modulo 864$ form an arithmetic progression of $24$ terms which will then repeat until infinity. Therefore Sylvester sequence , reduced $mod 864$, forms an arithmetic progression of $24$ terms, which will repeat until infinity.
QED










share|cite|improve this question
























  • Please see math.meta.stackexchange.com/questions/5020 Incidentally, have you an OEIS number for this sequence?
    – Lord Shark the Unknown
    Nov 16 at 21:00












  • It's easy to compute the Sylvester numbers modulo any small number, just use the recursive definition...$a_{n+1}=a_n^2-a_n+1$. $pmod {864}$ the sequence goes ${2,3,7,43,79,115,151,187,223,259,295,331,367,403,439,475,511,547,583,619,655,cdots }$ and so on.
    – lulu
    Nov 16 at 21:09












  • Sylvester's sequence at OEIS: oeis.org/A000058
    – nickgard
    Nov 16 at 21:15










  • I expanded my answer to show how it arises simply via the Binomial Theorem.
    – Bill Dubuque
    Nov 18 at 0:09










  • I don't see a question here.
    – Gerry Myerson
    Nov 18 at 0:11













up vote
1
down vote

favorite









up vote
1
down vote

favorite











The following observation has been made:



Numbers in Sylvester's sequence,when reduced $modulo 864$, form an arithmetic progression, namely $$7,43,79,115,151,187,223,259,295,331,.....$$



This has been checked for the first ten members of the sequence:



$$7≡7(mod864)$$
$$43≡43(mod864)$$
$$1807≡79(mod864)$$
$$3263443≡115(mod864)$$
$$10650056950807≡151(mod864)$$
$$113423713055421844361000443≡187(mod864)$$
$$12864938683278671740537145998360961546653259485195807≡223(mod864)$$



I have been unable to check other numbers in this sequence, due to the rapid growth of the sequence, the numbers become too large to handle.
However, we can use congruence relations, congruence arithmetic and arithmetic of residue classes to prove that Sylvester numbers ,when reduced $modulo 864$, form an arithmetic progression.
Consider the following:
One may define the sequence by the recurrence relation:



$$si=si−1(si−1−1)+1$$



Sylvester's sequence can also be defined by the formula:



$$sn=1+∏n−1i=0si$$



$$7≡7 (mod864)$$
$$7x6+1=43≡43 (mod864)$$
$$43x42+1=1807≡79 (mod 864)$$
$$79x78+1=6163≡115 (mod 864)$$
$$115x114+1=13111≡151 (mod 864)$$
$$151x150+1=22651≡187 (mod 864)$$
$$187x186+1=34783≡223 (mod 864)$$
$$223x222+1=49507≡259 (mod 864)$$
$$259x258+1=66823≡295 (mod 864)$$
$$295x294+1=86731≡331 (mod 864)$$
$$331x330+1=109231≡367 (mod 864)$$
$$367x366+1=134323≡403 (mod 864)$$
$$403x402+1=162007≡439 (mod 864)$$
$$439x438+1=192283≡475 (mod 864)$$
$$475x474+1=225151≡511 (mod 864)$$
$$511x510+1=260611≡547 (mod 864)$$
$$547x546+1=298663≡583 (mod 864)$$
$$583x582+1=339307≡619 (mod 864)$$
$$619x618+1=382543≡655 (mod 864)$$
$$655x654+1=428371≡691 (mod 864)$$
$$691x690+1=476791≡727 (mod 864)$$
$$727x726+1=527803≡763 (mod 864)$$
$$763x762+1=581407≡799 (mod 864)$$
$$799x798+1=637603≡835 (mod 864)$$
$$835x834+1=696391≡7 (mod 864)$$
$$7x6+1 =43≡43 (mod 864)$$
$$43x42+1 =1807≡79 (mod 864)$$
$$79x78+1=6163≡115 (mod 864)$$
etc.



Notice that after $24$ cycles , we get back to where we started. Hence Sylvester numbers , reduced $modulo 864$ form an arithmetic progression of $24$ terms which will then repeat until infinity. Therefore Sylvester sequence , reduced $mod 864$, forms an arithmetic progression of $24$ terms, which will repeat until infinity.
QED










share|cite|improve this question















The following observation has been made:



Numbers in Sylvester's sequence,when reduced $modulo 864$, form an arithmetic progression, namely $$7,43,79,115,151,187,223,259,295,331,.....$$



This has been checked for the first ten members of the sequence:



$$7≡7(mod864)$$
$$43≡43(mod864)$$
$$1807≡79(mod864)$$
$$3263443≡115(mod864)$$
$$10650056950807≡151(mod864)$$
$$113423713055421844361000443≡187(mod864)$$
$$12864938683278671740537145998360961546653259485195807≡223(mod864)$$



I have been unable to check other numbers in this sequence, due to the rapid growth of the sequence, the numbers become too large to handle.
However, we can use congruence relations, congruence arithmetic and arithmetic of residue classes to prove that Sylvester numbers ,when reduced $modulo 864$, form an arithmetic progression.
Consider the following:
One may define the sequence by the recurrence relation:



$$si=si−1(si−1−1)+1$$



Sylvester's sequence can also be defined by the formula:



$$sn=1+∏n−1i=0si$$



$$7≡7 (mod864)$$
$$7x6+1=43≡43 (mod864)$$
$$43x42+1=1807≡79 (mod 864)$$
$$79x78+1=6163≡115 (mod 864)$$
$$115x114+1=13111≡151 (mod 864)$$
$$151x150+1=22651≡187 (mod 864)$$
$$187x186+1=34783≡223 (mod 864)$$
$$223x222+1=49507≡259 (mod 864)$$
$$259x258+1=66823≡295 (mod 864)$$
$$295x294+1=86731≡331 (mod 864)$$
$$331x330+1=109231≡367 (mod 864)$$
$$367x366+1=134323≡403 (mod 864)$$
$$403x402+1=162007≡439 (mod 864)$$
$$439x438+1=192283≡475 (mod 864)$$
$$475x474+1=225151≡511 (mod 864)$$
$$511x510+1=260611≡547 (mod 864)$$
$$547x546+1=298663≡583 (mod 864)$$
$$583x582+1=339307≡619 (mod 864)$$
$$619x618+1=382543≡655 (mod 864)$$
$$655x654+1=428371≡691 (mod 864)$$
$$691x690+1=476791≡727 (mod 864)$$
$$727x726+1=527803≡763 (mod 864)$$
$$763x762+1=581407≡799 (mod 864)$$
$$799x798+1=637603≡835 (mod 864)$$
$$835x834+1=696391≡7 (mod 864)$$
$$7x6+1 =43≡43 (mod 864)$$
$$43x42+1 =1807≡79 (mod 864)$$
$$79x78+1=6163≡115 (mod 864)$$
etc.



Notice that after $24$ cycles , we get back to where we started. Hence Sylvester numbers , reduced $modulo 864$ form an arithmetic progression of $24$ terms which will then repeat until infinity. Therefore Sylvester sequence , reduced $mod 864$, forms an arithmetic progression of $24$ terms, which will repeat until infinity.
QED







sequences-and-series elementary-number-theory modular-arithmetic






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edited Nov 18 at 3:20









AryanSonwatikar

819




819










asked Nov 16 at 20:54









Derek

1845




1845












  • Please see math.meta.stackexchange.com/questions/5020 Incidentally, have you an OEIS number for this sequence?
    – Lord Shark the Unknown
    Nov 16 at 21:00












  • It's easy to compute the Sylvester numbers modulo any small number, just use the recursive definition...$a_{n+1}=a_n^2-a_n+1$. $pmod {864}$ the sequence goes ${2,3,7,43,79,115,151,187,223,259,295,331,367,403,439,475,511,547,583,619,655,cdots }$ and so on.
    – lulu
    Nov 16 at 21:09












  • Sylvester's sequence at OEIS: oeis.org/A000058
    – nickgard
    Nov 16 at 21:15










  • I expanded my answer to show how it arises simply via the Binomial Theorem.
    – Bill Dubuque
    Nov 18 at 0:09










  • I don't see a question here.
    – Gerry Myerson
    Nov 18 at 0:11


















  • Please see math.meta.stackexchange.com/questions/5020 Incidentally, have you an OEIS number for this sequence?
    – Lord Shark the Unknown
    Nov 16 at 21:00












  • It's easy to compute the Sylvester numbers modulo any small number, just use the recursive definition...$a_{n+1}=a_n^2-a_n+1$. $pmod {864}$ the sequence goes ${2,3,7,43,79,115,151,187,223,259,295,331,367,403,439,475,511,547,583,619,655,cdots }$ and so on.
    – lulu
    Nov 16 at 21:09












  • Sylvester's sequence at OEIS: oeis.org/A000058
    – nickgard
    Nov 16 at 21:15










  • I expanded my answer to show how it arises simply via the Binomial Theorem.
    – Bill Dubuque
    Nov 18 at 0:09










  • I don't see a question here.
    – Gerry Myerson
    Nov 18 at 0:11
















Please see math.meta.stackexchange.com/questions/5020 Incidentally, have you an OEIS number for this sequence?
– Lord Shark the Unknown
Nov 16 at 21:00






Please see math.meta.stackexchange.com/questions/5020 Incidentally, have you an OEIS number for this sequence?
– Lord Shark the Unknown
Nov 16 at 21:00














It's easy to compute the Sylvester numbers modulo any small number, just use the recursive definition...$a_{n+1}=a_n^2-a_n+1$. $pmod {864}$ the sequence goes ${2,3,7,43,79,115,151,187,223,259,295,331,367,403,439,475,511,547,583,619,655,cdots }$ and so on.
– lulu
Nov 16 at 21:09






It's easy to compute the Sylvester numbers modulo any small number, just use the recursive definition...$a_{n+1}=a_n^2-a_n+1$. $pmod {864}$ the sequence goes ${2,3,7,43,79,115,151,187,223,259,295,331,367,403,439,475,511,547,583,619,655,cdots }$ and so on.
– lulu
Nov 16 at 21:09














Sylvester's sequence at OEIS: oeis.org/A000058
– nickgard
Nov 16 at 21:15




Sylvester's sequence at OEIS: oeis.org/A000058
– nickgard
Nov 16 at 21:15












I expanded my answer to show how it arises simply via the Binomial Theorem.
– Bill Dubuque
Nov 18 at 0:09




I expanded my answer to show how it arises simply via the Binomial Theorem.
– Bill Dubuque
Nov 18 at 0:09












I don't see a question here.
– Gerry Myerson
Nov 18 at 0:11




I don't see a question here.
– Gerry Myerson
Nov 18 at 0:11










2 Answers
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up vote
1
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Hint:



If $$a_n=36n+7$$
then
$$a_n^2-a_n+1=(36n+7)^2-(36n+7)+1=432n^2+468n+43=36(12n^2+13n+1)+7.$$



Now remains to study



$$(12n^2+13n+1)bmodfrac{864}{36}.$$



(Final hint, $12n(n+1)bmod24=0$.)






share|cite|improve this answer






























    up vote
    1
    down vote













    Below I give a simple direct proof, then I explain how it boils down to the Binomial Theorem.



    Notice $bmod 864!: a_n = 7! +! 36j Longrightarrow color{#0a0}{a_{n+1}},equiv, color{#0a0}{a_n}!+color{#c00}{36}, $ by



    $underbrace{color{#0a0}{a_{n+1}!-!a_n} = (a_n!-!1)^2}_{Large a_{n+1} = a_n^2-a_n+1}! = (6!+!36j)^2! = 36+432,underbrace{j(1!+!3j)}_{large rm even}equiv color{#c00}{36}.,$



    Remark $ $ This is closely connected to the arithmetic progression that arises from the first two terms of the Binomial Theorem $, (1+b)^n = color{#0a0}{1+nb},pmod{!b^2}.,$ To make this clearer we examine the sequence $,b_n := a_n! - 1,,$ which satisfies $,b_{k+1} = b_k + b_k^2,, b_0 = 1.$



    Lemma $ $ Suppose a sequence $b_k$ satisfies the recurrence $,b_{k+1} = b_k(1+ b_k).,$ Then $$ b := b_{large k}, Rightarrow, b_{large k+n} equiv b(1+b)^nequiv b(color{#0a0}{1+nb}) pmod {!b^3}qquad $$



    Proof $ $ By induction on $n$. Clear if $n=0.,$ Suppose for induction it is true for $n$. Then



    $qquad begin{align}
    bmod color{#c00}{b^3}!: b_{large j+n+1}, = &qquad, b_{large j+n} (1+b_{large j+n})\
    = &, color{#c00}b(1+nb)(1+b+ncolor{#c00}{b^2)}\
    equiv &, b(1+nb)(1+b) {rm by} color{#c00}bcolor{#c00}{b^2}equiv 0\
    equiv &, b(1+(n!+!1)b)\
    equiv &, b(1+b)^{large n+1}qquadqquadquad {bf QED}
    end{align}qquadqquadqquad $



    When $,b = 2!+!4iequiv 2pmod{!4}$ we can enlarge the modulus to $,4b^3$ since, as above



    $quad begin{align}
    bmod color{#c00}{4b^3}!: b_{j+n+1}
    equiv &, b(1+(n!+!1)b) + color{#c00}2,{underbrace{n(1+nb/2}_{color{#c00}{rm even}})},color{#c00}{b^3}\[.2em]
    end{align}qquadqquadqquad $



    Here $,b_2 = 6,$ so we get $,b_{2+n}equiv 6 + 36npmod{!864},,$ just as you observed.






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      2 Answers
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      2 Answers
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      active

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      up vote
      1
      down vote













      Hint:



      If $$a_n=36n+7$$
      then
      $$a_n^2-a_n+1=(36n+7)^2-(36n+7)+1=432n^2+468n+43=36(12n^2+13n+1)+7.$$



      Now remains to study



      $$(12n^2+13n+1)bmodfrac{864}{36}.$$



      (Final hint, $12n(n+1)bmod24=0$.)






      share|cite|improve this answer



























        up vote
        1
        down vote













        Hint:



        If $$a_n=36n+7$$
        then
        $$a_n^2-a_n+1=(36n+7)^2-(36n+7)+1=432n^2+468n+43=36(12n^2+13n+1)+7.$$



        Now remains to study



        $$(12n^2+13n+1)bmodfrac{864}{36}.$$



        (Final hint, $12n(n+1)bmod24=0$.)






        share|cite|improve this answer

























          up vote
          1
          down vote










          up vote
          1
          down vote









          Hint:



          If $$a_n=36n+7$$
          then
          $$a_n^2-a_n+1=(36n+7)^2-(36n+7)+1=432n^2+468n+43=36(12n^2+13n+1)+7.$$



          Now remains to study



          $$(12n^2+13n+1)bmodfrac{864}{36}.$$



          (Final hint, $12n(n+1)bmod24=0$.)






          share|cite|improve this answer














          Hint:



          If $$a_n=36n+7$$
          then
          $$a_n^2-a_n+1=(36n+7)^2-(36n+7)+1=432n^2+468n+43=36(12n^2+13n+1)+7.$$



          Now remains to study



          $$(12n^2+13n+1)bmodfrac{864}{36}.$$



          (Final hint, $12n(n+1)bmod24=0$.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 17 at 0:07

























          answered Nov 16 at 21:17









          Yves Daoust

          122k668217




          122k668217






















              up vote
              1
              down vote













              Below I give a simple direct proof, then I explain how it boils down to the Binomial Theorem.



              Notice $bmod 864!: a_n = 7! +! 36j Longrightarrow color{#0a0}{a_{n+1}},equiv, color{#0a0}{a_n}!+color{#c00}{36}, $ by



              $underbrace{color{#0a0}{a_{n+1}!-!a_n} = (a_n!-!1)^2}_{Large a_{n+1} = a_n^2-a_n+1}! = (6!+!36j)^2! = 36+432,underbrace{j(1!+!3j)}_{large rm even}equiv color{#c00}{36}.,$



              Remark $ $ This is closely connected to the arithmetic progression that arises from the first two terms of the Binomial Theorem $, (1+b)^n = color{#0a0}{1+nb},pmod{!b^2}.,$ To make this clearer we examine the sequence $,b_n := a_n! - 1,,$ which satisfies $,b_{k+1} = b_k + b_k^2,, b_0 = 1.$



              Lemma $ $ Suppose a sequence $b_k$ satisfies the recurrence $,b_{k+1} = b_k(1+ b_k).,$ Then $$ b := b_{large k}, Rightarrow, b_{large k+n} equiv b(1+b)^nequiv b(color{#0a0}{1+nb}) pmod {!b^3}qquad $$



              Proof $ $ By induction on $n$. Clear if $n=0.,$ Suppose for induction it is true for $n$. Then



              $qquad begin{align}
              bmod color{#c00}{b^3}!: b_{large j+n+1}, = &qquad, b_{large j+n} (1+b_{large j+n})\
              = &, color{#c00}b(1+nb)(1+b+ncolor{#c00}{b^2)}\
              equiv &, b(1+nb)(1+b) {rm by} color{#c00}bcolor{#c00}{b^2}equiv 0\
              equiv &, b(1+(n!+!1)b)\
              equiv &, b(1+b)^{large n+1}qquadqquadquad {bf QED}
              end{align}qquadqquadqquad $



              When $,b = 2!+!4iequiv 2pmod{!4}$ we can enlarge the modulus to $,4b^3$ since, as above



              $quad begin{align}
              bmod color{#c00}{4b^3}!: b_{j+n+1}
              equiv &, b(1+(n!+!1)b) + color{#c00}2,{underbrace{n(1+nb/2}_{color{#c00}{rm even}})},color{#c00}{b^3}\[.2em]
              end{align}qquadqquadqquad $



              Here $,b_2 = 6,$ so we get $,b_{2+n}equiv 6 + 36npmod{!864},,$ just as you observed.






              share|cite|improve this answer



























                up vote
                1
                down vote













                Below I give a simple direct proof, then I explain how it boils down to the Binomial Theorem.



                Notice $bmod 864!: a_n = 7! +! 36j Longrightarrow color{#0a0}{a_{n+1}},equiv, color{#0a0}{a_n}!+color{#c00}{36}, $ by



                $underbrace{color{#0a0}{a_{n+1}!-!a_n} = (a_n!-!1)^2}_{Large a_{n+1} = a_n^2-a_n+1}! = (6!+!36j)^2! = 36+432,underbrace{j(1!+!3j)}_{large rm even}equiv color{#c00}{36}.,$



                Remark $ $ This is closely connected to the arithmetic progression that arises from the first two terms of the Binomial Theorem $, (1+b)^n = color{#0a0}{1+nb},pmod{!b^2}.,$ To make this clearer we examine the sequence $,b_n := a_n! - 1,,$ which satisfies $,b_{k+1} = b_k + b_k^2,, b_0 = 1.$



                Lemma $ $ Suppose a sequence $b_k$ satisfies the recurrence $,b_{k+1} = b_k(1+ b_k).,$ Then $$ b := b_{large k}, Rightarrow, b_{large k+n} equiv b(1+b)^nequiv b(color{#0a0}{1+nb}) pmod {!b^3}qquad $$



                Proof $ $ By induction on $n$. Clear if $n=0.,$ Suppose for induction it is true for $n$. Then



                $qquad begin{align}
                bmod color{#c00}{b^3}!: b_{large j+n+1}, = &qquad, b_{large j+n} (1+b_{large j+n})\
                = &, color{#c00}b(1+nb)(1+b+ncolor{#c00}{b^2)}\
                equiv &, b(1+nb)(1+b) {rm by} color{#c00}bcolor{#c00}{b^2}equiv 0\
                equiv &, b(1+(n!+!1)b)\
                equiv &, b(1+b)^{large n+1}qquadqquadquad {bf QED}
                end{align}qquadqquadqquad $



                When $,b = 2!+!4iequiv 2pmod{!4}$ we can enlarge the modulus to $,4b^3$ since, as above



                $quad begin{align}
                bmod color{#c00}{4b^3}!: b_{j+n+1}
                equiv &, b(1+(n!+!1)b) + color{#c00}2,{underbrace{n(1+nb/2}_{color{#c00}{rm even}})},color{#c00}{b^3}\[.2em]
                end{align}qquadqquadqquad $



                Here $,b_2 = 6,$ so we get $,b_{2+n}equiv 6 + 36npmod{!864},,$ just as you observed.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Below I give a simple direct proof, then I explain how it boils down to the Binomial Theorem.



                  Notice $bmod 864!: a_n = 7! +! 36j Longrightarrow color{#0a0}{a_{n+1}},equiv, color{#0a0}{a_n}!+color{#c00}{36}, $ by



                  $underbrace{color{#0a0}{a_{n+1}!-!a_n} = (a_n!-!1)^2}_{Large a_{n+1} = a_n^2-a_n+1}! = (6!+!36j)^2! = 36+432,underbrace{j(1!+!3j)}_{large rm even}equiv color{#c00}{36}.,$



                  Remark $ $ This is closely connected to the arithmetic progression that arises from the first two terms of the Binomial Theorem $, (1+b)^n = color{#0a0}{1+nb},pmod{!b^2}.,$ To make this clearer we examine the sequence $,b_n := a_n! - 1,,$ which satisfies $,b_{k+1} = b_k + b_k^2,, b_0 = 1.$



                  Lemma $ $ Suppose a sequence $b_k$ satisfies the recurrence $,b_{k+1} = b_k(1+ b_k).,$ Then $$ b := b_{large k}, Rightarrow, b_{large k+n} equiv b(1+b)^nequiv b(color{#0a0}{1+nb}) pmod {!b^3}qquad $$



                  Proof $ $ By induction on $n$. Clear if $n=0.,$ Suppose for induction it is true for $n$. Then



                  $qquad begin{align}
                  bmod color{#c00}{b^3}!: b_{large j+n+1}, = &qquad, b_{large j+n} (1+b_{large j+n})\
                  = &, color{#c00}b(1+nb)(1+b+ncolor{#c00}{b^2)}\
                  equiv &, b(1+nb)(1+b) {rm by} color{#c00}bcolor{#c00}{b^2}equiv 0\
                  equiv &, b(1+(n!+!1)b)\
                  equiv &, b(1+b)^{large n+1}qquadqquadquad {bf QED}
                  end{align}qquadqquadqquad $



                  When $,b = 2!+!4iequiv 2pmod{!4}$ we can enlarge the modulus to $,4b^3$ since, as above



                  $quad begin{align}
                  bmod color{#c00}{4b^3}!: b_{j+n+1}
                  equiv &, b(1+(n!+!1)b) + color{#c00}2,{underbrace{n(1+nb/2}_{color{#c00}{rm even}})},color{#c00}{b^3}\[.2em]
                  end{align}qquadqquadqquad $



                  Here $,b_2 = 6,$ so we get $,b_{2+n}equiv 6 + 36npmod{!864},,$ just as you observed.






                  share|cite|improve this answer














                  Below I give a simple direct proof, then I explain how it boils down to the Binomial Theorem.



                  Notice $bmod 864!: a_n = 7! +! 36j Longrightarrow color{#0a0}{a_{n+1}},equiv, color{#0a0}{a_n}!+color{#c00}{36}, $ by



                  $underbrace{color{#0a0}{a_{n+1}!-!a_n} = (a_n!-!1)^2}_{Large a_{n+1} = a_n^2-a_n+1}! = (6!+!36j)^2! = 36+432,underbrace{j(1!+!3j)}_{large rm even}equiv color{#c00}{36}.,$



                  Remark $ $ This is closely connected to the arithmetic progression that arises from the first two terms of the Binomial Theorem $, (1+b)^n = color{#0a0}{1+nb},pmod{!b^2}.,$ To make this clearer we examine the sequence $,b_n := a_n! - 1,,$ which satisfies $,b_{k+1} = b_k + b_k^2,, b_0 = 1.$



                  Lemma $ $ Suppose a sequence $b_k$ satisfies the recurrence $,b_{k+1} = b_k(1+ b_k).,$ Then $$ b := b_{large k}, Rightarrow, b_{large k+n} equiv b(1+b)^nequiv b(color{#0a0}{1+nb}) pmod {!b^3}qquad $$



                  Proof $ $ By induction on $n$. Clear if $n=0.,$ Suppose for induction it is true for $n$. Then



                  $qquad begin{align}
                  bmod color{#c00}{b^3}!: b_{large j+n+1}, = &qquad, b_{large j+n} (1+b_{large j+n})\
                  = &, color{#c00}b(1+nb)(1+b+ncolor{#c00}{b^2)}\
                  equiv &, b(1+nb)(1+b) {rm by} color{#c00}bcolor{#c00}{b^2}equiv 0\
                  equiv &, b(1+(n!+!1)b)\
                  equiv &, b(1+b)^{large n+1}qquadqquadquad {bf QED}
                  end{align}qquadqquadqquad $



                  When $,b = 2!+!4iequiv 2pmod{!4}$ we can enlarge the modulus to $,4b^3$ since, as above



                  $quad begin{align}
                  bmod color{#c00}{4b^3}!: b_{j+n+1}
                  equiv &, b(1+(n!+!1)b) + color{#c00}2,{underbrace{n(1+nb/2}_{color{#c00}{rm even}})},color{#c00}{b^3}\[.2em]
                  end{align}qquadqquadqquad $



                  Here $,b_2 = 6,$ so we get $,b_{2+n}equiv 6 + 36npmod{!864},,$ just as you observed.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 18 at 0:13









                  Gerry Myerson

                  145k8147298




                  145k8147298










                  answered Nov 16 at 21:59









                  Bill Dubuque

                  207k29189624




                  207k29189624






























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