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How to derive this sequence: $1^3+5^3+3^3=153,16^3+50^3+33^3=165033,166^3+500^3+333^3=166500333,cdots$?

Derive a General formula for each term of this periodic sequence?How to show this sequence of functions weakly converge?How to establish convergence and find limit of the sequence$(n+1)^1/ln(n+1)$challenge problem: show this sequence is convergent.The $n$th term of this infinitely nested radical sequenceHow to prove that a sequence is divergent?How to show this sequence is decreasingHow to prove two sequence have a common limit.Finding general term of given sequence.Is this sequence following some rule: $(1+3)^2 + (1+2)^2=(2+3)^2$ , $(1+3+5)^2 + (3+4+5)^2=(4+5+6)^2$ , …?

4

2

$begingroup$

I found it is interesting but I don't know how the R.H.S is coming from the L.H.S,

i.e, how to derive this sequence?

The sequence is as follows:
$$1^3+5^3+3^3=153$$
$$16^3+50^3+33^3=165033$$
$$166^3+500^3+333^3=166500333$$
$$1666^3+5000^3+3333^3=166650003333$$
$$.$$
$$.$$
$$.$$
$$textand so on$$
So,any help please...

sequences-and-series

share|cite|improve this question

edited Mar 17 at 11:20

J.G.

32.1k23250

asked Mar 17 at 9:23

SureshSuresh

369110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Small error : in the middle it is $50 , 500,5000$ and so on.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 17 at 9:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Sequence $A246057$ in OEIS is also funny ! Notice that your sequence was entered in year 2017.
  $endgroup$
  – Claude Leibovici
  Mar 17 at 9:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you write every term $a_k$ in the sequence as $a_k=fracb_k9$, you should be able to get a closed fromula. Eg: 1666= 14994/9 = (15*(10^n)-6) /9 ; 5000 = 5*10^n ; 3333*9 = 29997 = 30*(10^n) - 3 . And 166650003333 *9 = 1499850029997 = (150*(10^n) - 15) *(10^(2n+1)) + (30*(10^n) -3) . Then you just plug them in the formula and see if all the components match. It's tedious, but it works.
  $endgroup$
  – user3257842
  Mar 17 at 10:13
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For the fun, in the OEIS window, just type curious cubic identity
  $endgroup$
  – Claude Leibovici
  Mar 17 at 10:15
  

  
  
  
  

add a comment | 

4

2

$begingroup$

I found it is interesting but I don't know how the R.H.S is coming from the L.H.S,

i.e, how to derive this sequence?

The sequence is as follows:
$$1^3+5^3+3^3=153$$
$$16^3+50^3+33^3=165033$$
$$166^3+500^3+333^3=166500333$$
$$1666^3+5000^3+3333^3=166650003333$$
$$.$$
$$.$$
$$.$$
$$textand so on$$
So,any help please...

sequences-and-series

share|cite|improve this question

edited Mar 17 at 11:20

J.G.

32.1k23250

asked Mar 17 at 9:23

SureshSuresh

369110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Small error : in the middle it is $50 , 500,5000$ and so on.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Mar 17 at 9:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Sequence $A246057$ in OEIS is also funny ! Notice that your sequence was entered in year 2017.
  $endgroup$
  – Claude Leibovici
  Mar 17 at 9:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you write every term $a_k$ in the sequence as $a_k=fracb_k9$, you should be able to get a closed fromula. Eg: 1666= 14994/9 = (15*(10^n)-6) /9 ; 5000 = 5*10^n ; 3333*9 = 29997 = 30*(10^n) - 3 . And 166650003333 *9 = 1499850029997 = (150*(10^n) - 15) *(10^(2n+1)) + (30*(10^n) -3) . Then you just plug them in the formula and see if all the components match. It's tedious, but it works.
  $endgroup$
  – user3257842
  Mar 17 at 10:13
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For the fun, in the OEIS window, just type curious cubic identity
  $endgroup$
  – Claude Leibovici
  Mar 17 at 10:15
  

  
  
  
  

add a comment | 

$begingroup$

I found it is interesting but I don't know how the R.H.S is coming from the L.H.S,

i.e, how to derive this sequence?

The sequence is as follows:
$$1^3+5^3+3^3=153$$
$$16^3+50^3+33^3=165033$$
$$166^3+500^3+333^3=166500333$$
$$1666^3+5000^3+3333^3=166650003333$$
$$.$$
$$.$$
$$.$$
$$textand so on$$
So,any help please...

sequences-and-series

share|cite|improve this question

edited Mar 17 at 11:20

J.G.

32.1k23250

asked Mar 17 at 9:23

SureshSuresh

369110

$endgroup$

share|cite|improve this question

edited Mar 17 at 11:20

J.G.

32.1k23250

asked Mar 17 at 9:23

SureshSuresh

369110

share|cite|improve this question

edited Mar 17 at 11:20

J.G.

32.1k23250

asked Mar 17 at 9:23

SureshSuresh

369110

edited Mar 17 at 11:20

J.G.

32.1k23250

edited Mar 17 at 11:20

J.G.

32.1k23250

asked Mar 17 at 9:23

SureshSuresh

369110

asked Mar 17 at 9:23

SureshSuresh

369110

3 Answers
3

active

oldest

votes

2

$begingroup$

The $n$th such equation is $$left(tfrac5times 10^n-1-23right)^3+(5times 10^n-1)^3+left(tfrac10^n-13right)^3=10^3n-1+tfrac23(10^n-1-1)10^2n+5times 10^2n-1+tfrac10^n-13,$$or with $x=10^n$ we can write it as $$left(tfracx-46right)^3+left(tfracx2right)^3+left(tfracx-13right)^3=tfracx^310+tfrac23left(tfracx10-1right)x^2+tfracx^22+tfracx-13,$$which we can verify algebraically (both sides are $tfrac(x-1)(x^2+2)6$).

share|cite|improve this answer

answered Mar 17 at 11:35

J.G.J.G.

32.1k23250

$endgroup$

add a comment | 

2

$begingroup$

First some notations, let express the repunits $r_n=11cdots 1=dfrac(10^n-1)9$ in term of $k=10^n$

$begincases
a_n&=16cdots 6&=10^n-1+6 r_n-1&=frack-46\
b_n&=50cdots 0&=5times 10^n-1&=frack2\
c_n&=33cdots 3&=3,r_n&=frack-13endcases$

The sum of cubes

$f(n)=a_n^3+b_n^3+c_n^3=left(frack-46right)^3+left(frac k2right)^3+left(frack-13right)^3=dfrack^3-k^2+2k-26=dfrac(k-1)(k^2+2)6$

Can be expressed like this:
$$boxedf(n)=dfrac(10^n-1)(10^2n+2)6$$

But we are interested in the non-factored expression:

$beginalignf(n)&=dfrack^3-k^2+2k-26\&=dfrack^3-4k^2+3k^2+2k-26\&=left(dfrack^2-4k^26right)+left(dfrac 3k^26right)+left(dfrac2k-26right)=k^2,a_n+k,b_n+c_n=overlinea_nb_nc_nendalign$

Where

$$overlinea_nb_nc_n=underbrace16cdots 6_a_ntimes 10^2n;underbrace50cdots 0_b_ntimes 10^n;underbrace33cdots 3_c_n$$

share|cite|improve this answer

answered Mar 17 at 12:46

zwimzwim

12.6k831

$endgroup$

add a comment | 

1

$begingroup$

This is sequence A281857 in OEIS. They give the recurrence relation
$$a_n = 1111,a_n-1 - 112110,a_n-2 + 1111000,a_n-3 - 1000000,a_n-4 $$ for which which you already found the first terms.

share|cite|improve this answer

answered Mar 17 at 9:43

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

add a comment | 

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2

$begingroup$

The $n$th such equation is $$left(tfrac5times 10^n-1-23right)^3+(5times 10^n-1)^3+left(tfrac10^n-13right)^3=10^3n-1+tfrac23(10^n-1-1)10^2n+5times 10^2n-1+tfrac10^n-13,$$or with $x=10^n$ we can write it as $$left(tfracx-46right)^3+left(tfracx2right)^3+left(tfracx-13right)^3=tfracx^310+tfrac23left(tfracx10-1right)x^2+tfracx^22+tfracx-13,$$which we can verify algebraically (both sides are $tfrac(x-1)(x^2+2)6$).

share|cite|improve this answer

answered Mar 17 at 11:35

J.G.J.G.

32.1k23250

$endgroup$

add a comment | 

2

$begingroup$

The $n$th such equation is $$left(tfrac5times 10^n-1-23right)^3+(5times 10^n-1)^3+left(tfrac10^n-13right)^3=10^3n-1+tfrac23(10^n-1-1)10^2n+5times 10^2n-1+tfrac10^n-13,$$or with $x=10^n$ we can write it as $$left(tfracx-46right)^3+left(tfracx2right)^3+left(tfracx-13right)^3=tfracx^310+tfrac23left(tfracx10-1right)x^2+tfracx^22+tfracx-13,$$which we can verify algebraically (both sides are $tfrac(x-1)(x^2+2)6$).

share|cite|improve this answer

answered Mar 17 at 11:35

J.G.J.G.

32.1k23250

$endgroup$

add a comment | 

$begingroup$

The $n$th such equation is $$left(tfrac5times 10^n-1-23right)^3+(5times 10^n-1)^3+left(tfrac10^n-13right)^3=10^3n-1+tfrac23(10^n-1-1)10^2n+5times 10^2n-1+tfrac10^n-13,$$or with $x=10^n$ we can write it as $$left(tfracx-46right)^3+left(tfracx2right)^3+left(tfracx-13right)^3=tfracx^310+tfrac23left(tfracx10-1right)x^2+tfracx^22+tfracx-13,$$which we can verify algebraically (both sides are $tfrac(x-1)(x^2+2)6$).

share|cite|improve this answer

answered Mar 17 at 11:35

J.G.J.G.

32.1k23250

$endgroup$

share|cite|improve this answer

answered Mar 17 at 11:35

J.G.J.G.

32.1k23250

answered Mar 17 at 11:35

J.G.J.G.

32.1k23250

answered Mar 17 at 11:35

J.G.J.G.

32.1k23250

2

$begingroup$

First some notations, let express the repunits $r_n=11cdots 1=dfrac(10^n-1)9$ in term of $k=10^n$

$begincases
a_n&=16cdots 6&=10^n-1+6 r_n-1&=frack-46\
b_n&=50cdots 0&=5times 10^n-1&=frack2\
c_n&=33cdots 3&=3,r_n&=frack-13endcases$

The sum of cubes

$f(n)=a_n^3+b_n^3+c_n^3=left(frack-46right)^3+left(frac k2right)^3+left(frack-13right)^3=dfrack^3-k^2+2k-26=dfrac(k-1)(k^2+2)6$

Can be expressed like this:
$$boxedf(n)=dfrac(10^n-1)(10^2n+2)6$$

But we are interested in the non-factored expression:

$beginalignf(n)&=dfrack^3-k^2+2k-26\&=dfrack^3-4k^2+3k^2+2k-26\&=left(dfrack^2-4k^26right)+left(dfrac 3k^26right)+left(dfrac2k-26right)=k^2,a_n+k,b_n+c_n=overlinea_nb_nc_nendalign$

Where

$$overlinea_nb_nc_n=underbrace16cdots 6_a_ntimes 10^2n;underbrace50cdots 0_b_ntimes 10^n;underbrace33cdots 3_c_n$$

share|cite|improve this answer

answered Mar 17 at 12:46

zwimzwim

12.6k831

$endgroup$

add a comment | 

2

$begingroup$

First some notations, let express the repunits $r_n=11cdots 1=dfrac(10^n-1)9$ in term of $k=10^n$

$begincases
a_n&=16cdots 6&=10^n-1+6 r_n-1&=frack-46\
b_n&=50cdots 0&=5times 10^n-1&=frack2\
c_n&=33cdots 3&=3,r_n&=frack-13endcases$

The sum of cubes

$f(n)=a_n^3+b_n^3+c_n^3=left(frack-46right)^3+left(frac k2right)^3+left(frack-13right)^3=dfrack^3-k^2+2k-26=dfrac(k-1)(k^2+2)6$

Can be expressed like this:
$$boxedf(n)=dfrac(10^n-1)(10^2n+2)6$$

But we are interested in the non-factored expression:

$beginalignf(n)&=dfrack^3-k^2+2k-26\&=dfrack^3-4k^2+3k^2+2k-26\&=left(dfrack^2-4k^26right)+left(dfrac 3k^26right)+left(dfrac2k-26right)=k^2,a_n+k,b_n+c_n=overlinea_nb_nc_nendalign$

Where

$$overlinea_nb_nc_n=underbrace16cdots 6_a_ntimes 10^2n;underbrace50cdots 0_b_ntimes 10^n;underbrace33cdots 3_c_n$$

share|cite|improve this answer

answered Mar 17 at 12:46

zwimzwim

12.6k831

$endgroup$

add a comment | 

$begingroup$

First some notations, let express the repunits $r_n=11cdots 1=dfrac(10^n-1)9$ in term of $k=10^n$

$begincases
a_n&=16cdots 6&=10^n-1+6 r_n-1&=frack-46\
b_n&=50cdots 0&=5times 10^n-1&=frack2\
c_n&=33cdots 3&=3,r_n&=frack-13endcases$

The sum of cubes

$f(n)=a_n^3+b_n^3+c_n^3=left(frack-46right)^3+left(frac k2right)^3+left(frack-13right)^3=dfrack^3-k^2+2k-26=dfrac(k-1)(k^2+2)6$

Can be expressed like this:
$$boxedf(n)=dfrac(10^n-1)(10^2n+2)6$$

But we are interested in the non-factored expression:

$beginalignf(n)&=dfrack^3-k^2+2k-26\&=dfrack^3-4k^2+3k^2+2k-26\&=left(dfrack^2-4k^26right)+left(dfrac 3k^26right)+left(dfrac2k-26right)=k^2,a_n+k,b_n+c_n=overlinea_nb_nc_nendalign$

Where

$$overlinea_nb_nc_n=underbrace16cdots 6_a_ntimes 10^2n;underbrace50cdots 0_b_ntimes 10^n;underbrace33cdots 3_c_n$$

share|cite|improve this answer

answered Mar 17 at 12:46

zwimzwim

12.6k831

$endgroup$

share|cite|improve this answer

answered Mar 17 at 12:46

zwimzwim

12.6k831

answered Mar 17 at 12:46

zwimzwim

12.6k831

answered Mar 17 at 12:46

zwimzwim

12.6k831

1

$begingroup$

This is sequence A281857 in OEIS. They give the recurrence relation
$$a_n = 1111,a_n-1 - 112110,a_n-2 + 1111000,a_n-3 - 1000000,a_n-4 $$ for which which you already found the first terms.

share|cite|improve this answer

answered Mar 17 at 9:43

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

add a comment | 

1

$begingroup$

This is sequence A281857 in OEIS. They give the recurrence relation
$$a_n = 1111,a_n-1 - 112110,a_n-2 + 1111000,a_n-3 - 1000000,a_n-4 $$ for which which you already found the first terms.

share|cite|improve this answer

answered Mar 17 at 9:43

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

add a comment | 

$begingroup$

This is sequence A281857 in OEIS. They give the recurrence relation
$$a_n = 1111,a_n-1 - 112110,a_n-2 + 1111000,a_n-3 - 1000000,a_n-4 $$ for which which you already found the first terms.

share|cite|improve this answer

answered Mar 17 at 9:43

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

share|cite|improve this answer

answered Mar 17 at 9:43

Claude LeiboviciClaude Leibovici

125k1158135

answered Mar 17 at 9:43

Claude LeiboviciClaude Leibovici

125k1158135

answered Mar 17 at 9:43

Claude LeiboviciClaude Leibovici

125k1158135