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Template:Sequence of the Day for August 11

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Intended for: August 11, 2011

Timetable

  • First draft entered by Alonso del Arte on May 30, 2011 as a verbatim copy of the write-up from December 11, 2010 ✓
  • Draft reviewed by Alonso del Arte on August 10, 2011
  • Draft approved (sort of...) by Daniel Forgues on August 10, 2011
  • New draft entered by Alonso del Arte on August 10, 2011
  • New draft reviewed by Daniel Forgues on August 10, 2011
  • This draft approved for this year and next by Alonso del Arte on August 11, 2011 on account of Daniel's added insight of connecting dots I was only dimly aware needed connecting. ✓
Yesterday's SOTD * Tomorrow's SOTD

The line below marks the end of the <noinclude> ... </noinclude> section.



A001020:
11n, n   ≥   0
.
{ 1, 11, 121, 1331, 14641, 161051, 1771561, ... }

In the classic Star Trek episode “The Trouble with Tribbles,” Spock estimates the growth of the Tribble population aboard the Enterprise thus: “1,771,561. That’s assuming one tribble, multiplying with an average litter of 10, producing a new generation every 12 hours over a period of three days.”

Note that for
n = 0
to 4 you get numbers which are concatenated single digit (base 10) binomial coefficients, while for
n   ≥   5
some binomial coefficients are now more than single digit (base 10) and thus overlap, e.g.
1 
 5 
 10
  10
    5
     1
------
161051
Setting
x = 10
and
y = 1
in
(x + y)n =
n
k   = 0
  
(  nk  ) xk yn  − k

explains why this is so.

See also:

* A096884
101n, n   ≥   0
.
* A097659
1001n, n   ≥   0
.