Template:Sequence of the Day for August 11

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

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A001020:

11 n, 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  )  x k y n  − k

explains why this is so.

See also:

* A096884

101 n, n   ≥   0

.
* A097659

1001 n, n   ≥   0

.