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November 10th, 2014, 11:44 AM  #1 
Member Joined: Jan 2013 Posts: 47 Thanks: 0  Super simple sequence question
Right so I've got myself in a right pickle on this question: Find the nth term of 2,1,2,11 So I've tried using the standard an^2 + bn + c But that hasn't worked .. Any help?? Last edited by skipjack; November 10th, 2014 at 03:03 PM. 
November 10th, 2014, 11:52 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,878 Thanks: 2240 Math Focus: Mainly analysis and algebra 
You will be able to make a cubic equation fit.

November 10th, 2014, 12:02 PM  #3 
Member Joined: Jan 2013 Posts: 47 Thanks: 0 
How would you do that?

November 10th, 2014, 12:25 PM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
You can fit any n+1 points to an nth degree equation  this is the 'boring' solution. In your case I get (2n^3  9n^2 + 16n  15)/3 which you could get by solving an^3 + bn^2 + cn + d = y for each of the (x, y) pairs you have. It's a bit tiresome. 
November 10th, 2014, 12:33 PM  #5 
Member Joined: Jan 2013 Posts: 47 Thanks: 0 
got 4/6n^3 3n^2 + 14 and 2/3 times n  29/3 Any good? 
November 10th, 2014, 12:57 PM  #6 
Member Joined: Jan 2013 Posts: 47 Thanks: 0 
Anyone else got an answer?

November 10th, 2014, 01:35 PM  #7 
Member Joined: Jan 2013 Posts: 47 Thanks: 0 
Someone must have a solution and answer!

November 10th, 2014, 01:40 PM  #8 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,546 Thanks: 925 Math Focus: Elementary mathematics and beyond 
You run out of room with differences ... so assume it's a cubic! Solve the system a + b + c + d = 2 8a + 4b + 2c + d = 1 27a + 9b + 3c + d = 2 64a + 16b + 4c + d = 11 where a, b, c and d are the coefficients of the generating function. By the way, please stop bumping. 
November 10th, 2014, 02:08 PM  #9 
Math Team Joined: Apr 2010 Posts: 2,770 Thanks: 356 
Another way, compute the differences. Code: 2 1 2 11 1 3 9 2 6 4 $\displaystyle a\cdot {n \choose 3} + b\cdot {n \choose 2}+c \cdot {n \choose 1}+d\cdot {n \choose 0}$ Then, look at the left 'column', having numbers 2, 1, 2 and 4. These are the values for a, b, c, d; (a, b, c, d) = (4, 2, 1, 2) to get $\displaystyle 4\cdot {n \choose 3} + 2\cdot {n \choose 2}+1 \cdot {n \choose 1}2\cdot {n \choose 0}$, which can be proved by induction. You don't need to compute all difference to get these. Now, x in {0, 1, 2, 3} give the resulting elements from the sequence. An appropriate translation gives a desired offset (n>n1). Alternatively, the first differences are of the form 3^k, for k in {0, 1, 2}. Summing them, with adding an extra constant, gives (3^n1)/2  2. n in {0, 1, 2, 3} produce the sequence. Last edited by skipjack; November 10th, 2014 at 04:34 PM. 
November 10th, 2014, 04:17 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 17,739 Thanks: 1361 
Or 2n! + n  4, for n = 0, 1, 2 and 3.


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