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Subject: "Recursive Series"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #483
Reading Topic #483
Ramsey_KJ
Member since Sep-23-04
Nov-13-04, 10:38 AM (EST)
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"Recursive Series"
 
   This problem appears in the yahoo groups IntermediateNumberTheory
Group.
The following concerns the set of recursive series: R_0 = 0, R_1 = 2^
(n+1)-1 and R_i = 6*R_(i-1) - R_(i-2) + K where K = 4*(2^n -1) (for n
= 0, K = 0 and (R_i)^2 is a triangular number). The product any two
adjacent terms of these series is also a triangular number (can be
represented as the product x*(x+1)/2 where x is an integer) if n is
an positive interger (0,1,2,3 etc.). Otherwise the product can still
be put in the form x*(x+1)/2 where x_i are integers times a negative
power of two. A special case is K = -4 (limit as n -> minus infinity)
in that case the R series (where the product of any two adjacent
terms is a triangular number) is 5, 2, 3, 12, 65 etc, and x_i are
also integers.

Challenge
1. Find a non-recursive formula for the ith term of the R series for
a given K (other than 0).
2. Find the recursive series x_i for the triangular numbers x_i *
(x_i+1)/2 represented by R_i*R_(i-1). Hint it is of the form 6* x_(i-
1) - x_(i-2) + K2 where K2 is a constant dependent on K.
3. Find a non recursive formula for the ith term x_i corresponding to
a given K.


Have a Good Day
KJ Ramsey
Have a Good Day
KJ Ramsey


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