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SOLUTION TO MONTHLY PROBLEM PROPOSED BY D. E. KNUTH
 

Summary: SOLUTION TO MONTHLY PROBLEM
PROPOSED BY D. E. KNUTH
Tewodros Amdeberhan
Department of Mathematics, Temple University, Philadelphia PA 19122, USA
tewodros@euclid.math.temple.edu
Problem 10568: P Let n be a nonnegative integer. The sequence de ned by x0 = n and
xk+1 = xk ,dpxke for k 0 converges to 0. Let fn be the number of steps required; i.e. xfn = 0
but xfn,1 0. Find a closed form for fn.
Solution: Consider the following partition of the positive integers into intervals":
for each nonnegative integer m, de ne successively
f amk + 1;bmk ; bmk + 1;amk + 1 g;
for k = 1;2;:::;2m ,1, and augmented by am2m + 1;bm2m ; where
amk := 4m + 2k ,32m + kk ,1; and bmk := 4m + 2k ,22m + k2
:
Claim:
fn =

2m+1
,m,2 + 2k ,2; if n 2 amk + 1;bmk
2m+1

  

Source: Amdeberhan, Tewodros - Department of Mathematics, Tulane University

 

Collections: Mathematics