| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
(a) Let f(n) be the smallest positive integer which is not a factor
of n. Continue the series f(n), f(f(n)), f(f(f(n))), ... until
you reach 2. What is the maximum length of the series?
(b) Let g(n) be the second smallest positive integer which is not a
factor of n. Continue the series g(n), g(g(n)), g(g(g(n))), ...
until you reach 2. What is the maximum length of the series?
[This problem is due to David Grabiner of Claremont H.S., Claremont, CA]
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 578.1 | Answer to (a). | CHOVAX::YOUNG | I think we're all bozos on this BUS. | Sun Sep 14 1986 14:27 | 25 |
(a):
The maximum length of such a series is 3.
The smallest postive integer which is not a factor of n, must
be of the form:
m
P
Where (P) is any prime, and (m) is any
natural number. Now if (n) was 2 then we were done before we started,
so that comprises 0 steps. If (n) is any odd number then f(n)=2
so that we have 1 step. If (n) is any even number not divisible
by 3 then f(n)=3, f(f(n))=2, for 2 steps. All remaining (n) must
then be of the form:
a b
2 * 3 * ...
m Now f(n) must be of
the form P <> (2,3). Once again, if P is odd then f(f(n))=2
for 2 steps. However, if P is 2 then f(f(n))=3, and f(f(f(n)))=2
which is 3 steps.
This covers all possibilities.
-- Barry
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| 578.2 | Answer to (b). | CHOVAX::YOUNG | I think we're all bozos on this BUS. | Sun Sep 14 1986 14:33 | 5 |
(b):
This series can NEVER reach 2.
Reason: 2 is never the second smallest non-factor of any number.
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| 578.3 | CLT::GILBERT | eager like a child | Tue Sep 16 1986 18:25 | 1 | |
(c) Same as (b), but continue until you reach 3. | |||||