| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 2006.1 | n = 1 or 3 | JOBURG::BUCHANAN | dodging lions and wasting time | Wed Oct 25 1995 09:34 | 4 |
| because it's South African, I'll reply.
n>=4 => S[n] == S[4] == 3 mod 5
but 3 is a quadratic non-residue mod 5.
So we need only check n=<3.
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| 2006.2 | Not many | CHEFS::STRANGEWAYS | Andy Strangeways@REO DTN 830-3216 | Wed Oct 25 1995 09:39 | 7 |
| Only for n = 1 and n = 3.
S[4] = 1 + 2 + 6 + 24 = 33.
For n > 4, n! == 0 (10), so S[n] == 3 (10) and cannot be a square.
Andy.
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| 2006.3 | Re .1 | CHEFS::STRANGEWAYS | Andy Strangeways@REO DTN 830-3216 | Wed Oct 25 1995 09:41 | 1 |
| Notes collision!
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| 2006.4 | more puzzles | JOBURG::BUCHANAN | dodging lions and wasting time | Wed Oct 25 1995 12:30 | 18 |
| Some more puzzles. For which values of n:
(a) is S[n] "square-free"? (divisible by no square number > 1)
(b) is S[n] "square-less"? (Let arity(p,x) = a be the largest exponent
such that p^a divides x, but p^(a+1) does not divide x. Then say that
x is *square-less* if there is no prime p dividing x such that
arity(p,x) is even.)
Those two were easy, now a little harder...
(c) Let p be prime. Let a[p,n] = arity(p,S[n]). Conjecture: as n ->
infinity, the sequence a[p,n] is bounded above, by some value A[p].
(d) Conjecture: as p -> infinity, A[p] is bounded above, by some
value A*.
Andrew.
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| 2006.5 | | AUSSIE::GARSON | achtentachtig kacheltjes | Thu Oct 26 1995 07:38 | 12 |
| re .4
(a)
It is easy to see that
t|S[n] and t|(n+1)! => t|S[m] for all m >= n
Since 9|S[5] and 9|6!, 9|S[m] for all m >= 5
Hence the only possible square-free S[n] are those with n in {1,2,3,4}
and by inspection the solution is n=1 or n=2 or n=4.
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| 2006.6 | | AUSSIE::GARSON | achtentachtig kacheltjes | Thu Oct 26 1995 08:03 | 19 |
| re .4
(b)
It is nearly as easy to see that
arity(p,S[n]) = t and p^(t+1) | (n+1)! => arity(p,S[m]) = t for all m >= n
Since arity(3,S[8]) = 2 and 3^3 | 9!, arity(3,S[m]) = 2 for all m >= 8
Hence the only possible square-less S[n] are those with n in {1..7}
and by inspection the solution is n=1 or n=2 or n=4.
(c) looks harder. It would I think be necessary to show that eventually
the antecedent above is satisfied. Note though that t would not
necessarily be A[p]. For example A[3] >= 4 while t = 2 in the notation
above. By the way did you intend A[p] to be the least upper bound?
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| 2006.7 | I am unanimous with you | JOBURG::BUCHANAN | dodging lions and wasting time | Thu Oct 26 1995 10:54 | 6 |
| > (c) looks harder. It would I think be necessary to show that eventually
> the antecedent above is satisfied. Note though that t would not
> necessarily be A[p]. For example A[3] >= 4 while t = 2 in the notation
> above. By the way did you intend A[p] to be the least upper bound?
Yes to all 5 points.
|