sum{k>=m} floor(n/k) = n
Leroy Quet
Consider the sequence A145265 of the On-Line Encyclopedia Of Integer
Sequences.
A positive integer n is included if there exists a positive integer m
such that sum{k>=0} floor(n/(m+k)) = n.
(Terms calculated by Stefan Steinerberger :
1, 4, 5, 8, 15, 18, 19, 22, 23, 26, 33, 36, 37, 40, 41, 44, 51, 54,
55, 58, 59, 62, 69, 72, 73, 76, 77, 80, 87, 90, 91, 94, 95, 98, 105,
108, 109, 112, 113, 116, 123, 126, 127, 130, 131, 134, 141, 144, 145,
148, 149, 152, 159, 162, 163, 166, 167, 170, 177, 180, 181, 184,...)
Does this sequence contain all of those, and only those, positive
integers that are congruent to 0, 1, 4, 5, 8, 15 (mod 18)?
It sure seems like it.
Thanks,
Leroy Quet
Mensanator
> I posted this on sci.math, but no one responded.
There's are reasons no one responded.
>
> Consider the sequence A145265 of the On-Line Encyclopedia Of Integer
> Sequences.
The first reason is you need to post a link, so that _I_
don't have to do any work looking it up.
>
> A positive integer n is included if there exists a positive integer m
> such that sum{k>=0} floor(n/(m+k)) = n.
What the fuck is k? That's another reason. Explain the
problem completely here, don't depend on people going
to OLEI, especially if you didn't provide a link,
because they won't.
>
> (Terms calculated by Stefan Steinerberger :
> 1, 4, 5, 8, 15, 18, 19, 22, 23, 26, 33, 36, 37, 40, 41, 44, 51, 54,
> 55, 58, 59, 62, 69, 72, 73, 76, 77, 80, 87, 90, 91, 94, 95, 98, 105,
> 108, 109, 112, 113, 116, 123, 126, 127, 130, 131, 134, 141, 144, 145,
> 148, 149, 152, 159, 162, 163, 166, 167, 170, 177, 180, 181, 184,...)
>
> Does this sequence contain all of those, and only those, positive
> integers that are congruent to 0, 1, 4, 5, 8, 15 (mod 18)?
>
> It sure seems like it.
So? The first 6th Generation Type [1,2] Mersenne Hailstone
has 53338 _DIGITS_ !!! Your little sequence proves nothing.
>
> Thanks,
> Leroy Quet
Leroy Quet
Mensanator wrote:
> On Nov 2, 10:27�am, Leroy Quet <qqq...@mindspring.com> wrote:
> > I posted this on sci.math, but no one responded.
>
> There's are reasons no one responded.
>
> >
> > Consider the sequence A145265 of the On-Line Encyclopedia Of Integer
> > Sequences.
>
> The first reason is you need to post a link, so that _I_
> don't have to do any work looking it up.
>
http://www.research.att.com/~njas/sequences/A145265
> > A positive integer n is included if there exists a positive integer m
> > such that sum{k>=0} floor(n/(m+k)) = n.
>
> What the fuck is k? That's another reason. Explain the
> problem completely here, don't depend on people going
> to OLEI, especially if you didn't provide a link,
> because they won't.
I thought that was obvious. k is the index of summation. k equals 0
then 1 then 2, then...
>
> >
> > (Terms calculated by Stefan Steinerberger :
> > 1, 4, 5, 8, 15, 18, 19, 22, 23, 26, 33, 36, 37, 40, 41, 44, 51, 54,
> > 55, 58, 59, 62, 69, 72, 73, 76, 77, 80, 87, 90, 91, 94, 95, 98, 105,
> > 108, 109, 112, 113, 116, 123, 126, 127, 130, 131, 134, 141, 144, 145,
> > 148, 149, 152, 159, 162, 163, 166, 167, 170, 177, 180, 181, 184,...)
> >
> > Does this sequence contain all of those, and only those, positive
> > integers that are congruent to 0, 1, 4, 5, 8, 15 (mod 18)?
> >
> > It sure seems like it.
>
> So? The first 6th Generation Type [1,2] Mersenne Hailstone
> has 53338 _DIGITS_ !!! Your little sequence proves nothing.
>
Yes, I know that the pattern could break. I would hope that someone
could calculate a HUGE number of terms to see if the pattern is
maintained for that huge number of terms.
Thanks,
Leroy Quet
Leroy Quet
>...
> Yes, I know that the pattern could break. I would hope that someone
> could calculate a HUGE number of terms to see if the pattern is
> maintained for that huge number of terms.
I know that calculating a huge number of terms would prove nothing if
the pattern is maintained. But there would be a chance the pattern
could be broken, which would prove my conjecture wrong.
Also, obviously I know that the pattern could break. That is why I am
posting this thread. Maybe someone has a PROOF that the pattern is
maintained, IF it indeed is maintained. Other than that, I hope for a
counter-example.
Thanks,
Leroy Quet
Mensanator
> Mensanator wrote:
> > On Nov 2, 10:27 am, Leroy Quet <qqq...@mindspring.com> wrote:
> > > I posted this on sci.math, but no one responded.
>
> > There's are reasons no one responded.
>
> > > Consider the sequence A145265 of the On-Line Encyclopedia Of Integer
> > > Sequences.
>
> > The first reason is you need to post a link, so that _I_
> > don't have to do any work looking it up.
>
> http://www.research.att.com/~njas/sequences/A145265
>
> > > A positive integer n is included if there exists a positive integer m
> > > such that sum{k>=0} floor(n/(m+k)) = n.
>
> > What the fuck is k? That's another reason. Explain the
> > problem completely here, don't depend on people going
> > to OLEI, especially if you didn't provide a link,
> > because they won't.
>
> I thought that was obvious. k is the index of summation. k equals 0
> then 1 then 2, then...
And then what? k goes to infinity? Some limit you
haven't defined?
In your example:
EXAMPLE Checking n = 8: floor(8/3) + floor(8/4) + floor(8/5) +
floor(8/6) + floor(8/7) + floor(8/8) = 2 + 2 + 1 + 1 + 1 + 1 = 8. So 8
is included in the sequence. Checking n = 6: floor(6/2) + floor(6/3) +
floor(6/4) + floor(6/5) + floor(6/6) = 3 + 2 + 1 + 1 + 1 = 8, which is
> 6. But floor(6/3) + floor(6/4) + floor(6/5) + floor(6/6) = 2 + 1 + 1
+ 1 = 5, which is < 6. So, 6 is not included in the sequence.
The first case has 6 terms. That's k=0 to 5? And m=3?
Why k=0 to 5?
And for the 2nd you started with m=3 and then tried
m=2, one with 4 terms one with 5? You just add terms
until you hit or bust?
Those OLEI guys accept sloppy work, don't they?
>
>
>
>
>
>
>
> > > (Terms calculated by Stefan Steinerberger :
> > > 1, 4, 5, 8, 15, 18, 19, 22, 23, 26, 33, 36, 37, 40, 41, 44, 51, 54,
> > > 55, 58, 59, 62, 69, 72, 73, 76, 77, 80, 87, 90, 91, 94, 95, 98, 105,
> > > 108, 109, 112, 113, 116, 123, 126, 127, 130, 131, 134, 141, 144, 145,
> > > 148, 149, 152, 159, 162, 163, 166, 167, 170, 177, 180, 181, 184,...)
>
> > > Does this sequence contain all of those, and only those, positive
> > > integers that are congruent to 0, 1, 4, 5, 8, 15 (mod 18)?
>
> > > It sure seems like it.
>
> > So? The first 6th Generation Type [1,2] Mersenne Hailstone
> > has 53338 _DIGITS_ !!! Your little sequence proves nothing.
>
> Yes, I know that the pattern could break. I would hope that someone
> could calculate a HUGE number of terms to see if the pattern is
> maintained for that huge number of terms.
>
> Thanks,
> Leroy Quet- Hide quoted text -
>
> - Show quoted text -
Jake
> On Nov 2, 12:12 pm, Leroy Quet <qqq...@mindspring.com> wrote:
>
>
>
> > Mensanator wrote:
> > > On Nov 2, 10:27 am, Leroy Quet <qqq...@mindspring.com> wrote:
> > > > I posted this on sci.math, but no one responded.
>
> > > There's are reasons no one responded.
>
> > > > Consider the sequence A145265 of the On-Line Encyclopedia Of Integer
> > > > Sequences.
>
> > > The first reason is you need to post a link, so that _I_
> > > don't have to do any work looking it up.
>
> >http://www.research.att.com/~njas/sequences/A145265
>
> > > > A positive integer n is included if there exists a positive integer m
> > > > such that sum{k>=0} floor(n/(m+k)) = n.
>
> > > What the fuck is k? That's another reason. Explain the
> > > problem completely here, don't depend on people going
> > > to OLEI, especially if you didn't provide a link,
> > > because they won't.
>
> > I thought that was obvious. k is the index of summation. k equals 0
> > then 1 then 2, then...
>
> And then what? k goes to infinity? Some limit you
> haven't defined?
You're quite an ass. It's standard that sum{k>=0} means to sum over k
= 0, 1, 2, ... In this problem, n/(m+k) is less than one for k large
enough, so floor(n/(m+k)) is nonzero for only finitely many values of
k.
Jake
It's true. It's helpful to look at the sum in the reverse order, in
which case we want to know whether there is a p such that
(*) n = sum{n>=l>=p} floor(n/l)
The number of 1's in the sequence {floor(n/l) : n>=l>=1} is ceil(n/2);
the number of 2's is ceil(2n/3) - ceil(n/2); the number of 3's is
ceil(3n/4) - ceil(2n/3); and so on. If we add up just the terms equal
to 1 or 2, we get roughly n/2 + 2(2n/3 - n/2) = 5n/6, but if we add up
all the terms equal to 1, 2 or 3, we get roughly n/2 + 2(2n/3 - n/2) +
3(3n/4 - 2n/3) = 13n/12. This means that if n is sufficiently large
and satisfies (*), then the sum will be of 1's and 2's and some number
of 3's. Therefore, to check whether it's possible for n to satisfy
(*), we need only see whether n is congruent to ceil(n/2) + 2(ceil(2n/
3) - ceil(n/2)) modulo 3. For n = 18q + r, this is true exactly when r
= 0, 1, 4, 5, 8 or 15. (The equality can be checked by hand for
"insufficiently large" n.)
Patrick Hamlyn
Well spotted. I rate your post infinitely more useful than Insanator's.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms...@egroups.com)
Leroy Quet
Patrick Hamlyn wrote:
> Well spotted. I rate your post infinitely more useful than Insanator's.
Yes, thank you, Jake!
Thanks,
Leroy Quet
Mensanator
wrote:
I wasn't trying to be helpful. I was pointing out that the
deficiencies
in the original post, not that I had resolutions for them. Maybe if
Jake
had replied to Leroy's original post in sci.math, then this situation
wouldn't have developed.
> --
> Patrick Hamlyn posting from Perth, Western Australia
> Windsurfing capital of the Southern Hemisphere
> Moderator: polyforms group (polyforms-subscr...@egroups.com)- Hide quoted text -
Mensanator
> On Nov 2, 7:02 pm, Mensanator <mensana...@aol.com> wrote:
>
>
>
>
>
> > On Nov 2, 12:12 pm, Leroy Quet <qqq...@mindspring.com> wrote:
>
> > > Mensanator wrote:
> > > > On Nov 2, 10:27 am, Leroy Quet <qqq...@mindspring.com> wrote:
> > > > > I posted this on sci.math, but no one responded.
>
> > > > There's are reasons no one responded.
>
> > > > > Consider the sequence A145265 of the On-Line Encyclopedia Of Integer
> > > > > Sequences.
>
> > > > The first reason is you need to post a link, so that _I_
> > > > don't have to do any work looking it up.
>
> > >http://www.research.att.com/~njas/sequences/A145265
>
> > > > > A positive integer n is included if there exists a positive integer m
> > > > > such that sum{k>=0} floor(n/(m+k)) = n.
>
> > > > What the fuck is k? That's another reason. Explain the
> > > > problem completely here, don't depend on people going
> > > > to OLEI, especially if you didn't provide a link,
> > > > because they won't.
>
> > > I thought that was obvious. k is the index of summation. k equals 0
> > > then 1 then 2, then...
>
> > And then what? k goes to infinity? Some limit you
> > haven't defined?
>
> You're quite an ass.
Better to be a hated smart-ass than a beloved dumb-ass.
> It's standard that sum{k>=0} means to sum over k
> = 0, 1, 2, ... In this problem, n/(m+k) is less than one for k large
> enough, so floor(n/(m+k)) is nonzero for only finitely many values of
> k.
This is rec.puzzles, not a math textbook.
James Dow Allen
post my own more tedious derivation.
On Nov 2, 11:27 pm, Leroy Quet <qqq...@mindspring.com> wrote:
> A positive integer n is included if there exists a positive integer m
> such that sum{k>=0} floor(n/(m+k)) = n.
Let's adopt the shorthand
T_k = floor(n/(m+k))
> Does this sequence contain all of those, and only those, positive
> integers that are congruent to 0, 1, 4, 5, 8, 15 (mod 18)?
The problem breaks into three parts:
(1) Proof of "all of those".
(2) Proof of "only those".
(3) Explaining why 18 is special.
Proofs on finite sets are easier than proofs on infinite
sets. Here, fortunately, as exhibited next, the infinite
series on the integers defined above can be reduced to
finite inequalities on just finitely-many residue classes.
(1) Proof of "all of those" is straightforward.
Write
n = 18a + t
m = 5a + 1 + floor(5t/18)
Work out the details for each of the six satisfactory
values of t.
For example, when t = 15 we find
T_k = 3 for 0 <= k <= a
T_k = 2 for a < k <= 4a + 2
T_k = 1 for 4a + 2 < k <= 13a + 10
T_k = 0 for 13a + 10 < k
so sum T_k = 3*(a + 1) + 2*(3a + 2) + 1*(9a + 8)
= 18a + 15 = 18a + t = n
as was to be demonstrated.
(2) I suppose the proof of "only those" will be
similar though more tedious: instead of 6 cases,
we might need to consider 24 cases (12 values of t,
and both the m that's a little too big and
the m that's a little too small).
(3) Why do these solutions have a cycle of size 18?
(Only an informal explanation is needed, as the
"proof" would follow from (1) and (2).)
Repeat the work summarized in (1) above, but use
(x,y) instead of (18,5); that is write
n = xa + t
m = ya + t'
The inequalities become
T_k = 3 for 0 <= k <= (x/3 - y)a + s
T_k = 2 for ... < k <= (x/2 - y)a + s'
T_k = 1 for ... < k <= (x - y)a + s''
where s,s',s'' depend on t,t' but not a.
The equation
sum T_k = n = xa + t
turns out to be
((11/6)x - 3y) a + s''' = xa + t
Il est aisé à voir :-)
s''' = t
whence
5x = 18y
which is (more or less :-) what was to be demonstrated.
James Dow Allen