Infinite balls, infinite groups with finite number of balls of each value
Multi tool use
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You have N balls of value 0, N of value 1, N of value 2, etc., N of each value up to infinity.
They are separated into a group of X balls with total value 1, another group of X with total value 2, another of X with value 3, etc., one different group of X for each total value up to infinity.
What is the smallest N for X=4 to be possible? What is the general equation for any value X?
sequences-and-series combinatorics limits
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add a comment |
$begingroup$
You have N balls of value 0, N of value 1, N of value 2, etc., N of each value up to infinity.
They are separated into a group of X balls with total value 1, another group of X with total value 2, another of X with value 3, etc., one different group of X for each total value up to infinity.
What is the smallest N for X=4 to be possible? What is the general equation for any value X?
sequences-and-series combinatorics limits
$endgroup$
$begingroup$
Are the balls replaced back into the box?
$endgroup$
– Jam
Dec 10 '18 at 22:53
$begingroup$
I don't quite understand what your question is. If at each "step", you're just removing $1$ or $4$ balls then clearly doing this up to $k$ will just require $k$ or $4k$ steps. What are you defining a "step" to be, for it to depend on $N$?
$endgroup$
– Jam
Dec 10 '18 at 22:58
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Step k is the removal of 4 balls totalling k in value. The total removed is the sum of all the steps as no balls are returned.
$endgroup$
– Ben Franks
Dec 11 '18 at 8:02
add a comment |
$begingroup$
You have N balls of value 0, N of value 1, N of value 2, etc., N of each value up to infinity.
They are separated into a group of X balls with total value 1, another group of X with total value 2, another of X with value 3, etc., one different group of X for each total value up to infinity.
What is the smallest N for X=4 to be possible? What is the general equation for any value X?
sequences-and-series combinatorics limits
$endgroup$
You have N balls of value 0, N of value 1, N of value 2, etc., N of each value up to infinity.
They are separated into a group of X balls with total value 1, another group of X with total value 2, another of X with value 3, etc., one different group of X for each total value up to infinity.
What is the smallest N for X=4 to be possible? What is the general equation for any value X?
sequences-and-series combinatorics limits
sequences-and-series combinatorics limits
edited Dec 13 '18 at 8:29
Ben Franks
asked Dec 10 '18 at 22:45
Ben FranksBen Franks
263110
263110
$begingroup$
Are the balls replaced back into the box?
$endgroup$
– Jam
Dec 10 '18 at 22:53
$begingroup$
I don't quite understand what your question is. If at each "step", you're just removing $1$ or $4$ balls then clearly doing this up to $k$ will just require $k$ or $4k$ steps. What are you defining a "step" to be, for it to depend on $N$?
$endgroup$
– Jam
Dec 10 '18 at 22:58
$begingroup$
Step k is the removal of 4 balls totalling k in value. The total removed is the sum of all the steps as no balls are returned.
$endgroup$
– Ben Franks
Dec 11 '18 at 8:02
add a comment |
$begingroup$
Are the balls replaced back into the box?
$endgroup$
– Jam
Dec 10 '18 at 22:53
$begingroup$
I don't quite understand what your question is. If at each "step", you're just removing $1$ or $4$ balls then clearly doing this up to $k$ will just require $k$ or $4k$ steps. What are you defining a "step" to be, for it to depend on $N$?
$endgroup$
– Jam
Dec 10 '18 at 22:58
$begingroup$
Step k is the removal of 4 balls totalling k in value. The total removed is the sum of all the steps as no balls are returned.
$endgroup$
– Ben Franks
Dec 11 '18 at 8:02
$begingroup$
Are the balls replaced back into the box?
$endgroup$
– Jam
Dec 10 '18 at 22:53
$begingroup$
Are the balls replaced back into the box?
$endgroup$
– Jam
Dec 10 '18 at 22:53
$begingroup$
I don't quite understand what your question is. If at each "step", you're just removing $1$ or $4$ balls then clearly doing this up to $k$ will just require $k$ or $4k$ steps. What are you defining a "step" to be, for it to depend on $N$?
$endgroup$
– Jam
Dec 10 '18 at 22:58
$begingroup$
I don't quite understand what your question is. If at each "step", you're just removing $1$ or $4$ balls then clearly doing this up to $k$ will just require $k$ or $4k$ steps. What are you defining a "step" to be, for it to depend on $N$?
$endgroup$
– Jam
Dec 10 '18 at 22:58
$begingroup$
Step k is the removal of 4 balls totalling k in value. The total removed is the sum of all the steps as no balls are returned.
$endgroup$
– Ben Franks
Dec 11 '18 at 8:02
$begingroup$
Step k is the removal of 4 balls totalling k in value. The total removed is the sum of all the steps as no balls are returned.
$endgroup$
– Ben Franks
Dec 11 '18 at 8:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can show that $Nge X^2$. Indeed, for any natural $k$ consider a union $U$ of the first $kN$ groups. The sum $S$ of ball’s values of $U$ is $sum_{i=1}^{kN} i=frac 12 kN(kN+1).$
On the other hand, for each $j$, $U$ contains at most $N$ balls with value $j$. Since $|U|=kXcdot N$, we have $Sge Nsum_{j=0}^{kX-1} j=frac 12 N(kX-1)kX.$ Thus $$frac 12 kN(kN+1)ge frac 12 N(kX-1)kX$$ or $kN+1ge kX^2-X$. Since this inequality hold for all $k$, we have $Nge X^2$.
$endgroup$
$begingroup$
do you have an upper bound to go with this lower bound?
$endgroup$
– Ben Franks
Dec 18 '18 at 15:46
$begingroup$
@BenFranks Unortunately, no. I think it should be obtained by a different way, maybe by come combinatorial constructions.
$endgroup$
– Alex Ravsky
Dec 18 '18 at 15:48
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
We can show that $Nge X^2$. Indeed, for any natural $k$ consider a union $U$ of the first $kN$ groups. The sum $S$ of ball’s values of $U$ is $sum_{i=1}^{kN} i=frac 12 kN(kN+1).$
On the other hand, for each $j$, $U$ contains at most $N$ balls with value $j$. Since $|U|=kXcdot N$, we have $Sge Nsum_{j=0}^{kX-1} j=frac 12 N(kX-1)kX.$ Thus $$frac 12 kN(kN+1)ge frac 12 N(kX-1)kX$$ or $kN+1ge kX^2-X$. Since this inequality hold for all $k$, we have $Nge X^2$.
$endgroup$
$begingroup$
do you have an upper bound to go with this lower bound?
$endgroup$
– Ben Franks
Dec 18 '18 at 15:46
$begingroup$
@BenFranks Unortunately, no. I think it should be obtained by a different way, maybe by come combinatorial constructions.
$endgroup$
– Alex Ravsky
Dec 18 '18 at 15:48
add a comment |
$begingroup$
We can show that $Nge X^2$. Indeed, for any natural $k$ consider a union $U$ of the first $kN$ groups. The sum $S$ of ball’s values of $U$ is $sum_{i=1}^{kN} i=frac 12 kN(kN+1).$
On the other hand, for each $j$, $U$ contains at most $N$ balls with value $j$. Since $|U|=kXcdot N$, we have $Sge Nsum_{j=0}^{kX-1} j=frac 12 N(kX-1)kX.$ Thus $$frac 12 kN(kN+1)ge frac 12 N(kX-1)kX$$ or $kN+1ge kX^2-X$. Since this inequality hold for all $k$, we have $Nge X^2$.
$endgroup$
$begingroup$
do you have an upper bound to go with this lower bound?
$endgroup$
– Ben Franks
Dec 18 '18 at 15:46
$begingroup$
@BenFranks Unortunately, no. I think it should be obtained by a different way, maybe by come combinatorial constructions.
$endgroup$
– Alex Ravsky
Dec 18 '18 at 15:48
add a comment |
$begingroup$
We can show that $Nge X^2$. Indeed, for any natural $k$ consider a union $U$ of the first $kN$ groups. The sum $S$ of ball’s values of $U$ is $sum_{i=1}^{kN} i=frac 12 kN(kN+1).$
On the other hand, for each $j$, $U$ contains at most $N$ balls with value $j$. Since $|U|=kXcdot N$, we have $Sge Nsum_{j=0}^{kX-1} j=frac 12 N(kX-1)kX.$ Thus $$frac 12 kN(kN+1)ge frac 12 N(kX-1)kX$$ or $kN+1ge kX^2-X$. Since this inequality hold for all $k$, we have $Nge X^2$.
$endgroup$
We can show that $Nge X^2$. Indeed, for any natural $k$ consider a union $U$ of the first $kN$ groups. The sum $S$ of ball’s values of $U$ is $sum_{i=1}^{kN} i=frac 12 kN(kN+1).$
On the other hand, for each $j$, $U$ contains at most $N$ balls with value $j$. Since $|U|=kXcdot N$, we have $Sge Nsum_{j=0}^{kX-1} j=frac 12 N(kX-1)kX.$ Thus $$frac 12 kN(kN+1)ge frac 12 N(kX-1)kX$$ or $kN+1ge kX^2-X$. Since this inequality hold for all $k$, we have $Nge X^2$.
answered Dec 15 '18 at 11:53
Alex RavskyAlex Ravsky
40.5k32282
40.5k32282
$begingroup$
do you have an upper bound to go with this lower bound?
$endgroup$
– Ben Franks
Dec 18 '18 at 15:46
$begingroup$
@BenFranks Unortunately, no. I think it should be obtained by a different way, maybe by come combinatorial constructions.
$endgroup$
– Alex Ravsky
Dec 18 '18 at 15:48
add a comment |
$begingroup$
do you have an upper bound to go with this lower bound?
$endgroup$
– Ben Franks
Dec 18 '18 at 15:46
$begingroup$
@BenFranks Unortunately, no. I think it should be obtained by a different way, maybe by come combinatorial constructions.
$endgroup$
– Alex Ravsky
Dec 18 '18 at 15:48
$begingroup$
do you have an upper bound to go with this lower bound?
$endgroup$
– Ben Franks
Dec 18 '18 at 15:46
$begingroup$
do you have an upper bound to go with this lower bound?
$endgroup$
– Ben Franks
Dec 18 '18 at 15:46
$begingroup$
@BenFranks Unortunately, no. I think it should be obtained by a different way, maybe by come combinatorial constructions.
$endgroup$
– Alex Ravsky
Dec 18 '18 at 15:48
$begingroup$
@BenFranks Unortunately, no. I think it should be obtained by a different way, maybe by come combinatorial constructions.
$endgroup$
– Alex Ravsky
Dec 18 '18 at 15:48
add a comment |
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$begingroup$
Are the balls replaced back into the box?
$endgroup$
– Jam
Dec 10 '18 at 22:53
$begingroup$
I don't quite understand what your question is. If at each "step", you're just removing $1$ or $4$ balls then clearly doing this up to $k$ will just require $k$ or $4k$ steps. What are you defining a "step" to be, for it to depend on $N$?
$endgroup$
– Jam
Dec 10 '18 at 22:58
$begingroup$
Step k is the removal of 4 balls totalling k in value. The total removed is the sum of all the steps as no balls are returned.
$endgroup$
– Ben Franks
Dec 11 '18 at 8:02