Infinite balls, infinite groups with finite number of balls of each value












3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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










  • $begingroup$
    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


















3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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










  • $begingroup$
    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
















3












3








3


3



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $Nge X^2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
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$begingroup$

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 $Nge X^2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $Nge X^2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $Nge X^2$.






share|cite|improve this answer









$endgroup$



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 $Nge X^2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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