A sequence that avoids both arithmetic and geometric progressions
Sequences that avoid arithmetic progressions have been studied, e.g., "Sequences Containing No 3-Term Arithmetic Progressions," Janusz Dybizbański, 2012, journal link.
I started to explore sequences that avoid both arithmetic and geometric progressions,
i.e., avoid $x, x+c, x+2c$ and avoid $y, c y, c^2 y$ anywhere in the sequence
(not necessarily consecutively).
Starting with $(1,2)$, one cannot extend with $3$ because $(1,2,3)$ forms an
arithemtical progression, and one cannot extend with $4$ because $(1,2,4)$ is a geometric
progression. But $(1,2,5)$ is fine.
Continuing in the same manner leads to the following "greedy" sequence:
$$1, 2, 5, 6, 12, 13, 15, 16, 32, 33, 35, 39, 40, 42, 56, 81, 84, 85, 88,$$
$$90, 93, 94, 108, 109, 113, 115, 116, 159, 189, 207, 208, 222, ldots$$
This sequence is not in the OEIS.
Here are a few questions:
Q1. What is its growth rate?
Q2. Does $sum_{i=1}^infty 1/s_i$ converge? (Where $s_i$ is the $i$-th term of the above
sequence.)
Q3. If it does, does it converge to e?
Update: No. The sum appears to be approximately $2.73 > e$, as per
@MichaelStocker and @Turambar.
That is wild numerical speculation. The first 457 terms (the extent
of the graph above) sum to 2.70261.
Addendum. 11Jul2014. Starting with $(0,1)$ rather than $(1,2)$ renders
a direct hit on OEIS A225571.
sequences-and-series number-theory
asked Jul 7 '14 at 12:02
Joseph O'RourkeJoseph O'Rourke
17.9k348107
|
show 7 more comments
Sequences that avoid arithmetic progressions have been studied, e.g., "Sequences Containing No 3-Term Arithmetic Progressions," Janusz Dybizbański, 2012, journal link.
I started to explore sequences that avoid both arithmetic and geometric progressions,
i.e., avoid $x, x+c, x+2c$ and avoid $y, c y, c^2 y$ anywhere in the sequence
(not necessarily consecutively).
Starting with $(1,2)$, one cannot extend with $3$ because $(1,2,3)$ forms an
arithemtical progression, and one cannot extend with $4$ because $(1,2,4)$ is a geometric
progression. But $(1,2,5)$ is fine.
Continuing in the same manner leads to the following "greedy" sequence:
$$1, 2, 5, 6, 12, 13, 15, 16, 32, 33, 35, 39, 40, 42, 56, 81, 84, 85, 88,$$
$$90, 93, 94, 108, 109, 113, 115, 116, 159, 189, 207, 208, 222, ldots$$
This sequence is not in the OEIS.
Here are a few questions:
Q1. What is its growth rate?
Q2. Does $sum_{i=1}^infty 1/s_i$ converge? (Where $s_i$ is the $i$-th term of the above
sequence.)
Q3. If it does, does it converge to e?
Update: No. The sum appears to be approximately $2.73 > e$, as per
@MichaelStocker and @Turambar.
That is wild numerical speculation. The first 457 terms (the extent
of the graph above) sum to 2.70261.
Addendum. 11Jul2014. Starting with $(0,1)$ rather than $(1,2)$ renders
a direct hit on OEIS A225571.
sequences-and-series number-theory
asked Jul 7 '14 at 12:02
Joseph O'RourkeJoseph O'Rourke
17.9k348107
6
Expanding on your numerical data I conclude that Q3 can be answered in the negative. 457th term is 17933 with sum of reciprocals : 2.702607644337383 1000th term is 66102 : 2.718058753659135 but alas, somewhere shortly after that the sum excedes e. 2000 253701 2.7259164358931023. 3000 442429 2.7287027640882804.
– Michael Stocker
Jul 7 '14 at 13:16
3
The answer to Q2 is definitely Yes, it converges, because even just avoiding arithmetic progressions leads to convergence. Converges to what constant is unclear.
– Joseph O'Rourke
Jul 9 '14 at 0:33
2
@JosephO'Rourke Really? I may be ignorant here so going out on a limb, but I don't think $a_n=nln n$ is in an arithmetic progression, but the sum of its reciprocals diverges.
– John Molokach
Mar 25 '16 at 2:46
1
@JohnMolokach In fact, when you look at the difference in terms in the OEIS sequence, it looks a lot like OEIS A006519, which is $O(nln(n))$. The sequence talked about in the question is a bit different, but I think asymptotically they'd be the same, meaning the sum of reciprocals diverges. And of course, this is just a heuristic, not a proof.
– Turambar
Apr 28 '16 at 17:52
1
On the other hand, after looking at it out to $i = 10,000$, the chart of the differences doesn't seem to be as similar to A006519 as I thought. Its peaks grow faster and have a more bell curve distribution around them instead of the recursive characteristics of A006519. At $i = 10,000$, $ln(s_i)/ln(i) = 1.679258559$ and generally growing, with the sum of reciprocals around $2.734896156$. So if the "power" keeps generally growing (and doesn't start dropping to 1 after some point), then the sum is below $2.736095249$.
– Turambar
May 10 '16 at 16:14
|
show 7 more comments
Sequences that avoid arithmetic progressions have been studied, e.g., "Sequences Containing No 3-Term Arithmetic Progressions," Janusz Dybizbański, 2012, journal link.
I started to explore sequences that avoid both arithmetic and geometric progressions,
i.e., avoid $x, x+c, x+2c$ and avoid $y, c y, c^2 y$ anywhere in the sequence
(not necessarily consecutively).
Starting with $(1,2)$, one cannot extend with $3$ because $(1,2,3)$ forms an
arithemtical progression, and one cannot extend with $4$ because $(1,2,4)$ is a geometric
progression. But $(1,2,5)$ is fine.
Continuing in the same manner leads to the following "greedy" sequence:
$$1, 2, 5, 6, 12, 13, 15, 16, 32, 33, 35, 39, 40, 42, 56, 81, 84, 85, 88,$$
$$90, 93, 94, 108, 109, 113, 115, 116, 159, 189, 207, 208, 222, ldots$$
This sequence is not in the OEIS.
Here are a few questions:
Q1. What is its growth rate?
Q2. Does $sum_{i=1}^infty 1/s_i$ converge? (Where $s_i$ is the $i$-th term of the above
sequence.)
Q3. If it does, does it converge to e?
Update: No. The sum appears to be approximately $2.73 > e$, as per
@MichaelStocker and @Turambar.
That is wild numerical speculation. The first 457 terms (the extent
of the graph above) sum to 2.70261.
Addendum. 11Jul2014. Starting with $(0,1)$ rather than $(1,2)$ renders
a direct hit on OEIS A225571.
sequences-and-series number-theory
asked Jul 7 '14 at 12:02
Joseph O'RourkeJoseph O'Rourke
17.9k348107
Sequences that avoid arithmetic progressions have been studied, e.g., "Sequences Containing No 3-Term Arithmetic Progressions," Janusz Dybizbański, 2012, journal link.
I started to explore sequences that avoid both arithmetic and geometric progressions,
i.e., avoid $x, x+c, x+2c$ and avoid $y, c y, c^2 y$ anywhere in the sequence
(not necessarily consecutively).
Starting with $(1,2)$, one cannot extend with $3$ because $(1,2,3)$ forms an
arithemtical progression, and one cannot extend with $4$ because $(1,2,4)$ is a geometric
progression. But $(1,2,5)$ is fine.
Continuing in the same manner leads to the following "greedy" sequence:
$$1, 2, 5, 6, 12, 13, 15, 16, 32, 33, 35, 39, 40, 42, 56, 81, 84, 85, 88,$$
$$90, 93, 94, 108, 109, 113, 115, 116, 159, 189, 207, 208, 222, ldots$$
This sequence is not in the OEIS.
Here are a few questions:
Q1. What is its growth rate?
Q2. Does $sum_{i=1}^infty 1/s_i$ converge? (Where $s_i$ is the $i$-th term of the above
sequence.)
Q3. If it does, does it converge to e?
Update: No. The sum appears to be approximately $2.73 > e$, as per
@MichaelStocker and @Turambar.
That is wild numerical speculation. The first 457 terms (the extent
of the graph above) sum to 2.70261.
Addendum. 11Jul2014. Starting with $(0,1)$ rather than $(1,2)$ renders
a direct hit on OEIS A225571.
sequences-and-series number-theory
sequences-and-series number-theory
asked Jul 7 '14 at 12:02
Joseph O'RourkeJoseph O'Rourke
17.9k348107
asked Jul 7 '14 at 12:02
Joseph O'RourkeJoseph O'Rourke
17.9k348107
edited yesterday
Joseph O'Rourke
asked Jul 7 '14 at 12:02
Joseph O'RourkeJoseph O'Rourke
17.9k348107
asked Jul 7 '14 at 12:02
Joseph O'RourkeJoseph O'Rourke
17.9k348107
asked Jul 7 '14 at 12:02
Joseph O'RourkeJoseph O'Rourke
17.9k348107
17.9k348107
6
Expanding on your numerical data I conclude that Q3 can be answered in the negative. 457th term is 17933 with sum of reciprocals : 2.702607644337383 1000th term is 66102 : 2.718058753659135 but alas, somewhere shortly after that the sum excedes e. 2000 253701 2.7259164358931023. 3000 442429 2.7287027640882804.
– Michael Stocker
Jul 7 '14 at 13:16
3
The answer to Q2 is definitely Yes, it converges, because even just avoiding arithmetic progressions leads to convergence. Converges to what constant is unclear.
– Joseph O'Rourke
Jul 9 '14 at 0:33
2
@JosephO'Rourke Really? I may be ignorant here so going out on a limb, but I don't think $a_n=nln n$ is in an arithmetic progression, but the sum of its reciprocals diverges.
– John Molokach
Mar 25 '16 at 2:46
1
@JohnMolokach In fact, when you look at the difference in terms in the OEIS sequence, it looks a lot like OEIS A006519, which is $O(nln(n))$. The sequence talked about in the question is a bit different, but I think asymptotically they'd be the same, meaning the sum of reciprocals diverges. And of course, this is just a heuristic, not a proof.
– Turambar
Apr 28 '16 at 17:52
1
On the other hand, after looking at it out to $i = 10,000$, the chart of the differences doesn't seem to be as similar to A006519 as I thought. Its peaks grow faster and have a more bell curve distribution around them instead of the recursive characteristics of A006519. At $i = 10,000$, $ln(s_i)/ln(i) = 1.679258559$ and generally growing, with the sum of reciprocals around $2.734896156$. So if the "power" keeps generally growing (and doesn't start dropping to 1 after some point), then the sum is below $2.736095249$.
– Turambar
May 10 '16 at 16:14
|
show 7 more comments
6
Expanding on your numerical data I conclude that Q3 can be answered in the negative. 457th term is 17933 with sum of reciprocals : 2.702607644337383 1000th term is 66102 : 2.718058753659135 but alas, somewhere shortly after that the sum excedes e. 2000 253701 2.7259164358931023. 3000 442429 2.7287027640882804.
– Michael Stocker
Jul 7 '14 at 13:16
3
The answer to Q2 is definitely Yes, it converges, because even just avoiding arithmetic progressions leads to convergence. Converges to what constant is unclear.
– Joseph O'Rourke
Jul 9 '14 at 0:33
2
@JosephO'Rourke Really? I may be ignorant here so going out on a limb, but I don't think $a_n=nln n$ is in an arithmetic progression, but the sum of its reciprocals diverges.
– John Molokach
Mar 25 '16 at 2:46
1
@JohnMolokach In fact, when you look at the difference in terms in the OEIS sequence, it looks a lot like OEIS A006519, which is $O(nln(n))$. The sequence talked about in the question is a bit different, but I think asymptotically they'd be the same, meaning the sum of reciprocals diverges. And of course, this is just a heuristic, not a proof.
– Turambar
Apr 28 '16 at 17:52
1
On the other hand, after looking at it out to $i = 10,000$, the chart of the differences doesn't seem to be as similar to A006519 as I thought. Its peaks grow faster and have a more bell curve distribution around them instead of the recursive characteristics of A006519. At $i = 10,000$, $ln(s_i)/ln(i) = 1.679258559$ and generally growing, with the sum of reciprocals around $2.734896156$. So if the "power" keeps generally growing (and doesn't start dropping to 1 after some point), then the sum is below $2.736095249$.
– Turambar
May 10 '16 at 16:14
6
6
Expanding on your numerical data I conclude that Q3 can be answered in the negative. 457th term is 17933 with sum of reciprocals : 2.702607644337383 1000th term is 66102 : 2.718058753659135 but alas, somewhere shortly after that the sum excedes e. 2000 253701 2.7259164358931023. 3000 442429 2.7287027640882804.
– Michael Stocker
Jul 7 '14 at 13:16
Expanding on your numerical data I conclude that Q3 can be answered in the negative. 457th term is 17933 with sum of reciprocals : 2.702607644337383 1000th term is 66102 : 2.718058753659135 but alas, somewhere shortly after that the sum excedes e. 2000 253701 2.7259164358931023. 3000 442429 2.7287027640882804.
– Michael Stocker
Jul 7 '14 at 13:16
3
3
The answer to Q2 is definitely Yes, it converges, because even just avoiding arithmetic progressions leads to convergence. Converges to what constant is unclear.
– Joseph O'Rourke
Jul 9 '14 at 0:33
The answer to Q2 is definitely Yes, it converges, because even just avoiding arithmetic progressions leads to convergence. Converges to what constant is unclear.
– Joseph O'Rourke
Jul 9 '14 at 0:33
2
2
@JosephO'Rourke Really? I may be ignorant here so going out on a limb, but I don't think $a_n=nln n$ is in an arithmetic progression, but the sum of its reciprocals diverges.
– John Molokach
Mar 25 '16 at 2:46
@JosephO'Rourke Really? I may be ignorant here so going out on a limb, but I don't think $a_n=nln n$ is in an arithmetic progression, but the sum of its reciprocals diverges.
– John Molokach
Mar 25 '16 at 2:46
1
1
@JohnMolokach In fact, when you look at the difference in terms in the OEIS sequence, it looks a lot like OEIS A006519, which is $O(nln(n))$. The sequence talked about in the question is a bit different, but I think asymptotically they'd be the same, meaning the sum of reciprocals diverges. And of course, this is just a heuristic, not a proof.
– Turambar
Apr 28 '16 at 17:52
@JohnMolokach In fact, when you look at the difference in terms in the OEIS sequence, it looks a lot like OEIS A006519, which is $O(nln(n))$. The sequence talked about in the question is a bit different, but I think asymptotically they'd be the same, meaning the sum of reciprocals diverges. And of course, this is just a heuristic, not a proof.
– Turambar
Apr 28 '16 at 17:52
1
1
On the other hand, after looking at it out to $i = 10,000$, the chart of the differences doesn't seem to be as similar to A006519 as I thought. Its peaks grow faster and have a more bell curve distribution around them instead of the recursive characteristics of A006519. At $i = 10,000$, $ln(s_i)/ln(i) = 1.679258559$ and generally growing, with the sum of reciprocals around $2.734896156$. So if the "power" keeps generally growing (and doesn't start dropping to 1 after some point), then the sum is below $2.736095249$.
– Turambar
May 10 '16 at 16:14
On the other hand, after looking at it out to $i = 10,000$, the chart of the differences doesn't seem to be as similar to A006519 as I thought. Its peaks grow faster and have a more bell curve distribution around them instead of the recursive characteristics of A006519. At $i = 10,000$, $ln(s_i)/ln(i) = 1.679258559$ and generally growing, with the sum of reciprocals around $2.734896156$. So if the "power" keeps generally growing (and doesn't start dropping to 1 after some point), then the sum is below $2.736095249$.
– Turambar
May 10 '16 at 16:14
|
show 7 more comments
2 Answers
2
active
oldest
votes
$color{brown}{textbf{HINT}}$
Denote the target sequence ${F_3(n)}$ and let us try to estimate the probability $P(N)}$ that natural number belongs to ${F_3}.$
Suppose
$$F_3(1)=1,quad F_3(2)=2,quad P(1)=P(2)=P(5) = P(6) = 1,\ P(3)=P(4)=P(7)=P(8)=0,tag1$$
$$V(N)=sumlimits_{i=1}^{N}P(i),quad F_3(N) = left[V^{-1}(N)right].tag2$$
Let $P_a(N)$ is the probability that N does not belong to arithmetic progression and $P_g(N)$ is the similar probability for geometric progressions.
Suppose
$$P(N) = P_a(N)P_g(N)).tag3$$
$color{brown}{textbf{Arithmetic Probability estimation.}}$
Suppose
$$P_a(N)=prodlimits_{k=1}^{[N/2]}P_a(N,k),tag4$$
where $P_a(N,k)$ is the probability that arithmetic progression ${N-2k,N-k, N}$ does not exist for any $j.$
Suppose
$$P_a(N,k) = big(1-P(N-2k)big)big((1-P(N-k)big).tag5$$
$color{brown}{textbf{Geometric Probability estimation.}}$
Suppose
$$P_g(N)=prodlimits_{k=1}^{left[,sqrt Nn,right]}P_g(N,k),tag6$$
where $P_g(N,k)$ is the probability that geometric progression $left(dfrac{N}{k^2}, dfrac Nk, Nright}$ with the denominator $k$ does not exist for all $i,j.$
Taking in account that the geometric progression can exist only if $k^2,| N,$ suppose
$$P_g(N,k) = left(1-dfrac1{k^2}Pleft(dfrac{N}{k^2}right)right)left(1-Pleft(dfrac{N}{k}right)right).tag7$$
$color{brown}{textbf{Common model.}}$
Common model can be simplified to next one,
begin{cases}
P(N) = 1-prodlimits_{k=1}^{[N/2-1]}Big(1-big(1-P(N-2k)big)big(1-P(N-k)big)Big)\
timesprodlimits_{k=1}^{left[sqrt Nright]}left(1-left(1-dfrac1{k^2}Pleft(dfrac{N}{k^2}right)right)left(1-Pleft(dfrac{N}{k}right)right)right)\[4pt]
P(1)=P(2)=P(5) = P(6) = 1,quad P(3)=P(4)=P(7)=P(8)=0\
V(N)=sumlimits_{i=1}^{N}P(i),quad F_3(n) = left[V^{-1}(n)right].tag8
end{cases}
Looking the solution in the form of
$$left{begin{align}
&P(N)=P_v(N),quadtext{where}quad v=left[dfrac{N-1 mod 4}2right],\[4pt]
&P_0(N)=
begin{cases}
1,quadtext{if}quad N<9\
cN^{-s},quadtext{otherwise}
end{cases}\[4pt]
&P_1(N)=
begin{cases}
0,quadtext{if}quad N<9\
dN^{-t},quadtext{otherwise}
end{cases}
end{align}right.tag9$$
then
$$begin{cases}
&P_0(N) =1-&prodlimits_{k=1}^{[N/4-1]}Big(1-big(1-P_0(N-4k)big)big(1-P_1(N-2k)big)Big)\
&×Big(1-big(1-P_1(N-4k-2)big)big(1-P_0(N-2k-1)big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}P_0left(dfrac{N}{(2k-1)^2}right)right)
left(1-P_1left(dfrac{N}{2k-1}right)right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}P_1left(dfrac{N}{4k^2}right)right)
left(1-P_0left(dfrac{N}{2k}right)right)right)\[4pt]
&P_1(N) = 1- &prodlimits_{k=1}^{[N/4-1]}Big(1-big(1-P_1(N-4k)big)big(1-P_0(N-2k)big)Big)\
&&Big(1-big(1-P_0(N-4k-2)big)big(1-P_1(N-2k-1)big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}P_1left(dfrac{N}{(2k-1)^2}right)right)
left(1-P_0left(dfrac{N}{2k-1}right)right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}P_0left(dfrac{N}{4k^2}right)right)
left(1-P_1left(dfrac{N}{2k}right)right)right).
end{cases}tag{10}$$
Taking in account $(9),$ can be written
$$begin{cases}
&P_0(N) =1-&prodlimits_{k=[N/4-2]}^{[N/4-1]},c(N-2k-1)^{-s}prodlimits_{k=1}^{[N/4-3]}Big(1-big(1-c(N-4k)^{-s}big)big(1-d(N-2k)^{-t}big)Big)\
&×Big(1-big(1-d(N-4k-2)^{-t}big)big(1-c(N-2k-1)^{-s}big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}cleft(dfrac{N}{(2k-1)^2}right)^{-s}right)
left(1-dleft(dfrac{N}{2k-1}right)^{-t}right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}dleft(dfrac{N}{4k^2}right)^{-t}right)
left(1-cleft(dfrac{N}{2k}right)^{-s}right)right)\[4pt]
&P_1(N) = 1- &prodlimits_{k=[N/4-2]}^{[N/4-1]},c(N-2k)^{-s}prodlimits_{k=1}^{[N/4-3]}Big(1-big(1-d(N-4k)^{-t}big)big(1-(N-2k)^{-s}big)Big)\
&&Big(1-big(1-(N-4k-2)^{-s}big)big(1-d(N-2k-1)^{-t}big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}dleft(dfrac{N}{(2k-1)^2}right)^{-t}right)
left(1-cleft(dfrac{N}{2k-1}right)^{-s}right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}cleft(dfrac{N}{4k^2}right)^{-s}right)
left(1-dleft(dfrac{N}{2k}right)^{-t}right)right).
end{cases}tag{11}$$
Model $(11)$ should be checked theoretically and practically, but it gives the approach to the required estimations.
The next steps are estimation of parameters $$c,d,s,t$$ and using of obtained model.
answered Jan 2 at 19:52
Yuri NegometyanovYuri Negometyanov
10.9k1727
@Joseph O'Rourke Approximation can be changed, but parametrization is actual.
– Yuri Negometyanov
23 hours ago
add a comment |
Since both @awwalker and @mathworker21 mentioned Erdős' conjecture, and because
a paper discussing this conjecture was just published, I thought I would mention it:
Erdős Conjecture (1940s or 1950s). If $A subset mathbb{N}$
satisfies $sum_{n in A} frac{1}{n}= infty$, then
$A$ contains arbitrarily long arithmetic progressions.
In
- Grochow, Joshua. "New applications of the polynomial method: The cap
set conjecture and beyond." Bulletin of the American Mathematical
Society, Vol.56, No.1, Jan. 2019,
he says:
"It remains open even to prove that a set $A$ satisfying the hypothesis contains $3$-term arithmetic progressions."
add a comment |
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2 Answers
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$color{brown}{textbf{HINT}}$
Denote the target sequence ${F_3(n)}$ and let us try to estimate the probability $P(N)}$ that natural number belongs to ${F_3}.$
Suppose
$$F_3(1)=1,quad F_3(2)=2,quad P(1)=P(2)=P(5) = P(6) = 1,\ P(3)=P(4)=P(7)=P(8)=0,tag1$$
$$V(N)=sumlimits_{i=1}^{N}P(i),quad F_3(N) = left[V^{-1}(N)right].tag2$$
Let $P_a(N)$ is the probability that N does not belong to arithmetic progression and $P_g(N)$ is the similar probability for geometric progressions.
Suppose
$$P(N) = P_a(N)P_g(N)).tag3$$
$color{brown}{textbf{Arithmetic Probability estimation.}}$
Suppose
$$P_a(N)=prodlimits_{k=1}^{[N/2]}P_a(N,k),tag4$$
where $P_a(N,k)$ is the probability that arithmetic progression ${N-2k,N-k, N}$ does not exist for any $j.$
Suppose
$$P_a(N,k) = big(1-P(N-2k)big)big((1-P(N-k)big).tag5$$
$color{brown}{textbf{Geometric Probability estimation.}}$
Suppose
$$P_g(N)=prodlimits_{k=1}^{left[,sqrt Nn,right]}P_g(N,k),tag6$$
where $P_g(N,k)$ is the probability that geometric progression $left(dfrac{N}{k^2}, dfrac Nk, Nright}$ with the denominator $k$ does not exist for all $i,j.$
Taking in account that the geometric progression can exist only if $k^2,| N,$ suppose
$$P_g(N,k) = left(1-dfrac1{k^2}Pleft(dfrac{N}{k^2}right)right)left(1-Pleft(dfrac{N}{k}right)right).tag7$$
$color{brown}{textbf{Common model.}}$
Common model can be simplified to next one,
begin{cases}
P(N) = 1-prodlimits_{k=1}^{[N/2-1]}Big(1-big(1-P(N-2k)big)big(1-P(N-k)big)Big)\
timesprodlimits_{k=1}^{left[sqrt Nright]}left(1-left(1-dfrac1{k^2}Pleft(dfrac{N}{k^2}right)right)left(1-Pleft(dfrac{N}{k}right)right)right)\[4pt]
P(1)=P(2)=P(5) = P(6) = 1,quad P(3)=P(4)=P(7)=P(8)=0\
V(N)=sumlimits_{i=1}^{N}P(i),quad F_3(n) = left[V^{-1}(n)right].tag8
end{cases}
Looking the solution in the form of
$$left{begin{align}
&P(N)=P_v(N),quadtext{where}quad v=left[dfrac{N-1 mod 4}2right],\[4pt]
&P_0(N)=
begin{cases}
1,quadtext{if}quad N<9\
cN^{-s},quadtext{otherwise}
end{cases}\[4pt]
&P_1(N)=
begin{cases}
0,quadtext{if}quad N<9\
dN^{-t},quadtext{otherwise}
end{cases}
end{align}right.tag9$$
then
$$begin{cases}
&P_0(N) =1-&prodlimits_{k=1}^{[N/4-1]}Big(1-big(1-P_0(N-4k)big)big(1-P_1(N-2k)big)Big)\
&×Big(1-big(1-P_1(N-4k-2)big)big(1-P_0(N-2k-1)big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}P_0left(dfrac{N}{(2k-1)^2}right)right)
left(1-P_1left(dfrac{N}{2k-1}right)right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}P_1left(dfrac{N}{4k^2}right)right)
left(1-P_0left(dfrac{N}{2k}right)right)right)\[4pt]
&P_1(N) = 1- &prodlimits_{k=1}^{[N/4-1]}Big(1-big(1-P_1(N-4k)big)big(1-P_0(N-2k)big)Big)\
&&Big(1-big(1-P_0(N-4k-2)big)big(1-P_1(N-2k-1)big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}P_1left(dfrac{N}{(2k-1)^2}right)right)
left(1-P_0left(dfrac{N}{2k-1}right)right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}P_0left(dfrac{N}{4k^2}right)right)
left(1-P_1left(dfrac{N}{2k}right)right)right).
end{cases}tag{10}$$
Taking in account $(9),$ can be written
$$begin{cases}
&P_0(N) =1-&prodlimits_{k=[N/4-2]}^{[N/4-1]},c(N-2k-1)^{-s}prodlimits_{k=1}^{[N/4-3]}Big(1-big(1-c(N-4k)^{-s}big)big(1-d(N-2k)^{-t}big)Big)\
&×Big(1-big(1-d(N-4k-2)^{-t}big)big(1-c(N-2k-1)^{-s}big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}cleft(dfrac{N}{(2k-1)^2}right)^{-s}right)
left(1-dleft(dfrac{N}{2k-1}right)^{-t}right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}dleft(dfrac{N}{4k^2}right)^{-t}right)
left(1-cleft(dfrac{N}{2k}right)^{-s}right)right)\[4pt]
&P_1(N) = 1- &prodlimits_{k=[N/4-2]}^{[N/4-1]},c(N-2k)^{-s}prodlimits_{k=1}^{[N/4-3]}Big(1-big(1-d(N-4k)^{-t}big)big(1-(N-2k)^{-s}big)Big)\
&&Big(1-big(1-(N-4k-2)^{-s}big)big(1-d(N-2k-1)^{-t}big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}dleft(dfrac{N}{(2k-1)^2}right)^{-t}right)
left(1-cleft(dfrac{N}{2k-1}right)^{-s}right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}cleft(dfrac{N}{4k^2}right)^{-s}right)
left(1-dleft(dfrac{N}{2k}right)^{-t}right)right).
end{cases}tag{11}$$
Model $(11)$ should be checked theoretically and practically, but it gives the approach to the required estimations.
The next steps are estimation of parameters $$c,d,s,t$$ and using of obtained model.
answered Jan 2 at 19:52
Yuri NegometyanovYuri Negometyanov
10.9k1727
@Joseph O'Rourke Approximation can be changed, but parametrization is actual.
– Yuri Negometyanov
23 hours ago
add a comment |
$color{brown}{textbf{HINT}}$
Denote the target sequence ${F_3(n)}$ and let us try to estimate the probability $P(N)}$ that natural number belongs to ${F_3}.$
Suppose
$$F_3(1)=1,quad F_3(2)=2,quad P(1)=P(2)=P(5) = P(6) = 1,\ P(3)=P(4)=P(7)=P(8)=0,tag1$$
$$V(N)=sumlimits_{i=1}^{N}P(i),quad F_3(N) = left[V^{-1}(N)right].tag2$$
Let $P_a(N)$ is the probability that N does not belong to arithmetic progression and $P_g(N)$ is the similar probability for geometric progressions.
Suppose
$$P(N) = P_a(N)P_g(N)).tag3$$
$color{brown}{textbf{Arithmetic Probability estimation.}}$
Suppose
$$P_a(N)=prodlimits_{k=1}^{[N/2]}P_a(N,k),tag4$$
where $P_a(N,k)$ is the probability that arithmetic progression ${N-2k,N-k, N}$ does not exist for any $j.$
Suppose
$$P_a(N,k) = big(1-P(N-2k)big)big((1-P(N-k)big).tag5$$
$color{brown}{textbf{Geometric Probability estimation.}}$
Suppose
$$P_g(N)=prodlimits_{k=1}^{left[,sqrt Nn,right]}P_g(N,k),tag6$$
where $P_g(N,k)$ is the probability that geometric progression $left(dfrac{N}{k^2}, dfrac Nk, Nright}$ with the denominator $k$ does not exist for all $i,j.$
Taking in account that the geometric progression can exist only if $k^2,| N,$ suppose
$$P_g(N,k) = left(1-dfrac1{k^2}Pleft(dfrac{N}{k^2}right)right)left(1-Pleft(dfrac{N}{k}right)right).tag7$$
$color{brown}{textbf{Common model.}}$
Common model can be simplified to next one,
begin{cases}
P(N) = 1-prodlimits_{k=1}^{[N/2-1]}Big(1-big(1-P(N-2k)big)big(1-P(N-k)big)Big)\
timesprodlimits_{k=1}^{left[sqrt Nright]}left(1-left(1-dfrac1{k^2}Pleft(dfrac{N}{k^2}right)right)left(1-Pleft(dfrac{N}{k}right)right)right)\[4pt]
P(1)=P(2)=P(5) = P(6) = 1,quad P(3)=P(4)=P(7)=P(8)=0\
V(N)=sumlimits_{i=1}^{N}P(i),quad F_3(n) = left[V^{-1}(n)right].tag8
end{cases}
Looking the solution in the form of
$$left{begin{align}
&P(N)=P_v(N),quadtext{where}quad v=left[dfrac{N-1 mod 4}2right],\[4pt]
&P_0(N)=
begin{cases}
1,quadtext{if}quad N<9\
cN^{-s},quadtext{otherwise}
end{cases}\[4pt]
&P_1(N)=
begin{cases}
0,quadtext{if}quad N<9\
dN^{-t},quadtext{otherwise}
end{cases}
end{align}right.tag9$$
then
$$begin{cases}
&P_0(N) =1-&prodlimits_{k=1}^{[N/4-1]}Big(1-big(1-P_0(N-4k)big)big(1-P_1(N-2k)big)Big)\
&×Big(1-big(1-P_1(N-4k-2)big)big(1-P_0(N-2k-1)big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}P_0left(dfrac{N}{(2k-1)^2}right)right)
left(1-P_1left(dfrac{N}{2k-1}right)right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}P_1left(dfrac{N}{4k^2}right)right)
left(1-P_0left(dfrac{N}{2k}right)right)right)\[4pt]
&P_1(N) = 1- &prodlimits_{k=1}^{[N/4-1]}Big(1-big(1-P_1(N-4k)big)big(1-P_0(N-2k)big)Big)\
&&Big(1-big(1-P_0(N-4k-2)big)big(1-P_1(N-2k-1)big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}P_1left(dfrac{N}{(2k-1)^2}right)right)
left(1-P_0left(dfrac{N}{2k-1}right)right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}P_0left(dfrac{N}{4k^2}right)right)
left(1-P_1left(dfrac{N}{2k}right)right)right).
end{cases}tag{10}$$
Taking in account $(9),$ can be written
$$begin{cases}
&P_0(N) =1-&prodlimits_{k=[N/4-2]}^{[N/4-1]},c(N-2k-1)^{-s}prodlimits_{k=1}^{[N/4-3]}Big(1-big(1-c(N-4k)^{-s}big)big(1-d(N-2k)^{-t}big)Big)\
&×Big(1-big(1-d(N-4k-2)^{-t}big)big(1-c(N-2k-1)^{-s}big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}cleft(dfrac{N}{(2k-1)^2}right)^{-s}right)
left(1-dleft(dfrac{N}{2k-1}right)^{-t}right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}dleft(dfrac{N}{4k^2}right)^{-t}right)
left(1-cleft(dfrac{N}{2k}right)^{-s}right)right)\[4pt]
&P_1(N) = 1- &prodlimits_{k=[N/4-2]}^{[N/4-1]},c(N-2k)^{-s}prodlimits_{k=1}^{[N/4-3]}Big(1-big(1-d(N-4k)^{-t}big)big(1-(N-2k)^{-s}big)Big)\
&&Big(1-big(1-(N-4k-2)^{-s}big)big(1-d(N-2k-1)^{-t}big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}dleft(dfrac{N}{(2k-1)^2}right)^{-t}right)
left(1-cleft(dfrac{N}{2k-1}right)^{-s}right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}cleft(dfrac{N}{4k^2}right)^{-s}right)
left(1-dleft(dfrac{N}{2k}right)^{-t}right)right).
end{cases}tag{11}$$
Model $(11)$ should be checked theoretically and practically, but it gives the approach to the required estimations.
The next steps are estimation of parameters $$c,d,s,t$$ and using of obtained model.
answered Jan 2 at 19:52
Yuri NegometyanovYuri Negometyanov
10.9k1727
@Joseph O'Rourke Approximation can be changed, but parametrization is actual.
– Yuri Negometyanov
23 hours ago
add a comment |
$color{brown}{textbf{HINT}}$
Denote the target sequence ${F_3(n)}$ and let us try to estimate the probability $P(N)}$ that natural number belongs to ${F_3}.$
Suppose
$$F_3(1)=1,quad F_3(2)=2,quad P(1)=P(2)=P(5) = P(6) = 1,\ P(3)=P(4)=P(7)=P(8)=0,tag1$$
$$V(N)=sumlimits_{i=1}^{N}P(i),quad F_3(N) = left[V^{-1}(N)right].tag2$$
Let $P_a(N)$ is the probability that N does not belong to arithmetic progression and $P_g(N)$ is the similar probability for geometric progressions.
Suppose
$$P(N) = P_a(N)P_g(N)).tag3$$
$color{brown}{textbf{Arithmetic Probability estimation.}}$
Suppose
$$P_a(N)=prodlimits_{k=1}^{[N/2]}P_a(N,k),tag4$$
where $P_a(N,k)$ is the probability that arithmetic progression ${N-2k,N-k, N}$ does not exist for any $j.$
Suppose
$$P_a(N,k) = big(1-P(N-2k)big)big((1-P(N-k)big).tag5$$
$color{brown}{textbf{Geometric Probability estimation.}}$
Suppose
$$P_g(N)=prodlimits_{k=1}^{left[,sqrt Nn,right]}P_g(N,k),tag6$$
where $P_g(N,k)$ is the probability that geometric progression $left(dfrac{N}{k^2}, dfrac Nk, Nright}$ with the denominator $k$ does not exist for all $i,j.$
Taking in account that the geometric progression can exist only if $k^2,| N,$ suppose
$$P_g(N,k) = left(1-dfrac1{k^2}Pleft(dfrac{N}{k^2}right)right)left(1-Pleft(dfrac{N}{k}right)right).tag7$$
$color{brown}{textbf{Common model.}}$
Common model can be simplified to next one,
begin{cases}
P(N) = 1-prodlimits_{k=1}^{[N/2-1]}Big(1-big(1-P(N-2k)big)big(1-P(N-k)big)Big)\
timesprodlimits_{k=1}^{left[sqrt Nright]}left(1-left(1-dfrac1{k^2}Pleft(dfrac{N}{k^2}right)right)left(1-Pleft(dfrac{N}{k}right)right)right)\[4pt]
P(1)=P(2)=P(5) = P(6) = 1,quad P(3)=P(4)=P(7)=P(8)=0\
V(N)=sumlimits_{i=1}^{N}P(i),quad F_3(n) = left[V^{-1}(n)right].tag8
end{cases}
Looking the solution in the form of
$$left{begin{align}
&P(N)=P_v(N),quadtext{where}quad v=left[dfrac{N-1 mod 4}2right],\[4pt]
&P_0(N)=
begin{cases}
1,quadtext{if}quad N<9\
cN^{-s},quadtext{otherwise}
end{cases}\[4pt]
&P_1(N)=
begin{cases}
0,quadtext{if}quad N<9\
dN^{-t},quadtext{otherwise}
end{cases}
end{align}right.tag9$$
then
$$begin{cases}
&P_0(N) =1-&prodlimits_{k=1}^{[N/4-1]}Big(1-big(1-P_0(N-4k)big)big(1-P_1(N-2k)big)Big)\
&×Big(1-big(1-P_1(N-4k-2)big)big(1-P_0(N-2k-1)big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}P_0left(dfrac{N}{(2k-1)^2}right)right)
left(1-P_1left(dfrac{N}{2k-1}right)right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}P_1left(dfrac{N}{4k^2}right)right)
left(1-P_0left(dfrac{N}{2k}right)right)right)\[4pt]
&P_1(N) = 1- &prodlimits_{k=1}^{[N/4-1]}Big(1-big(1-P_1(N-4k)big)big(1-P_0(N-2k)big)Big)\
&&Big(1-big(1-P_0(N-4k-2)big)big(1-P_1(N-2k-1)big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}P_1left(dfrac{N}{(2k-1)^2}right)right)
left(1-P_0left(dfrac{N}{2k-1}right)right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}P_0left(dfrac{N}{4k^2}right)right)
left(1-P_1left(dfrac{N}{2k}right)right)right).
end{cases}tag{10}$$
Taking in account $(9),$ can be written
$$begin{cases}
&P_0(N) =1-&prodlimits_{k=[N/4-2]}^{[N/4-1]},c(N-2k-1)^{-s}prodlimits_{k=1}^{[N/4-3]}Big(1-big(1-c(N-4k)^{-s}big)big(1-d(N-2k)^{-t}big)Big)\
&×Big(1-big(1-d(N-4k-2)^{-t}big)big(1-c(N-2k-1)^{-s}big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}cleft(dfrac{N}{(2k-1)^2}right)^{-s}right)
left(1-dleft(dfrac{N}{2k-1}right)^{-t}right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}dleft(dfrac{N}{4k^2}right)^{-t}right)
left(1-cleft(dfrac{N}{2k}right)^{-s}right)right)\[4pt]
&P_1(N) = 1- &prodlimits_{k=[N/4-2]}^{[N/4-1]},c(N-2k)^{-s}prodlimits_{k=1}^{[N/4-3]}Big(1-big(1-d(N-4k)^{-t}big)big(1-(N-2k)^{-s}big)Big)\
&&Big(1-big(1-(N-4k-2)^{-s}big)big(1-d(N-2k-1)^{-t}big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}dleft(dfrac{N}{(2k-1)^2}right)^{-t}right)
left(1-cleft(dfrac{N}{2k-1}right)^{-s}right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}cleft(dfrac{N}{4k^2}right)^{-s}right)
left(1-dleft(dfrac{N}{2k}right)^{-t}right)right).
end{cases}tag{11}$$
Model $(11)$ should be checked theoretically and practically, but it gives the approach to the required estimations.
The next steps are estimation of parameters $$c,d,s,t$$ and using of obtained model.
answered Jan 2 at 19:52
Yuri NegometyanovYuri Negometyanov
10.9k1727
$color{brown}{textbf{HINT}}$
Denote the target sequence ${F_3(n)}$ and let us try to estimate the probability $P(N)}$ that natural number belongs to ${F_3}.$
Suppose
$$F_3(1)=1,quad F_3(2)=2,quad P(1)=P(2)=P(5) = P(6) = 1,\ P(3)=P(4)=P(7)=P(8)=0,tag1$$
$$V(N)=sumlimits_{i=1}^{N}P(i),quad F_3(N) = left[V^{-1}(N)right].tag2$$
Let $P_a(N)$ is the probability that N does not belong to arithmetic progression and $P_g(N)$ is the similar probability for geometric progressions.
Suppose
$$P(N) = P_a(N)P_g(N)).tag3$$
$color{brown}{textbf{Arithmetic Probability estimation.}}$
Suppose
$$P_a(N)=prodlimits_{k=1}^{[N/2]}P_a(N,k),tag4$$
where $P_a(N,k)$ is the probability that arithmetic progression ${N-2k,N-k, N}$ does not exist for any $j.$
Suppose
$$P_a(N,k) = big(1-P(N-2k)big)big((1-P(N-k)big).tag5$$
$color{brown}{textbf{Geometric Probability estimation.}}$
Suppose
$$P_g(N)=prodlimits_{k=1}^{left[,sqrt Nn,right]}P_g(N,k),tag6$$
where $P_g(N,k)$ is the probability that geometric progression $left(dfrac{N}{k^2}, dfrac Nk, Nright}$ with the denominator $k$ does not exist for all $i,j.$
Taking in account that the geometric progression can exist only if $k^2,| N,$ suppose
$$P_g(N,k) = left(1-dfrac1{k^2}Pleft(dfrac{N}{k^2}right)right)left(1-Pleft(dfrac{N}{k}right)right).tag7$$
$color{brown}{textbf{Common model.}}$
Common model can be simplified to next one,
begin{cases}
P(N) = 1-prodlimits_{k=1}^{[N/2-1]}Big(1-big(1-P(N-2k)big)big(1-P(N-k)big)Big)\
timesprodlimits_{k=1}^{left[sqrt Nright]}left(1-left(1-dfrac1{k^2}Pleft(dfrac{N}{k^2}right)right)left(1-Pleft(dfrac{N}{k}right)right)right)\[4pt]
P(1)=P(2)=P(5) = P(6) = 1,quad P(3)=P(4)=P(7)=P(8)=0\
V(N)=sumlimits_{i=1}^{N}P(i),quad F_3(n) = left[V^{-1}(n)right].tag8
end{cases}
Looking the solution in the form of
$$left{begin{align}
&P(N)=P_v(N),quadtext{where}quad v=left[dfrac{N-1 mod 4}2right],\[4pt]
&P_0(N)=
begin{cases}
1,quadtext{if}quad N<9\
cN^{-s},quadtext{otherwise}
end{cases}\[4pt]
&P_1(N)=
begin{cases}
0,quadtext{if}quad N<9\
dN^{-t},quadtext{otherwise}
end{cases}
end{align}right.tag9$$
then
$$begin{cases}
&P_0(N) =1-&prodlimits_{k=1}^{[N/4-1]}Big(1-big(1-P_0(N-4k)big)big(1-P_1(N-2k)big)Big)\
&×Big(1-big(1-P_1(N-4k-2)big)big(1-P_0(N-2k-1)big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}P_0left(dfrac{N}{(2k-1)^2}right)right)
left(1-P_1left(dfrac{N}{2k-1}right)right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}P_1left(dfrac{N}{4k^2}right)right)
left(1-P_0left(dfrac{N}{2k}right)right)right)\[4pt]
&P_1(N) = 1- &prodlimits_{k=1}^{[N/4-1]}Big(1-big(1-P_1(N-4k)big)big(1-P_0(N-2k)big)Big)\
&&Big(1-big(1-P_0(N-4k-2)big)big(1-P_1(N-2k-1)big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}P_1left(dfrac{N}{(2k-1)^2}right)right)
left(1-P_0left(dfrac{N}{2k-1}right)right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}P_0left(dfrac{N}{4k^2}right)right)
left(1-P_1left(dfrac{N}{2k}right)right)right).
end{cases}tag{10}$$
Taking in account $(9),$ can be written
$$begin{cases}
&P_0(N) =1-&prodlimits_{k=[N/4-2]}^{[N/4-1]},c(N-2k-1)^{-s}prodlimits_{k=1}^{[N/4-3]}Big(1-big(1-c(N-4k)^{-s}big)big(1-d(N-2k)^{-t}big)Big)\
&×Big(1-big(1-d(N-4k-2)^{-t}big)big(1-c(N-2k-1)^{-s}big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}cleft(dfrac{N}{(2k-1)^2}right)^{-s}right)
left(1-dleft(dfrac{N}{2k-1}right)^{-t}right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}dleft(dfrac{N}{4k^2}right)^{-t}right)
left(1-cleft(dfrac{N}{2k}right)^{-s}right)right)\[4pt]
&P_1(N) = 1- &prodlimits_{k=[N/4-2]}^{[N/4-1]},c(N-2k)^{-s}prodlimits_{k=1}^{[N/4-3]}Big(1-big(1-d(N-4k)^{-t}big)big(1-(N-2k)^{-s}big)Big)\
&&Big(1-big(1-(N-4k-2)^{-s}big)big(1-d(N-2k-1)^{-t}big)Big)\
&×prodlimits_{k=1}^{left[sqrt N/2right]-1}
left(1-left(1-dfrac1{(2k-1)^2}dleft(dfrac{N}{(2k-1)^2}right)^{-t}right)
left(1-cleft(dfrac{N}{2k-1}right)^{-s}right)right)\[4pt]
&×left(1-left(1-dfrac1{4k^2}cleft(dfrac{N}{4k^2}right)^{-s}right)
left(1-dleft(dfrac{N}{2k}right)^{-t}right)right).
end{cases}tag{11}$$
Model $(11)$ should be checked theoretically and practically, but it gives the approach to the required estimations.
The next steps are estimation of parameters $$c,d,s,t$$ and using of obtained model.
answered Jan 2 at 19:52
Yuri NegometyanovYuri Negometyanov
10.9k1727
edited 22 hours ago
answered Jan 2 at 19:52
Yuri NegometyanovYuri Negometyanov
10.9k1727
answered Jan 2 at 19:52
Yuri NegometyanovYuri Negometyanov
10.9k1727
answered Jan 2 at 19:52
Yuri NegometyanovYuri Negometyanov
10.9k1727
10.9k1727
@Joseph O'Rourke Approximation can be changed, but parametrization is actual.
– Yuri Negometyanov
23 hours ago
add a comment |
@Joseph O'Rourke Approximation can be changed, but parametrization is actual.
– Yuri Negometyanov
23 hours ago
@Joseph O'Rourke Approximation can be changed, but parametrization is actual.
– Yuri Negometyanov
23 hours ago
@Joseph O'Rourke Approximation can be changed, but parametrization is actual.
– Yuri Negometyanov
23 hours ago
add a comment |
Since both @awwalker and @mathworker21 mentioned Erdős' conjecture, and because
a paper discussing this conjecture was just published, I thought I would mention it:
Erdős Conjecture (1940s or 1950s). If $A subset mathbb{N}$
satisfies $sum_{n in A} frac{1}{n}= infty$, then
$A$ contains arbitrarily long arithmetic progressions.
In
- Grochow, Joshua. "New applications of the polynomial method: The cap
set conjecture and beyond." Bulletin of the American Mathematical
Society, Vol.56, No.1, Jan. 2019,
he says:
"It remains open even to prove that a set $A$ satisfying the hypothesis contains $3$-term arithmetic progressions."
add a comment |
Since both @awwalker and @mathworker21 mentioned Erdős' conjecture, and because
a paper discussing this conjecture was just published, I thought I would mention it:
Erdős Conjecture (1940s or 1950s). If $A subset mathbb{N}$
satisfies $sum_{n in A} frac{1}{n}= infty$, then
$A$ contains arbitrarily long arithmetic progressions.
In
- Grochow, Joshua. "New applications of the polynomial method: The cap
set conjecture and beyond." Bulletin of the American Mathematical
Society, Vol.56, No.1, Jan. 2019,
he says:
"It remains open even to prove that a set $A$ satisfying the hypothesis contains $3$-term arithmetic progressions."
add a comment |
Since both @awwalker and @mathworker21 mentioned Erdős' conjecture, and because
a paper discussing this conjecture was just published, I thought I would mention it:
Erdős Conjecture (1940s or 1950s). If $A subset mathbb{N}$
satisfies $sum_{n in A} frac{1}{n}= infty$, then
$A$ contains arbitrarily long arithmetic progressions.
In
- Grochow, Joshua. "New applications of the polynomial method: The cap
set conjecture and beyond." Bulletin of the American Mathematical
Society, Vol.56, No.1, Jan. 2019,
he says:
"It remains open even to prove that a set $A$ satisfying the hypothesis contains $3$-term arithmetic progressions."
Since both @awwalker and @mathworker21 mentioned Erdős' conjecture, and because
a paper discussing this conjecture was just published, I thought I would mention it:
Erdős Conjecture (1940s or 1950s). If $A subset mathbb{N}$
satisfies $sum_{n in A} frac{1}{n}= infty$, then
$A$ contains arbitrarily long arithmetic progressions.
In
- Grochow, Joshua. "New applications of the polynomial method: The cap
set conjecture and beyond." Bulletin of the American Mathematical
Society, Vol.56, No.1, Jan. 2019,
he says:
"It remains open even to prove that a set $A$ satisfying the hypothesis contains $3$-term arithmetic progressions."
edited yesterday
community wiki
2 revs
Joseph O'Rourke
add a comment |
add a comment |
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6
Expanding on your numerical data I conclude that Q3 can be answered in the negative. 457th term is 17933 with sum of reciprocals : 2.702607644337383 1000th term is 66102 : 2.718058753659135 but alas, somewhere shortly after that the sum excedes e. 2000 253701 2.7259164358931023. 3000 442429 2.7287027640882804.
– Michael Stocker
Jul 7 '14 at 13:16
3
The answer to Q2 is definitely Yes, it converges, because even just avoiding arithmetic progressions leads to convergence. Converges to what constant is unclear.
– Joseph O'Rourke
Jul 9 '14 at 0:33
2
@JosephO'Rourke Really? I may be ignorant here so going out on a limb, but I don't think $a_n=nln n$ is in an arithmetic progression, but the sum of its reciprocals diverges.
– John Molokach
Mar 25 '16 at 2:46
1
@JohnMolokach In fact, when you look at the difference in terms in the OEIS sequence, it looks a lot like OEIS A006519, which is $O(nln(n))$. The sequence talked about in the question is a bit different, but I think asymptotically they'd be the same, meaning the sum of reciprocals diverges. And of course, this is just a heuristic, not a proof.
– Turambar
Apr 28 '16 at 17:52
1
On the other hand, after looking at it out to $i = 10,000$, the chart of the differences doesn't seem to be as similar to A006519 as I thought. Its peaks grow faster and have a more bell curve distribution around them instead of the recursive characteristics of A006519. At $i = 10,000$, $ln(s_i)/ln(i) = 1.679258559$ and generally growing, with the sum of reciprocals around $2.734896156$. So if the "power" keeps generally growing (and doesn't start dropping to 1 after some point), then the sum is below $2.736095249$.
– Turambar
May 10 '16 at 16:14