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Is there any algorithm to identify the smallest example of gap size $n-1$ between consecutive prime numbers
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In contemplating Goldbach's conjecture, I became interested in gaps between successive primes. If $n<a<b<2n$ and the range $a$ to $b$ is a primeless gap, then one could ignore any primes in the range $2n-b$ to $2n-a$ for purposes of Goldbach solutions concerning the even number $2n$. That line of thought is not getting anywhere, but it did turn my attention to gaps.
There are several ways to achieve a gap between successive primes of at least $n-1$, viz: $kmid (n!+k)$ for $2le kle n$. If $N$ is defined as the least common multiple of $n$ (see OEIS A003418), then $kmid (N+k)$ for $2le kle n$. Finally gcd$(k,(n# +k))>1$ for $2le kle n$.
These methods all work. For a gap of $n-1=5$; $n=6$, the factorial method gives the five numbers $722,723,724,725,726$ without any primes. The lcm method gives the five numbers $62,63,64,65,66$ without any primes. The primorial method gives the five numbers $32,33,34,35,36$ without any primes. But the smallest sequence of five numbers without any primes is $24,25,26,27,28$. I see that this occurs because the two odd numbers have distinct small prime factors, but I cannot see any way (other than direct examination) to generate or identify this sequence.
Added by edit: It is interesting to note that since the factorial and the lcm methods both yield sequences whose members are divisible (seriatim) by every $k$ in the applicable range, and the lcm method generates a smaller answer by virtue of casting out 'superfluous' occurrences of prime factors, then intermediate sequences of $(n-1)$ numbers lacking a prime can be obtained by subtracting various multiples of the lcm $N$ from the factorial $n!$. In the context of the example given, this generates sequences of five primeless numbers starting at $662; 602; 542; 482; 422; 362; 302; 242; 182; 122$. Further to that thought: Any number $m(n#)$ can be the basis for such a sequence, i.e. $(m(n#)+k)$ for $2le kle n$.
I have searched MSE to see if this question has been asked before, and I did not find it (which surprised me a little): Is there an algorithmic way to generate or identify the first or smallest occurrence of a prime gap of size $(n-1)$?
prime-numbers prime-gaps goldbachs-conjecture
asked Jul 27 at 19:17
Keith Backman
39227
add a comment |Â
up vote
3
down vote
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In contemplating Goldbach's conjecture, I became interested in gaps between successive primes. If $n<a<b<2n$ and the range $a$ to $b$ is a primeless gap, then one could ignore any primes in the range $2n-b$ to $2n-a$ for purposes of Goldbach solutions concerning the even number $2n$. That line of thought is not getting anywhere, but it did turn my attention to gaps.
There are several ways to achieve a gap between successive primes of at least $n-1$, viz: $kmid (n!+k)$ for $2le kle n$. If $N$ is defined as the least common multiple of $n$ (see OEIS A003418), then $kmid (N+k)$ for $2le kle n$. Finally gcd$(k,(n# +k))>1$ for $2le kle n$.
These methods all work. For a gap of $n-1=5$; $n=6$, the factorial method gives the five numbers $722,723,724,725,726$ without any primes. The lcm method gives the five numbers $62,63,64,65,66$ without any primes. The primorial method gives the five numbers $32,33,34,35,36$ without any primes. But the smallest sequence of five numbers without any primes is $24,25,26,27,28$. I see that this occurs because the two odd numbers have distinct small prime factors, but I cannot see any way (other than direct examination) to generate or identify this sequence.
Added by edit: It is interesting to note that since the factorial and the lcm methods both yield sequences whose members are divisible (seriatim) by every $k$ in the applicable range, and the lcm method generates a smaller answer by virtue of casting out 'superfluous' occurrences of prime factors, then intermediate sequences of $(n-1)$ numbers lacking a prime can be obtained by subtracting various multiples of the lcm $N$ from the factorial $n!$. In the context of the example given, this generates sequences of five primeless numbers starting at $662; 602; 542; 482; 422; 362; 302; 242; 182; 122$. Further to that thought: Any number $m(n#)$ can be the basis for such a sequence, i.e. $(m(n#)+k)$ for $2le kle n$.
I have searched MSE to see if this question has been asked before, and I did not find it (which surprised me a little): Is there an algorithmic way to generate or identify the first or smallest occurrence of a prime gap of size $(n-1)$?
prime-numbers prime-gaps goldbachs-conjecture
asked Jul 27 at 19:17
Keith Backman
39227
2
no, it is an extremely difficult problem. See the table of maximal prime gaps at en.wikipedia.org/wiki/Prime_gap Evidently there is slow progress, they now have the first 78 maximal gaps. Last time I looked it was 75. They have (slightly) passed $10^19$
– Will Jagy
Jul 27 at 19:53
add a comment |Â
up vote
3
down vote
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up vote
3
down vote
favorite
In contemplating Goldbach's conjecture, I became interested in gaps between successive primes. If $n<a<b<2n$ and the range $a$ to $b$ is a primeless gap, then one could ignore any primes in the range $2n-b$ to $2n-a$ for purposes of Goldbach solutions concerning the even number $2n$. That line of thought is not getting anywhere, but it did turn my attention to gaps.
There are several ways to achieve a gap between successive primes of at least $n-1$, viz: $kmid (n!+k)$ for $2le kle n$. If $N$ is defined as the least common multiple of $n$ (see OEIS A003418), then $kmid (N+k)$ for $2le kle n$. Finally gcd$(k,(n# +k))>1$ for $2le kle n$.
These methods all work. For a gap of $n-1=5$; $n=6$, the factorial method gives the five numbers $722,723,724,725,726$ without any primes. The lcm method gives the five numbers $62,63,64,65,66$ without any primes. The primorial method gives the five numbers $32,33,34,35,36$ without any primes. But the smallest sequence of five numbers without any primes is $24,25,26,27,28$. I see that this occurs because the two odd numbers have distinct small prime factors, but I cannot see any way (other than direct examination) to generate or identify this sequence.
Added by edit: It is interesting to note that since the factorial and the lcm methods both yield sequences whose members are divisible (seriatim) by every $k$ in the applicable range, and the lcm method generates a smaller answer by virtue of casting out 'superfluous' occurrences of prime factors, then intermediate sequences of $(n-1)$ numbers lacking a prime can be obtained by subtracting various multiples of the lcm $N$ from the factorial $n!$. In the context of the example given, this generates sequences of five primeless numbers starting at $662; 602; 542; 482; 422; 362; 302; 242; 182; 122$. Further to that thought: Any number $m(n#)$ can be the basis for such a sequence, i.e. $(m(n#)+k)$ for $2le kle n$.
I have searched MSE to see if this question has been asked before, and I did not find it (which surprised me a little): Is there an algorithmic way to generate or identify the first or smallest occurrence of a prime gap of size $(n-1)$?
prime-numbers prime-gaps goldbachs-conjecture
asked Jul 27 at 19:17
Keith Backman
39227
In contemplating Goldbach's conjecture, I became interested in gaps between successive primes. If $n<a<b<2n$ and the range $a$ to $b$ is a primeless gap, then one could ignore any primes in the range $2n-b$ to $2n-a$ for purposes of Goldbach solutions concerning the even number $2n$. That line of thought is not getting anywhere, but it did turn my attention to gaps.
There are several ways to achieve a gap between successive primes of at least $n-1$, viz: $kmid (n!+k)$ for $2le kle n$. If $N$ is defined as the least common multiple of $n$ (see OEIS A003418), then $kmid (N+k)$ for $2le kle n$. Finally gcd$(k,(n# +k))>1$ for $2le kle n$.
These methods all work. For a gap of $n-1=5$; $n=6$, the factorial method gives the five numbers $722,723,724,725,726$ without any primes. The lcm method gives the five numbers $62,63,64,65,66$ without any primes. The primorial method gives the five numbers $32,33,34,35,36$ without any primes. But the smallest sequence of five numbers without any primes is $24,25,26,27,28$. I see that this occurs because the two odd numbers have distinct small prime factors, but I cannot see any way (other than direct examination) to generate or identify this sequence.
Added by edit: It is interesting to note that since the factorial and the lcm methods both yield sequences whose members are divisible (seriatim) by every $k$ in the applicable range, and the lcm method generates a smaller answer by virtue of casting out 'superfluous' occurrences of prime factors, then intermediate sequences of $(n-1)$ numbers lacking a prime can be obtained by subtracting various multiples of the lcm $N$ from the factorial $n!$. In the context of the example given, this generates sequences of five primeless numbers starting at $662; 602; 542; 482; 422; 362; 302; 242; 182; 122$. Further to that thought: Any number $m(n#)$ can be the basis for such a sequence, i.e. $(m(n#)+k)$ for $2le kle n$.
I have searched MSE to see if this question has been asked before, and I did not find it (which surprised me a little): Is there an algorithmic way to generate or identify the first or smallest occurrence of a prime gap of size $(n-1)$?
prime-numbers prime-gaps goldbachs-conjecture
asked Jul 27 at 19:17
Keith Backman
39227
edited Jul 28 at 15:58
asked Jul 27 at 19:17
Keith Backman
39227
asked Jul 27 at 19:17
Keith Backman
39227
asked Jul 27 at 19:17
Keith Backman
39227
39227
2
no, it is an extremely difficult problem. See the table of maximal prime gaps at en.wikipedia.org/wiki/Prime_gap Evidently there is slow progress, they now have the first 78 maximal gaps. Last time I looked it was 75. They have (slightly) passed $10^19$
– Will Jagy
Jul 27 at 19:53
add a comment |Â
2
no, it is an extremely difficult problem. See the table of maximal prime gaps at en.wikipedia.org/wiki/Prime_gap Evidently there is slow progress, they now have the first 78 maximal gaps. Last time I looked it was 75. They have (slightly) passed $10^19$
– Will Jagy
Jul 27 at 19:53
2
2
no, it is an extremely difficult problem. See the table of maximal prime gaps at en.wikipedia.org/wiki/Prime_gap Evidently there is slow progress, they now have the first 78 maximal gaps. Last time I looked it was 75. They have (slightly) passed $10^19$
– Will Jagy
Jul 27 at 19:53
no, it is an extremely difficult problem. See the table of maximal prime gaps at en.wikipedia.org/wiki/Prime_gap Evidently there is slow progress, they now have the first 78 maximal gaps. Last time I looked it was 75. They have (slightly) passed $10^19$
– Will Jagy
Jul 27 at 19:53
add a comment |Â
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2
no, it is an extremely difficult problem. See the table of maximal prime gaps at en.wikipedia.org/wiki/Prime_gap Evidently there is slow progress, they now have the first 78 maximal gaps. Last time I looked it was 75. They have (slightly) passed $10^19$
– Will Jagy
Jul 27 at 19:53