## Prime k-tuplets and their Reverse Gap Patterns

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• Has anyone noticed that the recent Chermoni-Wroblewski 17-tuplet (message #22949) has this left to right gap pattern namely (4, 6, 2, 4, 6, 2, 6, 6, 4, 2, 4,
Message 1 of 7 , Aug 19, 2011
Has anyone noticed that the recent Chermoni-Wroblewski 17-tuplet (message #22949) has this left to right gap pattern namely (4, 6, 2, 4, 6, 2, 6, 6, 4, 2, 4, 6, 2, 6, 4, 2) and that this set of 17 consecutive primes namely, (17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83) has a left to right gap pattern (2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4) that is the reverse of the 17-tuplet pattern?

Seventeen consecutive primes starting with 17 and with the same 17-tuplet gaps in exact reverse order, rather a neat coincidence what? Furthermore, the leftmost 3 digits of the initial term of the 17-tuplet "102" and the rightmost 2 digits "17" are both divisible by 17. Ranaan and Jarek outdid themselves with this one!

Using Tony Forbes list of permissible prime k-tuple patterns, I found that each k up to and including 23 has a pattern that is the reverse of a pattern of some other set of k consecutive primes. But none with a prime k had a corresponding reverse pattern set whose initial term equaled k.

Does this reverse pattern correspondence hold true for any admissible prime k-tuplet? i.e. for any k>2, there exists a permissible prime k-tuplet and an associated set S of k consecutive odd positive primes, whose left to right gap pattern is the reverse of the left to right gap pattern of the tuplet.

Thanks folks.

Bill Sindelar

____________________________________________________________
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• ... The k-tuple conjecture is unproven so it s not known whether a set of primes actually exists for each admissible tuplet. It is well-known and easily shown
Message 2 of 7 , Aug 20, 2011
Bill Sindelar wrote:
> Does this reverse pattern correspondence hold true for any admissible prime
> k-tuplet? i.e. for any k>2, there exists a permissible prime k-tuplet and an
> associated set S of k consecutive odd positive primes, whose left to right
> gap pattern is the reverse of the left to right gap pattern of the tuplet.

The k-tuple conjecture is unproven so it's not known whether a set of
primes actually exists for each admissible tuplet.

It is well-known and easily shown that a gap pattern is admissible iff the
In particular, (a_1, a_2, ..., a_k) is an admissible tuple iff
(-a_k, ..., -a_2, -a_1) is admissible.
The latter has the reverse gap pattern.

Raanan Chermoni & Jaroslaw Wroblewski found two 27-digit 19-tuplets:
They both have the reverse gap pattern of the 19 primes from 13 to 89.
As far as I know, this is the largest known sets of consecutive primes
with reverse gap patterns.
The largest known sets with the same (not reversed) gap patterns is
probably still the 21 I found in 2006:

--
Jens Kruse Andersen
• Hi all, I just read that it is known that the series of reciprical odd perfect numbers converges, however it is not known if any odd perfect numbers exist. Is
Message 3 of 7 , Aug 22, 2011
Hi all,

I just read that it is known that the series of reciprical odd perfect
numbers converges, however it is not known if any odd perfect numbers exist.
Is the convergence proof easy enough to deliver to the list? If so, I'd
love to learn it. How about *even* perfect numbers? They gotta converge,
right?

Roahn
• ... See http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.0259v1.pdf for a stronger result. David
Message 4 of 7 , Aug 23, 2011
"Roahn Wynar" <rwynar@...> wrote:

> I just read that it is known that the series of
> reciprical odd perfect numbers converges

See
http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.0259v1.pdf
for a stronger result.

David
• ... Indeed, they do. Thanks to Euler, we know that all even perfect numbers are of the form (2^p-1)*2^(p-1). This allows very simple (and very crude) bounding
Message 5 of 7 , Aug 23, 2011
> How about *even* perfect numbers? They gotta converge, right?

Indeed, they do. Thanks to Euler, we know that all even perfect numbers
are of the form (2^p-1)*2^(p-1). This allows very simple (and very crude)
bounding (recipsum denoting the sum of reciprocals):

recipsum( (2^p-1)*2^(p-1) | p is prime) <=
recipsum( (2^n-1)*2^(n-1) | n >= 2) <=
recipsum( 2^(n-1)*2^(n-1) | n >= 2) =
recipsum( 4^n | n >= 1) = 1/3.

In fact, thanks to the rapid growth of the perfect numbers, one only needs
to take first few terms of the sum to obtain very good approximation of
its actual value:

0.2045201428389264301781344290984555766773...

Peter
• ... Off-list, Professor Dimiter Skordev asked me why I regard the latter result as stronger . My reply was as follows: The asymptotic number of perfect
Message 6 of 7 , Aug 24, 2011

> "Roahn Wynar" <rwynar@> wrote:
>
> > I just read that it is known that the series of
> > reciprical odd perfect numbers converges
>
> See
> http://arxiv.org/PS_cache/arxiv/pdf/1101/1101.0259v1.pdf
> for a stronger result.

Off-list, Professor Dimiter Skordev asked me why I regard
the latter result as "stronger". My reply was as follows:

The asymptotic number of perfect numbers less than x was proven in

MR0090600 (19,837d)
Hornfeck, Bernhard; Wirsing, Eduard
Über die Häufigkeit vollkommener Zahlen. (German)
Math. Ann. 133 (1957), 431-438.

to be less than x^eps for all eps>0.
From this it trivially follows that the sum of the reciprocals
of the perfect numbers is finite.

I regard Carl's result in

MR0618552 (82m:10012)
Pomerance, Carl
On the distribution of amicable numbers. II.
J. Reine Angew. Math. 325 (1981), 183-188.

as "stronger" since it proves the finiteness of
of the sum of the reciprocals of the amicable numbers.

This was, to my mind, a real "tour de force".

David
• ... The really tough question is whether the sum of the reciprocals of the sociable numbers is finite. One might guess so, from the opinions in
Message 7 of 7 , Aug 24, 2011

> The asymptotic number of perfect numbers less than x was proven in
>
> MR0090600 (19,837d)
> Hornfeck, Bernhard; Wirsing, Eduard
> Über die Häufigkeit vollkommener Zahlen. (German)
> Math. Ann. 133 (1957), 431-438.
>
> to be less than x^eps for all eps>0.
> From this it trivially follows that the sum of the reciprocals
> of the perfect numbers is finite.
>
> I regard Carl's result in
>
> MR0618552 (82m:10012)
> Pomerance, Carl
> On the distribution of amicable numbers. II.
> J. Reine Angew. Math. 325 (1981), 183-188.
>
> as "stronger" since it proves the finiteness of
> of the sum of the reciprocals of the amicable numbers.

The really tough question is whether the sum of the
reciprocals of the sociable numbers is finite.

One might guess so, from the opinions in
http://www.math.dartmouth.edu/~carlp/sociable.pdf
but there seems little hope of a proof,
given that we do not know whether an integer as small
as 276 is sociable.

David
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