A005002 inconsistent with wikipedia entry about Stirling numbers of the second kind?
Olivier Gerard
olivier.gerard at gmail.com
Wed Jul 16 09:52:00 CEST 2008
On Tue, Jul 15, 2008 at 22:18, N. J. A. Sloane <njas at research.att.com> wrote:
> Maximilian said:
>
> (Anyway, it is not nice that
> reading the referenced paper would be necessary to know the definition
> of the sequence.)
>
As one of the implied principles of the OEIS could be phrased:
First needed, First edited (or commented, or extended, or illustrated).
> njas: you are forgetting what the OEIS is for: if you come across
> njas: a sequence in your work, the OEIS tells you where it
> njas: has appeared in the literatue.
>
> njas: It is certainly true that many entries do not have a
> njas: self-contained definition. That will have to wait
> njas: until I have a larger staff to help.
>
og: Neil, you have already seqfans (about 200 subscribers) !
One of the main uses of the seqfan mailing list is precisely to
allow collaborative enhancements about sequences one comes
accross in one's work and removes obscurities or ambiguities.
It has precisely worked that way for this case about Rhyming Schemes.
A suggestion I could make to rapid-fire submitters :
You want your sequences in ? Good ! Work at
enhancing the OEIS by looking at references and add or cross-check
definitions, links and meaningful examples in already existing sequences.
It should be at least half of your mail exchange with Neil,
and through the comment interface.
Neil: in the same spirit as the "more" keywords, you could
perhaps create one or several keywords to signal older sequences
needing examples or self-contained definitions (this would be
different from the "uned" keyword). This would help focusing teamwork
on these eventually.
Olivier
At 9:41 PM -0700 7/15/08, Jack Brennen wrote:
>To determine if an odd number N is flimsy, take the finite set of residues
>of 2^a (mod N).
>Assume that the number of 1's in the binary representation of N is equal to C.
>
>To show that the number is flimsy, find a way to construct zero by adding
>up some number
>of residues of 2^a (mod N) using less than C terms. To show that the
>number is sturdy,
>show that it's impossible to do so.
>
>For instance, for the number 37, which has 3 ones in its binary
>representation, the
>set of residues of 2^a (mod 37) is actually all of the numbers except
>zero. Clearly
>we can find two that sum to zero. :) So 37 is flimsy.
>
>On the other hand, for the number 15, which has 4 ones in its binary
>representation,
>the set of residues of 2^a (mod 15) is: 1, 2, 4, 8. It is impossible to
>sum 2 or 3
>of these to add up to zero.
I haven't been following this thread too closely. Has someone noted that
the current sequence is missing 37, 74 (because it is 2*37), 97, and 101?
97 has three bits and 97*172961 has two bits. 101 has four bits and
101*365 has three bits (and 101*11147523830125 has two bits!).
Tony
At 9:41 PM -0700 7/15/08, Jack Brennen wrote:
>To determine if an odd number N is flimsy, take the finite set of residues
>of 2^a (mod N).
>Assume that the number of 1's in the binary representation of N is equal to C.
>
>To show that the number is flimsy, find a way to construct zero by adding
>up some number
>of residues of 2^a (mod N) using less than C terms. To show that the
>number is sturdy,
>show that it's impossible to do so.
>
>For instance, for the number 37, which has 3 ones in its binary
>representation, the
>set of residues of 2^a (mod 37) is actually all of the numbers except
>zero. Clearly
>we can find two that sum to zero. :) So 37 is flimsy.
>
>On the other hand, for the number 15, which has 4 ones in its binary
>representation,
>the set of residues of 2^a (mod 15) is: 1, 2, 4, 8. It is impossible to
>sum 2 or 3
>of these to add up to zero.
I haven't been following this thread too closely. Has someone noted that
the current sequence is missing 37, 74 (because it is 2*37), 97, and 101?
97 has three bits and 97*172961 has two bits. 101 has four bits and
101*365 has three bits (and 101*11147523830125 has two bits!).
And 67 belongs to this sequence because 67*128207979 has two bits.
And 81 belongs because 81*1657009 has two bits.
Tony
At 9:41 PM -0700 7/15/08, Jack Brennen wrote:
>To determine if an odd number N is flimsy, take the finite set of residues
>of 2^a (mod N).
>Assume that the number of 1's in the binary representation of N is equal to C.
>
>To show that the number is flimsy, find a way to construct zero by adding
>up some number
>of residues of 2^a (mod N) using less than C terms. To show that the
>number is sturdy,
>show that it's impossible to do so.
>
>For instance, for the number 37, which has 3 ones in its binary
>representation, the
>set of residues of 2^a (mod 37) is actually all of the numbers except
>zero. Clearly
>we can find two that sum to zero. :) So 37 is flimsy.
>
>On the other hand, for the number 15, which has 4 ones in its binary
>representation,
>the set of residues of 2^a (mod 15) is: 1, 2, 4, 8. It is impossible to
>sum 2 or 3
>of these to add up to zero.
I haven't been following this thread too closely. Has someone noted that
the current sequence is missing 37, 74 (because it is 2*37), 97, and 101?
97 has three bits and 97*172961 has two bits. 101 has four bits and
101*365 has three bits (and 101*11147523830125 has two bits!).
And 67 belongs to this sequence because 67*128207979 has two bits.
Tony
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