This site is supported by donations to The OEIS Foundation.

Hello!

# Boundary values set right

In the section "A markup style study" of my user page I wrote:

"If some boundary value of a formula f(n) (for example for n=0) does not give an integer value (for example (1/2)) it is much more sensible to replace the formula by ceil(f(n)) or floor(f(n)) then to decapitate the sequence."

Below is a response of Jaume Oliver i Lafont to this remark.

Another trick to avoid non-integer boundary values.

At the very bottom of your user page on the OEIS wiki you suggest rounding upwards or downwards in order to avoid non-integer values at say a(0).

If that non-int value is a rational a/b, an interesting alternative is considering b*f(n), instead of ceil(f(n)) or floor(f(n)). Following your example, if a(0) is 1/2, we can consider the integer sequence 2*a(n), which will have initial value 1.

This keeps all the original information into the "integerized" sequence. For example, two sequences a(n) and b(n) that are identical for n>0 but differ only at non-integer values a(0) != b(0), will be still different after this (linear) transformation, while rounding (which is non-linear) might force them to appear equal.

Jaume Oliver i Lafont

Thank you Jaume Oliver! Clearly every such method is a substitute. The best way is of course to look at the meaning of the sequence and to embed the sequence in a broader setting.

A famous example for this are the Bernoulli number. Originally they were defined by Bernoulli only for n = 2, 4, 6,... (this enumeration is called the "archaic" on Weisstein's "Bernoulli Number" page at MathWorld). For odd n > 1 it is clear that Bn = 0. But what are the Bernoulli numbers for n = 0 and n = 1?

For all integer n > 1 Bn equals limx → n −xζ(1 − x). So let's check if this limit exists also for    n = 0 and n = 1. Well, it does. So isn't it then the most natural way to define B0 and B1 in this manner, Bn uniformly for all n? I think clearly it is. The Bernoulli number are the children of the zeta function!

Regrettably many mathematicians do follow this simple and coherent extension even at the price to be inconsistent with their own conceptual framework. For example Donald Knuth in his (et alia) Concrete Mathematics defines the Stirling polynomials in formula (6.52) giving rise to the combinatorial identity Bn = n! σn(1) for n ≥ 0 if this natural extension is chosen.

On the other hand Knuth does not follow his own framework and defines later B1 = −1/2 which looks completely absurd to me and breaks the simple connection with the Stirling polynomials.

Peter Luschny 23:33, 12 January 2010 (UTC)

In the gallery of new files Special:NewFiles, I see that you are making nifty figurate numbers images. I guess you are planning to insert them in the figurate numbers pages, should I just let you do it or do you want me to insert them?

and are both missing, by the way...

Thanks Daniel Forgues 20:49, 30 July 2010 (UTC)

Dear Daniel, I would not dare to add anything to your Gesamtkunstwerk. I use the images on my user page http://oeis.org/wiki/User:Peter_Luschny/FigurateNumber

Yes, there is something wrong with S16, it is located here but not accesible. I already informed David Applegate. If you like to insert them in the figurate numbers pages just go ahead.

Cheers Peter Luschny 22:26, 30 July 2010 (UTC)

## Notation matters

I very much enjoyed your 'column' "Notation Matters." For some strange reason, to me it kind of feels satisfying to be rebuked for obscure notation; after a long, busy day, it actually brightened my day. And obviously I am happy that it led to the discovery of a sequence that was inexplicably absent from the table and which you have made explicably present. I'm thinking it should be the Sequence of the Day a few days after it appears in the OEIS. Alonso del Arte 03:20, 5 November 2010 (UTC)

Just now realized this: My notation was not only "obscure," "cumbersome" and "not optimal," it was WRONG! When n is composite, the Kronecker delta doesn't just zero out an iteration through a divisor of n, but the entire product. This is obviously wrong and had to be changed immediately. However, I'm not ready to accept Knuth's notation for coprime, as the symbol seems rather overloaded. Thank you for allowing me to come to the realization of my error on my own. Alonso del Arte 05:31, 5 November 2010 (UTC)

### Dear Alonso,

it is very common in mathematics that notations are overloaded. To say 'a is prime to b' looks to me as a good number-theoretic overload for the geometrical 'a is orthogonal to b'.

And clearly the first and foremost value of a good notation is to reduce the chance of making errors. If it happened that I did not make an error in my formulas than because of this: I had not to think about them, I had just to put three symbols under the sum-sign.

Cheers Peter Peter Luschny 12:38, 5 November 2010 (UTC)

I created the page Orderings on which the different types of orderings appear as sections (if need be, each type of ordering could have their own page.)

It seems that we have all the following orderings, am I right?

Daniel Forgues 03:44, 13 January 2011 (UTC) — Daniel Forgues 04:33, 13 January 2011 (UTC)

#### Peter's respons: About encyclopaedism, the 'A-St-order' and the rules of the road.

Well, my sympathy for encyclopaedism is bounded. I think it is important to make the systematics clear; however individual notions should only be introduced on demand.

The above names are not very good as there are two different kinds of orders involved: the internal (a<b<c) versus (a>b>c)) and the external (a<b<c) < (a<c<b) versus (a<b<c) > (a<c<b). I do not know better ones, though.

The linguistic side is even more complicated as the names refer to relations which are relative. So first you have to define a reference point like 'lexicographic order'. Now it is clear what the 'reverse lexicographic order' is. However the 'reflected' in 'reflected lexicographic order' refers to the internal order. So what is the reference point here? Par ordre de Mufti MacMahon, Andrews and Knuth it is the descending order. This has to be fixed once and for all times like the sense of revolution: clockwise or counterclockwise; or like the rules of the road: right-hand traffic or left-hand-traffic.

So what K. Goldberg, M. Newman and E. Haynsworth have written in the chapter 'Combinatorial Analysis' of the blue HMF is a flagrant misuse of mathematical traffic rules: What they tabulate there on page 831 are no integer partitions. However, this conceptual crash was not stopped by the editors in charge Abramowitz and Stegun. Therefore this order is named after them: this is the infamous 'Abramowitz-Stegun-order'. (Or even 'A-St-order'; yes, even this inept name can be found 61 times on OEIS.)

Unfortunately the name 'Abramowitz-Stegun-order' is very popular on OEIS. Why unfortunately? Am I nitpicking? Because I have experienced the following respons: "Reflect the 'A-St-partitions' and you have the partitions in lexicographic order." However, this is wrong. The 'Abramowitz-Stegun-order' is not based on the lexicographic order; it is based on the colexicographic order. I believe this misunderstanding is promoted by the use of this ad hoc name. If the order would be called 'reflected colexicographic order' such misunderstandings would be obviated at the outset. And save generations of OEIS users the time to search for the exact meaning of 'A-St-order'. Peter Luschny 14:46, 13 January 2011 (UTC)

In Partitions#A comparison, there is now side by side your original table with 6 kinds of orderings and to the right a table including all 8 orderings. I would like to have this ordering comparison table on Orderings#A comparison and to have a link to it from Partitions#A comparison instead of having the table there. Do you agree in having the table with all 8 orderings (for the sake of completeness) on Orderings#A comparison? Since you did the bulk of the work on it (my adding the 2 columns was the easy part) you should get the credit for it in the history of orderings, while if I do it myself I could only credit it to you in a comment. — Daniel Forgues 04:28, 14 January 2011 (UTC)
Dear Daniel, just go ahead and do whatever you think is appropriate; all you have to fear is being mercilessly edited by jj :). When I write on the official wiki, which is collaborative writing, I do not want any 'credit'; and, by the way, I do not give any personal credit though I do give the reference. Only if something is cited from my user pages I would be pleased to see it handled according to the standards of academic behaviour. On the other hand I saw quite a few entries on the official pages of the wiki which are very subjective and would much better be placed in a personal blog or on a user page. Peter Luschny 15:54, 15 January 2011 (UTC)
I'll then copy the comparison table of orderings to Orderings and replace the table on Partitions with a link to the table on page Orderings. — Daniel Forgues 02:43, 16 January 2011 (UTC)

## "Abramowitz and Stegun" ordering of the partitions

I put back (as you originally said in your table, looking at the history) that it is the graded reflected colexicographic ordering of the partitions, I think it is right. (at some point I thought it might be the graded reverse reflected lexicographic ordering of the partitions or the reverse reflected colexicographic ordering of the partitions). — Daniel Forgues 01:35, 19 January 2011 (UTC)

What is confusing is that the partitions don't have the same number of parts (or equivalently have different number of parts of size 0, which are omitted) while in your comparison table every element has 4 parts. After having spent a huge amount of time on orderings of partitions and orderings of prime signatures, I still get confused about the orderings when the number of parts of the elements varies. I would appreciate if you could modify your comparison table so as to consider elements with different number of parts (or different number of parts of size 0, not omitted.)

P.S. I tried to rename (using the standard terminology)

A118914 Concatenation of the prime signatures (in sorted lists of exponents of distinct prime factors) of the positive integers.

as

Concatenation of the prime signatures (in lexicographic order of exponents of prime power components) of the positive integers.

but it was rejected.

As per your table it should be (Comment: The exponents can be read off Abramowitz and Stegun, p. 831, column labeled "pi".)

A036035 Least integer of each prime signature, in graded reflected colexicographic order of exponents.

The approved version as of now is (I made many flip-flopping edits between different ordering names, not good...):

A036035 Least integer of each prime signature, in graded reverse reflected colexicographic order of exponents.

So, which one is right?

Prime signatures have elements with different number of parts (or equivalently different number of parts of size 0) so the same confusion occurs... since your table always have elements with 4 parts.

It is confusing for me, maybe other people do not see it as confusing...? Those numerous ordering conventions are tedious to deal with...

I could try to modify your table myself to account for different number of part of size 0, problem is that I might mess things up since I still get confused, so I'd rather abstain to do so. — Daniel Forgues 02:17, 19 January 2011 (UTC)

#### re:

Dear Daniel,

"It is confusing [...] Those numerous ordering conventions are tedious to deal with... ". Yes, I agree, all that can be very confusing. Order and disorder is closely related. However, all I wanted to say I have already said. It is summarized in the three lines which I put on the partition page.

 Ref Lex reflected lexicographic A080576 "Maple" ordering of the partitions Rev Lex reverse lexicographic A080577 "Mathematica" ordering of the partitions Ref CoLex reflected colexicographic A036036 "Abramowitz and Stegun" ordering of the partitions

These cases cover, I think, 80% or more of what is referred to in the database. Clearly you can write big books on partially ordered sets and on lattice theory. However, I will leave this to the expert.

So what can I do for you? Looking at [1] and [2] I see that Matt Insall seems to be such an expert. Write him an email and invite him to our discussion group. This seems to be a good place to pursuit this topic further. Peter Luschny 09:23, 19 January 2011 (UTC)

I fixed your link to seqcomp (removed the |). — Daniel Forgues 04:31, 20 January 2011 (UTC)
I made a better comparison table (involving elements with different number of parts, using parts valued 0) and now I see things very clearly! I verified and corrected the related sequences. — Daniel Forgues 07:52, 20 January 2011 (UTC)

Caterpillars and "Peter's ordering". That's nice and helpful! I like it. Yesterday, when I was contemplating this subject again, I realized for the first time that I am using for many years a completely different ordering which is not even in your extended table (and which I also could not find elsewhere). I came up with this ordering the very first time I learned about integer partitions; I used it ever since because it enables me to quickly write down the partitions for small n with paper and pencil without pain. And in fact I also made use of it on my blog here before ( the caterpillar notation), again without realizing that I was using a fameless ordering. And a nameless. I will now formalize the definition and add it as a sequence to the OEIS. Juxtaposing it to the 'Maple', 'Mathematica' and 'A-St' orderings. Perhaps I will name it 'Peter's ordering'? I am quite sure that this time the name would be rejected by the editors :) As John C. Baez said: "... mathematical discoveries are never named after the people who made them." Neither Abramowitz nor Stegun ever wrote a single line about orderings of partitions. However, I am also sure that you will quickly find the systematic name. Peter Luschny 11:40, 20 January 2011 (UTC)

Your caterpillar notation is on your page of permutation trees, and is a notation for flattened trees. Are there many variants of your caterpillar notation. You are saying that your caterpillar notation corresponds to an ordering of partitions, a partition corresponding to a flattened tree, if I understand? I didn't figure things out yet. If it is another ordering of partitions, then please add it to the page Orderings of partitions. If it is an ordering of permutations or rooted trees, then it would go on the Permutations or Rooted trees page. — Daniel Forgues 03:56, 22 January 2011 (UTC)

Does a very hungry caterpillar eat integer partitions? The caterpillar notation is a notation related to binary trees. Here is another recent application inspired by a discussion on group/seqcomp on double enumerations: tree-traversal and displacement. The ordering of integer partitions related to this idea is now described on the page "Integer Partition Trees" as the flattening of the Fenner-Loizou tree of type [VRL, decreasing, forward] and will hopefully become "Peter's ordering" A182937. If we take the Fenner-Loizou tree as a canonical reference point "Peter's ordering" is dual to the "lexicographic ordering" in a very basic way, like the sense of revolution 'clockwise' is dual to 'counterclockwise', and refers to the sense of traversing the periphery of the tree. To apply or not apply the caterpillar notation in this case is a matter of convenience. It provides more information than the standard notation for total orderings as it refers to a partial ordering no longer seen after 'flattening'. Not everything I say on the page 'integer partition trees' will fit one-to-one into the nomenclature you give on other pages.Peter Luschny 15:58, 22 January 2011 (UTC)

## New sequence

Hello Peter. In my work on restricted partitions and compositions of sets are treated eight sequences one of them is missing from OEIS. Please go to my talk page--Adi Dani 00:06, 25 April 2011 (UTC)

# 1

Dear Peter after several attempts, following your instructions i allocate a new sequence in OEIS if you find a little time please comment on my sequence because I think that you are very competent in this area. Sequence take number A189886.

# 2

Dear Adi, I tried to write your formula as a Maple function and recieved the message: "Error, the argument to factorial should be non-negative". How did you compute your values? Then I tried Wolframs Alpha with your Mathematica code; for some reason Alpha failed to give me an answer. So I am in the moment stuck in reading of your formula properly. Did you check the hint that factorials should not be negative?

# 3

I use ,,Mathematica,, When i write formula using symbol ,,sigma,, formula works correctly and when i use ,,sum,, i get several error messages but finally i correct the formula and formula works. Try write formula with ,,sigma,, symbol. --Adi Dani 06:19, 30 April 2011 (UTC)

# 4

0+1/3*2!*1!*(-1)!+1/2*2!*0!^2+4/9*2!*(-4)!*4!+2/3*1!*(-2)!*3!+2!*0!^2+3/2*2!*1!*(-1)!+9/4*(-2)1*4!*0!

You see that this includes the factors (-1)!, (-4)!, (-2)!, (-1)! and (-2)!. As long as there are negative factorials this formula will not work.

# 5

I have done the formula on Mathematica program

Table[Sum[k!/(2^(k+j-2m)3^(m-j))m!/(j!(m-j)!)*j!/((k+2j-3m)!(3m-j-k)!),{m,0,k},{j,0,3m-k}],{k,0,20}]


i hop that will works--Adi Dani 08:57, 30 April 2011 (UTC)

# 6

Dear Adi, I do not see that you showed an interest in the problem of negative factorials. Did you trace your formula? Did you look at the explicit form for k=2 which I provided above? If you want to discuss your work with me this does only make sense if you answer to my observations; if not, it is wasted time and we should stop our conversation here.

# 7

Yes! I want to discuss now, I see that something interesting happen here From formula

${\displaystyle {\overline {c}}(k,N_{3})=\sum _{m=\lfloor k/3\rfloor }^{k}\sum _{i=0}^{3m-k}{\frac {m!~k!}{(m-i)!~(k+2i-3m)!~(3m-i-k)!~2^{k+i-2m}~3^{m-i}}}.\,}$

in case k=2 we gate

${\displaystyle {\overline {c}}(2,N_{3})=\sum _{m=\lfloor 2/3\rfloor }^{2}\sum _{i=0}^{3m-2}{\frac {m!~2!}{(m-i)!~(2+2i-3m)!~(3m-i-2)!~2^{2+i-2m}~3^{m-i}}}=\sum _{m=0}^{2}\sum _{i=0}^{3m-2}{\frac {m!~2!}{(m-i)!~(2+2i-3m)!~(3m-i-2)!~2^{2+i-2m}~3^{m-i}}}=\,}$
${\displaystyle \sum _{i=0}^{-2}{\frac {0!2!}{(0-i)!(2+2i)!(-i)!2^{2+i}~3^{-i}}}+\sum _{i=0}^{1}{\frac {2!}{(1-i)!~(2i-1)!~(1-i)!~2^{0}~3^{1-i}}}+}$
${\displaystyle +\sum _{i=0}^{4}{\frac {2!2!}{(2-i)!~(2i-4)!~(4-i)!~2^{i-2}~3^{2-i}}}}$

we can see that there appear negative factorial. How to explain that!! Your question is good. Is needed explanation, for the moment I can't explain.--Adi Dani 13:51, 30 April 2011 (UTC)

# 8

Here is explanation

${\displaystyle {\overline {c}}(2,N_{3})=\sum _{m=0}^{2}\sum _{i=0}^{m}{\binom {m}{i}}{\binom {i}{2+2i-3m}}{\frac {2!}{2^{2+i-2m}3^{m-i}}}\,}$
${\displaystyle \sum _{i=0}^{0}{\binom {0}{i}}{\binom {i}{2+2i}}{\frac {2!}{2^{2+i}3^{-i}}}={\binom {0}{0}}{\binom {0}{2}}{\frac {2!}{2^{2}3^{0}}}=0\,}$
${\displaystyle \sum _{i=0}^{1}{\binom {1}{i}}{\binom {i}{2i-1}}{\frac {2!}{2^{i}3^{1-i}}}={\binom {1}{0}}{\binom {0}{-1}}{\frac {2!}{2^{0}3^{1}}}+{\binom {1}{1}}{\binom {1}{1}}{\frac {2!}{2^{1}3^{0}}}=0+1=1\,}$
${\displaystyle \sum _{i=0}^{2}{\binom {2}{i}}{\binom {i}{2i-4}}{\frac {2!}{2^{i-2}3^{2-i}}}={\binom {2}{0}}{\binom {0}{-4}}{\frac {2!}{2^{-2}3^{2}}}+{\binom {2}{1}}{\binom {1}{-2}}{\frac {2!}{2^{-1}3^{1}}}+{\binom {2}{2}}{\binom {2}{0}}{\frac {2!}{2^{0}3^{0}}}=0+0+2=2\,}$

so ${\displaystyle {\overline {c}}(2,N_{3})=0+1+2=3}$ because ${\displaystyle {\binom {m}{k}}=0}$ for ${\displaystyle k>m\,}$--Adi Dani 15:47, 30 April 2011 (UTC)

# 9

Yes, this is the right direction. I told you this already on April 25 on the discussion page of 'set compositions', in my comment on your formula of the number of set compositions over set N2

${\displaystyle {\overline {c}}_{m}(k,N_{2})={\frac {m!k!}{(2m-k)!(k-m)!2^{k-m}}}}$

I wrote:

This formula has a problem: it is valid only for k = m..2m.
(Otherwise a factorial of a negative integer appears).
And this should be indicated! Therefore it is more advantageous to write

${\displaystyle {\overline {c}}_{m}(k,N_{2})=2^{m-k}k!{\binom {m}{2m-k}}}$
Similar remarks apply to the other formulas.


## Proposed sequence: Subgroups of powers of Z2

Hi,
as you recommended, I've created this description of the sequence I'd like to propose for admission to OEIS:
User:Tilman Piesk/Subgroups of powers of Z2
Greetings, Tilman Piesk 19:25, 19 May 2011 (UTC)

# 1

Hi Tilman, this looks interesting. So the next step is to bring things into the form required for a submission to OEIS.

The submission form looks like this: https://sites.google.com/site/seqcomp/seqsupp/oeissubmissionform

Now you can reflect on the entries of this form. I will help you if you want.

Let us look for instance at this sequence:

1, (until here for Z20) 3, (until here for Z21) 5, 9, 15, (until here for Z22) 17, 33, 51, 65, 85, 105, 129, 153, 165, 195, 255, (until here for Z23)

The numbers go into the DATA section. The hints say: Entries usually give at most 200 characters. This means that you should enter there the following segment:

1, 3, 5, 9, 15, 17, 33, 51, 65, 85, 105, 129, 153, 165, 195, 255, 257, 513, 771, 1025, 1285, 1545, 2049, 2313, 2565, 3075, 3855, 4097, 4369, 4641, 5185, 6273, 8193, 8481, 8721, 9345, 10305, 12291, 13107, 15555, 16385, 16705, 17025, 17425, 18465, 20485, 21845

The sequence is a list, so the OFFSET is 1.

You need only one KEYWORD: tabf, which means that it is a triangle of irregular shape. This by the way suggests another sequence. How long are the rows? As far as I can see this sequence starts 1,1,3,11,51,...

So next consider the other entries. Take your time. You can post your concept here, if you want, and I will discuss it with you. Cheers, Peter Luschny 00:31, 20 May 2011 (UTC)

Done so far: User:Tilman Piesk/Subgroups of powers of Z2/submission form
I'm not sure, if "tabf" is appropriate here, because the lines in this triangle wouldn't make sense on their own - as e.g. the lines in A022166 do.
Greetings, Tilman Piesk 22:15, 21 May 2011 (UTC)
Great! I suggest to continue this discussion on the discussion page of your submission concept. This also might invite comments from other users. Peter Luschny 08:24, 22 May 2011 (UTC)

## Seq. o' th' Day for July 15

Peter, could you look over Template:Sequence of the Day for July 15 (regarding phi-torials) and if it meets with your approval, mark it approved? Alonso del Arte 23:04, 10 July 2011 (UTC)

Thanks very much, Peter. Alonso del Arte 14:40, 14 July 2011 (UTC)

## Pochhammer symbols in sequence entries

Peter, how would you recommend notating Pochhammer symbols in OEIS sequence entries? Alonso del Arte 05:10, 5 October 2011 (UTC) P.S. It is specifically A196305 (draft) that has me wondering about this. Alonso del Arte 05:13, 5 October 2011 (UTC)

Unfortunately the 'pochhammer symbol' is used differently by different authors (including Pochhammer himself, who used his symbol to denote the binomial coefficient). So the best thing is not to use it all! There are good replacements; as always (with the notable exception of B_{1} = bernoulli(1)) I follow the advice of Don Knuth. Use the names falling and rising factorial powers instead. This disambiguates things and additionally give them more descriptive names.

Now the question is how to write falling and rising factorial powers with the typewriter :-) I have to confess that I have never done this.

If someone forces me to write with a typewriter I would not try to use a special symbol, but write something like

Let r_n(x) = x*(x+1)*...*(x+n-1) then etc...

Let f_n(x) = x*(x-1)*...*(x-n+1) then etc...

Often you can avoid all this; as in the case A196305: A much better way to write this name/formula simply is, with offset 0:

a(n) = 2^n*product(k-13,k=1..n)/n!

Not that thrilling this formula anyway. Peter Luschny 09:30, 5 October 2011 (UTC) [edited]

Thank you very much, Peter. I have told Alex and hopefully he will decide to eschew the confusing Pochhammer notations.
Maybe no one will ever force you to write about math on a typewriter, but there are sequences you've added to the OEIS containing things like sums, products and I think maybe even integrals. Alonso del Arte 18:06, 5 October 2011 (UTC)

## Brauche etwas Hilfe bei A195665

Hallo Peter,
vielleicht kannst du mir mal ein paar Hinweise zu A195665 geben. T.D.Noe schreibt immer nur, ich solle mir andere Folgen ansehen, ich bin mir aber nicht sicher, welche Schlüsse ich aus denen ziehen soll. Am ehesten Information entfernen, aber das ginge imho sehr auf Kosten der Verständlichkeit. Ich sehe auch den Unterschied zu meiner Folge A195467 nicht wirklich, die akzeptiert wurde. Mit A195663 bzw. A195664 habe ich das gleiche Problem. Grüße, Tilman Piesk 23:42, 17 October 2011 (UTC)

Hello Tilman, it is not easy for me to give advice. Perhaps: Reduce the information. Give a short but formal description, add a program. Don't try to explain too much. I would replace the examples by the sequence in table form. Delete the tables in the comments. Reduce the links to one or at most two. Try to write a short program. Sage is a very good free CAS and uses Phyton which is well suited to formalize your sequences. Peter Luschny 00:32, 18 October 2011 (UTC)

## Set Partitions

Hi. Maybe you want to take a look at this enumeration of partitions, which doesn't depend on the number of elements in the set: File:Set partitions 5; lexicographical.svg Do you know if there is any other (possibly better) way to enumerate the infinite set of finite set partitions? Frohe Restfeiertage, Tilman Piesk 02:04, 29 December 2011 (UTC)

Hello Tilman! I transferred your message from 'User talk:Peter Luschny/SetPartitions' back here. I watch only this page here so it is better to leave a notice here. Now to your question: Your enumeration of set partitions is nice. What could a 'better way' to enumerate be? This depends on what use you have in mind. Certainly there are other representations which have their good use. For instance I like to look at them in the following way:
Here the set partitions are divided into those with a free column (partitions with singletons) and those with only joined columns (multiton partitions). Thus the set partitions are the disjoint union (or direct sum) of partitions with singletons and multiton partitions. More on this you can find on my blog on SetPartitions. With regard to your plot I would certainly like to see also a column with the representations using Genji-Kô symbols. This would make it easier to bridge between your representation and more classical ones. Peter Luschny 20:55, 1 January 2012 (UTC)

Hi.

Greetings, Tilman Piesk 15:24, 3 January 2012 (UTC)

Well, I cannot point you to more enumerations than you can find on my user page. But why not ask your question on the seqfan list? You might get a good answer there. http://list.seqfan.eu/pipermail/seqfan/
"I don't get why you enumerate the bars of the Genji signs from right to left." I do not think this makes a significant difference. I just started to paint the signs and they turned out this way ;-) Peter Luschny 19:36, 3 January 2012 (UTC)

Tilman Piesk 23:17, 15 January 2012 (UTC)

(1) Tilman, I propose that you write a user page here on this wiki, User : Tilman_Piesk / SetPartitions, and present your nice diagrams on this page together with a description. Then you can link to that page from A000296 or from whatever seq which is appropriate. And I will link from my blog to this page. (2) I do not know. A question you could pose on the seqfan-list or on http://math.stackexchange.com (3) Yes, I did not notice your comment on nimbers. I will look into it. Peter Luschny 00:23, 16 January 2012 (UTC)

## 0^0

Maybe in Germany most non-science people know better than to say 0^0 = 0, but here in America you'll find lots of people who think that. That was a big part of the impetus for me to even start the page in the first place. I had asked a couple of people back then, but last week I went around asking a lot of people this question, and my suspicion was confirmed. Alonso del Arte 00:23, 24 January 2012 (UTC)

Alonso, I believe that when you ask the man on the street (in what country so ever) what 0^0 or 00 means, they will think this is an emoticon or something like that. Even when they recognize the power function they will have not the slightest idea. Most of them probably will not say 'I do not know' but they will give some guess. And this guess will be based on the totally irrelevant optical impression which the symbols 0^0 or 00 make. So I think your naïve question will not measure anything related to mathematics but only some sort of physiological/psychological reactions. If you want to find out the mathematical knowledge of some people with regard to the value of the power function at (0,0) you better ask: "What is the value of exp(s log z) when s=0 and z=0?" Peter Luschny 11:50, 24 January 2012 (UTC)

Peter, now I'm thinking that in Olivier's shoes I would also ban this topic from SeqFan just as he has. Alonso del Arte 15:56, 24 January 2012 (UTC)

## e in the Pascal Triangle

e in the Pascal Triangle

it would be great to add it to Pascal's triangle! — Daniel Forgues 02:02, 10 May 2013 (UTC)

It's nice, isn't it? It is entertaining, but it is not really surprising nor a deep result. But sure, it can be added to the wiki page. Why not just go ahead Daniel? By the way, here is OEIS on G+. Peter Luschny 06:04, 10 May 2013 (UTC)
I guess it's not surprising to find e in Pascal's triangle, since the binomial distribution is asymptotic to the Gaussian distribution. — Daniel Forgues 06:36, 10 May 2013 (UTC)

## Centralized Bernoulli

On your page https://oeis.org/wiki/User:Peter_Luschny/CentralizedBernoulli , after the code:

A064538 := n ->
denom((bernoulli(n+1,x+1) − bernoulli(n+1,1))/(n+1)):


you say, "Now observe that the quotient (n+1) can be pulled out of the denominator becoming a factor." and you give the code:

A064538 := n ->
(n+1)*denom(bernoulli(n+1,x+1) - bernoulli(n+1,1)):


You need to justify this step in this case, because it is not true in general that denom((a/b)/c) = c*denom(a/b) for integers a, b, c > 0. For example, 3 = denom(1/3) = denom((2/3)/2) != 2*denom(2/3) = 6. Thanks. - Jonathan Sondow 20:19, 18 November 2015 (UTC)

## Drafts A316349, A316387

Dear Dr. Luschny, where are you ? I think all OEIS is missing you :) In this message i'd like to invite you for edits or comments on sequences A316349, A316387. Note that this is not question to approve them faster, as clear invitation since your comments and suggestion are appreciated :)

Thank you, regards

Kolosov Petro (talk) 19:47, 6 July 2018 (EDT)

## Sparse rulers to length 10501

I've started https://oeis.org/A309407 and https://oeis.org/draft/A326499 and updated https://oeis.org/A046693 . I can send you sparse rulers to length 10501. A046693(n)-A309407(n) is 0 or 1 up to at least n=10567 if your conjecture is correct.

[This was written by Ed Pegg Jr, September 12, 2019 but not signed.]

## A242431

Hello. I have been looking at triangle A242431, and I have a question about the definition, which reads:

Triangle read by rows: T(n, k) = (k+1)*T(n-1, k) + sum(j=k..n, T(n-1, j)) for k < n, T(n, n) = 1; n>=0, 0<=k<=n.

The summation index j runs from k to n, so that its final iteration is T(n-1, n). However, T(n,k) is not expressly defined for k > n. The Maple and Sage code on the sequence page gets around this by setting T(n,k) to 0 if k>n, but the "official" terms of the sequence do not include any 0's. I wonder if it would be better to change "j=k..n" in the definition to "j=k..n-1".

Here are some formulas I found for A242431:

T(n,k) = 1 + Sum_{i = k+1..n} i*(i+1)^(n-i) for 0 <= k <= n.

T(n,k) = T(n,k+1) + (k+2)*(T(n-1,k) - T(n-1,k+1)) for 0 <= k <= n-2.

T(n,k) = T(n,k+1) + (k+1)*(k+2)^(n-k-1) for 0 <= k < n.

The closed formula could also be used to simplify the code at A047970:

A047970(n) = 1 + Sum_{i = 1..n} i*(i+1)^(n-i)