The correct response of the structure of a skyscraper to external stresses such as wind, depends on the rigidity of the pillars and their distribution on the basis of the skyscraper.

Well, from the analogies demonstrated with the 2-symmetric distribution we know that the maximum stiffness is at the tails of the permutations distribution, while the 2-symmetric distribution is in the center of the distribution and is above all if we are looking for the 2-supersymmetric case I talked in the 16 dec post.

The correct response to wind stress requires that the load-bearing pillars possess both characteristics, each of which has a very low probability of being obtained without design efforts. ]]>

could call the Identity permutation and its reverse as the configurations with maximum stiffness.

From the bell shaped distribution we know that the frequencies distribution on the extreme tail of the curve

are very low if compared to the center near to the vertex.

This is a good lesson because says that obtaining high stiffness does not arrive accidentally but

need geat efforts in search for proper design and calculations.

There are very important case of errors in stiffness calculation, one of them ia the Citicorp skyscraper the forth in

height in New York City which suffered from such a problem. ]]>

– The lines of Mac Mahon triangle can be thought as a frequency distribution of permutations.

– For Mac Mahon about one century ago, these numbers indicated the permutations with identical number of inversions,

we talk of subsets of size 2 order isomorphic to 12 and 21, the two definitions are equivalent. (need demonstration).

– From a statistics point of view we can see the Mac Mahon triangle with n growing as a new distribution I call it

“Mac Mahon distribution” in analogy to binomial and standard distribution.

– I conjecture that this distribution for n going to infinity approach the standard distribution.

– I have started investigation to extend these concept to 3-symmetric and k-symmetric subsets.

– I NEED YOUR HELP TO FOUND THE SUBSETS STRUCTURE AND BEHAVIOUR. SOMEONE WANT TO COOPERATE ? ]]>

also: http://www.mathematische-basteleien.de/macmahon.htm. ]]>

MacMahon was elected a fellow of the Royal Society in 1890. He received the Royal Society Royal Medal in 1900, the Sylvester Medal in 1919, and the Morgan Medal by the London Mathematical Society in 1923. MacMahon was the President of the London Mathematical Society from 1894 to 1896.

MacMahon is best known for his study of symmetric functions and enumeration of plane partitions; see MacMahon Master theorem. His two volume Combinatory analysis, published in 1915/16,[2] is the first major book in enumerative combinatorics.

In the movie The Man Who Knew Infinity Kevin McNally plays as MacMahon. The film accurately depicts the first meeting of MacMahon and Srinivasa Ramanujan, where Ramanujan successfully completes some mathematical calculations.

You can see Kevin McNally where https://en.wikipedia.org/wiki/Kevin_McNally.

Has someone of you seen the movie? ]]>

As an example take n=4 which is 2-symmetric.

The 24 permutations of this group are diveded into classes according to the content of permutation isomorphic to (12) or (21)

let us call the group using the number of permutations of each type, they are:

n=4 classes :(6,0) (5,1) (4,2) (3,3) (2,4) (1,5) (0,6)

n=4 elements: 1, 3, 5, 6, 5, 3, 1 the 4th line of Mac Mahon triangle tell us how many elements are in each class

I also modified the OEIS A008302 formula to produce one line of the Mac Mahon triangle

The modified formula is as follows:

{TL(n) = local(v=vector(n*(n-1)/2+1));my(A=1+x); for(i=1,n, A = 1 + intformal(A – q*subst(A,x,q*x +x^2*O(x^n)))/(1-q));

v=vector(#v,k,polcoeff(n!*polcoeff(A,n,x),k-1,q));}

The first 8 lines of the triangle.

line 1: [1]

line 2: [1, 1]

line 3: [1, 2, 2, 1]

line 4: [1, 3, 5, 6, 5, 3, 1]

line 5: [1, 4, 9, 15, 20, 22, 20, 15, 9, 4, 1]

line 6: [1, 5, 14, 29, 49, 71, 90, 101, 101, 90, 71, 49, 29, 14, 5, 1]

line 7: [1, 6, 20, 49, 98, 169, 259, 359, 455, 531, 573, 573, 531, 455, 359, 259, 169, 98, 49, 20, 6, 1]

line 8: [1, 7, 27, 76, 174, 343, 602, 961, 1415, 1940, 2493, 3017, 3450, 3736, 3836, 3736, 3450, 3017, 2493, 1940, 1415, 961, 602, 343, 174, 76, 27, 7, 1]

It is improper to use the name triangle because the lenth of line don’t increase in linear manner being equal to C(n,2)+1 the combinations.

]]>From my post of 5 december it is possible to derive a faormula for the direct calculation of the number of 2-symmetric permutation.

The formula is

if(n%4<2,print(T(n,n*(n-1)/4)))

A formula for T(n,k) can be found in OEIS A008302.

This is the result for n from 2 to 23

2 is not

3 is not

4 is 2-symmetric with 6 elements

5 is 2-symmetric with 22 elements

6 is not

7 is not

8 is 2-symmetric with 3836 elements

9 is 2-symmetric with 29228 elements

10 is not

11 is not

12 is 2-symmetric with 25598186 elements

13 is 2-symmetric with 296643390 elements

14 is not

15 is not

16 is 2-symmetric with 738680521142 elements

17 is 2-symmetric with 11501573822788 elements

18 is not

19 is not

20 is 2-symmetric with 62119523114983224 elements

21 is 2-symmetric with 1214967840930909302 elements

22 is not

23 is not

The formula works up to 100 and more Than the calculation time start to increase exponentially.

for n = 100 I got:

n=100;if(n%4<2,print(T(n,n*(n-1)/4)))

221162231799801437337491422676415187183834729960140576944399302421324901266274924973791449425648337816306724828465719503145180465615716643026616888287189208

this number has 156 digits.

the 2 tables are wrongly formatted because of tabs wich where not transferred correctly, change as follows with blanks.

var STD perm#

0 0 6

1 1 10

4 2 6

9 3 2

var STD perm#

0 0 3836

1 1 7472

4 2 6900

9 3 6034

16 4 4986

25 5 3880

36 6 2830

49 7 1922

64 8 1204

81 9 686

100 10 348

121 11 152

144 12 54

169 13 14

196 14 2

Than you

Angelo

]]>The sequence A008302 – OEIS – Triangle of Mahonian numbers T(n,k): coefficients in expansion of Product_{i=0..n-1} (1 + x + … + x^i), where k ranges from 0 to A000217(n-1), presents a connection to the density distribution of 2-symmetrical permutation.

If we take the following subsequences:

n=4 [T(8)..T(14)]

n=5 [T(15)..T(25)]

n=8 [T(64)..T(92)]

n=9 [T(93)..T(129)]

we can notice that these are exactly the number of permutations of order n which are order-isomorphic to (12) or (21) starting from the first, which is

the class (only one element) with only (12) type permutations, proceeding with increasing (21) and decreasing (12) densities, up to the 2-symmetric position wich is in trhe center of sequence. From this point the sequence proceeds always increasing (21) and decreasing (12) up to the final class (only 1 element) wich indicates the only permutation with all (21) type. ]]>

var STD perm#

0 0 22

1 1 40

4 2 30

9 3 18

16 4 8

25 5 2

There are 22 permutations with variance 0 perfectly balanced, 40 with variance 1, and so on….

If we form this sequence: 1, 4, 9, 15, 20, 22, 20, 15, 9, 4, 1 and make a search into OEIS

we found that this sequence formed by the number of permutations of each unbalanced class divided by 2 and in decreasing order

with respect to the variance followed by the number of balanced permutations followed again by the number of permutations of

each unbalanced class divided by 2 in ascending order is the sequence A008302 Triangle of Mahonian numbers T(n,k).

I did the same for n=8 with the following results:

var STD perm#

0 0 3836

1 1 7472

4 2 6900

9 3 6034

16 4 4986

25 5 3880

36 6 2830

49 7 1922

64 8 1204

81 9 686

100 10 348

121 11 152

144 12 54

169 13 14

196 14 2

We can form in the same manner the following sequence:

1, 7, 27, 76, 174, 343, 602, 961, 1415, 1940, 2493, 3017, 3450, 3736, 3836, 3736, 3450, ……. uo to 1

and we found that also this sequence is part of the sequence A008302 Triangle of Mahonian numbers T(n,k)

for different n and k.

Please Tanya give my e-mail address to Drake Thomas and let me know your comment. For your knowledge I did not have an education in

math and do it for fun, so I’m not able to proceed further without your help. But I will cooperate if you think this worth of attention,

have a lot of additional material and ideas.