Here is what I found for Beethoven’s 9th recordings.

Berlin Philharmonic 60:10 125 musicians

Vienna Philharmonic 63:15 134 musicians

Cleveland Symphony 66:01 105 musicians

San Francisco Symphony 70:41 100 musicians

Chicago Symphony 74:52 100 musicians

Using this dataset, I calculated a least-squared linear regression and found that time = 102 minutes – 0.31 * musicians (with a mean-squared-error of only 8.64 minutes).

Therefore, a 60 member orchestra should perform Beethoven’s 9th in 83:24, give or take a few minutes.

caveat: the number of musicians is correct for each group at some recent time, not necessarily at the time of the recording in question.

I started this analysis as a joke, but the correlation seems stronger than mere chance. I submit the following theory. Prominent orchestras not only have more musicians, but more accomplished musicians and conductors, allowing them to maintain a brisker tempo through the presto sections. I intended to include the one data point from the problem statement, but it does not correlate well, so I left it out. I question the authenticity of your data! ðŸ˜Ž

]]>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.