Conjecture: A293561(n) = StirlingS2[n + 1, 3] - Binomial[n, 2]

[Ven Popov (Ven)]

A293561(n) =  0,0,0,3,19,80,286,945...

I independently stumbled on the construction StirlingS2[n + 1, 3] - Binomial[n, 2], which seems to agree for all computable n (I tried up to 100).

The explicit formula is (3^n−2^{n+1}+1−n^2+n)​/2

Not sure what the connection is because A293561, which is a column of A142249, only has an analytic definition in terms of polylogarithms.

As for a(n) = StirlingS2[n + 1, 3] - Binomial[n, 2], here is what it counts:

The number of pairs of disjoint subsets of an n-element set that are non-empty and not both singletons.

For n=3:
1|23, 12|3, 13|2

For n=4:
1|23, 1|24, 1|34, 2|13, 2|14, 2|34, 3|12, 3|14, 3|24, 4|12, 4|13, 4|23
1|234, 2|134, 3|124, 4|123,
12|34, 13|24, 14|23, 

The StirlingS2[n + 1, 3] gives us the number of partitions of n-set into 3 subsets, 2 of them non-empty. Then Binomial[n, 2] removes the cases where the 2 main subsets only have 1 element each.

This construction arose when I was playing around with Erdos's Problem 1. Given a set S of n integers, you can obtain all such pairs of disjoint subsets (non-empty, not both singletons), then subtract the sum of one from the sum of the other. If any of the differences are 0, the set is not subset-sum distinct.

[Simon Plouffe]

Hello,  the sequence A293561 satisfies this
formula ; (5*x-3)/(3*x-1)/(2*x-1)/(x-1)^3 = f(x) 

the 300'th term is (computed from the formula) 
68445739529294187995663013691044157983231847812668716198704118
704698412320303849887521942026376077028188991584158547667025456603046141703840775

which agrees with the big b table. 

That formula appears in my big list of formulas here :
A293561 10.1363 [(5*x-3)/(3*x-1)/(2*x-1)/(x-1)^3 ogf]
on page 2127 of the document : 
https://plouffe.fr/articles/OEIS_conjectured_formulas%202023.pdf

Best regards, 
 Simon Plouffe

 

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[Ven Popov]

Thanks, Simon! It’s easy to confirm that OGF is exactly x^{-3} times the GF of a(n) = S(n+1,3) - \binom{n}{2}, so they’re fully consistent. The combinatorial interpretation I gave is therefore an interpretation of your OGF as well (up to index shift)

 

Cheers,

Ven

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