Some work I am doing is connected with a sequence 1, 3, 40, 1225, 67956, $\dots$ which agrees with http://oeis.org/A012250 for all eight terms. The only useful information in OEIS on this sequence is the reference D. N. Verma, Towards Classifying Finite PointSet Configurations, preprint, 1997. Does anyone know how to obtain this preprint? Verma himself passed away last year.

1$\begingroup$ For anyone thinking about Googling it: if you search for the title and subtract OEIS, you get two references to the fact that he gave a talk about it in 1997. $\endgroup$– Allen KnutsonFeb 28, 2013 at 2:54

$\begingroup$ @Richard: That's not one of the typewritten manuscripts I got from Verma, but I know that he had a tendency to leave such projects incomplete. His approach to mathematics was sometimes highly insightful, but was always quirky. As a person he was definitely one of a kind, but left us with too little completed (and reliable) work. $\endgroup$– Jim HumphreysMar 1, 2013 at 1:20

5$\begingroup$ I would like to thank Yannic Vargas, who located a copy of Verma's paper in the LACIM library and emailed me a scanned file. $\endgroup$– Richard StanleyMar 2, 2013 at 1:53

2$\begingroup$ The work referred to in the question is now available. See Theorem 4.6 and Remark 4.7 at math.mit.edu/~rstan/papers/distinctparts.pdf. $\endgroup$– Richard StanleyMay 25, 2013 at 0:04
2 Answers
I don't have Verma's preprint, but there are more modern references on the subject, which don't seem to be on OEIS. Basically one looks at the GIT quotient $(\mathbb{P^1})^n//\operatorname{SL}_2$, and your sequence corresponds to its degree under a certain embedding in projective space for values of $n=4,6,8,...$.
See section 2.11 in the paper "The moduli space of n points on the line is cut out by simple quadrics when n is not six", by B. Howard, J. Millson, A. Snowden and R. Vakil. An even more recent reference that also contains a sort of formula for these degrees is "The ring of evenly weighted points on the line", by M. Hering, B. Howard. Hope this helps.

2$\begingroup$ Thanks, this is very helpful. For $n=2m$ I get the formula $$ \frac 12\sum_{j=0}^{m1}(mj)^{n3}(1)^{j+1}{n\choose j} (2jn+1). $$ Assuming that my sequence and Verma's are the same (which must be true), then this gives a simpler formula than in the reference you provide. $\endgroup$ Feb 28, 2013 at 21:44


1$\begingroup$ The above formula can be simplified to $$ \frac 12\sum_{j=0}^{m1}(mj)^{n3}(1)^{j+1}\binom nj. $$ $\endgroup$ Apr 5, 2013 at 17:45
You might try to contact one of the organizers of the
Workshop on Algebraic Combinatorics, 920 June 1997
Org.: F. Bergeron (UQAM), N. Bergeron (CRM & York Univ.), C. Reutenauer (UQAM)
(or ask s/o in the Math Dept of IIT Bombay, http://www.math.iitb.ac.in/.)