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 Point-Set Configurations, preprint, 1997. Does anyone know how to obtain this preprint? Verma himself passed away last year.

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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. – Allen Knutson Feb 28 '13 at 2:54
@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. – Jim Humphreys Mar 1 '13 at 1:20
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. – Richard Stanley Mar 2 '13 at 1:53
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. – Richard Stanley May 25 '13 at 0:04

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.

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Thanks, this is very helpful. For $n=2m$ I get the formula $$\frac 12\sum_{j=0}^{m-1}(m-j)^{n-3}(-1)^{j+1}{n\choose j} (2j-n+1).$$ 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. – Richard Stanley Feb 28 '13 at 21:44
very nice answer – Abdelmalek Abdesselam Mar 1 '13 at 17:51
The above formula can be simplified to $$\frac 12\sum_{j=0}^{m-1}(m-j)^{n-3}(-1)^{j+1}\binom nj.$$ – Richard Stanley Apr 5 '13 at 17:45

You might try to contact one of the organizers of the
Workshop on Algebraic Combinatorics, 9-20 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/.)

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