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A006860
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Erroneous version of A223911: Tiered orders on n nodes.
(Formerly M2959)
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4
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1, 3, 13, 111, 1381, 25623, 678133, 26269735, 1447451707, 114973020921, 13034306495563
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OFFSET
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1,2
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COMMENTS
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WARNING: The currently listed value of a(8) is inconsistent with the result from Kreweras and Klarner quoted below, as pointed out by Michel Marcus. - M. F. Hasler, Nov 03 2012
Graded posets, i.e., those in which every maximal chain has the same length. (The terminology "graded" is also used to refer to a weaker notion; see A001833.)
Kreweras observed and Klarner proved that a(n) is congruent to 1 (resp. 3) modulo 6 when n is odd (resp. even). - Michel Marcus, Nov 03 2012
Using the formulas in the paper from Klarner (cf. PARI code), I get 1, 3, 13, 85, 801, 10231, 168253, 3437673, 85162465, 2511412651, 86805640461, 3469622549053, ... - M. F. Hasler, Nov 07 2012
The values currently in the sequence through 25623 are certainly correct (I've enumerated these posets by brute force and other methods). (...) Klarner's eq.(2) contains a typo: instead of f(m_1, m_h) it should be f(m_1, m_2). (The point here is that the Hasse diagram of each of these posets decomposes as a bunch of bipartite graphs layered on top of each other; there are f(m_1, m_2) ways to choose the bipartite graph between the first two ranks of vertices, then f(m_2, m_3) ways to choose the bipartite graph between the second and third ranks of vertices, etc.) (...). When I implement Klarner's eqs.(1) and (2) (corrected) I get the following sequence: 1, 3, 13, 111, 1381, 25623, 678133, 26169951, 1447456261, 114973232583, ... Now we get the right terms up as far as I personally have experience (...) and they agree with Kreweras (and the current OEIS sequence) until a(8), at which point there is disagreement. [Joel Brewster Lewis, Mar 06 2013; private communication to M. F. Hasler]
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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PROG
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(PARI) ee(n)={my(f(m, n)=sum(k=0, m, (-1)^(m-k)*binomial(m, k)*(2^k-1)^n), C(n, m)=n!/prod(i=1, #m, m[i]!), t(h, n)=my(s=0); forvec(m=vector(h, i, [if(i<h, 1, n-h+1), n-h+1]), if(0<m[h]=n-sum(i=1, h-1, m[i]), s+=C(n, m)*prod(i=1, h-1, f(m[i], m[h])))); s); sum(h=1, n, t(h, n))} \\ This implements the formula in Klarner's paper, where equation 2 contains a typo. It does NOT yield the correct terms. - M. F. Hasler, Nov 07 2012
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CROSSREFS
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KEYWORD
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dead
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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