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A000274
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Number of permutations of length n with 2 consecutive ascending pairs.
(Formerly M3048 N1236)
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10
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0, 0, 1, 3, 18, 110, 795, 6489, 59332, 600732, 6674805, 80765135, 1057289046, 14890154058, 224497707343, 3607998868005, 61576514013960, 1112225784377144, 21197714949305577, 425131949816628507, 8950146311929021210
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OFFSET
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1,4
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COMMENTS
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a(n) = number of excedances in all derangements of [n-1]. Example: a(5)=18 because the derangements of {1,2,3,4} are 4*123, 3*14*2, 3*4*12, 4*3*12, 2*14*3, 2*4*13, 2*3*4*1, 3*4*21, 4*3*21 with the 18 excedances marked. An excedance of a permutation p is a position i such that p(i)>i.
(End)
Appears to be the inverse binomial transform of A001286 (filling the two leading zeros in there), then shifting one place to the right. R. J. Mathar, Apr 04 2012
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REFERENCES
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F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 263.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 210 (divided by 2).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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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FORMULA
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a(n) = (1 + n) a(n - 1) + (3 + n) a(n - 2) + (3 - n) a(n - 3) + (2 - n) a(n - 4).
a(n) = (n-1)^2/(n-2)*a(n-1)-(-1)^n*(n-1)/2, n>2, a(2)=0. - Vladeta Jovovic, Aug 31 2003
a(n) = (1/2){[n!/e] - [(n-1)!/e]} (conjectured).
a(n) = (n-1)*GAMMA(n,-1)*exp(-1)/2 where GAMMA = incomplete Gamma function. [Mark van Hoeij, Nov 11 2009]
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MAPLE
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a:= n->sum(n!*sum((-1)^k/k!/2, j=1..n), k=0..n): seq(a(n), n=2..20); # Zerinvary Lajos, May 17 2007
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MATHEMATICA
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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