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A002464
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Hertzsprung's problem: ways to arrange n non-attacking kings on an n X n board, with 1 in each row and column. Also number of permutations of length n without rising or falling successions.
(Formerly M2070 N0818)
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46
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1, 1, 0, 0, 2, 14, 90, 646, 5242, 47622, 479306, 5296790, 63779034, 831283558, 11661506218, 175203184374, 2806878055610, 47767457130566, 860568917787402, 16362838542699862, 327460573946510746, 6880329406055690790, 151436547414562736234, 3484423186862152966838
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OFFSET
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0,5
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COMMENTS
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Permutations of 12...n such that none of the following occur: 12, 23, ..., (n-1)n, 21, 32, ..., n(n-1).
This sequence is also the solution to the 'toast problem' devised by my house-mates and me as maths undergraduates some 27 years ago: Given a toast rack with n slots, how many ways can the slices be removed so that no two consecutive slices are removed from adjacent slots? - David Jones (david.jones(AT)zetnet.co.uk), Oct 24 2003
This sequence was also derived by the late D. P. Robbins. - David Callan, Nov 04 2003
Another interpretation: number of permutations of n containing exactly n different patterns of size n-1. - Olivier Gérard, Nov 05 2007
Number of directed Hamiltonian paths in the complement of the n-path graph P_n. - Andrew Howroyd, Mar 16 2016
There is an obvious connection between the two descriptions of the sequence: Replace the chessboard with a n X n zero-matrix and each king with "1". This matrix will transform the vector (1,2,..,n) into a permutation such that adjacent components do not differ by 1. The reverse is also true: Any such transformation is a solution of the king problem. - Gerhard Kirchner, Feb 10 2017
A formula of Poulet (1919) relates this to A326411: a(n) = T(n+2,1)/(n+2) + 2*T(n+1,1)/(n+1) + T(n,1)/n, where T(i,j) = A326411(i,j). - N. J. A. Sloane, Mar 08 2022
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REFERENCES
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W. Ahrens, Mathematische Unterhaltungen und Spiele. Teubner, Leipzig, Vol. 1, 3rd ed., 1921; Vol. 2, 2nd ed., 1918. See Vol. 1, p. 271.
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 263.
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).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.40.
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LINKS
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Michael Han, Tanya Khovanova, Ella Kim, Evin Liang, Miriam Lubashev, Oleg Polin, Vaibhav Rastogi, Benjamin Taycher, Ada Tsui and Cindy Wei, Fun with Latin Squares, arXiv:2109.01530 [math.HO], 2021.
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FORMULA
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If n = 0 or 1 then a(n) = 1; if n = 2 or 3 then a(n) = 0; otherwise a(n) = (n+1)*a(n-1) - (n-2)*a(n-2) - (n-5)*a(n-3) + (n-3)*a(n-4).
G.f.: e^((1 + x)/((-1 + x) * x)) * (1 + x) * Gamma(0, (1 + x)/((-1 + x) * x))/((-1 + x) * x). - Eric W. Weisstein, May 16 2014
Let S_{n, k} = number of permutations of 12...n with exactly k rising or falling successions. Let S[n](t) = Sum_{k >= 0} S_{n, k}*t^k. Then S[0] = 1; S[1] = 1; S[2] = 2*t; S[3] = 4*t+2*t^2; for n >= 4, S[n] = (n+1-t)*S[n-1] - (1-t)*(n-2+3*t)*S[n-2] - (1-t)^2*(n-5+t)*S[n-3] + (1-t)^3*(n-3)*S[n-4].
a(n) = n! + Sum_{k=1..n} (-1)^k * Sum_{t=1..k} binomial(k-1,t-1) * binomial(n-k,t) * 2^t * (n-k)!. - Max Alekseyev, Jan 29 2006
a(n) = Sum_{k=0..n} (-1)^(n-k)*k!*b(n,k), where g.f. for b(n,k) is (1-x)/(1-(1+y)*x-y*x^2), cf. A035607. - Vladeta Jovovic, Nov 24 2007
Asymptotic (M. Abramson and W. Moser, 1966): a(n)/n! ~ (1 - 2/n^2 - 10/(3*n^3) - 6/n^4 - 154/(15*n^5) - 88/(9*n^6) + 5336/(105*n^7) + 1612/(3*n^8) + 2098234/(567*n^9) + 36500686/(1575*n^10) + ... )/e^2. - Vaclav Kotesovec, Apr 19 2011, extended Dec 27 2020
Conjecture: a(n) = Sum_{k=1..n} k!*A080246(n-1, k-1) for n > 0. - John Keith, Nov 02 2020
Proof: a(n) = Sum_{k=1..n} k!*A080246(n-1, k-1) for n > 0. Since a(n) = Sum_{k=0..n-1} (-1)^k*(n-k)!*Sum_{i=0..k} binomial(n-k,i)*binomial(n-1-i,k-i) (M. Abramson and W. Moser, 1966) which is Sum_{k=1..n} (-1)^(k-1)(n-k+1)!*Sum{i=0..k-1} binomial(n-k+1,i)*binomial(n-1-i,k-1-i) = Sum_{k=1..n} (-1)^(n-k)(k!)*Sum_{i=0..n-k} binomial(k,i)*binomial(n-1-i,n-k-i) = k!*A080246(n-1, k-1) as (-1)^(n-k) = (-1)^(n+k) and binomial(n-1-i,k-1) = binomial(n-1-i,n-k-i). - Alex McGaw, Apr 13 2023
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EXAMPLE
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a(4) = 2: 2413, 3142.
a(5) = 14 corresponds to these 14 permutations of length 5: 13524, 14253, 24135, 24153, 25314, 31425, 31524, 35142, 35241, 41352, 42513, 42531, 52413, 53142.
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MAPLE
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MATHEMATICA
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(* computation from the permutation class *)
g[ L_ ] := Apply[ And, Map[ #>1&, L ] ]; f[ n_ ] := Length[ Select[ Permutations[ Range[ n ] ], g[ Rest[ Abs[ RotateRight[ # ]-# ] ] ]& ] ]; Table[ f[ n ], {n, 1, 8} ] (* Erich Friedman *)
(* or direct computation of terms *)
Table[n! + Sum[(-1)^r*(n-r)!*Sum[2^c *Binomial[r-1, c-1] *Binomial[n-r, c], {c, 1, r}], {r, 1, n-1}], {n, 1, 30}] (* Vaclav Kotesovec, Mar 28 2011 *)
(* or from g.f. *)
M = 30; CoefficientList[Sum[n!*x^n*(1-x)^n/(1+x)^n, {n, 0, M}] + O[x]^M, x] (* Jean-François Alcover, Jul 07 2015 *)
CoefficientList[Series[Exp[(1 + x)/((-1 + x) x)] (1 + x) Gamma[0, (1 + x)/((-1 + x) x)]/((-1 + x) x), {x, 0, 20}], x] (* Eric W. Weisstein, Apr 11 2018 *)
RecurrenceTable[{a[n] == (n + 1) a[n - 1] - (n - 2) a[n - 2] - (n - 5) a[n - 3] + (n - 3) a[n - 4], a[0] == a[1] == 1, a[2] == a[3] == 0}, a, {n, 0, 20}] (* Eric W. Weisstein, Apr 11 2018 *)
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PROG
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(PARI)
N = 66; x = 'x + O('x^N);
gf = sum(n=0, N, n!*(x*(1-x))^n/(1+x)^n );
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CROSSREFS
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KEYWORD
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nonn,nice,easy,changed
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AUTHOR
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EXTENSIONS
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Merged with the old A001100, Aug 19 2003
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STATUS
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approved
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