A078603 -id:A078603 - cp's OEIS frontend

A078628 Number of ways of arranging the numbers 1..n in a circle so that there is no consecutive triple i, i+1, i+2 or i, i-1, i-2 (mod n).

1, 1, 0, 4, 12, 76, 494, 3662, 30574, 284398, 2918924, 32791604, 400400062, 5281683678, 74866857910, 1135063409918, 18330526475060, 314169905117860, 5695984717957246, 108921059813769710, 2190998123920252622, 46250325111346491694

Offset: 1

N. J. A. Sloane, Dec 12 2002

nonn

This sequence can be related to A165964 by the use of auxiliary sequences (and the auxiliary sequences can themselves be calculated by recurrence relations). So if we desire we can determine any value of this sequence. [From Isaac Lambert, Oct 07 2009]

			a(4) = 4: 4 2 1 3, 4 3 1 2, 4 1 3 2, 4 2 3 1.
a(5) = 12: 5 3 1 2 4, 5 2 3 1 4, 5 4 2 1 3, 5 2 4 1 3, 5 1 4 2 3, 5 2 1 4 3, 5 1 3 4 2, 5 3 1 4 2, 5 4 1 3 2, 5 3 4 1 2, 5 2 4 3 1, 5 3 2 4 1.

Wayne M. Dymacek, Isaac Lambert and Kyle Parsons, Arithmetic Progressions in Permutations, http://math.ku.edu/~ilambert/CN.pdf, 2012. - From N. J. A. Sloane, Sep 14 2012

Isaac Lambert, Table of n, a(n) for n = 1..50
Wayne M. Dymáček and Isaac Lambert, Circular Permutations Avoiding Runs of i, i+1, i+2 or i, i-1, i-2, Journal of Integer Sequences, Vol. 14 (2011), Article 11.1.6.
Sean A. Irvine, Java program (github)
N. J. A. Sloane, FORTRAN program
Index entries for sequences related to shoe lacings

Cf. A078673. See A002816, A078603 for analogous sequence with restrictions only on pairs.

Cf. A095816, A165963, A165964.

a(11)-a(13) from John W. Layman, Nov 15 2004

a(14) from Isaac Lambert, Oct 07 2009

A002816 Number of polygons that can be formed from n points on a circle, no two adjacent.

Original entry on oeis.org

1, 0, 0, 0, 1, 3, 23, 177, 1553, 14963, 157931, 1814453, 22566237, 302267423, 4340478951, 66541218865, 1084982173641, 18752743351339, 342523093859011, 6593167693927885, 133408305489947029, 2831112931136162775, 62878579846490149375, 1458746608689369440265

Offset: 1

N. J. A. Sloane

Also number of ways of arranging the numbers 1..n in a circle so that adjacent numbers do not differ by 1 mod n. Reversing the direction around the circle does not count as a different solution (cf. A078603).
Also number of ways of seating n people around a circular table so that no one sits next to any of his neighbors in a previous seating order.
Suppose n people are seated at random around a circular table for two meals. Then p(n) = a(n)/((n-1)!/2) is the probability that no two people sit together at both meals.
Number of Hamiltonian cycles in the complement of C_n where C_n is the n-cycle graph. - Andrew Howroyd, Mar 15 2016

			a(6)=3: 135264, 136425, 142635.

P. Poulet, Reply to Query 4750, Permutations, L'Intermédiaire des Mathématiciens, 26 (1919), 117-121.
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).

T. D. Noe, Table of n, a(n) for n = 1..100
B. Aspvall and F. M. Liang, The dinner table problem, Technical Report CS-TR-80-829, Computer Science Department, Stanford, California, 1980.
D. P. Robbins, The probability that neighbors remain neighbors after random rearrangements, Amer. Math. Monthly 87 (1980), 122-124.
Eric Weisstein's World of Mathematics, Cycle Complement Graph
Eric Weisstein's World of Mathematics, Hamiltonian Cycle
Index entries for sequences related to shoe lacings

Cf. A000179 (Ménage problem), A078603, A078630, A078631, A242522 (Hamiltonian cycles in complement of path), A006184, left column of A326411.

dinner := proc(n) local j,k,sum; sum := (n-1)!/2 + (-1)^n; for k from 1 to n-1 do for j from 1 to min(n-k,k) do sum := sum+(-1)^k*binomial(k-1,j-1)*binomial(n-k,j)*n/(n-k)*(n-k-1)!/2*2^j; od; od; end;

t = {1, 0, 0, 0, 1, 3, 23}; Do[AppendTo[t, ((n^3 - 8*n^2 + 18*n - 21) t[[-1]] + 4*n*(n - 5) t[[-2]] - 2*(n - 6) (n^2 - 5 n + 3) t[[-3]] + (n^2 - 7*n + 9)  t[[-4]] + (n - 5) (n^2 - 5*n + 3) t[[-5]])/(n^2 - 7*n + 9)], {n, 8, 25}]; t (* T. D. Noe, Jan 04 2012 *)
Join[{1, 0}, RecurrenceTable[{a[3] == 0, a[4] == 0, a[5] == 1, a[6] == 3, a[7] == 23, (n^2 - 7 n + 9) a[n] == (n^3 - 8 n^2 + 18 n - 21) a[n - 1] + 4 n (n - 5) a[n - 2] - 2 (n - 6) (n^2 - 5 n + 3) a[n - 3] + (n^2 - 7 n + 9) a[n - 4] + (n - 5) (n^2 - 5 n + 3) a[n - 5]}, a, {n, 3, 20}]] (* Eric W. Weisstein, Feb 26 2021 *)

D-finite with recurrence (n^2 - 7n + 9)a(n) = (n^3 - 8n^2 + 18n - 21)a(n - 1) + 4n(n - 5)a(n - 2) - 2(n - 6)(n^2 - 5n + 3)a(n - 3) + (n^2 - 7n + 9)a(n - 4) + (n - 5)(n^2 - 5n + 3)a(n - 5), for n >= 8. - Poulet.
p(n) = exp(-2)*(1 + O(1/n)). - Aspvall and Liang.
Asymptotic: a(n)/(n-1)! ~ 1/(2*e^2)*(1 - 4/n + 20/(3n^3) + 58/(3n^4) + 796/(15n^5) + 7858/(45n^6) + 40324/(63n^7) + 140194/(63n^8) + 2444744/(405n^9) + 40680494/(14175n^10) + ...). - Vaclav Kotesovec, Apr 10 2012

Entry improved by Michael Steyer (m.steyer(AT)osram.de), Aug 30 2001

More terms from Sascha Kurz, Mar 22 2002

A089222 Number of ways of seating n people around a table for the second time without anyone sitting next to the same person as they did the first time.

Original entry on oeis.org

1, 0, 0, 0, 0, 10, 36, 322, 2832, 27954, 299260, 3474482, 43546872, 586722162, 8463487844, 130214368530, 2129319003680, 36889393903794, 675098760648204, 13015877566642418, 263726707757115400, 5603148830577775218, 124568968969991162100, 2892414672938546871250

Offset: 0

Udi Hadad (somebody(AT)netvision.net.il), Dec 22 2003

A078603 counts these arrangements up to circular symmetry (i.e., two arrangements are the same if one can be rotated to give the other). A002816 counts them up to dihedral symmetry (i.e., two arrangements are the same if one can be rotated or reflected to give the other). - Joel B. Lewis, Jan 28 2010

			a(4)=0 because trying to arrange 1,2,3,4 around a table will always give a couple who is sitting next to each other and differ by 1.

J. Snell, Introduction to Probability, e-book, pp. 101 Q. 20.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Art of Problem Solving forum, Random neighbors. - _Joel B. Lewis_, Jan 28 2010
B. Aspvall and F. M. Liang, The dinner table problem, Technical Report CS-TR-80-829, Computer Science Department, Stanford, California, 1980.
Charles M. Grinstead and J. Laurie Snell, Introduction to Probability.
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 626.
Roberto Tauraso, The Dinner Table Problem: The Rectangular Case, INTEGERS: Electronic Journal of Combinatorial Number Theory, Vol. 6 (2006), #A11. Note that in this paper a(1) = 1. See Column 2 in the table on page 3.

Cf. A002464, A002816, A078603, A078630, A078631.

Same[cperm_, n_] := ( For[same = False; i = 2, (i <= n) && ! same, i++, same = ((Mod[cperm[[i - 1]], n] + 1) == cperm[[i]]) || ((Mod[cperm[[ i]], n] + 1) == cperm[[i - 1]])]; same = same || ((Mod[cperm[[n]], n] + 1) == cperm[[1]]) || ((Mod[ cperm[[1]], n] + 1) == cperm[[n]]); Return[same]); CntSame[n_] := (allPerms = Permutations[Range[n]]; count = 0; For[j = 1, j <= n!, j++, perm = allPerms[[j]]; If[ ! Same[perm, n], count++ ]]; Return[count]);
(* or direct computation of terms *)
Table[If[n<3, 0, n! + (-1)^n*2n + Sum[(-1)^r*(n/(n-r))^2 * (n-r)! * Sum[2^c * Binomial[r-1,c-1] * Binomial[n-r,c], {c,1,r}], {r,1,n-1}]], {n,1,25}] (* Vaclav Kotesovec, Apr 06 2012 *)

Inclusion-exclusion gives that for n > 2, we have a(n) = n! + 2*n*(-1)^n + Sum_{1 <= k <= m < n} (-1)^m * (n/k) * binomial(n-m-1, k-1) * binomial(m-1, k-1) * 2^k * n * (n-m-1)!. - Joel B. Lewis, Jan 28 2010
a(n) = (3*n-30)*a(n-11) + (6*n-45)*a(n-10) + (5*n+18)*a(n-9) - (8*n-139)*a(n-8) - (26*n-204)*a(n-7) - (4*n-30)*a(n-6) + (26*n-148)*a(n-5) + (8*n-74)*a(n-4) - (9*n-18)*a(n-3) - (2*n-15)*a(n-2) + (n+2)*a(n-1), n >= 14. - Vaclav Kotesovec, Apr 13 2010
The asymptotic expansion from article by Aspvall and Liang (also cited in article by Tauraso) is wrong. Bad terms are 736/(15*n^5) + 8428/(45*n^6) + 40174/(63*n^7). Right asymptotic formula is a(n) ~ (n!/e^2)*(1 - 4/n + 20/(3*n^3) + 58/(3*n^4) + 796/(15*n^5) + 7858/(45*n^6) + 40324/(63*n^7) + 140194/(63*n^8) + ...). Verified also numerically. For example, for n=200, exact/asymptotic results are 1.0000000000125542243 (Aspvall + Liang), 1.0000000000000008990 (Kotesovec 7 terms) or 1.0000000000000000121 (Kotesovec 8 terms). - Vaclav Kotesovec, Apr 06 2012
a(n) = 2*n*A002816(n) for n > 1. - Martin Renner, Apr 01 2022

Tauraso reference from Parthasarathy Nambi, Dec 21 2006
More terms from Vladeta Jovovic, Nov 29 2009
a(0)=1 prepended by Alois P. Heinz, Jul 31 2019

A326404 Triangle T(n,k) read by rows: T(n,k) = number of ways of seating n people around a table for the second time so that k pairs are maintained. Rotated sequences are counted as one.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 0, 2, 0, 0, 4, 0, 2, 2, 0, 10, 10, 0, 2, 6, 24, 30, 40, 18, 0, 2, 46, 140, 224, 182, 98, 28, 0, 2, 354, 1088, 1480, 1280, 604, 192, 40, 0, 2, 3106, 9000, 12006, 9450, 4878, 1494, 330, 54, 0, 2, 29926, 83480, 107170, 82840, 41620, 14152, 3100, 520, 70, 0, 2

Offset: 0

Witold Tatkiewicz, Aug 01 2019

Definition requires "pairs" and for n=0 it is assumed that there is 1 way of seating 0 people around a table for the second time so that 0 pairs are maintained and 1 person forms only one pair with him/herself. Therefore T(0,0)=1, T(1,0)=0 and T(1,1)=1.
Sum of row n is equal to (n-1)! for n > 1.
Conjecture: The weighted average of each row using k as weights converges to 2 for large n and is given by the following formula: (Sum_{k} T(n,k)*k)/(Sum_{k} T(n,k)) = 2/(n-1) + 2.

			Assuming the initial order was {1,2,3,4,5} (therefore 1 and 5 form a pair as the first and last persons are neighbors in the case of a round table) there are 5 sets of ways of seating them again so that 3 pairs are conserved: {1,2,3,5,4}, {2,3,4,1,5}, {3,4,5,2,1}, {4,5,1,3,2}, {5,1,2,4,3}. Since within each set we do not allow for circular symmetry (e.g., {1,2,3,5,4} and its rotation to form {2,3,5,4,1} are counted as one) but we allow reflection ({1,2,3,5,4} and {4,5,3,2,1} are considered distinct), the total number of ways is 5*2 and therefore T(5,3)=10.
Unfolded table with n individuals (rows) forming k pairs (columns):
    0    1    2    3    4    5    6    7
0   1
1   0    1
2   0    0    1
3   0    0    0    2
4   0    0    4    0    2
5   2    0   10   10    0    2
6   6   24   30   40   18    0    2
7  46  140  224  182   98   28    0    2

Witold Tatkiewicz, Rows n = 0..17 of triangle, flattened
Witold Tatkiewicz, Link for Java program.

Cf. A078603 (column k=0).
Sum of n-th row is A000142(n-1) for n > 0.
Cf. A326390 (accounting for rotation and reflection symmetry), A326397 (disregards reflection symmetry but allows rotation), A326411 (disregards both reflection and rotation symmetry).

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T(n,n) = 2 for n > 2;
T(n,n-1) = 0 for n > 1.
Conjectures:
T(n,n-2) = n^2 + n - 2 for n > 3;
T(n,n-3) = (4/3)*n^3 + 2*n^2 - (16/3)*n + 2 for n > 4;
T(n,n-4) = (25/12)*n^4 + (23/6)*n^3 - (169/12)*n^2 + (85/6)*n - 6 for n > 5;
T(n,n-5) = (52/15)*n^5 + (25/3)*n^4 - (83/3)*n^3 + (221/3)*n^2 - (299/5)*n + 26 for n > 5;
T(n,n-6) = (707/120)*n^6 + (2037/120)*n^5 - (413/8)*n^4 + (2233/8)*n^3 - (5554/15)*n^2 + (3739/10)*n - 114 for n > 6.

cp's OEIS Frontend

A078628 Number of ways of arranging the numbers 1..n in a circle so that there is no consecutive triple i, i+1, i+2 or i, i-1, i-2 (mod n).

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A002816 Number of polygons that can be formed from n points on a circle, no two adjacent.

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A089222 Number of ways of seating n people around a table for the second time without anyone sitting next to the same person as they did the first time.

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A326404 Triangle T(n,k) read by rows: T(n,k) = number of ways of seating n people around a table for the second time so that k pairs are maintained. Rotated sequences are counted as one.

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