A002816 -id:A002816 - cp's OEIS frontend

A078630 Numerators of coefficients of asymptotic expansion of probability p(n) (see A002816) in powers of 1/n.

1, -4, 0, 20, 58, 796, 7858, 40324, 140194, 2444744, 40680494, -7117319032, -149539443124, -223750776484, -4960419494993024, -46146161037854692, -689434674121075448, -132496988938839119444, -9686633414582239854958, -442788087926096759821484

Offset: 0

N. J. A. Sloane, Dec 13 2002

sign

			p(n) = exp(-2)*(1 - 4/n + 20/(3n^3) + 58/(3n^4) + ...).

Vaclav Kotesovec, Table of n, a(n) for n = 0..50
B. Aspvall and F. M. Liang, The dinner table problem, Technical Report CS-TR-80-829, Computer Science Department, Stanford, California, 1980.

Cf. A078631, A089222.

t = 15;
y[n_]:=(1+Sum[Subscript[p,k]/n^k,{k,1,t}]);
mul=1;start=9; If[t>9,mul=n^(t-9);start=t];
w=Apart[Expand[mul*Simplify[
y[n]*n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
-((3*n-30)*y[n-11]
+(6*n-45)*y[n-10]*(n-10)
+(5*n+18)*y[n-9]*(n-9)*(n-10)
-(8*n-139)*y[n-8]*(n-8)*(n-9)*(n-10)
-(26*n-204)*y[n-7]*(n-7)*(n-8)*(n-9)*(n-10)
-(4*n-30)*y[n-6]*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
+(26*n-148)*y[n-5]*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
+(8*n-74)*y[n-4]*(n-4)*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
-(9*n-18)*y[n-3]*(n-3)*(n-4)*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
-(2*n-15)*y[n-2]*(n-2)*(n-3)*(n-4)*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
+(n+2)*y[n-1]*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10))],n],n];
sol=Solve[Table[Coefficient[w,n,j]==0,{j,start,start-t+1,-1}]];
asympt=y[n]/.sol[[1]];
Table[Numerator[Coefficient[asympt,n,-j]],{j,0,t}] (* Vaclav Kotesovec, Apr 06 2012 *)

Terms a(5)-a(19) from Vaclav Kotesovec, Apr 06 2012 (terms a(5)-a(7) were wrong, see A089222 for more information)

A078631 Denominators of coefficients of asymptotic expansion of probability p(n) (see A002816) in powers of 1/n.

Original entry on oeis.org

1, 1, 1, 3, 3, 15, 45, 63, 63, 405, 14175, 51975, 93555, 15795, 42567525, 49116375, 91216125, 2170943775, 19538493975, 109185701625, 3093594879375, 10257709336875, 428772250281375, 281764621613475, 158210081654625, 160789593855515625

Offset: 0

N. J. A. Sloane, Dec 13 2002

nonn

			p(n) = exp(-2)*(1 - 4/n + 20/(3n^3) + 58/(3n^4) + ...).

Vaclav Kotesovec, Table of n, a(n) for n = 0..50
B. Aspvall and F. M. Liang, The dinner table problem, Technical Report CS-TR-80-829, Computer Science Department, Stanford, California, 1980.

Cf. A078630, A089222.

t = 15;
y[n_]:=(1+Sum[Subscript[p,k]/n^k,{k,1,t}]);
mul=1;start=9; If[t>9,mul=n^(t-9);start=t];
w=Apart[Expand[mul*Simplify[
y[n]*n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
-((3*n-30)*y[n-11]
+(6*n-45)*y[n-10]*(n-10)
+(5*n+18)*y[n-9]*(n-9)*(n-10)
-(8*n-139)*y[n-8]*(n-8)*(n-9)*(n-10)
-(26*n-204)*y[n-7]*(n-7)*(n-8)*(n-9)*(n-10)
-(4*n-30)*y[n-6]*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
+(26*n-148)*y[n-5]*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
+(8*n-74)*y[n-4]*(n-4)*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
-(9*n-18)*y[n-3]*(n-3)*(n-4)*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
-(2*n-15)*y[n-2]*(n-2)*(n-3)*(n-4)*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10)
+(n+2)*y[n-1]*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*(n-6)*(n-7)*(n-8)*(n-9)*(n-10))],n],n];
sol=Solve[Table[Coefficient[w,n,j]==0,{j,start,start-t+1,-1}]];
asympt=y[n]/.sol[[1]];
Table[Denominator[Coefficient[asympt,n,-j]],{j,0,t}] (* Vaclav Kotesovec, Apr 06 2012 *)

Terms a(8)-a(25) from Vaclav Kotesovec, Apr 06 2012

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).

Original entry on oeis.org

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

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

A326411 Triangle T(n,k) read by rows: T(n,k) = the number of ways of seating n people around a table for the second time so that k pairs are maintained. Reflected and rotated sequences are counted as one.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 2, 0, 1, 1, 0, 5, 5, 0, 1, 3, 12, 15, 20, 9, 0, 1, 23, 70, 112, 91, 49, 14, 0, 1, 177, 544, 740, 640, 302, 96, 20, 0, 1, 1553, 4500, 6003, 4725, 2439, 747, 165, 27, 0, 1, 14963, 41740, 53585, 41420, 20810, 7076, 1550, 260, 35, 0, 1

Offset: 0

Witold Tatkiewicz, Aug 07 2019

Poulet (1919) arrives at this triangle of numbers by considering n-sided polygons whose vertices lie on a circle. Call a side of such a polygon simple if its endpoints are adjacent on the circle. Then T(n,k) is the number of such polygons with k simple sides. There is also a connection with A002464 (see that entry). - N. J. A. Sloane, Mar 08 2022
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.
The weighted average of each row using k as weights converges to 2 for large n and appears to be given by (Sum_{k} k*T(n,k))/n! = 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) nor reflection ({1,2,3,5,4} and {4,5,3,2,1} are also counted as one), the total number of ways is 5 and therefore T(5,3)=5.
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    1
4   0    0    2    0    1
5   1    0    5    5    0    1
6   3   12   15   20    9    0   1
7  23   70  112   91   49   14   0   1

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50; rows 0..17 from Witold Tatkiewicz)
P. Poulet, Query 4750, L'Intermédiaire des Mathématiciens, 26 (1919), 117-121. (Page 117)
P. Poulet, Query 4750, L'Intermédiaire des Mathématiciens, 26 (1919), 117-121. (Pages 118, 119)
P. Poulet, Query 4750, L'Intermédiaire des Mathématiciens, 26 (1919), 117-121. (Pages 120, 121)
Witold Tatkiewicz, link for java program.

Cf. A002816 (column k=0).
Row sums: A001710(n-1) = Sum_k T(n,k).
Cf. also A326390 (accounting for rotation and reflection symmetry), A326397 (disregards reflection symmetry but allows rotation), A326407 (disregards rotation symmetry but allows reflection).
See in addition A002464.

Java
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See Links section
```

A326411 := proc(n,k)
    option remember;
    if k > n or k < 0 then
        0;
    elif k = n then
        1;
    elif k =0 then
        if n < 5 then
            0 ;
        elif n = 5 then
            1 ;
        elif n = 6 then
            3 ;
        elif n = 7 then
            23 ;
        else
            # Poulet eq (6) page 120, shifted n->n-2
            -(n^3-8*n^2+18*n-21)*procname(n-1,0)
            -4*(n^2-5*n)*procname(n-2,0)
            +2*(n^3-11*n^2+33*n-18)*procname(n-3,0)
            -(n^2-7*n+9)*procname(n-4,0)
            -(n^3-10*n^2+28*n-15)*procname(n-5,0) ;
            -%/(n^2-7*n+9) ;
        end if;
    elif n <= 3 then
        0;
    else
        # Poulet eq (3) page 119
        2*(n-k)*procname(n-1,k-1)/(n-1)+2*k*procname(n-1,k)/(n-1)
            +(k-2)*procname(n-2,k-2)/(n-2) - 2*(k-1)*procname(n-2,k-1)/(n-2) + k*procname(n-2,k)/(n-2) ;
        %*n/k ;
    end if;
end proc:
for n from 0 to 12 do
    for k from 0 to n do
        printf("%a ",A326411(n,k)) ;
    end do:
    printf("\n") ;
end do: # R. J. Mathar, Mar 17 2022

T[n_, k_] := T[n, k] = Which[k > n || k < 0, 0, k == n, 1, k == 0, Which[n<5, 0, n == 5, 1, n == 6, 3, n == 7, 23, True,
    pc = -(n^3 - 8*n^2 + 18*n - 21)*T[n-1, 0]
      - 4*(n^2 - 5*n)*T[n - 2, 0]
      + 2*(n^3 - 11*n^2 + 33*n - 18)*T[n-3, 0]
      - (n^2 - 7*n + 9)*T[n-4, 0]
      - (n^3 - 10*n^2 + 28*n - 15)*T[n-5, 0];
    -pc/(n^2 - 7*n + 9)], n <= 3, 0, True,
   pc = 2*(n-k)*T[n-1, k-1]/(n-1) + 2*k*T[n-1, k]/(n-1) +
     (k - 2)*T[n-2, k-2]/(n-2) -
     2*(k-1)*T[n-2, k-1]/(n-2) + k*T[n-2, k]/(n-2);
    pc*n/k];
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 17 2023, after R. J. Mathar *)

Q(n,k)={k*subst(serlaplace(polcoef((1 - 2*x -x^2)/((1 + x)*(1 + (1 - y)*x + y*x^2)) + O(x^n), n-1)), y, k)}
row(n)={Vec(if(n<3, 1, (Q(n,y/(y-1))/2 + (-1)^n)*(y-1)^n), -n-1)} \\ Andrew Howroyd, Mar 01 2024

It appears that Poulet gives recurrences that generate the whole triangle. - N. J. A. Sloane, Mar 09 2022
T(n,n) = 1;
T(n,n-1) = 0 for n >= 1;
T(n,n-2) = n*(n-3)/2 for n >= 4 [Poulet];
T(n,n-3) = n*(n-4)*(2*n-7)/3 for n >= 4 [Poulet, corrected by N. J. A. Sloane, Mar 09 2022]
T(n,n-4) = (25/24)*n^4 + (23/12)*n^3 - (169/24)*n^2 + (85/12)*n - 3 for n > 5 (conjectured); [see Poulet]
T(n,n-5) = (26/15)*n^5 + (25/6)*n^4 - (83/6)*n^3 + (221/6)*n^2 - (299/10)*n + 13 for n > 5 (conjectured); [see Poulet]
T(n,n-6) = (707/240)*n^6 + (2037/240)*n^5 - (413/16)*n^4 + (2233/16)*n^3 - (2777/15)*n^2 + (3739/20)*n - 57 for n > 6 (conjectured). [See Poulet]

A002493 Number of ways to arrange n non-attacking kings on an n X n board, with 2 sides identified to form a cylinder, with 1 in each row and column.

Original entry on oeis.org

1, 0, 0, 0, 10, 60, 462, 3920, 36954, 382740, 4327510, 53088888, 702756210, 9988248956, 151751644590, 2454798429600, 42130249479562, 764681923900260, 14636063499474054, 294639009867223880

Offset: 1

N. J. A. Sloane

nonn

Number of directed Hamiltonian paths in the complement of C_n where C_n is the n-cycle graph. - Andrew Howroyd, Mar 15 2016

Number of ways of arranging n consecutive integers in a circle such that no pair of adjacent integers differ by 1, rotations are distinct. - Graham Holmes, Sep 03 2020

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).

Vincenzo Librandi, Table of n, a(n) for n = 1..100
M. Abramson and W. O. J. Moser, Permutations without rising or falling w-sequences, Ann. Math. Stat., 38 (1967), 1245-1254.
Eli Bagno, Estrella Eisenberg, Shulamit Reches, and Moriah Sigron, Counting King Permutations on the Cylinder, arXiv:2001.02948 [math.CO], 2020.
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 627-8.
A. J. Schwenk, Letter to N. J. A. Sloane, Mar 24 1980 (with enclosure and notes)

Cf. A002464, A002816, A006184.

Right diagonal of A338838.

b1:= proc(n, r) local gu, x; if r=0 then RETURN(0): fi: gu := (x*diff(x*(1+x)/(1-x),x))* (x*(1 + x)/(1 - x))^(r-1); gu := taylor(gu, x = 0, n +1); coeff(gu, x, n ) end: b:=proc(n) local r: if n=1 then 1 elif n=2 then 0 else add((-1)^(n-r)*r!*b1(n,r),r=0..n): fi: end: # Doron Zeilberger, Nov 14 2007

b[n_]:=(If[n>0, 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}], 0]); Table[If[n>2, b[n]-2*Sum[b[n-1-2k], {k, 0, Floor[n/2]}], If[n==1, 1, 0]], {n, 1, 25}] (* Vaclav Kotesovec after Vladeta Jovovic, Apr 06 2012 *)

The linear recurrence operator annihilating this sequence is (N is the shift operator Na(n):=a(n + 1)) is - 3*(43*n + 197)*(n - 2)*(n + 1)/( - 1222 + 753*n + 349*n^2) - 5*(n - 1)*(44*n^2 + 477*n + 1222)/( - 1222 + 753*n + 349*n^2)*N + 2*(n + 1)*(239*n^2 + 873*n - 1232)/( - 1222 + 753*n + 349*n^2)*N^2 + 4*(394 - 259*n + 215*n^2 + 55*n^3)/( - 1222 + 753*n + 349*n^2)*N^3 - ( - 7342 + 3699*n + 2718*n^2 + 349*n^3)/( - 1222 + 753*n + 349*n^2)*N^4 + N^5. - Doron Zeilberger, Nov 14 2007
a(n) = Sum((-1)^(n-k)*k!*A102413(n,k),k=1..n), n>2. - Vladeta Jovovic, Nov 23 2007
a(n) = b(n+1) - 2*Sum_{k=0..floor(n/2)} b(n-2*k) for n>1, where b(n)=A002464(n) if n>0 else b(0)=0. - Vladeta Jovovic, Nov 24 2007
Asymptotic: a(n) ~ n!/e^2*(1 - 2/n - 2/n^2 - 4/(3n^3) + 8/(3n^4) + 326/(15n^5) + 4834/(45n^6) + 154258/(315n^7) + 232564/(105n^8) + ...). - Vaclav Kotesovec, Apr 06 2012
a(n) = n! + Sum_{i=1..n-1} ((-1)^i * (n-i-1)! * n * Sum_{j=0..i-1} (2^(j+1) * C(i-1,j) * C(n-i,j+1))), for n>=5. - Andrew Woods, Jan 08 2015

A231091 Number of distinct (modulo rotation) unicursal star polygons (not necessarily regular, no edge joins adjacent vertices) that can be formed by connecting the vertices of a regular n-gon.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 5, 27, 175, 1533, 14361, 151575, 1735869, 21594863, 289365383, 4158887007, 63822480809, 1041820050629, 18027531255745, 329658402237171, 6352776451924233, 128686951765990343, 2733851297673484765, 60781108703102022027, 1411481990523638719737

Offset: 1

Stewart Gordon, Nov 03 2013

For polygons in general see A000939 and A000949, and especially the Golomb-Welch reference. - N. J. A. Sloane, Nov 21 2013

			For n=5, only solution is the regular pentagram.
For n=6, only solution is the unicursal hexagram (see Wikipedia link).
For n=7, two regular heptagrams and three irregular forms are possible.

Andrew Howroyd, Table of n, a(n) for n = 1..200
Stewart Gordon, The five possible stars for n=7 (SVG file)
Wikipedia, Unicursal hexagram

Cf. A000939 (if edges may join adjacent vertices), A000940, A002816 (rotations and reflections counted separately), A326411, A370459 (up to rotations and reflections), A370068 (directed edges).

Cf. A283184.

\\ Requires a370068 from A370068, b(n) is A283184.
b(n)={subst(serlaplace(polcoef((1 - x)/(1 + (1 - 2*y)*x + 2*y*x^2) + O(x*x^n), n)), y, 1)}
a(n)={(if(n%2==0 && n > 2, b(n/2-1)/2) + a370068(n))/2} \\ Andrew Howroyd, Mar 01 2024

a(n) = (A370068(n) + A283184(n/2-1)/2)/2 for even n >= 4; a(n) = A370068(n)/2 for odd n. - Andrew Howroyd, Feb 24 2024

a(15) onwards from Andrew Howroyd, Feb 23 2024

A078603 Number of ways of arranging the numbers 1..n in a circle so that adjacent numbers do not differ by 1 mod n.

Original entry on oeis.org

1, 0, 0, 0, 2, 6, 46, 354, 3106, 29926, 315862, 3628906, 45132474, 604534846, 8680957902, 133082437730, 2169964347282, 37505486702678, 685046187718022, 13186335387855770, 266816610979894058, 5662225862272325550

Offset: 1

N. J. A. Sloane, Dec 11 2002

nonn

			a(5) = 2: 1 3 5 2 4, 1 4 2 5 3; a(6) = 6: 1 4 6 2 5 3, 1 5 2 4 6 3, 1 5 3 6 2 4, 1 3 6 4 2 5, 1 4 2 6 3 5, 1 3 5 2 6 4.

Index entries for sequences related to shoe lacings

Twice A002816.
The sequence n*a(n) is A089222.
See also A078628.

For n>1, a(n) = 2*A002816(n).

The sequence was missing a zero; also added a cross-reference Joel B. Lewis, Jan 28 2010

A370459 Number of unicursal stars with n vertices.

Original entry on oeis.org

0, 0, 1, 1, 5, 19, 112, 828, 7441, 76579, 871225, 10809051, 144730446, 2079635889, 31912025537, 520913578812, 9013780062785, 164829273635749, 3176388519597555, 64343477504391475, 1366925655386979893, 30390554390984325019, 705740995420852895453

Offset: 3

Adam M. Scherlis, Feb 19 2024

nonn

A unicursal star is a closed loop formed by diagonals of a regular n-gon.
These are Hamiltonian cycles on the graph complement of the n-cycle.
Allowing polygon diagonals, but not sides, is equivalent to requiring every edge to cross at least one other edge.
These are counted up to rotation and reflection, i.e., modulo dihedral symmetry of the n-gon.
Inspired by a unicursal dodecagram drawn by Gordon FitzGerald (see links).

			For n=5, there is only the regular pentagram {5/2}.
For n=6, there is only the unicursal hexagram.
For n=7, in addition to the two regular heptagrams {7/2} and {7/3}, there are three nontrivial unicursal heptagrams represented by:
 (0, 2, 4, 1, 6, 3, 5, 0)
 (0, 2, 5, 1, 3, 6, 4, 0)
 (0, 2, 5, 1, 4, 6, 3, 0).

Andrew Howroyd, Table of n, a(n) for n = 3..200
Gordon FitzGerald, illustration of a unicursal dodecagram.
Wikipedia, Unicursal hexagram.

Cf. A000940 (polygon sides allowed).
Cf. A055684 (cases with dihedral symmetry only).
Cf. A002816 (rotations and reflections counted separately).
Cf. A231091 (up to rotations only), A370769 (achiral).

\\ Requires a370068 from A370068.
Ro(n)=-(-1)^n + subst(serlaplace(polcoef(((1 - x)^2)/(2*(1 + x)*(1 + (1 - 2*y)*x + 2*y*x^2)) + O(x*x^n), n)), y, 1)
Re(n)=subst(serlaplace(polcoef((1 - x - 2*x^2)/(4*(1 + (1 - 2*y)*x + 2*y*x^2)) + O(x*x^n), n)), y, 1)
a(n)={if(n<3, 0, (if(n%2, 2*Ro(n\2), Re(n/2)) + a370068(n))/4)} \\ Andrew Howroyd, Mar 01 2024

a(n) = (A231091(n) + A370769(n))/2. - Andrew Howroyd, Mar 06 2024

a(14) onwards from Andrew Howroyd, Feb 26 2024

cp's OEIS Frontend

A078630 Numerators of coefficients of asymptotic expansion of probability p(n) (see A002816) in powers of 1/n.

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A078631 Denominators of coefficients of asymptotic expansion of probability p(n) (see A002816) in powers of 1/n.

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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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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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A326411 Triangle T(n,k) read by rows: T(n,k) = the number of ways of seating n people around a table for the second time so that k pairs are maintained. Reflected and rotated sequences are counted as one.

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A002493 Number of ways to arrange n non-attacking kings on an n X n board, with 2 sides identified to form a cylinder, with 1 in each row and column.

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A231091 Number of distinct (modulo rotation) unicursal star polygons (not necessarily regular, no edge joins adjacent vertices) that can be formed by connecting the vertices of a regular n-gon.

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