A242527 -id:A242527 - cp's OEIS frontend

A051252 Number of essentially different ways of arranging numbers 1 through 2n around a circle so that sum of each pair of adjacent numbers is prime.

1, 1, 1, 2, 48, 512, 1440, 40512, 385072, 3154650, 106906168, 3197817022, 82924866213, 4025168862425, 127854811616691

Offset: 1

Jud McCranie reports that he was able to find a solution for each n <= 225 (2n <= 450) in just a few seconds. - Jul 05 2002
Is there a proof that this can always be done?
The Mathematica program for this sequence uses backtracking to find all solutions for a given n. To verify that at least one solution exists for a given n, the backtracking function be made to stop when the first solution is found. Solutions have been found for n <= 48. - T. D. Noe, Jun 19 2002
This sequence is from the prime circle problem. There is no known proof that a(n) > 0 for all n. However, for many n (see A072618 and A072676), we can prove that a(n) > 0. Also, the sequence A072616 seems to imply that there are always solutions in which the odd (or even) numbers are in order around the circle. - T. D. Noe, Jul 01 2002
Prime circles can apparently be generated for any n using the Mathematica programs given in A072676 and A072184. - T. D. Noe, Jul 08 2002
The following seems to always produce a solution: Work around the circle starting with 1 but after that always choosing the largest remaining number that fits. For example, if n = 4 this gives 1, 6, 7, 4, 3, 8, 5, 2. See A088643 for a sequence on a related idea. - Paul Boddington, Oct 30 2007
See A228917 for a similar conjecture on twin primes. - Zhi-Wei Sun, Sep 08 2013
See A242527 for a similar problem on the set of numbers {0 through (n-1)}. - Stanislav Sykora, May 30 2014
James Tilley and Stan Wagon report that all terms up to n = 10^6 are nonzero. Charles R Greathouse IV, Feb 05 2016

			One arrangement for 2n=6 is 1,4,3,2,5,6 and this is essentially unique, so a(3)=1.

R. K. Guy, Unsolved Problems in Number Theory, second edition, Springer, 1994. See section C1.

S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.
Eric Weisstein's World of Mathematics, Prime Circle.

Cf. A000341, A070897, A072616, A072617, A072618, A072676, A072184, A103839, A227050, A228917, A242527, A242528.

$RecursionLimit=500; try[lev_] := Module[{t, j}, If[lev>2n, (*then make sure the sum of the first and last is prime*) If[PrimeQ[soln[[1]]+soln[[2n]]]&&soln[[2]]<=soln[[2n]], (*Print[soln]; *) cnt++ ], (*else append another number to the soln list*) t=soln[[lev-1]]; For[j=1, j<=Length[s[[t]]], j++, If[ !MemberQ[soln, s[[t]][[j]]], soln[[lev]]=s[[t]][[j]]; try[lev+1]; soln[[lev]]=0]]]]; For[lst={}; n=1, n<=7, n++, s=Table[{}, {2n}]; For[i=1, i<=2n, i++, For[j=1, j<=2n, j++, If[i!=j&&PrimeQ[i+j], AppendTo[s[[i]], j]]]]; soln=Table[0, {2n}]; soln[[1]]=1; cnt=0; try[2]; AppendTo[lst, cnt]]; lst (* T. D. Noe *)

a(14)-a(15) from Max Alekseyev, Sep 19 2013

A242522 Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is at least 2.

Original entry on oeis.org

0, 0, 0, 0, 1, 5, 33, 245, 2053, 19137, 196705, 2212037, 27029085, 356723177, 5058388153, 76712450925, 1239124984693, 21241164552785, 385159565775633, 7365975246680597, 148182892455224845, 3128251523599365177, 69149857480654157545, 1597343462243140957757

Offset: 1

Stanislav Sykora, May 27 2014

nonn

a(n)=NPC(n;S;P) is the count of all neighbor-property cycles for a specific set S of n elements and a specific pair-property P. For more details, see the link and A242519.
Number of Hamiltonian cycles in the complement of P_n, where P_n is the n-path graph. - Andrew Howroyd, Mar 15 2016
a(n) also agrees with the number of optimal fundamentally distinct radio labelings of the wheel graph on (n+1) nodes for n = 5 up to at least n = 10 (and likely all larger n). - Eric W. Weisstein, Jan 12 2021
a(n) also agrees with the number of optimal fundamentally distinct radio labelings of the n-dipyramidal graph for n = 5 up to at least n = 9 (and likely all larger n). - Eric W. Weisstein, Jan 14 2021

			The 5 cycles of length n=6 having this property are {1,3,5,2,4,6}, {1,3,5,2,6,4}, {1,3,6,4,2,5}, {1,4,2,5,3,6}, {1,4,2,6,3,5}.

Andrew Woods, Table of n, a(n) for n = 1..100 (terms up to a(24) from _Hiroaki Yamanouchi_, Aug 28 2014)
F. C. Holroyd and W. J. G. Wingate, Cycles in the complement of a tree or other graph, Discrete Math., 55 (1985), 267-282.
S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.
Eric Weisstein's World of Mathematics, Dipyramidal Graph
Eric Weisstein's World of Mathematics, Hamiltonian Cycle
Eric Weisstein's World of Mathematics, Path Complement Graph
Eric Weisstein's World of Mathematics, Radio Labeling
Eric Weisstein's World of Mathematics, Wheel Graph

Cf. A242519, A242520, A242521, A242523, A242524, A242525, A242526, A242527, A242528, A242529, A242530, A242531, A242532, A242533, A242534.

Cf. A006184.

a[n_ /; n < 5] = 0; a[5] = 1; a[6] = 5; a[n_] := a[n] = n a[n - 1] - (n - 5) a[n - 2] - (n - 4) a[n - 3] + (n - 4) a[n - 4]; Array[a, 24] (* Jean-François Alcover, Oct 07 2017 *)
Join[{0, 0}, RecurrenceTable[{a[n] == n a[n - 1] - (n - 5) a[n - 2] - (n - 4) a[n - 3] + (n - 4) a[n - 4], a[3] == a[4] == 0, a[5] == 1, a[6] == 5}, a, {n, 20}]] (* Eric W. Weisstein, Apr 12 2018 *)

a(n) = A002493(n)/(2*n), n>1. - Andrew Woods, Dec 08 2014
a(n) = Sum_{k=1..n} (-1)^(n-k)*k!*A102413(n,k) / (2*n), n>2. - Andrew Woods after Vladeta Jovovic, Dec 08 2014
a(n) = (n-1)!/2 + sum_{i=1..n-1} ((-1)^i * (n-i-1)! * sum_{j=0..i-1} (2^j * C(i-1,j) * C(n-i,j+1))), for n>=5. - Andrew Woods, Jan 08 2015
a(n) = n a(n-1) - (n-5) a(n-2) - (n-4) a(n-3) + (n-4) a(n-4), for n>6. - Jean-François Alcover, Oct 07 2017

A242528 Number of cyclic arrangements of {0,1,...,n-1} such that both the difference and the sum of any two neighbors are prime.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 4, 18, 13, 62, 8, 133, 225, 209, 32, 2644, 4462, 61341, 113986, 750294, 176301, 7575912, 3575686, 7705362, 36777080, 108638048, 97295807

Offset: 1

Stanislav Sykora, May 30 2014

a(n)=NPC(n;S;P) is the count of all neighbor-property cycles for a specific set S of n elements and a specific pair-property P. For more details, see the link and A242519.

In this case the set is S={0 through n-1}. For the same pair-property P but the set S={1 through n}, see A227050.

			For n=12 (the first n for which a(n)>0) there are two such cycles:
C_1={0, 5, 2, 9, 4, 1, 6, 11, 8, 3, 10, 7},
C_2={0, 7, 10, 3, 8, 5, 2, 9, 4, 1, 6, 11}.

S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.

Cf. A227050, A242519, A242520, A242521, A242522, A242523, A242524, A242525, A242526, A242527, A242529, A242530, A242531, A242532, A242533, A242534.

A242528[n_] :=
Count[Map[lpf, Map[j0f, Permutations[Range[n - 1]]]], 0]/2;
j0f[x_] := Join[{0}, x, {0}];
lpf[x_] := Length[
   Join[Select[asf[x], ! PrimeQ[#] &],
    Select[Differences[x], ! PrimeQ[#] &]]];
asf[x_] := Module[{i}, Table[x[[i]] + x[[i + 1]], {i, Length[x] - 1}]];
Table[A242528[n], {n, 1, 8}]
(* OR, a less simple, but more efficient implementation. *)
A242528[n_, perm_, remain_] := Module[{opt, lr, i, new},
   If[remain == {},
     If[PrimeQ[First[perm] - Last[perm]] &&
       PrimeQ[First[perm] + Last[perm]], ct++];
     Return[ct],
     opt = remain; lr = Length[remain];
     For[i = 1, i <= lr, i++,
      new = First[opt]; opt = Rest[opt];
      If[! (PrimeQ[Last[perm] - new] && PrimeQ[Last[perm] + new]),
       Continue[]];
      A242528[n, Join[perm, {new}],
       Complement[Range[n - 1], perm, {new}]];
      ];
     Return[ct];
     ];
   ];
Table[ct = 0; A242528[n, {0}, Range[n - 1]]/2, {n, 1, 18}]
(* Robert Price, Oct 22 2018 *)

a(29)-a(33) from Fausto A. C. Cariboni, May 20 2017

A242519 Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is 2^k for some k=0,1,2,...

Original entry on oeis.org

0, 1, 1, 1, 4, 8, 14, 32, 142, 426, 1204, 3747, 9374, 26306, 77700, 219877, 1169656, 4736264, 17360564, 69631372, 242754286, 891384309, 3412857926, 12836957200, 42721475348, 152125749587, 549831594988

Offset: 1

Stanislav Sykora, May 27 2014

a(n)=NPC(n;S;P) is the count of all neighbor-property cycles for a specific set S of n elements and a specific pair-property P. Evaluating this sequence for n>=3 is equivalent to counting Hamiltonian cycles in a pair-property graph with n vertices and is often quite hard. For more details, see the link.

			The four such cycles of length 5 are:
C_1={1,2,3,4,5}, C_2={1,2,4,3,5}, C_3={1,2,4,5,3}, C_4={1,3,2,4,5}.
The first and the last of the 426 such cycles of length 10 are:
C_1={1,2,3,4,5,6,7,8,10,9}, C_426={1,5,7,8,6,4,3,2,10,9}.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..27 (first 21 terms from Stanislav Sykora)
S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.

Cf. A001710, A236602, A242520, A242521, A242522, A242523, A242524, A242525, A242526, A242527, A242528, A242529, A242530, A242531, A242532, A242533, A242534.

A242519[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
lpf[x_] := Length[Select[Abs[Differences[x]], ! MemberQ[t, #] &]];
t = Table[2^k, {k, 0, 10}];
Join[{0, 1}, Table[A242519[n], {n, 3, 10}]]
(* OR, a less simple, but more efficient implementation. *)
A242519[n_, perm_, remain_] := Module[{opt, lr, i, new},
   If[remain == {},
     If[MemberQ[t, Abs[First[perm] - Last[perm]]], ct++];
     Return[ct],
     opt = remain; lr = Length[remain];
     For[i = 1, i <= lr, i++,
      new = First[opt]; opt = Rest[opt];
      If[! MemberQ[t, Abs[Last[perm] - new]], Continue[]];
      A242519[n, Join[perm, {new}],
       Complement[Range[2, n], perm, {new}]];
      ];
     Return[ct];
     ];
   ];
t = Table[2^k, {k, 0, 10}];
Join[{0, 1}, Table[ct = 0; A242519[n, {1}, Range[2, n]]/2, {n, 3, 12}]] (* Robert Price, Oct 22 2018 *)

For any S and any P, and for n>=3, NPC(n;S;P)<=A001710(n-1).

a(22)-a(27) from Hiroaki Yamanouchi, Aug 29 2014

A242520 Number of cyclic arrangements of S={1,2,...,2n} such that the difference between any two neighbors is 3^k for some k=0,1,2,...

Original entry on oeis.org

1, 1, 2, 3, 27, 165, 676, 3584, 19108, 80754, 386776, 1807342, 8218582, 114618650, 1410831012, 12144300991, 126350575684

Offset: 1

Stanislav Sykora, May 27 2014

a(n)=NPC(2n;S;P) is the count of all neighbor-property cycles for a specific set S of 2n elements and a specific pair-property P. For more details, see the link and A242519.

In this particular instance of NPC(n;S;P), all the terms with odd cycle lengths are necessarily zero.

			The two such cycles of length n=6 are:
C_1={1,2,3,6,5,4}, C_2={1,2,5,6,3,4}.
The first and last of the 27 such cycles of length n=10 are:
C_1={1,2,3,4,5,6,7,8,9,10}, C_27={1,4,7,8,5,2,3,6,9,10}.

S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.

Cf. A242519, A242521, A242522, A242523, A242524, A242525, A242526, A242527, A242528, A242529, A242530, A242531, A242532, A242533, A242534.

A242520[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, 2 n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
lpf[x_] := Length[Select[Abs[Differences[x]], ! MemberQ[t, #] &]];
t = Table[3^k, {k, 0, 10}];
Join[{1}, Table[A242520[n], {n, 2, 5}]]
(* OR, a less simple, but more efficient implementation. *)
A242520[n_, perm_, remain_] := Module[{opt, lr, i, new},
   If[remain == {},
     If[MemberQ[t, Abs[First[perm] - Last[perm]]], ct++];
     Return[ct],
     opt = remain; lr = Length[remain];
     For[i = 1, i <= lr, i++,
      new = First[opt]; opt = Rest[opt];
      If[! MemberQ[t, Abs[Last[perm] - new]], Continue[]];
      A242520[n, Join[perm, {new}],
       Complement[Range[2, 2 n], perm, {new}]];
      ];
     Return[ct];
     ];
   ];
t = Table[3^k, {k, 0, 10}];
Join[{1}, Table[ct = 0; A242520[n, {1}, Range[2, 2 n]]/2, {n, 2, 8}]] (* Robert Price, Oct 22 2018 *)

a(14)-a(17) from Andrew Howroyd, Apr 05 2016

A242521 Number of cyclic arrangements (up to direction) of {1,2,...,n} such that the difference between any two neighbors is b^k for some b>1 and k>1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 4, 6, 9, 42, 231, 1052, 3818, 10086, 27892, 90076, 310301, 993680, 4663558, 22038882, 162588454, 1246422151, 8655752023, 58951670318, 347675502245

Offset: 1

Stanislav Sykora, May 27 2014

a(n)=NPC(n;S;P) is the count of all neighbor-property cycles for a specific set S={1,2,...,n} of n elements and a specific pair-property P. For more details, see the link and A242519.

			The two cycles of length n=13 (the smallest n such that a(n)>0) are: C_1={1,5,9,13,4,8,12,3,7,11,2,6,10}, C_2={1,9,5,13,4,8,12,3,7,11,2,6,10}.

S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.

Cf. A242519, A242520, A242522, A242523, A242524, A242525, A242526, A242527, A242528, A242529, A242530, A242531, A242532, A242533, A242534.

A242521[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
lpf[x_] := Length[Select[Abs[Differences[x]], ! MemberQ[t, #] &]];
t = Flatten[Table[b^k, {k, 2, 5}, {b, 2, 5}]];
Table[A242521[n], {n, 1, 10}]
(* OR, a less simple, but more efficient implementation. *)
A242521[n_, perm_, remain_] := Module[{opt, lr, i, new},
   If[remain == {},
     If[MemberQ[t, Abs[First[perm] - Last[perm]]], ct++];
     Return[ct],
     opt = remain; lr = Length[remain];
     For[i = 1, i <= lr, i++,
      new = First[opt]; opt = Rest[opt];
      If[! MemberQ[t, Abs[Last[perm] - new]], Continue[]];
      A242521[n, Join[perm, {new}],
       Complement[Range[2, n], perm, {new}]];
      ];
     Return[ct];
     ];
   ];
t = Flatten[Table[b^k, {k, 2, 5}, {b, 2, 5}]];
Table[ct = 0; A242521[n, {1}, Range[2, n]]/2, {n, 1, 18}] (* Robert Price, Oct 24 2018 *)

a(27)-a(30) from Max Alekseyev, Jul 12 2014

a(31)-a(32) from Fausto A. C. Cariboni, May 17 2017, May 24 2017

A242523 Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is at least 3.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 11, 125, 1351, 15330, 184846, 2382084, 32795170, 481379278, 7513591430, 124363961357, 2176990766569, 40199252548280, 781143277669538, 15937382209774353, 340696424417421213, 7616192835573406931, 177723017354688250713, 4321711817908214684734

Offset: 1

Stanislav Sykora, May 27 2014

a(n)=NPC(n;S;P) is the count of all neighbor-property cycles for a specific set S of n elements and a specific pair-property P. For more details, see the link and A242519.

			The shortest cycle with this property has length n=7: {1,4,7,3,6,2,5}.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..27 (terms a(1)-a(15) from _Stanislav Sykora_)
S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.

Cf. A242519, A242520, A242521, A242522, A242524, A242525, A242526, A242527, A242528, A242529, A242530, A242531, A242532, A242533, A242534.

A242523[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
lpf[x_] := Length[Select[Abs[Differences[x]], # < 3 &]];
Table[A242523[n], {n, 1, 10}]
 (* OR, a less simple, but more efficient implementation. *)
A242523[n_, perm_, remain_] := Module[{opt, lr, i, new},
   If[remain == {},
     If[Abs[First[perm] - Last[perm]] >= 3, ct++];
     Return[ct],
     opt = remain; lr = Length[remain];
     For[i = 1, i <= lr, i++,
      new = First[opt]; opt = Rest[opt];
      If[Abs[Last[perm] - new] < 3, Continue[]];
      A242523[n, Join[perm, {new}],
       Complement[Range[2, n], perm, {new}]];
      ];
     Return[ct];
     ];
   ];
Table[ct = 0; A242523[n, {1}, Range[2, n]]/2, {n, 1, 11}] (* Robert Price, Oct 24 2018 *)

a(16)-a(25) from Hiroaki Yamanouchi, Aug 28 2014

A242524 Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is at least 4.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 24, 504, 8320, 131384, 2070087, 33465414, 561681192, 9842378284, 180447203232, 3462736479324, 69517900171056, 1458720714556848, 31955023452174314, 729874911380470641, 17359562438053760533, 429391730229931885360

Offset: 1

Stanislav Sykora, May 27 2014

a(n)=NPC(n;S;P) is the count of all neighbor-property cycles for a specific set S of n elements and a specific pair-property P. For more details, see the link and A242519.

			The shortest such cycle has length n=9: {1,5,9,4,8,3,7,2,6}.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..27 (terms a(1)-a(16) from _Stanislav Sykora_)
S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.

Cf. A242519, A242520, A242521, A242522, A242523, A242525, A242526, A242527, A242528, A242529, A242530, A242531, A242532, A242533, A242534.

A242524[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
lpf[x_] := Length[Select[Abs[Differences[x]], # < 4 &]];
Table[A242524[n], {n, 1, 10}]
 (* OR, a less simple, but more efficient implementation. *)
A242524[n_, perm_, remain_] := Module[{opt, lr, i, new},
   If[remain == {},
     If[Abs[First[perm] - Last[perm]] >= 4, ct++];
     Return[ct],
     opt = remain; lr = Length[remain];
     For[i = 1, i <= lr, i++,
      new = First[opt]; opt = Rest[opt];
      If[Abs[Last[perm] - new] < 4, Continue[]];
      A242524[n, Join[perm, {new}],
       Complement[Range[2, n], perm, {new}]];
      ];
     Return[ct];
     ];
   ];
Table[ct = 0; A242524[n, {1}, Range[2, n]]/2, {n, 1, 12}] (* Robert Price, Oct 24 2018 *)

a(17)-a(25) from Hiroaki Yamanouchi, Aug 29 2014

A242525 Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is at most 3.

Original entry on oeis.org

1, 1, 1, 3, 6, 10, 17, 31, 57, 104, 188, 340, 616, 1117, 2025, 3670, 6651, 12054, 21847, 39596, 71764, 130065, 235730, 427238, 774328, 1403395, 2543518, 4609881, 8354965, 15142569, 27444447, 49740415, 90149708, 163387657, 296124381, 536696900

Offset: 1

Stanislav Sykora, May 27 2014

nonn

a(n) = NPC(n;S;P) is the count of all neighbor-property cycles for a specific set S of n elements and a specific pair-property P. For more details, see the link and A242519.

			For n=4, The three cycles are: C_1={1,2,3,4}, C_2={1,2,4,3}, C_3={1,3,2,4}.
The first and the last of the 104 such cycles of length n=10 are: C_1={1,2,3,5,6,8,9,10,7,4}, C_104={1,3,6,9,10,8,7,5,2,4}.

Yifan Xie, Table of n, a(n) for n = 1..3000 (First 100 terms from Andrew Howroyd)
Stanislav Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.
Yifan Xie, Proof of the formulas

Cf. A242519, A242520, A242521, A242522, A242523, A242524, A242526, A242527, A242528, A242529, A242530, A242531, A242532, A242533, A242534.

A242525[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
lpf[x_] := Length[Select[Abs[Differences[x]], # > 3 &]];
Join[{1, 1}, Table[A242525[n], {n, 3, 10}]]
(* OR, a less simple, but more efficient implementation. *)
A242525[n_, perm_, remain_] := Module[{opt, lr, i, new},
   If[remain == {},
     If[Abs[First[perm] - Last[perm]] <= 3, ct++];
     Return[ct],
     opt = remain; lr = Length[remain];
     For[i = 1, i <= lr, i++,
      new = First[opt]; opt = Rest[opt];
      If[Abs[Last[perm] - new] > 3, Continue[]];
      A242525[n, Join[perm, {new}],
       Complement[Range[2, n], perm, {new}]];
      ];
     Return[ct];
     ];
   ];
Join[{1, 1},
Table[ct = 0; A242525[n, {1}, Range[2, n]]/2, {n, 3, 12}] ](* Robert Price, Oct 24 2018 *)

lista(nn) = {my(v=[1, 1, 1, 3, 6, 10, 17]); for(n=8, nn, v = concat(v, v[n-1] + v[n-2] + v[n-4] + v[n-5])); v}; \\ Yifan Xie, Mar 20 2025

Empirical: a(n) = a(n-1)+a(n-2)+a(n-4)+a(n-5) for n>7. - Andrew Howroyd, Apr 08 2016
Empirical G.f.: x^2 + ((1-x)^2*(1+x)^2)/(1-x-x^2-x^4-x^5). - Andrew Howroyd, Apr 08 2016
Empirical first differences of A185265. - Sean A. Irvine, Jun 26 2022
See link for proofs of the above formulas. - Yifan Xie, Mar 19 2025

a(28)-a(35) from Andrew Howroyd, Apr 08 2016

A242526 Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is at most 4.

Original entry on oeis.org

1, 1, 1, 3, 12, 36, 90, 214, 521, 1335, 3473, 9016, 23220, 59428, 152052, 389636, 999776, 2566517, 6586825, 16899574, 43352560, 111213798, 285319258, 732016006, 1878072638, 4818362046, 12361809384, 31714901077, 81366445061, 208750870961

Offset: 1

Stanislav Sykora, May 27 2014

nonn

a(n)=NPC(n;S;P) is the count of all neighbor-property cycles for a specific set S of n elements and a specific pair-property P. For more details, see the link and A242519.

			The 3 cycles of length n=4 are: {1,2,3,4},{1,2,4,3},{1,3,2,4}.
The first and the last of the 1335 such cycles of length n=10 are:
C_1={1,2,3,4,6,7,8,10,9,5}, C_1335={1,4,8,10,9,7,6,3,2,5}.

Andrew Howroyd, Table of n, a(n) for n = 1..100
S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.

Cf. A242519, A242520, A242521, A242522, A242523, A242524, A242525, A242527, A242528, A242529, A242530, A242531, A242532, A242533, A242534.

A242526[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
lpf[x_] := Length[Select[Abs[Differences[x]], # > 4 &]];
Join[{1, 1}, Table[A242526[n], {n, 3, 10}]]
(* OR, a less simple, but more efficient implementation. *)
A242526[n_, perm_, remain_] := Module[{opt, lr, i, new},
   If[remain == {},
     If[Abs[First[perm] - Last[perm]] <= 4, ct++];
     Return[ct],
     opt = remain; lr = Length[remain];
     For[i = 1, i <= lr, i++,
      new = First[opt]; opt = Rest[opt];
      If[Abs[Last[perm] - new] > 4, Continue[]];
      A242526[n, Join[perm, {new}],
       Complement[Range[2, n], perm, {new}]];
      ];
     Return[ct];
     ];
   ];
Join[{1, 1}, Table[ct = 0; A242526[n, {1}, Range[2, n]]/2, {n, 3, 12}] ](* Robert Price, Oct 25 2018 *)

From Andrew Howroyd, Apr 08 2016: (Start)
Empirical: a(n) = 2*a(n-1) + a(n-2) - a(n-4) + 9*a(n-5) + 5*a(n-6) - a(n-7) - 7*a(n-8) - 10*a(n-9) + 2*a(n-10) + 2*a(n-11) + 2*a(n-12) + 4*a(n-13) - 2*a(n-17) - a(n-18) for n>20.
Empirical g.f.: x + (3 - 6*x - 2*x^2 - x^3 + 3*x^4 - 22*x^5 - 5*x^6 + x^7 + 8*x^8 + 14*x^9 - 6*x^10 + 2*x^11 - 6*x^12 - 6*x^13 - 3*x^15 + x^16 + 3*x^17) / (1 - 2*x - x^2 + x^4 - 9*x^5 - 5*x^6 + x^7 + 7*x^8 + 10*x^9 - 2*x^10 - 2*x^11 - 2*x^12 - 4*x^13 + 2*x^17 + x^18). (End)

a(22)-a(30) from Andrew Howroyd, Apr 08 2016

cp's OEIS Frontend

A051252 Number of essentially different ways of arranging numbers 1 through 2n around a circle so that sum of each pair of adjacent numbers is prime.

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A242522 Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is at least 2.

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A242528 Number of cyclic arrangements of {0,1,...,n-1} such that both the difference and the sum of any two neighbors are prime.

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A242519 Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is 2^k for some k=0,1,2,...

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A242520 Number of cyclic arrangements of S={1,2,...,2n} such that the difference between any two neighbors is 3^k for some k=0,1,2,...

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A242521 Number of cyclic arrangements (up to direction) of {1,2,...,n} such that the difference between any two neighbors is b^k for some b>1 and k>1.

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A242523 Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is at least 3.

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