A000255 -id:A000255 - cp's OEIS frontend

A132647 Number of permutations of [n] having no substring [k,k+1,k+2,k+3].

1, 1, 2, 6, 23, 117, 706, 4962, 39817, 359171, 3597936, 39630372, 476066277, 6194080387, 86776390796, 1302376048620, 20847721870931, 354549730559949, 6384006047649910, 121330369923079290, 2427196999663678987, 50981866833670160201, 1121806937829102793662

Offset: 0

Ivana Jovovic (ivana121(AT)EUnet.yu), Nov 14 2007

nonn

Andrew Howroyd, Table of n, a(n) for n = 0..200
D. M. Jackson and R. C. Read, A note on permutations without runs of given length, Aequationes Math. 17 (1978), no. 2-3, 336-343.

Cf. A000255, A002628.

seq(n)={Vec(sum(k=0, n, k!*((x^4-x)/(x^4-1) + O(x*x^n))^k))} \\ Andrew Howroyd, Aug 31 2018

G.f.: Sum_{n>=0} n!*((x^m-x)/(x^m-1))^n where m = 4.

a(n) ~ n! * (1 - 1/n^2 + 1/n^3 + 9/(2*n^4) + 7/n^5 - 55/(6*n^6) - 114/n^7 - 11419/(24*n^8) - 970/n^9 + 345199/(120*n^10) + ...). - Vaclav Kotesovec, Feb 17 2024

Terms a(16) and beyond from Andrew Howroyd, Aug 31 2018

A136301 Frequency of occurrence for each possible "probability of derangement" for a Secret Santa drawing in which each person draws a name in sequence and the only person who does not draw someone else's name is the one who draws the final name.

Original entry on oeis.org

1, 1, 1, 1, 5, 2, 1, 1, 13, 6, 13, 2, 6, 2, 1, 1, 29, 14, 73, 6, 42, 18, 29, 2, 18, 8, 14, 2, 6, 2, 1, 1, 61, 30, 301, 14, 186, 86, 301, 6, 102, 48, 186, 18, 102, 42, 61, 2, 42, 20, 86, 8, 48, 20, 30, 2, 18, 8, 14, 2, 6, 2, 1, 1, 125, 62, 1081, 30, 690, 330, 2069, 14, 414, 200, 1394

Offset: 3

Brian Parsonnet, Mar 22 2008

The sequence is best represented as a series of columns 1..n, where each column j has 2^(j-1) rows (see Example). For more details, see A136300.

The first column represents the case for 3 people (offset 3).

			Represented as a series of columns, where column j has 2^(j-1) rows, the sequence begins:
  row |j = 1   2   3   4   5 ...
  ----+-------------------------
    1 |    1   1   1   1   1 ...
    2 |        1   5  13  29 ...
    3 |        2   6  14  30 ...
    4 |        1  13  73 301 ...
    5 |            2   6  14 ...
    6 |            6  42 186 ...
    7 |            2  18  86 ...
    8 |            1  29 301 ...
    9 |                2   6 ...
   10 |               18 102 ...
   11 |                8  48 ...
   12 |               14 186 ...
   13 |                2  18 ...
   14 |                6 102 ...
   15 |                2  42 ...
   16 |                1  61 ...
   17 |                    2 ...
  ... |                  ... ...
.
If there are 5 people, numbered 1-5 according to the order in which they draw a name, and person #5 draws name #5, the first four people must draw 1-4 as a proper derangement, and there are 9 ways of doing so: 21435 / 23415 / 24135 / 31425 / 34125 / 34215 / 41235 / 43125 / 43215.
But the probability of each derangement depends on how many choices exist at each successive draw. The first person can draw from 4 possibilities (2,3,4,5). The second person nominally has 3 to choose from, unless the first person drew number 2, in which case person 2 may draw 4 possibilities (1,3,4,5), and so on. The probabilities of 21435 and 24135 are both then
        1/4 * 1/4 * 1/2 * 1/2 = 1/64.
More generally, if there are n people, at the i-th turn (i = 1..n), person i has either (n-i) or (n-i+1) choices, depending on whether the name of the person who is drawing has been chosen yet. A way to represent the two cases above is 01010, where a 0 indicates that the person's number is not yet drawn, and a 1 indicates it is.
For the n-th person to be forced to choose his or her own name, the last digit of this pattern must be 0, by definition. Similarly, the 1st digit must be a 0, and the second to last digit must be a 1. So all the problem patterns start with 0 and end with 10. For 5 people, that leaves 4 target patterns which cover all 9 derangements. By enumeration, that distribution can be shown to be (for the 3rd column = 5 person case):
        0-00-10 1 occurrences
        0-01-10 5 occurrences
        0-10-10 2 occurrences
        0-11-11 1 occurrences
1;
1, 1;
1, 5, 2, 1;
1, 13, 6, 13, 2, 6, 2, 1;
1, 29, 14, 73, 6, 42, 18, 29, 2, 18, 8, 14, 2, 6, 2, 1;

Brian Parsonnet, Table of n, a(n) for n = 3..257
Brian Parsonnet, Probability of Derangements

The application of this table towards final determination of the probabilities of derangements leads to sequence A136300, which is the sequence of numerators. The denominators are in A001044.
A048144 represents the peak value of all odd-numbers columns.
A000255 equals the sum of the bottom half of each column.
A000166 equals the sum of each column.
A047920 represents the frequency of replacements by person drawing at position n.
A008277, Triangle of Stirling numbers of 2nd kind, can be derived from A136301 through a series of transformations (see "Probability of Derangements.pdf").
Cf. A371761.

maxP = 15;
rows = Range[1, 2^(nP = maxP - 3)];
pasc = Table[
   Binomial[p + 1, i] - If[i >= p, 1, 0], {p, nP}, {i, 0, p}];
sFreq = Table[0, {maxP - 1}, {2^nP}]; sFreq[[2 ;; maxP - 1, 1]] = 1;
For[p = 1, p <= nP, p++,
  For[s = 1, s <= p, s++, rS = Range[2^(s - 1) + 1, 2^s];
        sFreq[[p + 2, rS]] = pasc[[p + 1 - s, 1 ;; p + 2 - s]] .
            sFreq[[s ;; p + 1, 1 ;; 2^(s - 1)]]]];
TableForm[ Transpose[ sFreq ] ]
(* Code snippet to illustrate the conjectured connection with A371761: *)
R[n_] := Table[Transpose[sFreq][[2^n]][[r]], {r, n + 1, maxP - 1}]
For[n = 0, n <= 6, n++, Print[n + 1, ": ", R[n]]] (* Peter Luschny, Apr 10 2024 *)

H(r,c) = Sum_{j=0..c-L(r)-1} H(T(r), L(r)+j) * M(c-T(r)-1, j) where M(y,z) = binomial distribution (y,z) when y - 1 > z and (y,z)-1 when y-1 <= z and T(r) = A053645 and L(r) = A000523.

Conjecture: Assume the table represented as in the Example section. Then row 2^n is row n + 1 of A371761. - Peter Luschny, Apr 10 2024

Edited by Brian Parsonnet, Mar 01 2011

A176734 a(n) = (n+7)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

Original entry on oeis.org

1, 8, 73, 746, 8425, 104084, 1395217, 20157542, 312129649, 5155334720, 90449857081, 1679650774658, 32908313146393, 678322072223756, 14672571587601985, 332293083938376254, 7862829504396683617, 194024597448534426872, 4984283037788104293289, 133083801736564331309210

Offset: 0

Wolfdieter Lang, Jul 14 2010

a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=8 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A049388(n) = (n+7)!/7!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs holds. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010).

			Necklaces and 8 cords problem. For n=4 one considers the following weak 2 part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively !4*1,binomial(4,3)*!3*c8(1), (binomial(4,2)*!2)*c8(2), and 1*c8(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c8(n):=A049388(n) numbers for the pure 8-cord problem (see the remark on the e.g.f. for the k cords problem in A000153; here for k=8: 1/(1-x)^8). This adds up as 9 + 4*2*8 + (6*1)*72 + 7920 = 8425 = a(4).

Cf. A176733 (necklaces and k=7 cords).

nxt[{n_,a_,b_}]:={n+1,b,(n+8)b+n*a}; Transpose[NestList[nxt,{1,1,8},20]][[2]] (* Harvey P. Dale, Mar 19 2013 *)
Table[(-1)^n HypergeometricPFQ[{9, -n}, {}, 1], {n, 0, 20}] (* Benedict W. J. Irwin, May 29 2016 *)

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^8) = exp(-x)/(1-x)^9, equivalent to the given recurrence.
a(n) = A086764(n+8,8).
a(n) = (-1)^n*2F0(9,-n;;1). - Benedict W. J. Irwin, May 29 2016

A176736 a(n) = (n+9)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

Original entry on oeis.org

1, 10, 111, 1352, 17909, 256134, 3931555, 64441684, 1123029513, 20730064706, 403978495031, 8286870547680, 178468044946621, 4025739435397822, 94912091598455979, 2334250550458513004, 59779945135439664785, 1591626582328767492474, 43990176790179196598143, 1260374228606935319612536

Offset: 0

Wolfdieter Lang, Jul 14 2010

a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=10 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A049398(n) = (n+9)!/9!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs holds. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010).

			Necklaces and 10 cords problem. For n=4 one considers the following weak 2-part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively !4*1,binomial(4,3)*!3*c10(1), (binomial(4,2)*! 2)*c10(2), and 1*c10(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c10(n):=A049398(n) numbers for the pure 10-cord problem (see the remark on the e.g.f. for the k-cord problem in A000153; here for k=10: 1/(1-x)^10). This adds up as 9 + 4*2*10 + (6*1)*110 + 17160 = 17909 = a(4).

Cf. A176735 (necklaces and k=9 cords).

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^10) = exp(-x)/(1-x)^11, equivalent to the given recurrence.

a(n) = A086764(n+10,10).

A189283 Number of permutations p of 1,2,...,n satisfying p(i+4)-p(i)<>4 for all 1<=i<=n-4.

Original entry on oeis.org

1, 1, 2, 6, 24, 114, 628, 4062, 30360, 255186, 2414292, 25350954, 292378968, 3673917102, 49928069188, 729534877758, 11403682481112, 189862332575658, 3354017704180052, 62654508729565554, 1233924707891272728, 25550498290562247438

Offset: 0

Vaclav Kotesovec, Apr 19 2011

a(n) is also number of ways to place n nonattacking pieces rook + semi-leaper[4,4] on an n X n chessboard.

Vaclav Kotesovec, Table of n, a(n) for n = 0..27 (Updated Jan 19 2019)
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 644.
Vaclav Kotesovec, Mathematica program for this sequence
George Spahn and Doron Zeilberger, Counting Permutations Where The Difference Between Entries Located r Places Apart Can never be s (For any given positive integers r and s), arXiv:2211.02550 [math.CO], 2022.

Cf. A000255, A189281, A189282, A189255.

Asymptotics (V. Kotesovec, Mar 2011): a(n)/n! ~ (1 + 7/n + 12/n^2)/e.

Terms a(26)-a(27) from Vaclav Kotesovec, Apr 20 2012

A209326 Number of permutations of [n] with a fixed point but no succession.

Original entry on oeis.org

0, 1, 0, 3, 7, 39, 207, 1437, 11203, 99041, 975645, 10601377, 125905445, 1622349059, 22539777113, 335845307359, 5341990288103, 90340567900583, 1618553943500599, 30623660893656205, 610152486797080443, 12769086757046132625, 280037186109883699885, 6422309829486480886809, 153727262708736577446741, 3833789797689152809143363

Offset: 0

Jon Perry, Jan 19 2013

nonn

A succession of a permutation p is a position i such that p(i+1)-p(i) = 1.

			For n=4 we have 1324, 1432, 2431, 3214, 3241, 4132 and 4213.

Max Alekseyev, Table of n, a(n) for n = 0..30

Cf. A000166, A000255, A002467, A180191, A207819, A207821, A209322, A209325.

a(n) = A000255(n-1) - A209322(n). - Max Alekseyev, Apr 03 2025

a(11)-a(14) from Alois P. Heinz, Jan 20 2013
a(15)-a(21) from Alois P. Heinz, Jul 04 2021
Terms a(22) onward from Max Alekseyev, Apr 03 2025

A288953 Number of relaxed compacted binary trees of right height at most one with minimal sequences between branch nodes except after the last branch node on level 0.

Original entry on oeis.org

1, 1, 3, 10, 51, 280, 1995, 15120, 138075, 1330560, 14812875, 172972800, 2271359475, 31135104000, 471038042475, 7410154752000, 126906349444875, 2252687044608000, 43078308695296875, 851515702861824000, 17984171447178811875, 391697223316439040000

Offset: 0

Michael Wallner, Jun 20 2017

nonn

A relaxed compacted binary tree of size n is a directed acyclic graph consisting of a binary tree with n internal nodes, one leaf, and n pointers. It is constructed from a binary tree of size n, where the first leaf in a post-order traversal is kept and all other leaves are replaced by pointers. These links may point to any node that has already been visited by the post-order traversal. The right height is the maximal number of right-edges (or right children) on all paths from the root to any leaf after deleting all pointers. A branch node is a node with a left and right edge (no pointer). See the Genitrini et al. link. - Michael Wallner, Apr 20 2017

a(n) is the number of plane increasing trees with n+1 nodes where in the growth process induced by the labels maximal young leaves and non-maximal young leaves alternate except for a sequence of maximal young leaves at the beginning. A young leaf is a leaf with no left sibling. A maximal young leaf is a young leaf with maximal label. See the Wallner link. - Michael Wallner, Apr 20 2017

			Denote by L the leaf and by o nodes. Every node has exactly two out-going edges or pointers. Internal edges are denoted by - or |. Pointers are omitted and may point to any node further right. The root is at level 0 at the very left.
The general structure is
  L-o-o-o-o-o-o-o-o
          | | | | |
          o o o o o.
For n=0 the a(0)=1 solution is L.
For n=1 the a(1)=1 solution is L-o.
For n=2 the a(2)=3 solutions are
L-o-o     L-o
            |
            o
  2    +   1    solutions of this shape with pointers.

Antoine Genitrini, Bernhard Gittenberger, Manuel Kauers and Michael Wallner, Asymptotic Enumeration of Compacted Binary Trees, arXiv:1703.10031 [math.CO], 2017
Michael Wallner, A bijection of plane increasing trees with relaxed binary trees of right height at most one, arXiv:1706.07163 [math.CO], 2017

Cf. A288954 (variation with additional initial sequence).
Cf. A177145 (variation without final sequence).
Cf. A001147 (relaxed compacted binary trees of right height at most one).
Cf. A082161 (relaxed compacted binary trees of unbounded right height).
Cf. A000032, A000246, A001879, A051577, A213527, A288950, A288952, A288954 (subclasses of relaxed compacted binary trees of right height at most one, see the Wallner link).
Cf. A000166, A000255, A000262, A052852, A123023, A130905, A176408, A201203 (variants of relaxed compacted binary trees of right height at most one, see the Wallner link).

E.g.f.: (2-z)/(3*(1-z)^2) + 1/(3*sqrt(1-z^2)).

A346204 a(n) is the number of permutations on [n] with at least one strong fixed point and at least one small descent.

Original entry on oeis.org

0, 0, 2, 5, 24, 128, 795, 5686, 46090, 418519, 4213098, 46595650, 561773033, 7333741536, 103065052300, 1551392868821, 24902155206164, 424588270621876, 7663358926666175, 145967769353476594, 2926073829112697318, 61577929208485406331, 1357369100658321844470, 31276096500003460511422

Offset: 1

Eugene Fiorini, Jared Glassband, Garrison Lee Koch, Sophia Lebiere, Xufei Liu, Evan Sabini, Nathan B. Shank, Andrew Woldar, Jul 10 2021

A small descent in a permutation p is a position i such that p(i)-p(i+1)=1.

A strong fixed point is a fixed point (or splitter) p(k)=k such that p(i) < k for i < k and p(j) > k for j > k.

			For n=4, the a(4)=5 permutations on [4] with strong fixed points and small descents: {(1*, 2*, [4, 3]), (1*, [3, 2], 4*), (1*, <4, 3, 2>), ([2, 1], 3*, 4*), (<3, 2, 1>, 4*)}. *strong fixed point, []small descent, <>consecutive small descents.

E. R. Berlekamp, J. H. Conway, and R. K. Guy, Winning Ways For Your Mathematical Plays, Vol. 1, CRC Press, 2001.

M. Lind, E. Fiorini, A. Woldar, and W. H. T. Wong, On Properties of Pebble Assignment Graphs, Journal of Integer Sequences, 24(6), 2020.

Cf. A000255, A000166, A000153, A000261, A001909, A001910, A055790, A346189, A346198, A346199.

import math
bn = [1,1,1]
wn = [0,0,0]
kn = [1,1,1]
def summation(n):
    final = bn[n] - bn[n-1]
    for k in range(4,n+1):
        final -= wn[k-1]*bn[n-k]
    return final
def smallsum(n):
    final = bn[n-1]
    for k in range(4,n+1):
        final += wn[k-1]*bn[n-k]
    return final
def derrangement(n):
    finalsum = 0
    for i in range(n+1):
        if i%2 == 0:
            finalsum += math.factorial(n)*1//math.factorial(i)
        else:
            finalsum -= math.factorial(n)*1//math.factorial(i)
    if finalsum != 0:
        return finalsum
    else:
        return 1
def fixedpoint(n):
    finalsum = math.factorial(n-1)
    for i in range(2,n):
        finalsum += math.factorial(i-i)*math.factorial(n-i-1)
        print(math.factorial(i-i)*math.factorial(n-i-1))
    return finalsum
def no_cycles(n):
    goal = n
    cycles = [0, 1]
    current = 2
    while current<= goal:
        new = 0
        k = 1
        while k<=current:
            new += (math.factorial(k-1)-cycles[k-1])*(math.factorial(current-k))
            k+=1
        cycles.append(new)
        current+=1
    return cycles
def total_func(n):
    for i in range(3,n+1):
        bn.append(derrangement(i+1)//(i))
        kn.append(smallsum(i))
        wn.append(summation(i))
    an = no_cycles(n)
    tl = [int(an[i]-kn[i]) for i in range(n+1)]
    factorial = [math.factorial(x) for x in range(0,n+1)]
    print("A346189 :" + str(wn[1:]))
    print("A346198 :" + str([factorial[i]-wn[i]-tl[i]-kn[i] for i in range(n+1)][1:]))
    print("A346199 :" + str(kn[1:]))
    print("A346204 :" + str(tl[1:]))
total_func(20)

a(n) = A006932(n) - A346199(n).

A369596 Number T(n,k) of permutations of [n] whose fixed points sum to k; triangle T(n,k), n>=0, 0<=k<=A000217(n), read by rows.

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 1, 2, 1, 1, 1, 0, 0, 1, 9, 2, 2, 3, 3, 2, 1, 1, 0, 0, 1, 44, 9, 9, 11, 11, 13, 5, 5, 4, 4, 2, 1, 1, 0, 0, 1, 265, 44, 44, 53, 53, 62, 64, 29, 22, 24, 16, 16, 8, 6, 5, 4, 2, 1, 1, 0, 0, 1, 1854, 265, 265, 309, 309, 353, 362, 406, 150, 159, 126, 126, 93, 86, 44, 36, 29, 19, 19, 9, 7, 5, 4, 2, 1, 1, 0, 0, 1

Offset: 0

Alois P. Heinz, Mar 02 2024

			T(3,0) = 2: 231, 312.
T(3,1) = 1: 132.
T(3,2) = 1: 321.
T(3,3) = 1: 213.
T(3,6) = 1: 123.
T(4,0) = 9: 2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321.
Triangle T(n,k) begins:
   1;
   0, 1;
   1, 0, 0,  1;
   2, 1, 1,  1,  0,  0, 1;
   9, 2, 2,  3,  3,  2, 1, 1, 0, 0, 1;
  44, 9, 9, 11, 11, 13, 5, 5, 4, 4, 2, 1, 1, 0, 0, 1;
  ...

Alois P. Heinz, Rows n = 0..50, flattened
Wikipedia, Permutation

Column k=0 gives A000166.
Column k=3 gives A000255(n-2) for n>=2.
Row sums give A000142.
Row lengths give A000124.
Reversed rows converge to A331518.
T(n,n) gives A369796.
Cf. A000217, A001710, A008289, A008290, A038720, A053632, A062869, A124327, A143946, A143947, A161680, A263753, A263756, A317527, A367955, A368338.

b:= proc(s) option remember; (n-> `if`(n=0, 1, add(expand(
      `if`(j=n, x^j, 1)*b(s minus {j})), j=s)))(nops(s))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b({$1..n})):
seq(T(n), n=0..7);
# second Maple program:
g:= proc(n) option remember; `if`(n=0, 1, n*g(n-1)+(-1)^n) end:
b:= proc(n, i, m) option remember; `if`(n>i*(i+1)/2, 0,
     `if`(n=0, g(m), b(n, i-1, m)+b(n-i, min(n-i, i-1), m-1)))
    end:
T:= (n, k)-> b(k, min(n, k), n):
seq(seq(T(n, k), k=0..n*(n+1)/2), n=0..7);

g[n_] := g[n] = If[n == 0, 1, n*g[n - 1] + (-1)^n];
b[n_, i_, m_] := b[n, i, m] = If[n > i*(i + 1)/2, 0,
   If[n == 0, g[m], b[n, i-1, m] + b[n-i, Min[n-i, i-1], m-1]]];
T[n_, k_] := b[k, Min[n, k], n];
Table[Table[T[n, k], {k, 0, n*(n + 1)/2}], {n, 0, 7}] // Flatten (* Jean-François Alcover, May 24 2024, after Alois P. Heinz *)

Sum_{k=0..A000217(n)} k * T(n,k) = A001710(n+1) for n >= 1.
Sum_{k=0..A000217(n)} (1+k) * T(n,k) = A038720(n) for n >= 1.
Sum_{k=0..A000217(n)} (n*(n+1)/2-k) * T(n,k) = A317527(n+1).
T(n,A161680(n)) = A331518(n).
T(n,A000217(n)) = 1.

A001887 Number of permutations p of {1,2,...,n} such that p(i) - i < 0 or p(i) - i > 2 for all i.

Original entry on oeis.org

1, 0, 0, 0, 1, 5, 33, 236, 1918, 17440, 175649, 1942171, 23396353, 305055960, 4280721564, 64330087888, 1030831875953, 17545848553729, 316150872317105, 6012076099604308, 120330082937778554

Offset: 0

N. J. A. Sloane

nonn

Previous name was: Hit polynomials.

J. Riordan, The enumeration of permutations with three-ply staircase restrictions, unpublished memorandum, Bell Telephone Laboratories, Murray Hill, NJ, Oct 1963. (See A001883)
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).

Alois P. Heinz, Table of n, a(n) for n = 0..300
E. R. Canfield, N. C. Wormald, Menage numbers, bijections and P-recursiveness, Discr. Math. 63 (1987) 117, table Section 7.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 373
V. Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 224.
N. J. A. Sloane, Annotated copy of Riordan's Three-Ply Staircase paper (unpublished memorandum, Bell Telephone Laboratories, Murray Hill, NJ, Oct 1963)
D. Zeilberger, Automatic Enumeration of Generalized Menage Numbers, arXiv preprint arXiv:1401.1089 [math.CO], 2014.

Cf. A000255, A000271, A001883-A001891, A075851, A075852.

First column of A080061.

nmax = 21;
gf = 1/(x^2-1)(x-Sum[n! (x(x-1)/(x^3-2x-1))^n + O[x]^nmax, {n, 0, nmax}]);
CoefficientList[gf, x] (* Jean-François Alcover, Aug 19 2018 *)

G.f.: (1/(x^2-1))*(x-Sum_{n>=0} n!*(x*(x-1)/(x^3-2*x-1))^n). - Vladeta Jovovic, Jun 30 2007
D-finite with recurrence (P. Flajolet, 1997): a(n) = (n-1)*a(n-1) + (n+2)*a(n-2) - (3*n-13)*a(n-3) - (2*n-8)*a(n-4) + (3*n-15)*a(n-5) + (n-4)*a(n-6) - (n-7)*a(n-7) - a(n-8), n>8.
a(n) ~ exp(-3) * n!. - Vaclav Kotesovec, Sep 10 2014

More terms from Vladimir Baltic and Vladeta Jovovic, Jan 05 2003

New name from Vaclav Kotesovec using a former comment by Vladimir Baltic and Vladeta Jovovic, Sep 16 2014

cp's OEIS Frontend

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A176734 a(n) = (n+7)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1.

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A176736 a(n) = (n+9)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1.

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A209326 Number of permutations of [n] with a fixed point but no succession.

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A288953 Number of relaxed compacted binary trees of right height at most one with minimal sequences between branch nodes except after the last branch node on level 0.

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A346204 a(n) is the number of permutations on [n] with at least one strong fixed point and at least one small descent.

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A176734 a(n) = (n+7)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

A176736 a(n) = (n+9)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.