cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A002866 a(0) = 1; for n > 0, a(n) = 2^(n-1)*n!.

Original entry on oeis.org

1, 1, 4, 24, 192, 1920, 23040, 322560, 5160960, 92897280, 1857945600, 40874803200, 980995276800, 25505877196800, 714164561510400, 21424936845312000, 685597979049984000, 23310331287699456000, 839171926357180416000, 31888533201572855808000, 1275541328062914232320000
Offset: 0

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Author

Keywords

Comments

Consider the set of n-1 odd numbers from 3 to 2n-1, i.e., {3, 5, ..., 2n-1}. There are 2^(n-1) subsets from {} to {3, 5, 7, ..., 2n-1}; a(n) = the sum of the products of terms of all the subsets. (Product for empty set = 1.) a(4) = 1 + 3 + 5 + 7 + 3*5 + 3*7 + 5*7 + 3*5*7 = 192. - Amarnath Murthy, Sep 06 2002
Also, a(n-1) is the number of ways to lace a shoe that has n pairs of eyelets such that there is a straight (horizontal) connection between all adjacent eyelet pairs. - Hugo Pfoertner, Jan 27 2003
This is also the denominator of the integral of ((1-x^2)^(n-1/2))/(Pi/4) where x ranges from 0 to 1. The numerator is (2*x)!/(x!*2^x). In both cases n starts at 1. E.g., the denominator when n=3 is 24 and the numerator is 15. - Al Hakanson (hawkuu(AT)excite.com), Oct 17 2003
Number of ways to use the elements of {1,...,n} once each to form a sequence of nonempty lists. - Bob Proctor, Apr 18 2005
Row sums of A131222. - Paul Barry, Jun 18 2007
Number of rotational symmetries of an n-cube. The number of all symmetries of an n-cube is A000165. See Egan for signed cycle notation, other notes, tables and animation. - Jonathan Vos Post, Nov 28 2007
1, 4, 24, 192, 1920, ... is the exponential (or binomial) convolution of 1, 1, 3, 15, 105, ... and 1, 3, 15, 105, 945 (A001147). - David Callan, Jul 25 2008
The n-th term of this sequence is the number of regions into which n-dimensional space is partitioned by the 2n hyperplanes of the form x_i=x_j and x_i=-x_j (for 1 <= i < j <= n). - Edward Scheinerman (ers(AT)jhu.edu), May 04 2008
a(n) is the number of ways to seat n churchgoers into pews and then linearly order the nonempty pews. - Geoffrey Critzer, Mar 16 2009
Equals the row sums of A156992. - Geoffrey Critzer, Mar 05 2010
From Gary W. Adamson, May 17 2010: (Start)
Next term in the series = (1, 3, 5, 7, ...) dot (1, 1, 4, 24, ...);
e.g., a(5) = 1920 = (1, 3, 5, 7, 9) dot (1, 1, 4, 24, 192) = (1 + 3 + 20 + 168 + 1728). (End)
a(n) is the number of ways to represent the permutations of {1,2,...,n} in cycle notation, taking into account that we can permute the order of all cycles and also have k ways to write a length-k cycle.
For positive n, a(n) equals the permanent of the n X n matrix with consecutive integers 1 to n along the main diagonal, consecutive integers 2 to n along the subdiagonal, and 1's everywhere else. - John M. Campbell, Jul 10 2011
From Dennis P. Walsh, Nov 26 2011: (Start)
Number of ways to arrange n books on consecutive bookshelves.
To derive a(n) = n!2^(n-1), we note that there are n! ways to arrange the books in a row. Then there are 2^(n-1) ways to place the arranged books on consecutive shelves since there are 2^(n-1) ordered partitions of n. Hence a(n) = n!2^(n-1).
Also, a(n) is the number of ways to stack n different alphabet blocks in contiguous stacks.
Furthermore, a(n) is the number of labeled, rooted forests that have (i) each root labeled larger than any nonroot, (ii) each root having exactly one child node, (iii) n non-root nodes, and (iv) each node in the forest with at most one child node.
Example: a(3)=24 since there are 24 arrangements of books b1, b2, and b3 on consecutive shelves, namely, |b1 b2 b3|, |b1 b3 b2|, |b2 b1 b3|, |b2 b3 b1|, |b3 b1 b2|, |b3 b2 b1|, |b1 b2||b3|, |b2 b1| |b3|, |b1 b3||b2|, |b3 b1||b2|, |b2 b3||b1|, |b3 b2||b1|, |b1||b2 b3|,|b1||b3 b2|, |b2||b1 b3|, |b2||b3 b1|, |b3||b1 b2|, |b3||b2 b1|, |b1||b2||b3|, |b1||b3||b2|, |b2||b1||b3|, |b2||b3||b1|, |b3||b1||b2|, and |b3||b2||b1|.
(End)
For n > 3, a(n) is the order of the Coxeter group (also called Weyl group) of type D_n. - Tom Edgar, Nov 05 2013

Examples

			For the shoe lacing: with the notation introduced in A078602 the a(3-1) = 4 "straight" lacings for 3 pairs of eyelets are: 125346, 125436, 134526, 143526. Their mirror images 134256, 143256, 152346, 152436 are not counted.
a(3) = 24 because the 24 rotations of a three-dimensional cube fall into four distinct classes:
(i) the identity, which leaves everything fixed;
(ii) 9 rotations which leave the centers of two faces fixed, comprising rotations of 90, 180 and 270 degrees for each of 3 pairs of faces;
(iii) 6 rotations which leave the centers of two edges fixed, comprising rotations of 180 degrees for each of 6 pairs of edges;
(iv) 8 rotations which leave two vertices fixed, comprising rotations of 120 and 240 degrees for each of 4 pairs of vertices. For an n-cube, rotations can be more complex. For example, in 4 dimensions a rotation can either act in a single plane, such as the x-y plane, while leaving any vectors orthogonal to that plane unchanged, or it can act in two orthogonal planes, performing rotations in both and leaving no vectors fixed. In higher dimensions, there will be room for more planes and more choices as to the number of planes in which a given rotation acts.
		

References

  • N. Bourbaki, Groupes et alg. de Lie, Chap. 4, 5, 6, p. 257.
  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992, Eq. (4.2.2.26)
  • 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).

Crossrefs

Bisections give A002671 and A274304.
Appears in A167584 (n >= 1); equals the row sums of A167594 (n >= 1). - Johannes W. Meijer, Nov 12 2009

Programs

  • FORTRAN
    See Pfoertner link.
    
  • Magma
    [1] cat [2^(n-1)*Factorial(n): n in [1..25]]; // G. C. Greubel, Jun 13 2019
    
  • Maple
    A002866 := n-> `if`(n=0,1,2^(n-1)*n!):
    with(combstruct); SeqSeqL := [S, {S=Sequence(U,card >= 1), U=Sequence(Z,card >=1)},labeled];
    seq(ceil(count(Subset(n))*count(Permutation(n))/2),n=0..17); # Zerinvary Lajos, Oct 16 2006
    G(x):=(1-x)/(1-2*x): f[0]:=G(x): for n from 1 to 26 do f[n]:=diff(f[n-1],x) od:x:=0:seq(f[n],n=0..17); # Zerinvary Lajos, Apr 04 2009
  • Mathematica
    Join[{1},Table[2^(n-1) n!,{n,25}]] (* Harvey P. Dale, Sep 27 2013 *)
    a[n_] := (-1)^n Hypergeometric2F1Regularized[1, -n, 2 - n, 2];
    Table[a[n], {n, 0, 20}]  (* Peter Luschny, Apr 26 2024 *)
  • PARI
    a(n)=if(n,n!<<(n-1),1) \\ Charles R Greathouse IV, Jan 13 2012
    
  • PARI
    a(n) = if(n == 0, 1, 2^(n-1)*n!);
    vector(25, n, a(n-1)) \\ Altug Alkan, Oct 18 2015
    
  • Sage
    [1] + [2^(n-1)*factorial(n) for n in (1..25)] # G. C. Greubel, Jun 13 2019

Formula

E.g.f.: (1 - x)/(1 - 2*x). - Paul Barry, May 26 2003, corrected Jun 18 2007
a(n) = n! * A011782(n).
For n >= 1, a(n) = Sum_{i=0..m/2} (-1)^i * binomial(n, i) * (n-2*i)^n. - Yong Kong (ykong(AT)curagen.com), Dec 28 2000
a(n) ~ 2^(1/2) * Pi^(1/2) * n^(3/2) * 2^n * e^(-n) * n^n*{1 + 13/12*n^(-1) + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 23 2001
E.g.f. is B(A(x)), where B(x) = 1/(1 - x) and A(x) = x/(1 - x). - Geoffrey Critzer, Mar 16 2009
a(n) = Sum_{k=1..n} A156992(n,k). - Dennis P. Walsh, Nov 26 2011
a(n+1) = Sum_{k=0..n} A132393(n,k)*2^(n+k), n>0. - Philippe Deléham, Nov 28 2011
G.f.: 1 + x/(1 - 4*x/(1 - 2*x/(1 - 6*x/(1 - 4*x/(1 - 8*x/(1 - 6*x/(1 - 10*x/(1 - ... (continued fraction). - Philippe Deléham, Nov 29 2011
a(n) = 2*n*a(n-1) for n >= 2. - Dennis P. Walsh, Nov 29 2011
G.f.: (1 + 1/G(0))/2, where G(k) = 1 + 2*x*k - 2*x*(k + 1)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 02 2012
G.f.: 1 + x/Q(0), m=4, where Q(k) = 1 - m*x*(2*k + 1) - m*x^2*(2*k + 1)*(2*k + 2)/(1 - m*x*(2*k + 2) - m*x^2*(2*k + 2)*(2*k + 3)/Q(k+1)) ; (continued fraction). - Sergei N. Gladkovskii, Sep 23 2013
G.f.: 1 + x/(G(0) - x), where G(k) = 1 + x*(k+1) - 4*x*(k + 1)/(1 - x*(k + 2)/G(k+1)); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
a(n) = Sum_{k=0..n} L(n,k)*k!; L(n,k) are the unsigned Lah numbers. - Peter Luschny, Oct 18 2014
a(n) = round(Sum_{k >= 1} log(k)^n/k^(3/2))/4, for n >= 1, which is related to the n-th derivative of zeta(x) evaluated at x = 3/2. - Richard R. Forberg, Jan 02 2015
a(n) = n!*hypergeom([-n+1], [], -1) for n>=1. - Peter Luschny, Apr 08 2015
From Amiram Eldar, Aug 04 2020: (Start)
Sum_{n >= 0} 1/a(n) = 2*sqrt(e) - 1.
Sum_{n >= 0} (-1)^n/a(n) = 2/sqrt(e) - 1. (End)

A078698 Number of ways to lace a shoe that has n pairs of eyelets such that each eyelet has at least one direct connection to the opposite side.

Original entry on oeis.org

1, 2, 20, 396, 14976, 907200, 79315200, 9551001600, 1513528934400, 305106949324800, 76296489615360000, 23175289163980800000, 8404709419090575360000, 3587225703492542791680000, 1779970753996760560435200000, 1016036270188884847558656000000, 661106386935312429191528448000000
Offset: 1

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Author

Hugo Pfoertner, Dec 18 2002

Keywords

Comments

The lace is "directed": reversing the order of eyelets along the path counts as a different solution. It must begin and end at the extreme pair of eyelets,

Examples

			a(3) = 20: label the eyelets 1,2,3 from front to back on the left side then 4,5,6 from back to front on the right side. The lacings are: 124356 154326 153426 142536 145236 135246 and the following and their mirror images: 125346 124536 125436 152346 153246 152436 154236.
Examples for n=2,3,4 can be found following the FORTRAN program at given link.
		

References

  • C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 494.

Crossrefs

Programs

  • Fortran
    c Program provided at Pfoertner link
  • Mathematica
    a[n_] := (n-1)!^2 Sum[Binomial[n-k, k]^2, {k, 0, n/2}];
    Array[a, 17] (* Jean-François Alcover, Jul 20 2018 *)

Formula

Conjecture: a(n) = (n-1)!^2*A051286(n). - Vladeta Jovovic, Sep 14 2005 (correct, see the Khrabrov/Kokhas reference, Joerg Arndt, May 26 2015)

Extensions

Terms a(9) and beyond (using A051286) from Joerg Arndt, May 26 2015

A078700 Number of symmetric ways to lace a shoe that has n pairs of eyelets such that each eyelet has at least one direct connection to the opposite side.

Original entry on oeis.org

1, 2, 6, 30, 192, 1560, 15120, 171360, 2217600, 32296320, 522547200, 9300614400, 180583603200, 3798482688000, 86044973414400, 2088355965696000, 54064489070592000, 1487129136869376000, 43312058119249920000
Offset: 1

Views

Author

Hugo Pfoertner, Dec 18 2002

Keywords

Comments

The lace must pass through each eyelet exactly one and must begin and end at the extreme pair of eyelets.

Examples

			a(3) = 6: label the eyelets 1,2,3 from front to back on the left side then 4,5,6 from back to front on the right side. The symmetric lacings are: 124356 154326 153426 142536 145236 135246.
		

Crossrefs

Programs

  • Maple
    ZL:=[S, {a = Atom, b = Atom, S = Prod(X,Sequence(Prod(X,b))), X = Sequence(b,card <= 1)}, labelled]: seq(combstruct[count](ZL, size=n), n=0..18); # Zerinvary Lajos, Mar 26 2008
  • Mathematica
    Table[Fibonacci[n + 1]*(n - 1)!, {n, 1, 19}] (* Zerinvary Lajos, Jul 09 2009 *)
  • Scheme
    (define (A078700 n) (* (A000142 (- n 1)) (A000045 (+ n 1)))) ;; Antti Karttunen, Jan 02 2007

Formula

a(n) = (n-1)!*Fibonacci(n+1) = A000142(n-1)*A000045(n+1). - Conjectured by Vladeta Jovovic, Jan 23 2005, proved by Antti Karttunen, Jan 06 2007.
Proof of Jovovic's conjecture from Antti Karttunen, Jan 06 2007: (Start)
Because of the symmetry and the beginning and ending conditions, we need to consider only n-1 eyelets on the other side. If considering only the distance from the starting and ending eyelets (the "level" of each eyelet-pair) through which the lace is traveling (but ignoring on which side it is), the lace will induce some permutation of {1..(n-1)} after it has visited half of the eyelets (and the remaining half of its route is wholly determined by the symmetry). This gives the factor (n-1)!. Independently of this, the condition that each eyelet has at least one direct connection to the opposite side, means that there is a simple bijection with binary strings of length n with no two consecutive 0's. I.e. we mark 0 when the lace stays on the same side and 1 when it crosses to the other side.
From the starting eyelet the lace can either cross to the other side (but not to the top one) or stay on the same side. However, after visiting half of the eylets, the lace MUST cross to the other side (on the same level), so this leaves n-1 eyelets where the choice is free, except that there can be no two consecutive 0's on the route. This gives the factor Fibonacci(n+1). (End)

Extensions

More terms added by Antti Karttunen, Jan 02 2007

A072503 Number of ways to lace a shoe with n eyelet pairs such that there is no direct "horizontal" connection between any adjacent eyelet pair.

Original entry on oeis.org

3, 45, 1824, 109560, 9520560, 1145057760, 181091917440
Offset: 3

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Author

Hugo Pfoertner, Jan 27 2003

Keywords

Comments

The lacing must not have any "straight connections" between adjacent eyelet pairs (e.g. 2<->2*n-1, 3<->2*n-2, 4<->2*n-3,....). There are no symmetric solutions.
From Sean A. Irvine, Oct 06 2024: (Start)
The lacing must begin and end with the top eyelet pair (1 and 2n in Pfoertner's numbering).
Every eyelet must be used.
Under versus over crossings are not considered distinct.
Three consecutive eyelets on the same side are not permitted.
Because there are no symmetric solutions, mirrored solutions can be handled by simply dividing the total solutions by 2. (End)

Examples

			The 6 non-straight lacings for n=3 are: 124536, 135426, 142356, 145326, 153246, 154236. Not counting mirror images we get a(3)=3.
		

Crossrefs

Programs

  • Fortran
    See Links.

Extensions

a(9) from Sean A. Irvine, Oct 07 2024
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