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.

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A000670 Fubini numbers: number of preferential arrangements of n labeled elements; or number of weak orders on n labeled elements; or number of ordered partitions of [n].

Original entry on oeis.org

1, 1, 3, 13, 75, 541, 4683, 47293, 545835, 7087261, 102247563, 1622632573, 28091567595, 526858348381, 10641342970443, 230283190977853, 5315654681981355, 130370767029135901, 3385534663256845323, 92801587319328411133, 2677687796244384203115, 81124824998504073881821
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

Number of ways n competitors can rank in a competition, allowing for the possibility of ties.
Also number of asymmetric generalized weak orders on n points.
Also called the ordered Bell numbers.
A weak order is a relation that is transitive and complete.
Called Fubini numbers by Comtet: counts formulas in Fubini theorem when switching the order of summation in multiple sums. - Olivier Gérard, Sep 30 2002 [Named after the Italian mathematician Guido Fubini (1879-1943). - Amiram Eldar, Jun 17 2021]
If the points are unlabeled then the answer is a(0) = 1, a(n) = 2^(n-1) (cf. A011782).
For n>0, a(n) is the number of elements in the Coxeter complex of type A_{n-1}. The corresponding sequence for type B is A080253 and there one can find a worked example as well as a geometric interpretation. - Tim Honeywill and Paul Boddington, Feb 10 2003
Also number of labeled (1+2)-free posets. - Detlef Pauly, May 25 2003
Also the number of chains of subsets starting with the empty set and ending with a set of n distinct objects. - Andrew Niedermaier, Feb 20 2004
From Michael Somos, Mar 04 2004: (Start)
Stirling transform of A007680(n) = [3,10,42,216,...] gives [3,13,75,541,...].
Stirling transform of a(n) = [1,3,13,75,...] is A083355(n) = [1,4,23,175,...].
Stirling transform of A000142(n) = [1,2,6,24,120,...] is a(n) = [1,3,13,75,...].
Stirling transform of A005359(n-1) = [1,0,2,0,24,0,...] is a(n-1) = [1,1,3,13,75,...].
Stirling transform of A005212(n-1) = [0,1,0,6,0,120,0,...] is a(n-1) = [0,1,3,13,75,...].
(End)
Unreduced denominators in convergent to log(2) = lim_{n->infinity} n*a(n-1)/a(n).
a(n) is congruent to a(n+(p-1)p^(h-1)) (mod p^h) for n >= h (see Barsky).
Stirling-Bernoulli transform of 1/(1-x^2). - Paul Barry, Apr 20 2005
This is the sequence of moments of the probability distribution of the number of tails before the first head in a sequence of fair coin tosses. The sequence of cumulants of the same probability distribution is A000629. That sequence is twice the result of deletion of the first term of this sequence. - Michael Hardy (hardy(AT)math.umn.edu), May 01 2005
With p(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j,i) = the j-th part of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i=1..p(n)} (n!/(Product_{j=1..p(i)} p(i,j)!)) * (p(i)!/(Product_{j=1..d(i)} m(i,j)!)). - Thomas Wieder, May 18 2005
The number of chains among subsets of [n]. The summed term in the new formula is the number of such chains of length k. - Micha Hofri (hofri(AT)wpi.edu), Jul 01 2006
Occurs also as first column of a matrix-inversion occurring in a sum-of-like-powers problem. Consider the problem for any fixed natural number m>2 of finding solutions to the equation Sum_{k=1..n} k^m = (k+1)^m. Erdős conjectured that there are no solutions for n, m > 2. Let D be the matrix of differences of D[m,n] := Sum_{k=1..n} k^m - (k+1)^m. Then the generating functions for the rows of this matrix D constitute a set of polynomials in n (for varying n along columns) and the m-th polynomial defining the m-th row. Let GF_D be the matrix of the coefficients of this set of polynomials. Then the present sequence is the (unsigned) first column of GF_D^-1. - Gottfried Helms, Apr 01 2007
Assuming A = log(2), D is d/dx and f(x) = x/(exp(x)-1), we have a(n) = (n!/2*A^(n+1)) Sum_{k=0..n} (A^k/k!) D^n f(-A) which gives Wilf's asymptotic value when n tends to infinity. Equivalently, D^n f(-a) = 2*( A*a(n) - 2*a(n-1) ). - Martin Kochanski (mjk(AT)cardbox.com), May 10 2007
List partition transform (see A133314) of (1,-1,-1,-1,...). - Tom Copeland, Oct 24 2007
First column of A154921. - Mats Granvik, Jan 17 2009
A slightly more transparent interpretation of a(n) is as the number of 'factor sequences' of N for the case in which N is a product of n distinct primes. A factor sequence of N of length k is of the form 1 = x(1), x(2), ..., x(k) = N, where {x(i)} is an increasing sequence such that x(i) divides x(i+1), i=1,2,...,k-1. For example, N=70 has the 13 factor sequences {1,70}, {1,2,70}, {1,5,70}, {1,7,70}, {1,10,70}, {1,14,70}, {1,35,70}, {1,2,10,70}, {1,2,14,70}, {1,5,10,70}, {1,5,35,70}, {1,7,14,70}, {1,7,35,70}. - Martin Griffiths, Mar 25 2009
Starting (1, 3, 13, 75, ...) = row sums of triangle A163204. - Gary W. Adamson, Jul 23 2009
Equals double inverse binomial transform of A007047: (1, 3, 11, 51, ...). - Gary W. Adamson, Aug 04 2009
If f(x) = Sum_{n>=0} c(n)*x^n converges for every x, then Sum_{n>=0} f(n*x)/2^(n+1) = Sum_{n>=0} c(n)*a(n)*x^n. Example: Sum_{n>=0} exp(n*x)/2^(n+1) = Sum_{n>=0} a(n)*x^n/n! = 1/(2-exp(x)) = e.g.f. - Miklos Kristof, Nov 02 2009
Hankel transform is A091804. - Paul Barry, Mar 30 2010
It appears that the prime numbers greater than 3 in this sequence (13, 541, 47293, ...) are of the form 4n+1. - Paul Muljadi, Jan 28 2011
The Fi1 and Fi2 triangle sums of A028246 are given by the terms of this sequence. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
The modified generating function A(x) = 1/(2-exp(x))-1 = x + 3*x^2/2! + 13*x^3/3! + ... satisfies the autonomous differential equation A' = 1 + 3*A + 2*A^2 with initial condition A(0) = 0. Applying [Bergeron et al., Theorem 1] leads to two combinatorial interpretations for this sequence: (A) a(n) gives the number of plane-increasing 0-1-2 trees on n vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 2 colors. (B) a(n) gives the number of non-plane-increasing 0-1-2 trees on n vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 4 colors. Examples are given below. - Peter Bala, Aug 31 2011
Starting with offset 1 = the eigensequence of A074909 (the beheaded Pascal's triangle), and row sums of triangle A208744. - Gary W. Adamson, Mar 05 2012
a(n) = number of words of length n on the alphabet of positive integers for which the letters appearing in the word form an initial segment of the positive integers. Example: a(2) = 3 counts 11, 12, 21. The map "record position of block containing i, 1<=i<=n" is a bijection from lists of sets on [n] to these words. (The lists of sets on [2] are 12, 1/2, 2/1.) - David Callan, Jun 24 2013
This sequence was the subject of one of the earliest uses of the database. Don Knuth, who had a computer printout of the database prior to the publication of the 1973 Handbook, wrote to N. J. A. Sloane on May 18, 1970, saying: "I have just had my first real 'success' using your index of sequences, finding a sequence treated by Cayley that turns out to be identical to another (a priori quite different) sequence that came up in connection with computer sorting." A000670 is discussed in Exercise 3 of Section 5.3.1 of The Art of Computer Programming, Vol. 3, 1973. - N. J. A. Sloane, Aug 21 2014
Ramanujan gives a method of finding a continued fraction of the solution x of an equation 1 = x + a2*x^2 + ... and uses log(2) as the solution of 1 = x + x^2/2 + x^3/6 + ... as an example giving the sequence of simplified convergents as 0/1, 1/1, 2/3, 9/13, 52/75, 375/541, ... of which the sequence of denominators is this sequence, while A052882 is the numerators. - Michael Somos, Jun 19 2015
For n>=1, a(n) is the number of Dyck paths (A000108) with (i) n+1 peaks (UD's), (ii) no UUDD's, and (iii) at least one valley vertex at every nonnegative height less than the height of the path. For example, a(2)=3 counts UDUDUD (of height 1 with 2 valley vertices at height 0), UDUUDUDD, UUDUDDUD. These paths correspond, under the "glove" or "accordion" bijection, to the ordered trees counted by Cayley in the 1859 reference, after a harmless pruning of the "long branches to a leaf" in Cayley's trees. (Cayley left the reader to infer the trees he was talking about from examples for small n and perhaps from his proof.) - David Callan, Jun 23 2015
From David L. Harden, Apr 09 2017: (Start)
Fix a set X and define two distance functions d,D on X to be metrically equivalent when d(x_1,y_1) <= d(x_2,y_2) iff D(x_1,y_1) <= D(x_2,y_2) for all x_1, y_1, x_2, y_2 in X.
Now suppose that we fix a function f from unordered pairs of distinct elements of X to {1,...,n}. Then choose positive real numbers d_1 <= ... <= d_n such that d(x,y) = d_{f(x,y)}; the set of all possible choices of the d_i's makes this an n-parameter family of distance functions on X. (The simplest example of such a family occurs when n is a triangular number: When that happens, write n = (k 2). Then the set of all distance functions on X, when |X| = k, is such a family.) The number of such distance functions, up to metric equivalence, is a(n).
It is easy to see that an equivalence class of distance functions gives rise to a well-defined weak order on {d_1, ..., d_n}. To see that any weak order is realizable, choose distances from the set of integers {n-1, ..., 2n-2} so that the triangle inequality is automatically satisfied. (End)
a(n) is the number of rooted labeled forests on n nodes that avoid the patterns 213, 312, and 321. - Kassie Archer, Aug 30 2018
From A.H.M. Smeets, Nov 17 2018: (Start)
Also the number of semantic different assignments to n variables (x_1, ..., x_n) including simultaneous assignments. From the example given by Joerg Arndt (Mar 18 2014), this is easily seen by replacing
"{i}" by "x_i := expression_i(x_1, ..., x_n)",
"{i, j}" by "x_i, x_j := expression_i(x_1, .., x_n), expression_j(x_1, ..., x_n)", i.e., simultaneous assignment to two different variables (i <> j),
similar for simultaneous assignments to more variables, and
"<" by ";", i.e., the sequential constructor. These examples are directly related to "Number of ways n competitors can rank in a competition, allowing for the possibility of ties." in the first comment.
From this also the number of different mean definitions as obtained by iteration of n different mean functions on n initial values. Examples:
the AGM(x1,x2) = AGM(x2,x1) is represented by {arithmetic mean, geometric mean}, i.e., simultaneous assignment in any iteration step;
Archimedes's scheme (for Pi) is represented by {geometric mean} < {harmonic mean}, i.e., sequential assignment in any iteration step;
the geometric mean of two values can also be observed by {arithmetic mean, harmonic mean};
the AGHM (as defined in A319215) is represented by {arithmetic mean, geometric mean, harmonic mean}, i.e., simultaneous assignment, but there are 12 other semantic different ways to assign the values in an AGHM scheme.
By applying power means (also called Holder means) this can be extended to any value of n. (End)
Total number of faces of all dimensions in the permutohedron of order n. For example, the permutohedron of order 3 (a hexagon) has 6 vertices + 6 edges + 1 2-face = 13 faces, and the permutohedron of order 4 (a truncated octahedron) has 24 vertices + 36 edges + 14 2-faces + 1 3-face = 75 faces. A001003 is the analogous sequence for the associahedron. - Noam Zeilberger, Dec 08 2019
Number of odd multinomial coefficients N!/(a_1!*a_2!*...*a_k!). Here each a_i is positive, and Sum_{i} a_i = N (so 2^{N-1} multinomial coefficients in all), where N is any positive integer whose binary expansion has n 1's. - Richard Stanley, Apr 05 2022 (edited Oct 19 2022)
From Peter Bala, Jul 08 2022: (Start)
Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 16 we obtain the sequence [1, 1, 3, 13, 11, 13, 11, 13, 11, 13, ...], with an apparent period of 2 beginning at a(4). Cf. A354242.
More generally, we conjecture that the same property holds for integer sequences having an e.g.f. of the form G(exp(x) - 1), where G(x) is an integral power series. (End)
a(n) is the number of ways to form a permutation of [n] and then choose a subset of its descent set. - Geoffrey Critzer, Apr 29 2023
This is the Akiyama-Tanigawa transform of A000079, the powers of two. - Shel Kaphan, May 02 2024

Examples

			Let the points be labeled 1,2,3,...
a(2) = 3: 1<2, 2<1, 1=2.
a(3) = 13 from the 13 arrangements: 1<2<3, 1<3<2, 2<1<3, 2<3<1, 3<1<2, 3<2<1, 1=2<3 1=3<2, 2=3<1, 1<2=3, 2<1=3, 3<1=2, 1=2=3.
Three competitors can finish in 13 ways: 1,2,3; 1,3,2; 2,1,3; 2,3,1; 3,1,2; 3,2,1; 1,1,3; 2,2,1; 1,3,1; 2,1,2; 3,1,1; 1,2,2; 1,1,1.
a(3) = 13. The 13 plane increasing 0-1-2 trees on 3 vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 2 colors, are:
........................................................
........1 (x3 colors).....1(x2 colors)....1(x2 colors)..
........|................/.\............./.\............
........2 (x3 colors)...2...3...........3...2...........
........|...............................................
........3...............................................
......====..............====............====............
.Totals 9......+..........2....+..........2....=..13....
........................................................
a(4) = 75. The 75 non-plane increasing 0-1-2 trees on 4 vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 4 colors, are:
...............................................................
.....1 (x3).....1(x4).......1(x4).....1(x4)........1(x3).......
.....|........./.\........./.\......./.\...........|...........
.....2 (x3)...2...3.(x3)..3...2(x3).4...2(x3)......2(x4).......
.....|.............\...........\.........\......../.\..........
.....3.(x3).........4...........4.........3......3...4.........
.....|.........................................................
.....4.........................................................
....====......=====........====......====.........====.........
Tots 27....+....12......+...12....+...12.......+...12...=...75.
From _Joerg Arndt_, Mar 18 2014: (Start)
The a(3) = 13 strings on the alphabet {1,2,3} containing all letters up to the maximal value appearing and the corresponding ordered set partitions are:
01:  [ 1 1 1 ]     { 1, 2, 3 }
02:  [ 1 1 2 ]     { 1, 2 } < { 3 }
03:  [ 1 2 1 ]     { 1, 3 } < { 2 }
04:  [ 2 1 1 ]     { 2, 3 } < { 1 }
05:  [ 1 2 2 ]     { 1 } < { 2, 3 }
06:  [ 2 1 2 ]     { 2 } < { 1, 3 }
07:  [ 2 2 1 ]     { 3 } < { 1, 2 }
08:  [ 1 2 3 ]     { 1 } < { 2 } < { 3 }
09:  [ 1 3 2 ]     { 1 } < { 3 } < { 2 }
00:  [ 2 1 3 ]     { 2 } < { 1 } < { 3 }
11:  [ 2 3 1 ]     { 3 } < { 1 } < { 2 }
12:  [ 3 1 2 ]     { 2 } < { 3 } < { 1 }
13:  [ 3 2 1 ]     { 3 } < { 2 } < { 1 }
(End)

References

  • Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.
  • Norman Biggs, E. Keith Lloyd and Robin J. Wilson, Graph Theory 1736-1936, Oxford, 1976, p. 44 (P(x)).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 183 (see R_n).
  • Kenneth S. Brown, Buildings, Springer-Verlag, 1988.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 228.
  • Jean-Marie De Koninck, Ces nombres qui nous fascinent, Entry 13, pp 4, Ellipses, Paris 2008.
  • P. J. Freyd, On the size of Heyting semi-lattices, preprint, 2002.
  • Ian P. Goulden and David M. Jackson, Combinatorial Enumeration, John Wiley and Sons, N.Y., 1983.
  • Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd Ed., 1994, exercise 7.44 (pp. 378, 571).
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • Donald E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, 1973, Section 5.3.1, Problem 3.
  • M. Muresan, Generalized Fubini numbers, Stud. Cerc. Mat., Vol. 37, No. 1 (1985), pp. 70-76.
  • Paul Peart, Hankel determinants via Stieltjes matrices. Proceedings of the Thirty-first Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 2000). Congr. Numer. 144 (2000), 153-159.
  • S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay 1957 Vol. 1, see page 19.
  • Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nuernberg, Jul 27 1994.
  • 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).
  • Richard P. Stanley, Enumerative Combinatorics, Wadsworth, Vol. 1, 1986; see Example 3.15.10, p. 146.
  • Jack van der Elsen, Black and White Transformations, Shaker Publishing, Maastricht, 2005, p. 18.

Links

Crossrefs

See A240763 for a list of the actual preferential arrangements themselves.
A000629, this sequence, A002050, A032109, A052856, A076726 are all more-or-less the same sequence. - N. J. A. Sloane, Jul 04 2012
Binomial transform of A052841. Inverse binomial transform of A000629.
Asymptotic to A034172.
Cf. A002144, A002869, A004121, A004122, A007047, A007318, A048144, A053525, A080253, A080254, A011782, A154921, A162312, A163204, A242280, A261959, A290376, A074206.
Row r=1 of A094416. Row 0 of array in A226513. Row n=1 of A262809.
Main diagonal of: A135313, A261781, A276890, A327245, A327583, A327584.
Row sums of triangles A019538, A131689, A208744 and A276891.
A217389 and A239914 give partial sums.
Column k=1 of A326322.

Programs

  • Haskell
    a000670 n = a000670_list !! n
a000670_list = 1 : f [1] (map tail $ tail a007318_tabl) where
   f xs (bs:bss) = y : f (y : xs) bss where y = sum $ zipWith (*) xs bs
-- Reinhard Zumkeller, Jul 26 2014
  • Magma
    R:=PowerSeriesRing(Rationals(), 40);
Coefficients(R!(Laplace( 1/(2-Exp(x)) ))); // G. C. Greubel, Jun 11 2024
  • Maple
    A000670 := proc(n) option remember; local k; if n <=1 then 1 else add(binomial(n,k)*A000670(n-k),k=1..n); fi; end;
with(combstruct); SeqSetL := [S, {S=Sequence(U), U=Set(Z,card >= 1)},labeled]; seq(count(SeqSetL,size=j),j=1..12);
with(combinat): a:=n->add(add((-1)^(k-i)*binomial(k, i)*i^n, i=0..n), k=0..n): seq(a(n), n=0..18); # Zerinvary Lajos, Jun 03 2007
a := n -> add(combinat:-eulerian1(n,k)*2^k,k=0..n): # Peter Luschny, Jan 02 2015
a := n -> (polylog(-n, 1/2)+`if`(n=0,1,0))/2: seq(round(evalf(a(n),32)), n=0..20); # Peter Luschny, Nov 03 2015
# next Maple program:
b:= proc(n, k) option remember;
     `if`(n=0, k!, k*b(n-1, k)+b(n-1, k+1))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..20);  # Alois P. Heinz, Aug 04 2021
  • Mathematica
    Table[(PolyLog[-z, 1/2] + KroneckerDelta[z])/2, {z, 0, 20}] (* Wouter Meeussen *)
a[0] = 1; a[n_]:= a[n]= Sum[Binomial[n, k]*a[n-k], {k, 1, n}]; Table[a[n], {n, 0, 30}] (* Roger L. Bagula and Gary W. Adamson, Sep 13 2008 *)
t = 30; Range[0, t]! CoefficientList[Series[1/(2 - Exp[x]), {x, 0, t}], x] (* Vincenzo Librandi, Mar 16 2014 *)
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1 / (2 - Exp@x), {x, 0, n}]]; (* Michael Somos, Jun 19 2015 *)
Table[Sum[k^n/2^(k+1),{k,0,Infinity}],{n,0,20}] (* Vaclav Kotesovec, Jun 26 2015 *)
Table[HurwitzLerchPhi[1/2, -n, 0]/2, {n, 0, 20}] (* Jean-François Alcover, Jan 31 2016 *)
Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*((i+r)^(n-r)/(i!*(k-i-r)!)), {i, 0, k-r}], {k, r, n}]; Fubini[0, 1] = 1; Table[Fubini[n, 1], {n, 0, 20}] (* Jean-François Alcover, Mar 31 2016 *)
Eulerian1[0, 0] = 1; Eulerian1[n_, k_] := Sum[(-1)^j (k-j+1)^n Binomial[n+1, j], {j, 0, k+1}]; Table[Sum[Eulerian1[n, k] 2^k, {k, 0, n}], {n, 0, 20}] (* Jean-François Alcover, Jul 13 2019, after Peter Luschny *)
Prepend[Table[-(-1)^k HurwitzLerchPhi[2, -k, 0]/2, {k, 1, 50}], 1] (* Federico Provvedi,Sep 05 2020 *)
Table[Sum[k!*StirlingS2[n,k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 22 2020 *)
  • Maxima
    makelist(sum(stirling2(n,k)*k!,k,0,n),n,0,12); /* Emanuele Munarini, Jul 07 2011 */
  • Maxima
    a[0]:1$ a[n]:=sum(binomial(n,k)*a[n-k],k,1,n)$ A000670(n):=a[n]$ makelist(A000670(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( subst( 1 / (1 - y), y, exp(x + x*O(x^n)) - 1), n))}; /* Michael Somos, Mar 04 2004 */
  • PARI
    Vec(serlaplace(1/(2-exp('x+O('x^66))))) /* Joerg Arndt, Jul 10 2011 */
  • PARI
    {a(n)=polcoeff(sum(m=0,n,m!*x^m/prod(k=1,m,1-k*x+x*O(x^n))),n)} /* Paul D. Hanna, Jul 20 2011 */
  • PARI
    {a(n) = if( n<1, n==0, sum(k=1, n, binomial(n, k) * a(n-k)))}; /* Michael Somos, Jul 16 2017 */
  • Python
    from math import factorial
from sympy.functions.combinatorial.numbers import stirling
def A000670(n): return sum(factorial(k)*stirling(n,k) for k in range(n+1)) # Chai Wah Wu, Nov 08 2022
  • Sage
    @CachedFunction
def A000670(n) : return 1 if n == 0 else add(A000670(k)*binomial(n,k) for k in range(n))
[A000670(n) for n in (0..20)] # Peter Luschny, Jul 14 2012

Formula

a(n) = Sum_{k=0..n} k! * StirlingS2(n,k) (whereas the Bell numbers A000110(n) = Sum_{k=0..n} StirlingS2(n,k)).
E.g.f.: 1/(2-exp(x)).
a(n) = Sum_{k=1..n} binomial(n, k)*a(n-k), a(0) = 1.
The e.g.f. y(x) satisfies y' = 2*y^2 - y.
a(n) = A052856(n) - 1, if n>0.
a(n) = A052882(n)/n, if n>0.
a(n) = A076726(n)/2.
a(n) is asymptotic to (1/2)*n!*log_2(e)^(n+1), where log_2(e) = 1.442695... [Barthelemy80, Wilf90].
For n >= 1, a(n) = (n!/2) * Sum_{k=-infinity..infinity} of (log(2) + 2 Pi i k)^(-n-1). - Dean Hickerson
a(n) = ((x*d/dx)^n)(1/(2-x)) evaluated at x=1. - Karol A. Penson, Sep 24 2001
For n>=1, a(n) = Sum_{k>=1} (k-1)^n/2^k = A000629(n)/2. - Benoit Cloitre, Sep 08 2002
Value of the n-th Eulerian polynomial (cf. A008292) at x=2. - Vladeta Jovovic, Sep 26 2003
First Eulerian transform of the powers of 2 [A000079]. See A000142 for definition of FET. - Ross La Haye, Feb 14 2005
a(n) = Sum_{k=0..n} (-1)^k*k!*Stirling2(n+1, k+1)*(1+(-1)^k)/2. - Paul Barry, Apr 20 2005
a(n) + a(n+1) = 2*A005649(n). - Philippe Deléham, May 16 2005 - Thomas Wieder, May 18 2005
Equals inverse binomial transform of A000629. - Gary W. Adamson, May 30 2005
a(n) = Sum_{k=0..n} k!*( Stirling2(n+2, k+2) - Stirling2(n+1, k+2) ). - Micha Hofri (hofri(AT)wpi.edu), Jul 01 2006
Recurrence: 2*a(n) = (a+1)^n where superscripts are converted to subscripts after binomial expansion - reminiscent of Bernoulli numbers' B_n = (B+1)^n. - Martin Kochanski (mjk(AT)cardbox.com), May 10 2007
a(n) = (-1)^n * n! * Laguerre(n,P((.),2)), umbrally, where P(j,t) are the polynomials in A131758. - Tom Copeland, Sep 27 2007
Formula in terms of the hypergeometric function, in Maple notation: a(n) = hypergeom([2,2...2],[1,1...1],1/2)/4, n=1,2..., where in the hypergeometric function there are n upper parameters all equal to 2 and n-1 lower parameters all equal to 1 and the argument is equal to 1/2. Example: a(4) = evalf(hypergeom([2,2,2,2],[1,1,1],1/2)/4) = 75. - Karol A. Penson, Oct 04 2007
a(n) = Sum_{k=0..n} A131689(n,k). - Philippe Deléham, Nov 03 2008
From Peter Bala, Jul 01 2009: (Start)
Analogy with the Bernoulli numbers.
We enlarge upon the above comment of M. Kochanski.
The Bernoulli polynomials B_n(x), n = 0,1,..., are given by the formula
(1)... B_n(x) := Sum_{k=0..n} binomial(n,k)*B(k)*x^(n-k),
where B(n) denotes the sequence of Bernoulli numbers B(0) = 1,
B(1) = -1/2, B(2) = 1/6, B(3) = 0, ....
By analogy, we associate with the present sequence an Appell sequence of polynomials {P_n(x)} n >= 0 defined by
(2)... P_n(x) := Sum_{k=0..n} binomial(n,k)*a(k)*x^(n-k).
These polynomials have similar properties to the Bernoulli polynomials.
The first few values are P_0(x) = 1, P_1(x) = x + 1,
P_2(x) = x^2 + 2*x + 3, P_3(x) = x^3 + 3*x^2 + 9*x + 13 and
P_4(x) = x^4 + 4*x^3 + 18*x^2 + 52*x + 75. See A154921 for the triangle of coefficients of these polynomials.
The e.g.f. for this polynomial sequence is
(3)... exp(x*t)/(2 - exp(t)) = 1 + (x + 1)*t + (x^2 + 2*x + 3)*t^2/2! + ....
The polynomials satisfy the difference equation
(4)... 2*P_n(x - 1) - P_n(x) = (x - 1)^n,
and so may be used to evaluate the weighted sums of powers of integers
(1/2)*1^m + (1/2)^2*2^m + (1/2)^3*3^m + ... + (1/2)^(n-1)*(n-1)^m
via the formula
(5)... Sum_{k=1..n-1} (1/2)^k*k^m = 2*P_m(0) - (1/2)^(n-1)*P_m(n),
analogous to the evaluation of the sums 1^m + 2^m + ... + (n-1)^m in terms of Bernoulli polynomials.
This last result can be generalized to
(6)... Sum_{k=1..n-1} (1/2)^k*(k+x)^m = 2*P_m(x)-(1/2)^(n-1)*P_m(x+n).
For more properties of the polynomials P_n(x), refer to A154921.
For further information on weighted sums of powers of integers and the associated polynomial sequences, see A162312.
The present sequence also occurs in the evaluation of another sum of powers of integers. Define
(7)... S_m(n) := Sum_{k=1..n-1} (1/2)^k*((n-k)*k)^m, m = 1,2,....
Then
(8)... S_m(n) = (-1)^m *[2*Q_m(-n) - (1/2)^(n-1)*Q_m(n)],
where Q_m(x) are polynomials in x given by
(9)... Q_m(x) = Sum_{k=0..m} a(m+k)*binomial(m,k)*x^(m-k).
The first few values are Q_1(x) = x + 3, Q_2(x) = 3*x^2 + 26*x + 75
and Q_3(x) = 13*x^3 + 225*x^2 + 1623*x + 4683.
For example, m = 2 gives
(10)... S_2(n) := Sum_{k=1..n-1} (1/2)^k*((n-k)*k)^2
= 2*(3*n^2 - 26*n + 75) - (1/2)^(n-1)*(3*n^2 + 26*n + 75).
(End)
G.f.: 1/(1-x/(1-2*x/(1-2*x/(1-4*x/(1-3*x/(1-6*x/(1-4*x/(1-8*x/(1-5*x/(1-10*x/(1-6*x/(1-... (continued fraction); coefficients of continued fraction are given by floor((n+2)/2)*(3-(-1)^n)/2 (A029578(n+2)). - Paul Barry, Mar 30 2010
G.f.: 1/(1-x-2*x^2/(1-4*x-8*x^2/(1-7*x-18*x^2/(1-10*x-32*x^2/(1../(1-(3*n+1)*x-2*(n+1)^2*x^2/(1-... (continued fraction). - Paul Barry, Jun 17 2010
G.f.: A(x) = Sum_{n>=0} n!*x^n / Product_{k=1..n} (1-k*x). - Paul D. Hanna, Jul 20 2011
a(n) = A074206(q_1*q_2*...*q_n), where {q_i} are distinct primes. - Vladimir Shevelev, Aug 05 2011
The adjusted e.g.f. A(x) := 1/(2-exp(x))-1, has inverse function A(x)^-1 = Integral_{t=0..x} 1/((1+t)*(1+2*t)). Applying [Dominici, Theorem 4.1] to invert the integral yields a formula for a(n): Let f(x) = (1+x)*(1+2*x). Let D be the operator f(x)*d/dx. Then a(n) = D^(n-1)(f(x)) evaluated at x = 0. Compare with A050351. - Peter Bala, Aug 31 2011
a(n) = D^n*(1/(1-x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A052801. - Peter Bala, Nov 25 2011
From Sergei N. Gladkovskii, from Oct 2011 to Oct 2013: (Start)
Continued fractions:
G.f.: 1+x/(1-x+2*x*(x-1)/(1+3*x*(2*x-1)/(1+4*x*(3*x-1)/(1+5*x*(4*x-1)/(1+... or 1+x/(U(0)-x), U(k) = 1+(k+2)*(k*x+x-1)/U(k+1).
E.g.f.: 1 + x/(G(0)-2*x) where G(k) = x + k + 1 - x*(k+1)/G(k+1).
E.g.f. (2 - 2*x)*(1 - 2*x^3/(8*x^2 - 4*x + (x^2 - 4*x + 2)*G(0)))/(x^2 - 4*x + 2) where G(k) = k^2 + k*(x+4) + 2*x + 3 - x*(k+1)*(k+3)^2 /G(k+1).
G.f.: 1 + x/G(0) where G(k) = 1 - 3*x*(k+1) - 2*x^2*(k+1)*(k+2)/G(k+1).
G.f.: 1/G(0) where G(k) = 1 - x*(k+1)/( 1 - 2*x*(k+1)/G(k+1) ).
G.f.: 1 + x/Q(0), where Q(k) = 1 - 3*x*(2*k+1) - 2*x^2*(2*k+1)*(2*k+2)/( 1 - 3*x*(2*k+2) - 2*x^2*(2*k+2)*(2*k+3)/Q(k+1) ).
G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)^2/( 2*x^2*(k+1)^2 - (1-x-3*x*k)*(1-4*x-3*x*k)/T(k+1) ). (End)
a(n) is always odd. For odd prime p and n >= 1, a((p-1)*n) = 0 (mod p). - Peter Bala, Sep 18 2013
a(n) = log(2)* Integral_{x>=0} floor(x)^n * 2^(-x) dx. - Peter Bala, Feb 06 2015
For n > 0, a(n) = Re(polygamma(n, i*log(2)/(2*Pi))/(2*Pi*i)^(n+1)) - n!/(2*log(2)^(n+1)). - Vladimir Reshetnikov, Oct 15 2015
a(n) = Sum_{k=1..n} (k*b2(k-1)*(k)!*Stirling2(n, k)), n>0, a(0)=1, where b2(n) is the n-th Bernoulli number of the second kind. - Vladimir Kruchinin, Nov 21 2016
Conjecture: a(n) = Sum_{k=0..2^(n-1)-1} A284005(k) for n > 0 with a(0) = 1. - Mikhail Kurkov, Jul 08 2018
a(n) = A074206(k) for squarefree k with n prime factors. In particular a(n) = A074206(A002110(n)). - Amiram Eldar, May 13 2019
For n > 0, a(n) = -(-1)^n / 2 * PHI(2, -n, 0), where PHI(z, s, a) is the Lerch zeta function. - Federico Provvedi, Sep 05 2020
a(n) = Sum_{s in S_n} Product_{i=1..n} binomial(i,s(i)-1), where s ranges over the set S_n of permutations of [n]. - Jose A. Rodriguez, Feb 02 2021
Sum_{n>=0} 1/a(n) = 2.425674839121428857970063350500499393706641093287018840857857170864211946122664... - Vaclav Kotesovec, Jun 17 2021
From Jacob Sprittulla, Oct 05 2021: (Start)
The following identities hold for sums over Stirling numbers of the second kind with even or odd second argument:
a(n) = 2 * Sum_{k=0..floor(n/2)} ((2k)! * Stirling2(n,2*k) ) - (-1)^n = 2*A052841-(-1)^n
a(n) = 2 * Sum_{k=0..floor(n/2)} ((2k+1)!* Stirling2(n,2*k+1))+ (-1)^n = 2*A089677+(-1)^n
a(n) = Sum_{k=1..floor((n+1)/2)} ((2k-1)!* Stirling2(n+1,2*k))
a(n) = Sum_{k=0..floor((n+1)/2)} ((2k)! * Stirling2(n+1,2*k+1)). (End)

A010050 a(n) = (2n)!.

Original entry on oeis.org

1, 2, 24, 720, 40320, 3628800, 479001600, 87178291200, 20922789888000, 6402373705728000, 2432902008176640000, 1124000727777607680000, 620448401733239439360000, 403291461126605635584000000, 304888344611713860501504000000, 265252859812191058636308480000000
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

Denominators in the expansion of cos(x): cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! - ...
Contribution from Peter Bala, Feb 21 2011: (Start)
We may compare the representation a(n) = Product_{k = 0..n-1} (n*(n+1)-k*(k+1)) with n! = Product_{k = 0..n-1} (n-k). Thus we may view a(n) as a generalized factorial function associated with the oblong numbers A002378. Cf. A000680.
The associated generalized binomial coefficients a(n)/(a(k)*a(n-k)) are triangle A086645, cf. A186432. (End)
Also, this sequence is the denominator of cosh(x) = (e^x + e^(-x))/2 = 1 + x^2/2! + x^4/4! + x^6/6! + ... - Mohammad K. Azarian, Jan 19 2012
Also (2n+1)-th derivative of arccoth(x) at x = 0. - Michel Lagneau, Aug 18 2012
Product of the partition parts of 2n+1 into exactly two positive integer parts, n > 0. Example: a(3) = 720, since 2(3)+1 = 7 has 3 partitions with exactly two positive integer parts: (6,1), (5,2), (4,3). Multiplying the parts in these partitions gives: 6! = 720. - Wesley Ivan Hurt, Jun 03 2013

Examples

			G.f. = 1 + 2*x + 24*x^2 + 720*x^3 + 40320*x^4 + 3628800*x^5 + ...

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 110.
  • H. B. Dwight, Tables of Integrals and Other Mathematical Data, Macmillan, NY, 1968, p. 88.
  • Isaac Newton, De analysi, 1669; reprinted in D. Whiteside, ed., The Mathematical Works of Isaac Newton, vol. 1, Johnson Reprint Co., 1964; see p. 20.
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapters 32 and 33, equations 32:6:1 and 33:6:1 at pages 300 and 314.

Links

Crossrefs

Cf. A000142, A000165, A009445, A073743.
Bisection of A005359, |A012251|, A012254, A070734.

Programs

Formula

a(n) = 2^n*A000680(n).
E.g.f.: arctanh(x) = Sum_{k>=0} a(k) * x^(2*k+1)/ (2*k+1)!.
E.g.f.: 1/(1-x^2) = Sum_{k>=0} a(k) * x^(2*k) / (2*k)!. - Paul Barry, Sep 14 2004
D-finite with recurrence: a(n+1) = a(n)*(2*n+1)*(2*n+2) = a(n)*A002939(n-1). - Lekraj Beedassy, Apr 29 2005
a(n) = Product_{k = 1..n} (2*k*n-k*(k-1)). - Peter Bala, Feb 21 2011
G.f.: G(0) where G(k) = 1 + 2*x*(2*k+1)*(4*k+1)/(1 - 4*x*(k+1)*(4*k+3)/(4*x*(k+1)*(4*k+3) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2012
a(n) = 2*A002674(n), n > 0. - Wesley Ivan Hurt, Jun 05 2013
From Ilya Gutkovskiy, Jan 20 2017: (Start)
a(n) ~ 2*sqrt(Pi)*4^n*n^(2*n+1/2)/exp(2*n).
Sum_{n>=0} 1/a(n) = cosh(1) = A073743. (End)

Extensions

Third line of data from M. F. Hasler, Apr 22 2015

A052841 Expansion of e.g.f.: 1/(exp(x)*(2-exp(x))).

Original entry on oeis.org

1, 0, 2, 6, 38, 270, 2342, 23646, 272918, 3543630, 51123782, 811316286, 14045783798, 263429174190, 5320671485222, 115141595488926, 2657827340990678, 65185383514567950, 1692767331628422662, 46400793659664205566, 1338843898122192101558, 40562412499252036940910
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

From Michael Somos, Mar 04 2004: (Start)
Stirling transform of A005359(n)=[0,2,0,24,0,720,...] is a(n)=[0,2,6,38,270,...].
Stirling transform of -(-1)^n*A052657(n-1)=[0,0,2,-6,48,-240,...] is a(n-1)=[0,0,2,6,38,270,...].
Stirling transform of -(-1)^n*A052558(n-1)=[1,-1,4,-12,72,-360,...] is a(n-1)=[1,0,2,6,38,270,...].
Stirling transform of 2*A052591(n)=[2,4,24,96,...] is a(n+1)=[2,6,38,270,...].
(End)
Also the central moments of a Geometric(1/2) random variable (for example the number of coin tosses until the first head). - Svante Janson, Dec 10 2012
Also the number of ordered set partitions of {1..n} with no cyclical adjacencies (successive elements in the same block, where 1 is a successor of n). - Gus Wiseman, Feb 13 2019
Also the number of ordered set partitions of {1..n} with an even number of blocks. - Geoffrey Critzer, Jul 04 2020

Examples

			From _Gus Wiseman_, Feb 13 2019: (Start)
The a(4) = 38 ordered set partitions with no cyclical adjacencies:
  {{1}{2}{3}{4}}  {{1}{24}{3}}  {{13}{24}}
  {{1}{2}{4}{3}}  {{1}{3}{24}}  {{24}{13}}
  {{1}{3}{2}{4}}  {{13}{2}{4}}
  {{1}{3}{4}{2}}  {{13}{4}{2}}
  {{1}{4}{2}{3}}  {{2}{13}{4}}
  {{1}{4}{3}{2}}  {{2}{4}{13}}
  {{2}{1}{3}{4}}  {{24}{1}{3}}
  {{2}{1}{4}{3}}  {{24}{3}{1}}
  {{2}{3}{1}{4}}  {{3}{1}{24}}
  {{2}{3}{4}{1}}  {{3}{24}{1}}
  {{2}{4}{1}{3}}  {{4}{13}{2}}
  {{2}{4}{3}{1}}  {{4}{2}{13}}
  {{3}{1}{2}{4}}
  {{3}{1}{4}{2}}
  {{3}{2}{1}{4}}
  {{3}{2}{4}{1}}
  {{3}{4}{1}{2}}
  {{3}{4}{2}{1}}
  {{4}{1}{2}{3}}
  {{4}{1}{3}{2}}
  {{4}{2}{1}{3}}
  {{4}{2}{3}{1}}
  {{4}{3}{1}{2}}
  {{4}{3}{2}{1}}
(End)

Links

Crossrefs

Main diagonal of A122101.
Inverse binomial transform of A000670.
Cf. A000126, A000296, A000670, A001610, A005359, A032032, A052558.
Cf. A052591, A052657, A052841, A124323, A156365, A169985, A173018.
Cf. A187784, A324011, A344037, A346208.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 40);
Coefficients(R!(Laplace( Exp(-x)/(2-Exp(x)) ))); // G. C. Greubel, Jun 11 2024
  • Maple
    spec := [S,{B=Prod(C,C),C=Set(Z,1 <= card),S=Sequence(B)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
P := proc(n,x) option remember; if n = 0 then 1 else
(n*x+2*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x); expand(%) fi end:
A052841 := n -> subs(x=2, P(n,x)):
seq(A052841(n), n=0..21); # Peter Luschny, Mar 07 2014
h := n -> add(combinat:-eulerian1(n, k)*2^k, k=0..n):
a := n -> (h(n)+(-1)^n)/2: seq(a(n), n=0..21); # Peter Luschny, Sep 19 2015
b := proc(n, m) option remember; if n = 0 then 1 else
     (m - 1)*b(n - 1, m) + (m + 1)*b(n - 1, m + 1) fi end:
a := n -> b(n, 0): seq(a(n), n = 0..21); # Peter Luschny, Jun 23 2023
  • Mathematica
    a[n_] := If[n == 0, 1, (PolyLog[-n, 1/2]/2 + (-1)^n)/2]; (* or *)
a[n_] := HurwitzLerchPhi[1/2, -n, -1]/2; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Feb 19 2016, after Vladeta Jovovic *)
With[{nn=30},CoefficientList[Series[1/(Exp[x](2-Exp[x])),{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Apr 08 2019 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(subst(1/(1-y^2),y,exp(x+x*O(x^n))-1),n))
  • PARI
    {a(n)=polcoeff(sum(m=0,n,(2*m)!*x^(2*m)/prod(k=1,2*m,1-k*x+x*O(x^n))),n)} /* Paul D. Hanna, Jul 20 2011 */
  • SageMath
    def A052841_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( exp(-x)/(2-exp(x)) ).egf_to_ogf().list()
A052841_list(40) # G. C. Greubel, Jun 11 2024

Formula

O.g.f.: Sum_{n>=0} (2*n)! * x^(2*n) / Product_{k=1..2*n} (1-k*x). - Paul D. Hanna, Jul 20 2011
a(n) = (A000670(n) + (-1)^n)/2 = Sum_{k>=0} (k-1)^n/2^(k+1). - Vladeta Jovovic, Feb 02 2003
Also, a(n) = Sum_{k=0..[n/2]} (2k)!*Stirling2(n, 2k). - Ralf Stephan, May 23 2004
a(n) = D^n*(1/(1-x^2)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A000670 and A005649. - Peter Bala, Nov 25 2011
E.g.f.: 1/(2*G(0)), where G(k) = 1 - 2^k/(2 - 4*x/(2*x - 2^k*(k+1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 22 2012
a(n) ~ n!/(4*(log(2))^(n+1)). - Vaclav Kotesovec, Aug 10 2013
a(n) = (h(n)+(-1)^n)/2 where h(n) = Sum_{k=0..n} E(n,k)*2^k and E(n,k) the Eulerian numbers A173018 (see also A156365). - Peter Luschny, Sep 19 2015
a(n) = (-1)^n + Sum_{k=0..n-1} binomial(n,k) * a(k). - Ilya Gutkovskiy, Jun 11 2020

Extensions

Edited by N. J. A. Sloane, Sep 06 2013

A002050 Number of simplices in barycentric subdivision of n-simplex.

Original entry on oeis.org

0, 1, 5, 25, 149, 1081, 9365, 94585, 1091669, 14174521, 204495125, 3245265145, 56183135189, 1053716696761, 21282685940885, 460566381955705, 10631309363962709, 260741534058271801, 6771069326513690645
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Stirling transform of A052849(n)=[1,4,12,48,240,...] is a(n)=[1,5,25,149,1081,...]. - Michael Somos, Mar 04 2004
Stirling transform of A000142(n-1)=[0,1,2,6,24,...] is a(n-1)=[0,1,5,25,149,...]. - Michael Somos, Mar 04 2004
Stirling transform of 2*A005359(n-1)=[1,0,4,0,48,0,...] is a(n-1)=[1,1,5,25,149,...]. - Michael Somos, Mar 04 2004
"Stirling-Bernoulli transform" of A000225. - Paul Barry, Apr 20 2005
a(n) is the number of nonempty words that can be formed from an alphabet of nonempty subsets of [n] so that the letters in each word are pairwise disjoint. - Geoffrey Critzer, Apr 12 2009
Row sums of A053440. - Peter Bala, Jul 12 2014
Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 9 we obtain the sequence [0, 1, 5, 7, 5, 1, 5, 4, 5, 7, 5, 1, 5, 4, 5, 7, 5, 1, 5, 4, 5, 7, 5, ...], with an apparent period of 6 = phi(9) beginning at a(5). - Peter Bala, Aug 03 2023

References

  • R. Austin, R. K. Guy, and R. Nowakowski, unpublished notes, circa 1987.
  • Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nuernberg, Jul 27 1994
  • 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).

Links

Crossrefs

A000629, A000670, A002050, A052856, A076726 are all more-or-less the same sequence. - N. J. A. Sloane, Jul 04 2012
A diagonal of the triangle in A241168. Row sums of A053440.

Programs

  • Mathematica
    Table[Sum[Binomial[n, i]*Sum[StirlingS2[i, k]*k!, {k, 1, i}], {i, 1, n}], {n, 0, 20}] (* Geoffrey Critzer, Apr 12 2009 *)
With[{nn=20},CoefficientList[Series[(Exp[2x]-Exp[x])/(2-Exp[x]),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, May 28 2013 *)
a[0] = 0; a[n_] := 2*Sum[k!*StirlingS2[n, k], {k, 2, n}] + 1; Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Sep 27 2013, after Vladimir Kruchinin *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(subst((y+y^2)/(1-y),y,exp(x+x*O(x^n))-1),n));

Formula

E.g.f.: (exp(2x)-exp(x))/(2-exp(x)).
a(n) = A000629(n) - 1.
a(n) = Sum_{k=0..n} (-1)^(n-k)k!*S2(n, k)(2^k-1). - Paul Barry, Apr 20 2005
a(n) = Sum_{k=1...n} binomial(n,k)*A000670(k). - Geoffrey Critzer, Apr 12 2009
a(n) ~ n!/log(2)^(n+1). - Vaclav Kotesovec, Jul 29 2013
a(n) = 1 + 2*Sum_{k=2..n} k!*Stirling2(n,k), n > 0, a(0)=1. - Vladimir Kruchinin, Sep 27 2013
G.f.: T(0)/(1-2*x) - 1/(1-x), where T(k) = 1 - 2*x^2*(k+1)^2/(2*x^2*(k+1)^2 - (1 - 2*x - 3*x*k)*(1 - 5*x - 3*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 29 2013
G.f.: Sum_{j>=1} j!*x^j / Product_{k=0..j} (1 - (k + 1)*x). - Ilya Gutkovskiy, Apr 04 2019

Extensions

More terms from James Sellers, Aug 22 2000

A111594 Triangle of arctanh numbers.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 0, 8, 0, 1, 0, 24, 0, 20, 0, 1, 0, 0, 184, 0, 40, 0, 1, 0, 720, 0, 784, 0, 70, 0, 1, 0, 0, 8448, 0, 2464, 0, 112, 0, 1, 0, 40320, 0, 52352, 0, 6384, 0, 168, 0, 1, 0, 0, 648576, 0, 229760, 0, 14448, 0, 240, 0, 1
Offset: 0

Views

Author

Wolfdieter Lang, Aug 23 2005

Keywords

Comments

Sheffer triangle associated to Sheffer triangle A060524.
For Sheffer triangles (matrices) see the explanation and S. Roman reference given under A048854.
The inverse matrix of A with elements a(n,m), n,m>=0, is given in A111593.
In the umbral calculus notation (see the S. Roman reference) this triangle would be called associated to (1,tanh(y)).
The row polynomials p(n,x):=sum(a(n,m)*x^m,m=0..n), together with the row polynomials s(n,x) of A060524 satisfy the exponential (or binomial) convolution identity s(n,x+y) = sum(binomial(n,k)*s(k,x)*p(n-k,y),k=0..n), n>=0.
Without the n=0 row and m=0 column and signed, this will become the Jabotinsky triangle A049218 (arctan numbers). For Jabotinsky matrices see the Knuth reference under A039692.
The row polynomials p(n,x) (defined above) have e.g.f. exp(x*arctanh(y)).
Exponential Riordan array [1, arctanh(x)] = [1, log(sqrt((1+x)/(1-x)))]. - Paul Barry, Apr 17 2008
Also the Bell transform of A005359. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Binomial convolution of row polynomials:
p(3,x)= 2*x+x^3; p(2,x)=x^2, p(1,x)= x, p(0,x)= 1,
together with those from A060524:
s(3,x)= 5*x+x^3; s(2,x)= 1+x^2, s(1,x)= x, s(0,x)= 1; therefore:
5*(x+y)+(x+y)^3 = s(3,x+y) = 1*s(0,x)*p(3,y) + 3*s(1,x)*p(2,y) + 3*s(2,x)*p(1,y) +1*s(3,x)*p(0,y) = 2*y+y^3 + 3*x*y^2 + 3*(1+x^2)*y + (5*x+x^3).
Triangle begins:
  1;
  0,   1;
  0,   0,    1;
  0,   2,    0,   1;
  0,   0,    8,   0,    1;
  0,  24,    0,  20,    0,  1;
  0,   0,  184,   0,   40,  0,   1;
  0, 720,    0, 784,    0, 70,   0, 1;
  0,   0, 8448,   0, 2464,  0, 112, 0, 1;
...

Links

Crossrefs

Row sums: A000246.
Cf. A005359, A049218, A060524, A111593.

Programs

  • Maple
    # The function BellMatrix is defined in A264428.
BellMatrix(n -> `if`(n::even, n!, 0), 10); # Peter Luschny, Jan 27 2016
  • Mathematica
    rows = 10;
t = Table[If[EvenQ[n], n!, 0], {n, 0, rows}];
T[n_, k_] := BellY[n, k, t];
Table[T[n, k], {n, 0, rows}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 22 2018, after Peter Luschny *)
  • Sage
    # uses[riordan_array from A256893]
riordan_array(1, atanh(x), 9, exp=true) # Peter Luschny, Apr 19 2015

Formula

E.g.f. for column m>=0: ((arctanh(x))^m)/m!.
a(n, m) = coefficient of x^n of ((arctanh(x))^m)/m!, n>=m>=0, else 0.
a(n, m) = a(n-1, m-1) + (n-2)*(n-1)*a(n-2, m), a(n, -1):=0, a(0, 0)=1, a(n, m)=0 for n

A052875 E.g.f.: (exp(x)-1)^2/(2-exp(x)).

Original entry on oeis.org

0, 0, 2, 12, 74, 540, 4682, 47292, 545834, 7087260, 102247562, 1622632572, 28091567594, 526858348380, 10641342970442, 230283190977852, 5315654681981354, 130370767029135900, 3385534663256845322, 92801587319328411132, 2677687796244384203114, 81124824998504073881820
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Previous name was: A simple grammar.
Stirling transform of A005359(n-1)=[0,0,2,0,24,0,...] is a(n-1)=[0,0,2,12,74,...]. - Michael Somos, Mar 04 2004
Stirling transform of -(-1)^n*A052566(n-1)=[1,-1,4,-6,48,...] is a(n-1)=[1,0,2,12,74,...]. - Michael Somos, Mar 04 2004
Stirling transform of A000142(n)=[0,2,6,24,120,...] is a(n)=[0,2,12,74,...]. - Michael Somos, Mar 04 2004
Stirling transform of A007680(n)=[2,10,42,216,...] is a(n+1)=[2,12,74,...]. - Michael Somos, Mar 04 2004
a(n) is the number of chains in the power set of {1,2,...,n} that do not contain the empty set and do not contain {1,2,...,n}. Equivalently, a(n) is the number of ordered set partitions of {1,2,...,n} into at least 2 classes. - Geoffrey Critzer, Sep 01 2014

Examples

			a(3) = 12 because we have: {{1}}, {{2}}, {{3}}, {{1,2}}, {{1,3}}, {{2,3}}, {{1}, {1,2}}, {{1}, {1,3}}, {{2}, {1,2}}, {{2}, {2,3}}, {{3}, {1,3}}, {{3}, {2,3}}. - _Geoffrey Critzer_, Sep 01 2014

Links

Crossrefs

Cf. A007047, A038719.

Programs

  • Maple
    spec := [S, {B = Set(Z, 1 <= card), C = Sequence(B, 1 <= card), S=Prod(B, C)}, labeled]: seq(combstruct[count](spec, size=n),  n=0..20);
  • Mathematica
    CoefficientList[Series[(E^x-1)^2/(2-E^x), {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Feb 25 2014 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(subst(y^2/(1-y),y,exp(x+x*O(x^n))-1),n))
  • Sage
    def A052875(n):
    return add(add((-1)^(j-i)*binomial(j,i)*i^n for i in range(n+1)) for j in range(n+1)) - 1
[A052875(n) for n in range(19)] # Peter Luschny, Jul 22 2014

Formula

Second column of A084416: Sum_{i=2..n} i!*Stirling2(n, i) = A000670(n)-1. - Vladeta Jovovic, Sep 15 2003
E.g.f.: (exp(x)-1)^2/(2-exp(x)).
a(n) ~ n! / (2 * (log(2))^(n+1)). - Vaclav Kotesovec, Feb 25 2014
E.g.f.: A(x)*(1/(1 - A(x)) - 1) where A(x)=exp(x)-1. - Geoffrey Critzer, Sep 01 2014

Extensions

New name using e.g.f., Vaclav Kotesovec, Feb 25 2014

A049218 Triangle T(n,k) of arctangent numbers: expansion of arctan(x)^n/n!.

Original entry on oeis.org

1, 0, 1, -2, 0, 1, 0, -8, 0, 1, 24, 0, -20, 0, 1, 0, 184, 0, -40, 0, 1, -720, 0, 784, 0, -70, 0, 1, 0, -8448, 0, 2464, 0, -112, 0, 1, 40320, 0, -52352, 0, 6384, 0, -168, 0, 1, 0, 648576, 0, -229760, 0, 14448, 0, -240, 0, 1, -3628800, 0, 5360256, 0, -804320, 0, 29568, 0, -330, 0, 1
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

|T(n,k)| gives the sum of the M_2 multinomial numbers (A036039) for those partitions of n with exactly k odd parts. E.g.: |T(6,2)| = 144 + 40 = 184 from the partitions of 6 with exactly two odd parts, namely (1,5) and (3,3), with M_2 numbers 144 and 40. Proof via the general Jabotinsky triangle formula for |T(n,k)| using partitions of n into k parts and their M_3 numbers (A036040). Then with the special e.g.f. of the (unsigned) k=1 column, f(x):= arctanh(x), only odd parts survive and the M_3 numbers are changed into the M_2 numbers. For the Knuth reference on Jabotinsky triangles see A039692. - Wolfdieter Lang, Feb 24 2005 [The first two sentences have been corrected thanks to the comment by José H. Nieto S. given below. - Wolfdieter Lang, Jan 16 2012]
|T(n,k)| gives the number of permutations of {1,2,...,n} (degree n permutations) with the number of odd cycles equal to k. E.g.: |T(5,3)|= 20 from the 20 degree 5 permutations with cycle structure (.)(.)(...). Proof: Use the cycle index polynomial for the symmetric group S_n (see the M_2 array A036039 or A102189) together with the partition interpretation of |T(n,k)| given above. - Wolfdieter Lang, Feb 24 2005 [See the following José H. Nieto S. correction. - Wolfdieter Lang, Jan 16 2012]
The first sentence of the above comment is inexact, it should be "|T(n,k)| gives the number of degree n permutations which decompose into exactly k odd cycles". The number of degree n permutations with k odd cycles (and, possibly, other cycles of even length) is given by A060524. - José H. Nieto S., Jan 15 2012
The unsigned triangle with e.g.f. exp(x*arctanh(z)) is the associated Jabotinsky type triangle for the Sheffer type triangle A060524. See the comments there. - Wolfdieter Lang, Feb 24 2005
Also the Bell transform of the sequence (-1)^(n/2)*A005359(n) without column 0. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 28 2016

Examples

			Triangle begins:
   1;
   0,   1;
  -2,   0,   1;
   0,  -8,   0,   1;
  24,   0, -20,   0,   1;
   0, 184,   0, -40,   0,   1;
  ...
O.g.f. for fifth subdiagonal: (24*t+16*t^2)/(1-t)^7 = 24*t + 184*t^2 + 784*t^3 + 2404*t^4 + ....

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 260.

Links

Crossrefs

Essentially same as A008309, which is the main entry for this sequence.
Row sums (unsigned) give A000246(n); signed row sums give A002019(n), n>=1. A137513.

Programs

  • Maple
    A049218 := proc(n,k)(-1)^((3*n+k)/2) *add(2^(j-k)*n!/j! *stirling1(j,k) *binomial(n-1,j-1),j=k..n) ; end proc: # R. J. Mathar, Feb 14 2011
# The function BellMatrix is defined in A264428.
# Adds (1,0,0,0, ..) as column 0.
BellMatrix(n -> `if`(n::odd, 0, (-1)^(n/2)*n!), 10); # Peter Luschny, Jan 28 2016
  • Mathematica
    t[n_, k_] := (-1)^((3n+k)/2)*Sum[ 2^(j-k)*n!/j!*StirlingS1[j, k]*Binomial[n-1, j-1], {j, k, n}]; Flatten[ Table[ t[n, k], {n, 1, 11}, {k, 1, n}]] (* Jean-François Alcover, Dec 06 2011, after Vladimir Kruchinin *)
BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
rows = 12;
M = BellMatrix[If[OddQ[#], 0, (-1)^(#/2)*#!]&, rows];
Table[M[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 23 2018, after Peter Luschny *)
  • PARI
    T(n,k)=polcoeff(serlaplace(atan(x)^k/k!), n)

Formula

E.g.f.: arctan(x)^k/k! = Sum_{n>=0} T(n, k) x^n/n!.
T(n,k) = ((-1)^((3*n+k)/2)*n!/2^k)*Sum_{i=k..n} 2^i*binomial(n-1,i-1)*Stirling1(i,k)/i!. - Vladimir Kruchinin, Feb 11 2011
E.g.f.: exp(t*arctan(x)) = 1 + t*x + t^2*x^2/2! + t*(t^2-2)*x^3/3! + .... The unsigned row polynomials are the Mittag-Leffler polynomials M(n,t/2). See A137513. The compositional inverse (with respect to x) (x-t/2*log((1+x)/(1-x)))^(-1) = x/(1-t) + 2*t/(1-t)^4*x^3/3!+ (24*t+16*t^2)/(1-t)^7*x^5/5! + .... The rational functions in t generate the (unsigned) diagonals of the table. See the Bala link. - Peter Bala, Dec 04 2011

Extensions

Additional comments from Michael Somos

A052811 A simple grammar: sequences of pairs of cycles.

Original entry on oeis.org

1, 0, 2, 6, 46, 340, 3308, 36288, 460752, 6551424, 103685232, 1803956880, 34247483664, 704301934752, 15598712592864, 370149922235520, 9369093828260736, 251968378971718656, 7174943434198029312
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Stirling transform of (-1)^n*a(n)=[0,2,-6,46,-340,...] is A005359(n)=[0,2,0,24,0,...]. - Michael Somos, Mar 04 2004

Links

Crossrefs

Cf. A346921.

Programs

  • Maple
    spec := [S,{B=Cycle(Z),C=Prod(B,B),S=Sequence(C)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    CoefficientList[Series[1/(1-Log[1-x]^2), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Sep 30 2013 *)
  • Maxima
    makelist((-1)^n*sum(stirling1(n, 2*k)*(2*k)!,k,0,floor(n/2)), n, 0, 18); /* Bruno Berselli, May 25 2011 */
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(1/(1-log(1-x)^2),n))
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=2*sum(j=1, i, binomial(i, j)*abs(stirling(j, 2, 1))*v[i-j+1])); v; \\ Seiichi Manyama, May 06 2022
  • PARI
    a(n) = sum(k=0, n\2, (2*k)!*abs(stirling(n, 2*k, 1))); \\ Seiichi Manyama, May 06 2022

Formula

a(n) = (-1)^n*Sum_{k=0..floor(n/2)} Stirling1(n, 2*k)*(2*k)!. - Vladeta Jovovic, Sep 22 2003
E.g.f.: 1/(1-log(1-x)^2).
a(n) = D^n(1/(1-x^2)) evaluated at x = 0, where D is the operator exp(x)*d/dx. Cf. A006252. - Peter Bala, Nov 25 2011
a(n) ~ n!/2 * exp(n)/(exp(1)-1)^(n+1). - Vaclav Kotesovec, Sep 30 2013
a(0) = 1; a(n) = 2 * Sum_{k=1..n} binomial(n,k) * |Stirling1(k,2)| * a(n-k). - Seiichi Manyama, May 06 2022

A167565 A triangle related to the a(n) formulas for the rows of the ED2 array A167560.

Original entry on oeis.org

1, 2, 0, 3, 1, 2, 4, 4, 16, 0, 5, 10, 67, 14, 24, 6, 20, 202, 124, 368, 0, 7, 35, 497, 601, 2736, 444, 720, 8, 56, 1064, 2120, 13712, 6464, 16896, 0, 9, 84, 2058, 6096, 53121, 48876, 186732, 25584, 40320, 10, 120, 3684, 15168, 171258, 257640, 1350296
Offset: 1

Views

Author

Johannes W. Meijer, Nov 10 2009

Keywords

Comments

The a(n) formulas given below correspond to the first ten rows of the ED2 array A167560.
The recurrence relations for the a(n) formulas for the left hand triangle columns, see the cross-references below, lead to the sequences A003148 and A007318.

Examples

			Row 1: a(n) = 1.
Row 2: a(n) = 2*n + 0.
Row 3: a(n) = 3*n^2 + 1*n + 2.
Row 4: a(n) = 4*n^3 + 4*n^2 + 16*n + 0.
Row 5: a(n) = 5*n^4 + 10*n^3 + 67*n^2 + 14*n + 24.
Row 6: a(n) = 6*n^5 + 20*n^4 + 202*n^3 + 124*n^2 + 368*n + 0.
Row 7: a(n) = 7*n^6 + 35*n^5 + 497*n^4 + 601*n^3 + 2736*n^2 + 444*n + 720.
Row 8: a(n) = 8*n^7 + 56*n^6 + 1064*n^5 + 2120*n^4 + 13712*n^3 + 6464*n^2 + 16896*n + 0.
Row 9: a(n) = 9*n^8 + 84*n^7 + 2058*n^6 + 6096*n^5 + 53121*n^4 + 48876*n^3 + 186732*n^2 + 25584*n + 40320.
Row 10: a(n) = 10*n^9 + 120*n^8 + 3684*n^7 + 15168*n^6 + 171258*n^5 + 257640*n^4 + 1350296*n^3 + 533472*n^2 + 1297152*n + 0.

Crossrefs

A167560 is the ED2 array.
A000012, A005843 (n=>1), 2*A104249 (n=>1), A167561, A167562 and A167563 equal the first sixth rows of the array.
A005359 equals the first right hand triangle column.
A000027, A000292, A167566, A167567 and A168304 equal the first five left hand triangle columns.
A000142 equals the row sums.
Cf. A003148 and A007318.

Extensions

Comment and links added by Johannes W. Meijer, Nov 23 2009

A320959 The exponential limit of arctanh (odd indices only).

Original entry on oeis.org

1, 10, 1248, 631440, 852647040, 2462394816000, 13241729554099200, 120563962753538304000, 1733764260005567741952000, 37343395325946891151466496000, 1155311729350231354981936496640000, 49626886713956000390638096497377280000, 2878005957927359237424925417166882734080000
Offset: 0

Views

Author

Peter Luschny, Nov 08 2018

Keywords

Comments

See A320956 for definitions and comments.

Examples

			Illustration of the convergence in the sense of A320956:
   [0] 0, 0, 0,  0, 0,    0, 0,      0, 0,         0, ...
   [1] 0, 1, 0,  2, 0,   24, 0,    720, 0,     40320, ... A005359
   [2] 0, 1, 0,  8, 0,  384, 0,  46080, 0,  10321920, ... A067624
   [3] 0, 1, 0, 10, 0,  984, 0, 262800, 0, 132289920, ...
   [4] 0, 1, 0, 10, 0, 1224, 0, 514800, 0, 445576320, ...
   [5] 0, 1, 0, 10, 0, 1248, 0, 615600, 0, 725840640, ...
   [6] 0, 1, 0, 10, 0, 1248, 0, 630720, 0, 832527360, ...
   [7] 0, 1, 0, 10, 0, 1248, 0, 631440, 0, 851155200, ...
   [8] 0, 1, 0, 10, 0, 1248, 0, 631440, 0, 852606720, ...
   [9] 0, 1, 0, 10, 0, 1248, 0, 631440, 0, 852647040, ...

Crossrefs

Cf. A320955 (exp), A320962 (log(x+1)), A320956 (sec+tan), A320958 (arcsin), this sequence (arctanh).
Cf. A005359, A067624.

Programs

  • Maple
    # The function ExpLim is defined in A320956.
L := [ExpLim(28, arctanh)]: seq(L[2*n], n=1..13);
  • Mathematica
    m = 13; CoefficientList[ArcTanh[x] + O[x]^(2 m + 1), x]*Range[0, 2 m - 1]!*BellB[Range[0, 2 m - 1]] // DeleteCases[#, 0]& (* Jean-François Alcover, Jul 23 2019 *)

Formula

For n >= 3 and odd, -a(m)*Zeta(m) = g(n), where g denotes the exponential limit of log(Gamma(x + 1)) and m = (n-1)/2.
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