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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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.

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.
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)

A000629 Number of necklaces of partitions of n+1 labeled beads.

Original entry on oeis.org

1, 2, 6, 26, 150, 1082, 9366, 94586, 1091670, 14174522, 204495126, 3245265146, 56183135190, 1053716696762, 21282685940886, 460566381955706, 10631309363962710, 260741534058271802, 6771069326513690646, 185603174638656822266, 5355375592488768406230
Offset: 0

Views

Author

N. J. A. Sloane, Don Knuth, Nick Singer (nsinger(AT)eos.hitc.com)

Keywords

Comments

Also the number of logically distinct strings of first order quantifiers in which n variables occur (C. S. Peirce, c. 1903). - Stephen Pollard (spollard(AT)truman.edu), Jun 07 2002
Stirling transform of A052849(n) = [2, 4, 12, 48, 240, ...] is a(n) = [2, 6, 26, 150, 1082, ...]. - Michael Somos, Mar 04 2004
Stirling transform of A000142(n-1) = [1, 1, 2, 6, 24, ...] is a(n-1) = [1, 2, 6, 26, ...]. - Michael Somos, Mar 04 2004
Stirling transform of (-1)^n * A024167(n-1) = [0, 1, -1, 5, -14, 94, ...] is a(n-2) = [0, 1, 2, 6, 26, ...]. - Michael Somos, Mar 04 2004
The asymptotic expansion of 2*log(n) - (2^1*log(1) + 2^2*log(2) + ... + 2^n*log(n))/2^n is (a(1)/1)/n + (a(2)/2)/n^2 + (a(3)/3)/n^3 + ... - Michael Somos, Aug 22 2004
This is the sequence of cumulants of the probability distribution of the number of tails before the first head in a sequence of fair coin tosses. - Michael Hardy (hardy(AT)math.umn.edu), May 01 2005
Appears to be row sums of A154921. - Mats Granvik, Jan 18 2009
This is the number of cyclically ordered partitions of n+1 labeled points. The ordered version is A000670. - Michael Somos, Jan 08 2011
A000670(n+1) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, Apr 27 2012
Row sums of A154921 as conjectured above by Granvik. a(n) gives the number of outcomes of a race between n horses H1,...,Hn, where if a horse falls it is not ranked. For example, when n = 2 the 6 outcomes are a dead heat, H1 wins H2 second, H2 wins H1 second, H1 wins H2 falls, H2 wins H1 falls or both fall. - Peter Bala, May 15 2012
Also the number of disjoint areas of a Venn diagram for n multisets. - Aurelian Radoaca, Jun 27 2016
Also the number of ways of ordering n nonnegative integers, allowing for the possibility of ties, and also comparing the smallest integers with 0. Each comparison with 0 gives two possibilities, x > 0 or x=0. As such, without comparison with 0, we get A000670, the number of ways of ordering n nonnegative integers, allowing for the possibility of ties, or the number of ways n competitors can rank in a competition, allowing for the possibility of ties. For instance, for 2 nonnegative integers x,y, there are the following 6 ways of ordering them: x = y = 0, x = y > 0, x > y = 0, x > y > 0, y > x = 0, y > x > 0. - Aurelian Radoaca, Jul 09 2016
Also the number of ordered set partitions of subsets of {1,...,n}. Also the number of chains of distinct nonempty subsets of {1,...,n}. - Gus Wiseman, Feb 01 2019
Number of combinations of a Simplex lock having n buttons.
Row sums of the unsigned cumulant expansion polynomials A127671 and logarithmic polynomials A263634. - Tom Copeland, Jun 04 2021
Also the number of vertices in the axis-aligned polytope consisting of all vectors x in R^n where, for all k in {1,...,n}, the k-th smallest coordinate of x lies in the interval [0, k]. - Adam P. Goucher, Jan 18 2023
Number of idempotent Boolean relation matrices whose complement is also idempotent. See Rosenblatt link. - Geoffrey Critzer, Feb 26 2023

Examples

			a(2)=6: the necklace representatives on 1,2,3 are ({123}), ({12},{3}), ({13},{2}), ({23},{1}), ({1},{2},{3}), ({1},{3},{2})
G.f. = 1 + 2*x + 6*x^2 + 26*x^3 + 150*x^4 + 1082*x^5 + 9366*x^6 + 94586*x^7 + ...
From _Gus Wiseman_, Feb 01 2019: (Start)
The a(3) = 26 ordered set partitions of subsets of {1,2,3} are:
  {}  {{1}}  {{2}}  {{3}}  {{12}}    {{13}}    {{23}}    {{123}}
                           {{1}{2}}  {{1}{3}}  {{2}{3}}  {{1}{23}}
                           {{2}{1}}  {{3}{1}}  {{3}{2}}  {{12}{3}}
                                                         {{13}{2}}
                                                         {{2}{13}}
                                                         {{23}{1}}
                                                         {{3}{12}}
                                                         {{1}{2}{3}}
                                                         {{1}{3}{2}}
                                                         {{2}{1}{3}}
                                                         {{2}{3}{1}}
                                                         {{3}{1}{2}}
                                                         {{3}{2}{1}}
(End)
		

References

  • R. Austin, R. K. Guy, and R. Nowakowski, unpublished notes, circa 1987.
  • N. G. de Bruijn, Asymptotic Methods in Analysis, Dover, 1981, p. 36.
  • Eric Hammer, The Calculations of Peirce's 4.453, Transactions of the Charles S. Peirce Society, Vol. 31 (1995), pp. 829-839.
  • D. E. Knuth, personal communication.
  • J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 174.
  • Charles Sanders Peirce, Collected Papers, eds. C. Hartshorne and P. Weiss, Harvard University Press, Cambridge, Vol. 4, 1933, pp. 364-365. (CP 4.453 in the electronic edition of The Collected Papers of Charles Sanders Peirce.)
  • Dawidson Razafimahatolotra, Number of Preorders to Compute Probability of Conflict of an Unstable Effectivity Function, Preprint, Paris School of Economics, University of Paris I, Nov 23 2007.

Crossrefs

Same as A076726 except for a(0). Cf. A008965, A052861, A008277.
Binomial transform of A000670, also double of A000670. - Joe Keane (jgk(AT)jgk.org)
A002050(n) = a(n) - 1.
A000629, A000670, A002050, A052856, A076726 are all more-or-less the same sequence. - N. J. A. Sloane, Jul 04 2012
Row sums of A028246.
A diagonal of the triangular array in A241168.
Row sums of unsigned A127671 and A263634.

Programs

  • Maple
    spec := [ B, {B=Cycle(Set(Z,card>=1))}, labeled ]; [seq(combstruct[count](spec, size=n), n=0..20)];
    a:=n->add(Stirling2(n+1,k)*(k-1)!,k=1..n+1); # Mike Zabrocki, Feb 05 2005
  • Mathematica
    a[ 0 ] = 1; a[ n_ ] := (a[ n ] = 1 + Sum[ Binomial[ n, k ] a[ n-k ], {k, 1, n} ])
    Table[ PolyLog[n, 1/2], {n, 0, -18, -1}] (* Robert G. Wilson v, Aug 05 2010 *)
    a[ n_] := If[ n<0, 0, PolyLog[ -n, 1/2]]; (* Michael Somos, Mar 07 2011 *)
    Table[Sum[(-1)^(n-k) StirlingS2[n,k]k! 2^k,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Oct 21 2011 *)
    Join[{1}, Rest[t=30; Range[0, t]! CoefficientList[Series[2/(2 - Exp[x]), {x, 0, t}], x]]] (* Vincenzo Librandi, Jan 02 2016 *)
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff(subst( (1 + y) / (1 - y), y, exp(x + x * O(x^n)) - 1), n))} /* Michael Somos, Mar 04 2004 */
    
  • PARI
    {a(n)=polcoeff(sum(m=0, n, 2^m*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)} \\ Paul D. Hanna, Jul 20 2011
    
  • Python
    from math import comb
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A000629(n): return 1+sum(comb(n,j)*A000629(j) for j in range(n)) if n else 1 # Chai Wah Wu, Sep 25 2023

Formula

a(n) = 2*A000670(n) - 0^n. - Michael Somos, Jan 08 2011
O.g.f.: Sum_{n>=0} 2^n*n!*x^n / Product_{k=0..n} (1+k*x). - Paul D. Hanna, Jul 20 2011
E.g.f.: exp(x) / (2 - exp(x)) = d/dx log(1 / (2 - exp(x))).
a(n) = Sum_{k>=1} k^n/2^k.
a(n) = 1 + Sum_{j=0..n-1} C(n, j)*a(j).
a(n) = round(n!/log(2)^(n+1)) (just for n <= 15). - Henry Bottomley, Jul 04 2000
a(n) is asymptotic to n!/log(2)^(n+1). - Benoit Cloitre, Oct 20 2002
a(n) = Sum_{k=0..n} (-1)^(n-k)*Stirling2(n, k)*k!*2^k. - Vladeta Jovovic, Sep 29 2003
a(n) = Sum_{k=1..n} A008292(n, k)*2^k; A008292: triangle of Eulerian numbers. - Philippe Deléham, Jun 05 2004
a(1) = 1, a(n) = 2*Sum_{k=1..n-1} k!*A008277(n-1, k) for n>1 or a(n) = Sum_{k=1..n} (k-1)!*A008277(n, k). - Mike Zabrocki, Feb 05 2005
a(n) = Sum_{k=0..n} Stirling2(n+1, k+1)*k!. - Paul Barry, Apr 20 2005
A000629 = binomial transform of this sequence. a(n) = sum of terms in n-th row of A028246. - Gary W. Adamson, May 30 2005
a(n) = 2*(-1)^n * n!*Laguerre(n,P((.),2)), umbrally, where P(j,t) are the polynomials in A131758. - Tom Copeland, Sep 28 2007
a(n) = 2^n*A(n,1/2); A(n,x) the Eulerian polynomials. - Peter Luschny, Aug 03 2010
a(n) = (-1)^n*b(n), where b(n) = -2*Sum_{k=0..n-1} binomial(n,k)*b(k), b(0)=1. - Vladimir Kruchinin, Jan 29 2011
Row sums of A028246. Let f(x) = x+x^2. Then a(n+1) = (f(x)*d/dx)^n f(x) evaluated at x = 1. - Peter Bala, Oct 06 2011
O.g.f.: 1+2*x/(U(0)-2*x) where U(k)=1+3*x+3*x*k-2*x*(k+2)*(1+x+x*k)/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 14 2011
E.g.f.: exp(x)/(2 - exp(x)) = 2/(2-Q(0))-1; Q(k)=1+x/(2*k+1-x*(2*k+1)/(x+(2*k+2)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 14 2011
G.f.: 1 / (1 - 2*x / (1 - 1*x / (1 - 4*x / (1 - 2*x / (1 - 6*x / ...))))). - Michael Somos, Apr 27 2012
PSUM transform of A162509. BINOMIAL transform is A007047. - Michael Somos, Apr 27 2012
G.f.: 1/G(0) where G(k) = 1 - x*(2*k+2)/( 1 - x*(k+1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 23 2013
E.g.f.: 1/E(0) where E(k) = 1 - x/(k+1)/(1 - 1/(1 + 1/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Mar 27 2013
G.f.: T(0)/(1-2*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, Oct 29 2013
a(n) = log(2)*integral_{x>=0} (ceiling(x))^n * 2^(-x) dx. - Peter Bala, Feb 06 2015

Extensions

a(19) from Michael Somos, Mar 07 2011

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

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).

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

A122704 a(n) = Sum_{k=0..n} 3^(n-k)*A123125(n, k).

Original entry on oeis.org

1, 1, 4, 22, 160, 1456, 15904, 202672, 2951680, 48361216, 880405504, 17630351872, 385148108800, 9114999832576, 232311251144704, 6343764407375872, 184778982658539520, 5718564661248065536, 187389427488113557504, 6481629887083387420672, 235993351028007334051840
Offset: 0

Views

Author

Philippe Deléham, Oct 22 2006

Keywords

Comments

a(n+1) = [1,4,22,160,1456,...] is the first Eulerian transform of A000244 (powers of 3), it is also the Stirling transform of A080599(n+1) = [1,3,12,66,450,...].

Examples

			G.f. = 1 + x + 4*x^2 + 22*x^3 + 160*x^4 + 1456*x^5 + 15904*x^6 + ... - _Michael Somos_, Jun 05 2021
		

Crossrefs

Cf. A076726.

Programs

  • Maple
    # From Peter Luschny, Jun 27 2019: (Start)
    seq(subs(x=3, add(combinat:-eulerian1(n,k)*x^k, k=0..n)), n=0..20);
    # Alternative:
    gf := -2/(exp(2*x) - 3): series(gf, x, 21): seq(n!*coeff(%, x, n), n=0..20);
    # (End)
    # Or via the recurrence of the Fubini polynomials:
    F := proc(n) option remember; if n = 0 then return 1 fi;
    expand(add(binomial(n, k)*F(n - k)*x, k = 1..n)) end:
    a := n -> 2^n*subs(x = 1/2, F(n)):
    seq(a(n), n = 0..20); # Peter Luschny, May 21 2021
    # next Maple program:
    a:= proc(n) option remember; `if`(n=0, 1, add(
          a(n-j)*binomial(n, j)*2^(j-1), j=1..n))
        end:
    seq(a(n), n=0..23);  # Alois P. Heinz, May 30 2021
  • Mathematica
    CoefficientList[Series[(Exp[x]-2*Cosh[x])/(2*Exp[x]-3*Cosh[x]), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 24 2013 *)
    Table[Sum[2^(n+1)*k^n/3^(k+1), {k, 0, Infinity}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 28 2013 *)
    Round@Table[(-1)^(n+1) (PolyLog[-n, Sqrt[3]] + PolyLog[-n, -Sqrt[3]])/3, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 31 2015 *)
    Table[Sum[StirlingS2[n, k]*2^(n-k)*k!, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jul 13 2018 *)
    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]*3^k, {k, 0, n}], {n, 0, 20}] (* Jean-François Alcover, Jul 13 2019, after Peter Luschny *)
    a[n_] := (-2)^(n + 1) PolyLog[-n, 3] / 3;
    Table[a[n], {n, 0, 20}] (* Peter Luschny, Aug 20 2021 *)
  • PARI
    {a(n)=polcoeff(sum(m=0,n,m!*x^m/prod(k=1,m,1-2*k*x+x*O(x^n))),n)} /* Paul D. Hanna, Jul 20 2011 */
    
  • PARI
    {a(n) = if(n<0, 0, n!*polcoeff( 2/(3 - exp(2*x + x*O(x^n))), n))}; /* Michael Somos, Jun 05 2021 */

Formula

O.g.f.: Sum_{n>=0} n! * x^n / Product_{k=1..n} (1-2*k*x). - Paul D. Hanna, Jul 20 2011
a(n) = Sum_{k=0..n} A131689(n,k)*2^(n-k). - Philippe Deléham, Oct 09 2007
a(n) = A_{n}(3) where A_{n}(x) are the Eulerian polynomials. - Peter Luschny, Aug 03 2010
E.g.f.: (exp(x) - 2*cosh(x))/(2*exp(x) - 3*cosh(x)) =1 + x/(U(0)-x) where U(k)= 4*k+1 - x/(1 + x/(4*k+3 - x/(1 + x/U(k+1)))); (continued fraction, 4-step). - Sergei N. Gladkovskii, Nov 08 2012
G.f.: 1 + x/G(0) where G(k) = 1 - x*2*(2*k+2) + x^2*(k+1)*(k+2)*(1-2^2)/G(k+1); (continued fraction due to T. J. Stieltjes). - Sergei N. Gladkovskii, Jan 11 2013
a(n) ~ n!/3 * (2/log(3))^(n+1). - Vaclav Kotesovec, Jun 24 2013
G.f.: 1/Q(0), where Q(k) = 1 - x*(4*k+1) - 3*x^2*(k+1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 30 2013
a(n) = Sum_{k>=0} 2^(n+1)*k^n/3^(k+1). - Vaclav Kotesovec, Nov 28 2013
a(n) = 2^n*log(3)* Integral_{x >= 0} (floor(x))^n * 3^(-x) dx. - Peter Bala, Feb 14 2015
From Karol A. Penson, Sep 04 2015: (Start)
E.g.f.: 2/(3-exp(2*x)).
Special values of the generalized hypergeometric function n_F_(n-1):
a(n) = (2^(n+1)/9) * hypergeom([2,2,..2],[1,1,..1],1/3), where the sequence in the first square bracket ("upper" parameters) has n elements all equal to 2 whereas the sequence in the second square bracket ("lower" parameters) has n-1 elements all equal to 1.
Example: a(4) = (2^5/9) * hypergeom([2,2,2,2],[1,1,1],1/3) = 16. (End)
a(n) = (-1)^(n+1)*(Li_{-n}(sqrt(3)) + Li_{-n}(-sqrt(3)))/3, where Li_n(x) is the polylogarithm. - Vladimir Reshetnikov, Oct 31 2015
a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * 2^(k - 1) * a(n-k). - Ilya Gutkovskiy, Jan 16 2020
a(n) = 2^n*F_{n}(1/2), where F_{n}(x) is the Fubini polynomial. This is another way to state the above formula from Ilya Gutkovskiy. - Peter Luschny, May 21 2021
a(n+1) = -2*a(n) + 3*Sum_{k=0..n} binomial(n, k)*a(k)*a(n-k). - Michael Somos, Jun 05 2021
a(n) = (-2)^(n + 1)*PolyLog(-n, 3)/3. - Peter Luschny, Aug 20 2021

Extensions

a(7) corrected (was 206672) and more terms from Peter Luschny, Aug 03 2010
More terms from Vaclav Kotesovec, Jul 13 2018

A052856 E.g.f.: (1-3*exp(x)+exp(2*x))/(exp(x)-2).

Original entry on oeis.org

1, 2, 4, 14, 76, 542, 4684, 47294, 545836, 7087262, 102247564, 1622632574, 28091567596, 526858348382, 10641342970444, 230283190977854, 5315654681981356, 130370767029135902, 3385534663256845324
Offset: 0

Views

Author

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

Keywords

Comments

Previous name was: A simple grammar.
Stirling transform of A005212(n-1)=[1,1,0,6,0,120,0,...] is a(n-1)=[1,2,4,14,76,...]. - Michael Somos, Mar 04 2004
Stirling transform of (-1)^n*A052612(n-1)=[0,2,-2,12,-24,...] is a(n-1)=[0,2,4,14,76,...]. - Michael Somos, Mar 04 2004
Stirling transform of A000142(n)=[2,2,6,24,120,...] is a(n)=[2,2,4,14,76,...]. - Michael Somos, Mar 04 2004

Crossrefs

A000670(n)=a(n)-1, if n>0. A032109(n)=a(n)/2, if n>0.
A000629, A000670, A002050, A052856, A076726 are all more-or-less the same sequence. - N. J. A. Sloane, Jul 04 2012

Programs

  • Maple
    spec := [S,{B=Sequence(C),C=Set(Z,1 <= card),S=Union(B,C)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    With[{nn=20},CoefficientList[Series[(1-3Exp[x]+Exp[x]^2)/(-2+Exp[x]),{x,0,nn}],x]Range[0,nn]!] (* Harvey P. Dale, Nov 24 2012 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(subst(y+1/(1-y),y,exp(x+x*O(x^n))-1),n))

Formula

E.g.f.: (1-3*exp(x)+exp(x)^2)/(-2+exp(x))
a(n) ~ n!/(2*(log(2))^(n+1)). - Vaclav Kotesovec, Oct 05 2013

Extensions

New name using e.g.f., Vaclav Kotesovec, Oct 05 2013

A195204 Triangle of coefficients of a sequence of binomial type polynomials.

Original entry on oeis.org

2, 2, 4, 6, 12, 8, 26, 60, 48, 16, 150, 380, 360, 160, 32, 1082, 2940, 3120, 1680, 480, 64, 9366, 26908, 31080, 19040, 6720, 1344, 128, 94586, 284508, 351344, 236880, 96320, 24192, 3584, 256
Offset: 1

Views

Author

Peter Bala, Sep 13 2011

Keywords

Comments

Define a polynomial sequence P_n(x) by means of the recursion
P_(n+1)(x) = x*(P_n(x)+ P_n(x+1)), with P_0(x) = 1.
The first few polynomials are
P_1(x) = 2*x, P_2(x) = 2*x*(2*x + 1),
P_3(x) = 2*x*(4*x^2 + 6*x + 3), P_4(x) = 2*x*(8*x^3+24*x^2+30*x+13).
The present table shows the coefficients of these polynomials (excluding P_0(x)) in ascending powers of x. The P_n(x) are a polynomial sequence of binomial type. In particular, if we denote P_n(x) by x^[n] then we have the analog of the binomial expansion
(x+y)^[n] = Sum_{k = 0..n} binomial(n,k)*x^[n-k]*y^[k].
There are further analogies between the x^[n] and the monomials x^n.
1) Dobinski-type formula
exp(-x)*Sum_{k >= 0} (-k)^[n]*x^k/k! = (-1)^n*Bell(n,2*x),
where the Bell (or exponential) polynomials are defined as
Bell(n,x) := Sum_{k = 1..n} Stirling2(n,k)*x^k.
Equivalently, the connection constants associated with the polynomial sequences {x^[n]} and {x^n} are (up to signs) the same as the connection constants associated with the polynomial sequences {Bell(n,2*x)} and {Bell(n,x)}. For example, the list of coefficients of x^[4] is [26,60,48,16] and a calculation gives
Bell(4,2*x) = -26*Bell(1,x) + 60*Bell(2,x) - 48*Bell(3,x) + 16*Bell(4,x).
2) Analog of Bernoulli's summation formula
Bernoulli's formula for the sum of the p-th powers of the first n positive integers is
Sum_{k = 1..n} k^p = (1/(p+1))*Sum_{k = 0..p} (-1)^k * binomial(p+1,k)*B_k*n^(p+1-k), where B_k = [1,-1/2,1/6,0,-1/30,...] is the sequence of Bernoulli numbers.
This generalizes to
2*Sum_{k = 1..n} k^[p] = 1/(p+1)*Sum_{k = 0..p} (-1)^k * binomial(p+1,k)*B_k*n^[p+1-k].
The polynomials P_n(x) belong to a family of polynomial sequences P_n(x,t) of binomial type, dependent on a parameter t, and defined recursively by P_(n+1)(x,t)= x*(P_n(x,t)+ t*P_n(x+1,t)), with P_0(x,t) = 1. When t = 0 we have P_n(x,0) = x^n, the monomial polynomials. The present table is the case t = 1. The case t = -2 is (up to signs) A079641. See also A195205 (case t = 2).
Triangle T(n,k) (1 <= k <= n), read by rows, given by (0, 1, 2, 2, 4, 3, 6, 4, 8, 5, 10, ...) DELTA (2, 0, 2, 0, 2, 0, 2, 0, 2, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 22 2011
T(n,k) is the number of binary relations R on [n] with index = 1 containing exactly k strongly connected components (SCC's) and satisfying the condition that if (x,y) is in R then x and y are in the same SCC. - Geoffrey Critzer, Jan 17 2024

Examples

			Triangle begins
n\k|....1......2......3......4......5......6......7
===================================================
..1|....2
..2|....2......4
..3|....6.....12......8
..4|...26.....60.....48.....16
..5|..150....380....360....160.....32
..6|.1082...2940...3120...1680....480.....64
..7|.9366..26908..31080..19040...6720...1344....128
...
Relation with rising factorials for row 4:
x^[4] = 16*x^4+48*x^3+60*x^2+26*x = 2^4*x*(x+1)*(x+2)*(x+3)-6*2^3*x*(x+1)*(x+2)+7*2^2*x*(x+1)-2*x, where [1,7,6,1] is the fourth row of the triangle of Stirling numbers of the second kind A008277.
Generalized Dobinski formula for row 4:
exp(-x)*Sum_{k >= 1} (-k)^[4]*x^k/k! = exp(-x)*Sum_{k >= 1} (16*k^4-48*k^3+60*k^2-26*k)*x^k/k! = 16*x^4+48*x^3+28*x^2+2*x = Bell(4,2*x).
Example of generalized Bernoulli summation formula:
2*(1^[2]+2^[2]+...+n^[2]) = 1/3*(B_0*n^[3]-3*B_1*n^[2]+3*B_2*n^[1]) =
n*(n+1)*(4*n+5)/3, where B_0 = 1, B_1 = -1/2, B_2 = 1/6 are Bernoulli numbers.
From _Philippe Deléham_, Dec 22 2011: (Start)
Triangle (0, 1, 2, 2, 4, 3, 6, ...) DELTA (2, 0, 2, 0, 2, ...) begins:
  1;
  0,    2;
  0,    2,     4;
  0,    6,    12,     8;
  0,   26,    60,    48,    16;
  0,  150,   380,   360,   160,   32;
  0, 1082,  2940,  3120,  1680,  480,   64;
  0, 9366, 26908, 31080, 19040, 6720, 1344, 128;
  ... (End)
		

Crossrefs

Cf. A000629 (row sums), A000670 (one half row sums), A014307 (row polys. at x = 1/2), A079641, A195205, A209849.

Programs

  • Maple
    # The function BellMatrix is defined in A264428.
    # Adds (1,0,0,0, ..) as column 0.
    BellMatrix(n -> (-1)^(n+1)*polylog(-n, 2), 10); # Peter Luschny, Jan 29 2016
  • Mathematica
    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[(-1)^(#+1) PolyLog[-#, 2]&, rows];
    Table[M[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 24 2018, after Peter Luschny *)

Formula

E.g.f.: F(x,z) := (exp(z)/(2-exp(z)))^x = Sum_{n>=0} P_n(x)*z^n/n!
= 1 + 2*x*z + (2*x+4*x^2)*z^2/2! + (6*x+12*x^2+8*x^3)*z^3/3! + ....
The generating function F(x,z) satisfies the partial differential equation d/dz(F(x,z)) = x*F(x,z) + x*F(x+1,z) and hence the row polynomials P_n(x) satisfy the recurrence relation
P_(n+1)(x)= x*(P_n(x) + P_n(x+1)), with P_0(x) = 1.
In what follows we change notation and write x^[n] for P_n(x).
Relation with the factorial polynomials:
For n >= 1,
x^[n] = Sum_{k = 1..n} (-1)^(n-k)*Stirling2(n,k)*2^k*x^(k),
and its inverse formula
2^n*x^(n) = Sum_{k = 1..n} |Stirling1(n,k)|*x^[k],
where x^(n) denotes the rising factorial x*(x+1)*...*(x+n-1).
Relation with the Bell polynomials:
The alternating n-th row entries (-1)^(n+k)*T(n,k) are the connection coefficients expressing the polynomial Bell(n,2*x) as a linear combination of Bell(k,x), 1 <= k <= n.
The delta operator:
The sequence of row polynomials is of binomial type. If D denotes the derivative operator d/dx then the delta operator D* for this sequence of binomial type polynomials is given by
D* = D/2 - log(cosh(D/2)) = log(2*exp(D)/(exp(D)+1))
= (D/2) - (D/2)^2/2! + 2*(D/2)^4/4! - 16*(D/2)^6/6! + 272*(D/2)^8/8! - ...,
where [1,2,16,272,...] is the sequence of tangent numbers A000182.
D* is the lowering operator for the row polynomials
(D*)x^[n] = n*x^[n-1].
Associated Bernoulli polynomials:
Generalized Bernoulli polynomial GB(n,x) associated with the polynomials x^[n] may be defined by
GB(n,x) := ((D*)/(exp(D)-1))x^[n].
They satisfy the difference equation
GB(n,x+1) - GB(n,x) = n*x^[n-1]
and have the expansion
GB(n,x) = -(1/2)*n*x^[n-1] + (1/2)*Sum_{k = 0..n} binomial(n,k) * B_k * x^[n-k], where B_k denotes the ordinary Bernoulli numbers.
The first few polynomials are
GB(0,x) = 1/2, GB(1,x) = x-3/4, GB(2,x) = 2*x^2-2*x+1/12,
GB(3,x) = 4*x^3-3*x^2-x, GB(4,x) = 8*x^4-4*x^2-4*x-1/60.
It can be shown that
1/(n+1)*(d/dx)(GB(n+1,x)) = Sum_{i = 0..n} 1/(i+1) * Sum_{k = 0..i} (-1)^k *binomial(i,k)*(x+k)^[n].
This generalizes a well-known formula for Bernoulli polynomials.
Relations with other sequences:
Row sums: A000629(n) = 2*A000670(n). Column 1: 2*A000670(n-1). Row polynomials evaluated at x = 1/2: {P_n(1/2)}n>=0 = [1,1,2,7,35,226,...] = A014307.
T(n,k) = A184962(n,k)*2^k. - Philippe Deléham, Feb 17 2013
Also the Bell transform of A076726. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 29 2016
Conjecture: o.g.f. as a continued fraction of Stieltjes type: 1/(1 - 2*x*z/(1 - z/(1 - 2*(x + 1)*z/(1 - 2*z/(1 - 2*(x + 2)*z/(1 - 3*z/(1 - 2*(x + 3)*z/(1 - 4*z/(1 - ... ))))))))). - Peter Bala, Dec 12 2024

Extensions

a(1) added by Philippe Deléham, Dec 22 2011

A032109 "BIJ" (reversible, indistinct, labeled) transform of 1,1,1,1,...

Original entry on oeis.org

1, 1, 2, 7, 38, 271, 2342, 23647, 272918, 3543631, 51123782, 811316287, 14045783798, 263429174191, 5320671485222, 115141595488927, 2657827340990678, 65185383514567951, 1692767331628422662, 46400793659664205567, 1338843898122192101558, 40562412499252036940911
Offset: 0

Views

Author

Keywords

Comments

Starting (1, 2, 7, 38, 271, ...) = A008292 * [1, 1, 2, 4, 8, ...]. - Gary W. Adamson, Dec 25 2008
The inverse binomial transform is 1, 0, 1, 3, 19, 135, 1171, 11823, 136459, ..., see A091346. - R. J. Mathar, Oct 17 2012
Stirling transform of A001710. - Anton Zakharov, Aug 06 2016
For n even (resp. n odd), a(n) counts the ordered partitions of {1,2,...,n} with an even (resp. odd) number of blocks, and a(n)-1 counts the ordered partitions of {1,2,...,n} with an odd (resp. even) number of blocks. - Jose A. Rodriguez, Feb 04 2021

Crossrefs

A000629, A000670, A002050, A032109, A052856, A076726 are all more-or-less the same sequence. - N. J. A. Sloane, Jul 04 2012
A052856(n)=2*a(n), if n>0.

Programs

  • Maple
    A032109 := proc(n)
        (A000670(n)+1)/2 ;
    end proc: # R. J. Mathar, Oct 17 2012
    a := n -> (polylog(-n, 1/2)+`if`(n=0,3,2))/4:
    seq(round(evalf(a(n), 32)), n=0..18); # Peter Luschny, Nov 03 2015
    # alternative Maple program:
    b:= proc(n, m) option remember; `if`(n=0, m!,
          add(b(n-1, max(m, j)), j=1..m+1))
        end:
    a:= n-> (b(n,0)+1)/2:
    seq(a(n), n=0..23);  # Alois P. Heinz, Sep 29 2017
  • Mathematica
    Table[(PolyLog[-n, 1/2] + 2 + KroneckerDelta[n])/4, {n, 0, 20}] (* Vladimir Reshetnikov, Nov 02 2015 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(subst((1-y^2/2)/(1-y),y,exp(x+x*O(x^n))-1),n))
    
  • PARI
    list(n)=my(v=Vec(subst((1-y^2/2)/(1-y),y,exp(x+x*O(x^n))-1)));vector(n+1,i,v[i]*(i-1)!) \\ Charles R Greathouse IV, Oct 17 2012

Formula

E.g.f.: (e^(2*x)-2*e^x-1)/(2*e^x-4).
a(n) = (A000670(n)+1)/2. - Vladeta Jovovic, Apr 13 2003
a(n) = A052875(n)/2 + 1. - Max Alekseyev, Jan 31 2021
a(n) ~ sqrt(Pi/2)*n^(n+1/2)/(2*log(2)^(n+1)*exp(n)). - Ilya Gutkovskiy, Aug 06 2016
a(n) = Sum_{s in S_n^even} Product_{i=1..n} binomial(i,s(i)-1), where s ranges over the set S_n^even of even permutations of [n]. - Jose A. Rodriguez, Feb 02 2021

A163204 Triangle read by rows, A095989 convolved with A000670.

Original entry on oeis.org

1, 1, 2, 3, 2, 8, 13, 6, 8, 48, 75, 26, 24, 48, 368, 541, 150, 104, 144, 368, 3376, 4683, 1082, 600, 624, 1104, 3376, 35824, 47293, 9366, 4328, 3600, 4784, 10128, 35824, 430512, 545835, 94586, 37464, 25968, 27600, 43888, 107472, 430512, 5773936, 7087261, 1091670, 378344, 224784, 199088, 253200, 465712, 1291536, 5773936, 85482032
Offset: 1

Views

Author

Gary W. Adamson, Jul 23 2009

Keywords

Comments

Row sums = A000670 starting with offset 1: (1, 3, 13, 75, 541, 4683,...).
Left border = A000670, right border = A095989.
Second column: A076726. - Michel Marcus, Mar 31 2016

Examples

			First few rows of the triangle are:
1;
1, 2;
3, 2, 8;
13, 6, 8, 48;
75, 26, 24, 48, 368;
541, 150, 104, 144, 368, 3376;
4683, 1082, 600, 624, 1104, 3376, 35824;
47293, 9366, 4328, 3600, 4784, 10128, 35824, 430512;
...
		

Crossrefs

Programs

  • Mathematica
    max = 10; 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; A000670 = Table[ Fubini[n, 1], {n, 0, max}]; s = 1 - 1/Sum[Fubini[k, 1] q^k, {k, 0, max}] + O[q]^max; A095989 = CoefficientList[s/q, q]; row[n_] := A095989[[1 ;; n]]*Reverse[A000670[[1 ;; n]]]; Table[row[n], {n, 1, max-1}] // Flatten (* Jean-François Alcover, Mar 31 2016 *)

Formula

Descending antidiagonals of a multiplication table formed by convolving A095989 with A000670, where A095989 is the INVERTi transform of A000670 starting (1, 3, 13, 75,...).

Extensions

a(23) corrected by Jean-François Alcover, Mar 31 2016
Terms a(37) onward added by G. C. Greubel, Dec 10 2016

A298668 Number T(n,k) of set partitions of [n] into k blocks such that the absolute difference between least elements of consecutive blocks is always > 1; triangle T(n,k), n>=0, 0<=k<=ceiling(n/2), read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 3, 0, 1, 7, 2, 0, 1, 15, 12, 0, 1, 31, 50, 6, 0, 1, 63, 180, 60, 0, 1, 127, 602, 390, 24, 0, 1, 255, 1932, 2100, 360, 0, 1, 511, 6050, 10206, 3360, 120, 0, 1, 1023, 18660, 46620, 25200, 2520, 0, 1, 2047, 57002, 204630, 166824, 31920, 720
Offset: 0

Views

Author

Alois P. Heinz, Jan 24 2018

Keywords

Examples

			T(5,1) = 1: 12345.
T(5,2) = 7: 1234|5, 1235|4, 123|45, 1245|3, 124|35, 125|34, 12|345.
T(5,3) = 2: 124|3|5, 12|34|5.
T(7,4) = 6: 1246|3|5|7, 124|36|5|7, 124|3|56|7, 126|34|5|7, 12|346|5|7, 12|34|56|7.
T(9,5) = 24: 12468|3|5|7|9, 1246|38|5|7|9, 1246|3|58|7|9, 1246|3|5|78|9, 1248|36|5|7|9, 124|368|5|7|9, 124|36|58|7|9, 124|36|5|78|9, 1248|3|56|7|9, 124|38|56|7|9, 124|3|568|7|9, 124|3|56|78|9, 1268|34|5|7|9, 126|348|5|7|9, 126|34|58|7|9, 126|34|5|78|9, 128|346|5|7|9, 12|3468|5|7|9, 12|346|58|7|9, 12|346|5|78|9, 128|34|56|7|9, 12|348|56|7|9, 12|34|568|7|9, 12|34|56|78|9.
Triangle T(n,k) begins:
  1;
  0, 1;
  0, 1;
  0, 1,    1;
  0, 1,    3;
  0, 1,    7,     2;
  0, 1,   15,    12;
  0, 1,   31,    50,     6;
  0, 1,   63,   180,    60;
  0, 1,  127,   602,   390,    24;
  0, 1,  255,  1932,  2100,   360;
  0, 1,  511,  6050, 10206,  3360,  120;
  0, 1, 1023, 18660, 46620, 25200, 2520;
  ...
		

Crossrefs

Columns k=0-11 give (offsets may differ): A000007, A057427, A168604, A028243, A028244, A028245, A032180, A228909, A228910, A228911, A228912, A228913.
Row sums give A229046(n-1) for n>0.
T(2n+1,n+1) gives A000142.
T(2n,n) gives A001710(n+1).

Programs

  • Maple
    b:= proc(n, m, t) option remember; `if`(n=0, x^m, add(
          b(n-1, max(m, j), `if`(j>m, 1, 0)), j=1..m+1-t))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
    seq(T(n), n=0..14);
    # second Maple program:
    T:= (n, k)-> `if`(k=0, `if`(n=0, 1, 0), (k-1)!*Stirling2(n-k+1, k)):
    seq(seq(T(n, k), k=0..ceil(n/2)), n=0..14);
    # third Maple program:
    T:= proc(n, k) option remember; `if`(k<2, `if`(n=0 xor k=0, 0, 1),
          `if`(k>ceil(n/2), 0, add((k-j)*T(n-1-j, k-j), j=0..1)))
        end:
    seq(seq(T(n, k), k=0..ceil(n/2)), n=0..14);
  • Mathematica
    T[n_, k_] := T[n, k] = If[k < 2, If[Xor[n == 0, k == 0], 0, 1],
         If[k > Ceiling[n/2], 0, Sum[(k-j) T[n-1-j, k-j], {j, 0, 1}]]];
    Table[Table[T[n, k], {k, 0, Ceiling[n/2]}], {n, 0, 14}] // Flatten (* Jean-François Alcover, Mar 08 2021, after third Maple program *)

Formula

T(n,k) = (k-1)! * Stirling2(n-k+1,k) for k>0, T(n,0) = A000007(n).
T(n,k) = Sum_{j=0..k-1} (-1)^j*C(k-1,j)*(k-j)^(n-k) for k>0, T(n,0) = A000007(n).
T(n,k) = (k-1)! * A136011(n,k) for n, k >= 1.
Sum_{j>=0} T(n+j,j) = A076726(n) = 2*A000670(n) = A000629(n) + A000007(n).
Showing 1-9 of 9 results.