cp's OEIS Frontend

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

Showing 1-10 of 14 results. Next

A154931 Third column of A154921.

Original entry on oeis.org

1, 3, 18, 130, 1125, 11361, 131124, 1702548, 24562575, 389799355, 6748339158, 126565340694, 2556332651145, 55320126580005, 1276961156453160, 31318513972988008, 813295166343147315, 22293401161982239071, 643251586018800611370, 19488333337058966337930
Offset: 2

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Author

Mats Granvik, Jan 17 2009

Keywords

Crossrefs

Cf. A154921.

Programs

  • Maple
    b:= proc(n, p) option remember; `if`(n=0, p!,
          add(b(n-j, p+j)/j!, j=1..n))
        end:
    a:= n-> b(n-2, 2)/2:
    seq(a(n), n=2..23);  # Alois P. Heinz, Feb 03 2019
  • Mathematica
    With[{nn=20},Drop[CoefficientList[Series[(x^2/2)/(2-Exp[x]),{x,0,nn}],x] Range[0,nn]!,2]] (* Harvey P. Dale, Jan 16 2013 *)

Formula

E.g.f.: 1/2!*x^2/(2-exp(x)) = x^2/2! + 3*x^3/3! + 18*x^4/4! + .... - Peter Bala, Apr 20 2012

Extensions

a(9) through a(18) from Peter Bala, Apr 20 2012
a(19)-a(21) from Alois P. Heinz, Feb 03 2019

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

Views

Author

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.

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

A263634 Irregular triangle read by rows: row n gives coefficients of n-th logarithmic polynomial L_n(x_1, x_2, ...) with monomials sorted into standard order.

Original entry on oeis.org

1, -1, 1, 2, -3, 1, -6, 12, -4, -3, 1, 24, -60, 20, 30, -5, -10, 1, -120, 360, -120, -270, 30, 120, 30, -6, -15, -10, 1, 720, -2520, 840, 2520, -210, -1260, -630, 42, 210, 140, 210, -7, -21, -35, 1
Offset: 1

Views

Author

N. J. A. Sloane, Oct 29 2015

Keywords

Comments

"Standard order" here means as produced by Maple's "sort" command.
From Petros Hadjicostas, May 27 2020: (Start)
According to the Maple help files for the "sort" command, polynomials in multiple variables are "sorted in total degree with ties broken by lexicographic order (this is called graded lexicographic order)."
Thus for example, x_1^2*x_3 = x_1*x_1*x_3 > x_1*x_2*x_2 = x_1*x_2^2, while x_1^2*x_4 = x_1*x_1*x_4 > x_1*x_2*x_3. (End)
Row sums are 0 (for n > 1). Numbers of terms in rows are partition numbers A000041.
From Tom Copeland, Nov 06 2015: (Start)
With the formal Taylor series f(x) = 1 + x[1] x + x[2] x^2/2! + ... , the partition polynomials of this entry give d[log(f(x))]/dx = L_1(x[1]) + L_2(x[1], x[2]) x + L_3(...) x^2/2! + ..., and the coefficients of the reduced polynomials with x[n] = t are signed A028246.
The raising operator R = x + d[log(f(D)]/dD = x + L_1(x[1]) + L_2[x[1], x[2]) D + L_3(x[1], x[2], x[3]) D^2/2! + ... with D = d/dx generates an Appell sequence of polynomials, given umbrally by P_n(x[1], ..., x[n]; x) = (x[.] + x)^n = Sum_{k=0..n} binomial(n,k) x[k] * x^(n-k) = R^n 1 with the e.g.f. f(t)*e^(x*t) = exp[t P.(x[1], ..., x[.]; x)]. P_0 = x[0] = 1.
The umbral compositional inverse Appell sequence is generated by R = x - d[log(f(D))]/dD with e.g.f. e^(x*t)/f(t) = exp[t IP.(x[1], ..., x[.]; x)], so umbrally IP_n(x[1], ..., x[n]; P.(x[1], ..., x[n]; x)) = x^n = P_n(x[1], ..., x[n]; IP.(x[1], ..., x[n]; x)). An unsigned array for the reduced IP_n(x[1], ..., x[n]; x) polynomials with IP_0 = x[0] = 1 and x[n] = -1 for n > 0 is A154921, for which f(t) = 2 - e^t. (End)
From Tom Copeland, Sep 08 2016: (Start)
The Appell formalism allows a matrix representation in the power basis x^n of the raising operator R that incorporates this array's partition polynomials L_n(x[1], ..., x[n]):
VP_(n+1) = VP_n * R = VP_n * XPS^(-1) * MX * XPS, where XPS is the matrix formed from multiplying the n-th diagonal of the Pascal matrix PS of A007318 by the indeterminate x[n], with x[0] = 1 for the main diagonal of ones, i.e., XPS[n,k] = PS[n,k] * x[n-k]; the matrix MX is A129185; the matrix XPS^(-1) is the inverse of XPS, which can be formed by multiplying the diagonals of the Pascal matrix by the partition polynomials IPT(n, x[1], ..., x[n]) of A133314, i.e., XPS^(-1)[n,k] = PS[n,k] * IPT(n-k, x[1], ...); and VP_n is the row vector in the power basis representing the Appell polynomial P_n(x) formed from the basic sequence of moments 1, x[1], x[2], ..., i.e., umbrally P_n(x) = (x[.] + x)^n = Sum_{k=0..n} binomial(n,k) * x[k] * x^(n-k).
Then R = XPS^(-1) * MX * XPS is the Pascal matrix PS with an additional first superdiagonal of ones and the other lower diagonals multiplied by the partition polynomials of this array, i.e., R[n,k] = PS[n,k] * L_{n+1-k}(x[1], ..., x[n+1-k]) except for the first superdiagonal of ones.
Consistently, VP_n = (1, 0, 0, ...) * R^n = (1, 0, 0, ...) * XPS^(-1) * MX^n * XPS = (1, 0, 0, ...) * MX^n * XPS = the n-th row vector of XPS, which is the vector representation of P_n(x) = (x[.] + x)^n with x[0] = 1.
See the Copeland link for the umbral representation R = exp[g.*D] * x * exp[h.*D] that reflects the matrix representations.
The Stirling partition polynomials of the first kind St1_n(a[1], a[2], ..., a[n]) of A036039, the Stirling partition polynomials of the second kind St2_n(b[1], b[2], ..., b[n]) of A036040, and the refined Lah polynomials Lah_n[c[1], c[2], ..., c[n]) of A130561 are Appell sequences in the respective distinguished indeterminates a[1], b[1], and c[1]. Comparing the formulas for their raising operators with that in this entry, L_n(x[1], x[2], ..., x[n]) evaluates to
A) (n-1)! * a[n] for x[n] = St1_n(a[1], a[2], ..., a[n]);
B) b[n] for x[n] = St2_n(b[1], b[2], ..., b[n]);
C) n! * c[n] for x[n] = Lah_n(c[1], c[2], ..., c[n]).
Conversely, from the respective e.g.f.s (added Sep 12 2016)
D) x[n] = St1_n(L_1(x[1])/0!, ..., L_n(x[1], ..., x[n])/(n-1)!);
E) x[n] = St2_n(L_1(x[1]), ..., L_n(x[1], ..., x[n]));
F) x[n] = Lah_n(L_1(x[1])/1!, ..., L_n(x[1], ..., x[n])/n!).
Given only the Appell sequence with no closed form for the e.g.f., the raising operator can be generated using this formalism, as has been partially done for A134264. (End)
For the Appell sequences above, the raising operator is related to the recursion P_(n+1)(x) = x * P_n(x) + Sum_{k=0..n} binomial(n,k) * L_(n-k+1)(x[1], ..., x[n+k-1]) * P_k(x). For a derivation and connections to formal cumulants (c_n = L_n(x[1], ...)) and moments (m_n = x[n]), see the Copeland link on noncrossing partitions. With x = 0, the recursion reduces to x[n+1] = Sum_{k = 0..n} binomial(n,k) * L_(n-k+1)(x[1], ..., x[n+k-1]) * x[k] with x[0] = 1. This array is a differently ordered version of A127671. - Tom Copeland, Sep 13 2016
With x[n] = x^(n-1), a signed version of A130850 is obtained. - Tom Copeland, Nov 14 2016
See p. 2 of Getzler for a relation to stable graphs called necklaces used in computations for Deligne-Mumford-Knudsen moduli spaces of stable curves of genus 1. - Tom Copeland, Nov 15 2019
For a relation to a combinatorial Faa di Bruno Hopf algebra related to functional composition, as presented by Connes and Moscovici, see Figueroa et al. - Tom Copeland, Jan 17 2020
From Tom Copeland, May 17 2020: (Start)
The e.g.f. of an Appell sequence is f(t) e^(x*t) with f(0) = 1. Given the Laguerre-Polya class function f(t) = e^(-a*t^2 + b*t) Product_m (1 - t/z_m) e^(t/z_m) with a = 0 for simplicity (more generally a >= 0) and b real and where the product runs over all the zeros z_m of f(t) with all zeros real and Sum_m 1/(z_m)^2 convergent, the raising operator of the Appell polynomials is R = x + b - Sum_{k > 0} c_(k+1) D^k with c_k = Sum_m (1/(z_m)^k), i.e., traces of powers of the reciprocals of the zeros. From R in earlier comments, b = L_1(x_1) and otherwise c_k = -L_k(x_1, ..., x_k).
The Laguerre / Turan / de Gua inequalities (Csordas and Williamson and Skovgaard) imply that all the zeros of each Appell polynomial are real and simple and its extrema are local maxima above the x-axis and local minima below and are located above or below the zeros of the next lower degree Appell polynomial. (End)
From Tom Copeland, Oct 15 2020: (Start)
With a_n = n! * b_n = (n-1)! * c_n for n > 0, represent a function with f(0) = a_0 = b_0 = 1 as an
A) exponential generating function (e.g.f), or formal Taylor series: f(x) = e^{a.x} = 1 + Sum_{n > 0} a_n * x^n/n!
B) ordinary generating function (o.g.f.), or formal power series: f(x) = 1/(1-b.x) = 1 + Sum_{n > 0} b_n * x^n
C) logarithmic generating function (l.g.f): f(x) = 1 - log(1 - c.x) = 1 + Sum_{n > 0} c_n * x^n /n.
Expansions of log(f(x)) are given in
I) A127671 and A263634 for the e.g.f: log[ e^{a.*x} ] = e^{L.(a_1,a_2,...)x} = Sum_{n > 0} L_n(a_1,...,a_n) * x^n/n!, the logarithmic polynomials, cumulant expansion polynomials
II) A263916 for the o.g.f.: log[ 1/(1-b.x) ] = log[ 1 - F.(b_1,b_2,...)x ] = -Sum_{n > 0} F_n(b_1,...,b_n) * x^n/n, the Faber polynomials.
Expansions of exp(f(x)-1) are given in
III) A036040 for an e.g.f: exp[ e^{a.x} - 1 ] = e^{BELL.(a_1,...)x}, the Bell/Touchard/exponential partition polynomials, a.k.a. the Stirling partition polynomials of the second kind
IV) A130561 for an o.g.f.: exp[ b.x/(1-b.x) ] = e^{LAH.(b.,...)x}, the Lah partition polynomials
V) A036039 for an l.g.f.: exp[ -log(1-c.x) ] = e^{CIP.(c_1,...)x}, the cycle index polynomials of the symmetric groups S_n, a.k.a. the Stirling partition polynomials of the first kind.
Since exp and log are a compositional inverse pair, one can extract the indeterminates of the log set of partition polynomials from the exp set and vice versa. For a discussion of the relations among these polynomials and the combinatorics of connected and disconnected graphs/maps, see Novak and LaCroix on classical moments and cumulants and the two books on statistical mechanics referenced in A036040. (End)
Ignoring signs, these polynomials appear in Schröder in the set of equations (II) on p. 343 and in Stewart's translation on p. 31. - Tom Copeland, Aug 25 2021

Examples

			The first few polynomials are:
(1) x[1].
(2) -x[1]^2 + x[2].
(3) 2*x[1]^3 - 3*x[1]*x[2] + x[3].
(4) -6*x[1]^4 + 12*x[1]^2*x[2] - 4*x[1]*x[3] - 3*x[2]^2 + x[4].
(5) 24*x[1]^5 - 60*x[1]^3*x[2] + 20*x[1]^2*x[3] + 30*x[1]*x[2]^2 - 5*x[1]*x[4] - 10*x[2]*x[3] + x[5].
(6) -120*x[1]^6 + 360*x[1]^4*x[2] - 120*x[1]^3*x[3] - 270*x[1]^2*x[2]^2 + 30*x[1]^2*x[4] + 120*x[1]*x[2]*x[3] + 30*x[2]^3 - 6*x[1]*x[5] - 15*x[2]*x[4] - 10*x[3]^2 + x[6].
...
[1]    1
[2]   -1,    1
[3]    2,   -3,     1
[4]   -6,   12,    -4,    -3,   1
[5]   24,  -60,    20,    30,  -5,  -10,   1
[6] -120,  360,  -120,  -270,  30,  120,  30, -6, -15, -10, 1
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 140, 156, 308.

Crossrefs

Programs

  • Maple
    triangle := proc(numrows) local E, s, Q;
    E := add(x[i]*t^i/i!, i=1..numrows);
    s := series(log(1 + E), t, numrows+1);
    Q := k -> sort(expand(k!*coeff(s, t, k)));
    seq(print(coeffs(Q(k))), k=1..numrows) end:
    triangle(6); # updated by Peter Luschny, May 27 2020

Formula

G.f.: Log(1 + Sum_{i >= 1} x_i*t^i/i!) = Sum_{n >= 1} L_n(x_1, x_2, ...)*t^n/n!. [Comtet, p. 140, Eq. [5a]. - corrected by Tom Copeland, Sep 08 2016]
Conjecture: row polynomials are R(n,1) for n > 0 where R(n,k) = R(n-1,k+1) - Sum_{j=1..n-1} binomial(n-2,j-1)*R(j,k)*R(n-j,1) for n > 1, k > 0 with R(1,k) = x_k for k > 0. - Mikhail Kurkov, Mar 30 2025

A208744 Triangle relating to ordered Bell numbers, A000670.

Original entry on oeis.org

1, 1, 2, 1, 3, 9, 1, 4, 18, 52, 1, 5, 30, 130, 375, 1, 6, 45, 260, 1125, 3246, 1, 7, 63, 455, 2625, 11361, 32781, 1, 8, 84, 728, 5250, 30296, 131124, 378344, 1, 9, 108, 1092, 9450, 68166, 393372, 1702548, 4912515, 1, 10, 135, 1560, 15750, 136332, 983430, 5675160, 24562575, 70872610
Offset: 1

Views

Author

Gary W. Adamson, Mar 05 2012

Keywords

Comments

Row sums = A000670 starting (1, 3, 13, 75,...).
Right border = A052882 starting (1, 2, 9, 52, 375,...).
A000670 is the eigensequence of triangle A074909, deleting the first "1".
Triangle A074909 is the "beheaded" Pascal's triangle: (1; 1,2; 1,3,3;...).

Examples

			Row 4 (nonzero terms) = (1, 4, 18, 52) = termwise product of (1, 4, 6, 4) and (1, 1, 3, 13).
First few rows of the triangle:
1;
1, 2;
1, 3, 9;
1, 4, 18, 52;
1, 5, 30, 130, 375;
1, 6, 45, 260, 1125, 3246;
1, 7, 63, 455, 2625, 11361, 32781;
1, 8, 84, 728, 5250, 30296, 131124, 378344;
...
		

Crossrefs

Programs

  • Mathematica
    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;
    a[n_, k_] := Binomial[n, k] Fubini[k, 1];
    Table[a[n, k], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, Mar 30 2016 *)

Formula

As infinite lower triangular matrices, A074909 * A000670 as the main diagonal and the rest zeros.
E.g.f. (exp(x) - 1)/(2 - exp(x*t)) = x + (1 + 2*t)*x^2/2! + (1 + 3*t + 9*t^2)*x^3/3! + .... Cf. A154921. - Peter Bala, Aug 31 2016

Extensions

a(36) corrected by Jean-François Alcover, Mar 30 2016

A162312 Triangular array, inverse of 2*P - I, where P is Pascal's triangle and I is the identity matrix.

Original entry on oeis.org

1, -2, 1, 6, -4, 1, -26, 18, -6, 1, 150, -104, 36, -8, 1, -1082, 750, -260, 60, -10, 1, 9366, -6492, 2250, -520, 90, -12, 1, -94586, 65562, -22722, 5250, -910, 126, -14, 1, 1091670, -756688, 262248, -60592, 10500, -1456, 168, -16, 1, -14174522, 9825030
Offset: 0

Views

Author

Peter Bala, Jul 01 2009

Keywords

Comments

We make a few remarks about the general array M(a) := (I - a*P)^-1, where a <> 1, and its connection with weighted sums of powers of positive integers. The present case corresponds to -M(2).
The array M(a) begins
/
| 1/(1-a)
| a/(1-a)^2............... 1/(1-a)
| (a+a^2)/(1-a)^3......... 2*a/(1-a)^2........ 1/(1-a)
| (a+4*a^2+a^3)/(1-a)^4... 3*(a+a^2)/(1-a)^3.. 3*a/(1-a)^2... 1/(1-a)
| ...
\
In the first column the numerator polynomials are the Eulerian polynomials A_n(a). See A008292.
The e.g.f. for this array is
(1)... exp(x*t)/(1-a*exp(t)) = 1/(1-a) + [a/(1-a)^2 + x/(1-a)]*t
+ [(a+a^2)/(1-a)^3 + 2*a*x/(1-a)^2 + x^2/(1-a)]*t^2/2! + ....
The row generating polynomials P_m(x) of the array M(a), which, of course, depend on a, have properties similar to those of the Bernoulli polynomials. They form an Appell sequence and may be expressed in terms of the Eulerian polynomials as
(2)... P_m(x) = sum {k=0..m} binomial(m,k) * A_k(a) / (1-a)^(k+1) * x^(m-k).
As a Newton series we have
(3)... P_m(x) = 1/(1-a)*sum {j = 0..m} sum {k = j..m}(a/(1-a))^j * k! * Stirling2(m,k) * binomial(x,k-j).
The proof of this result in the particular case a = -1 given in [Roman, p. 100] can be easily generalized to a proof of (3).
A result equivalent to (3) is
(4)... P_m(x) = 1/(1-a)*sum {j = 0..m} sum {k = 0..j} (a/(1-a))^j * (-1)^(j-k) * comb(j,k) * (x + k)^m,
which in turn leads to the infinite series expansion
(5)... P_m(x) = sum {k = 0..inf} a^k * (x + k)^m,
provided |a| < 1. See [Nelsen].
The polynomials P_m(x) satisfy the difference equation
(6)... P_m(x) - a*P_m(x + 1) = x^m (recall a <> 1),
which leads easily to the evaluation of the weighted sums of powers of integers
(7)... sum {k = 0..n-1} a^k * k^m = P_m(0) - a^n * P_m(n).
for m = 0,1,2,... and a <> 1.
More generally we have
(8)... sum {k = 0..n-1} a^k * (x + k)^m = P_m(x) - a^n * P_m(x + n).
for m = 0,1,2,... and a <> 1.
In the remaining case a = 1 the sums are evaluated in terms of the Bernoulli polynomials.
The most well-studied case is when a = -1. The row polynomials of the array M(-1) are then one half of the Euler polynomials E_m(x), which may be used to evaluate the alternating sums of powers of integers
(9)... 2*sum {k = 1..n-1} (-1)^k * k^m = E_m(0) - (-1)^n * E_m(n).

Examples

			Triangle begins
====================================================
n\k|.....0......1......2......3......4......5......6
====================================================
0..|.....1
1..|....-2......1
2..|.....6.....-4......1
3..|...-26.....18.....-6......1
4..|...150...-104.....36.....-8......1
5..|.-1082....750...-260.....60....-10......1
6..|..9366..-6492...2250...-520.....90....-12......1
...
		

References

  • S. Roman, The Umbral Calculus, Dover Publications.

Crossrefs

Programs

  • Maple
    #A162312
    with(combinat):
    T := (n,k) -> (-1)^(n+k)*binomial(n,k)
    *add( j!*stirling2(n-k+1,j+1),j = 0..n):
    for n from 0 to 9 do
    seq(T(n,k), k = 0..n);
    end do;
  • Mathematica
    Table[(-1)^(n+k) Binomial[n, k] PolyLog[k-n, 1/2], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2019 *)
  • PARI
    matrix(10, 10, n, k, 2*binomial(n-1,k-1) - (n==k))^(-1) \\ Michel Marcus, Jul 12 2018

Formula

TABLE ENTRIES
(1)... T(n,k) = (-1)^(n+k) * binomial(n,k) * A000629(n-k).
(2)... T(n,k) = (-1)^(n+k) * binomial(n,k) * sum {j = 0..n} j! * Stirling2(n-k+1,j+1).
GENERATING FUNCTION
(3)... exp(x*t)/(2*exp(t)-1) = 1 + (-2+x)*t + (6-4*x+x^2)*t^2/2!
+ ....
PROPERTIES OF ROW POLYNOMIALS
The row generating polynomials R_n(x) form an Appell sequence. The first few values are R_0(x) = 1, R_1(x) = x-2, R_2(x) = x^2-4*x+6 and R_3(x) = x^3-6*x^2+18*x-26.
They may be recursively computed by means of
(4)... R_n(x) = x^n - 2*sum {k = 0..n-1} binomial(n,k) * R_k(x).
Explicit formulas are
(5)... R_n(x) = sum {j = 0..n} sum {k = j..n} (-2)^j * k! * Stirling2(n,k) * binomial(x,k-j),
(6)... R_n(x) = (-1)^n * sum {j = 0..n} sum {k = j..n} k! * Stirling2(n,k) * binomial(-x+1,k-j),
and
(7)... R_n(x) = sum {j = 0..n} sum {k = 0..j} 2^j * (-1)^k * comb(j,k) * (x + k)^n.
Other expansions include
(8)... R_n(x) = sum {k = 0..n} binomial(n,k) * (-1)^k * A000670(k) * (x-1)^(n-k),
(9)... R_n(x) = sum {k = 0..n} binomial(n,k) * (-1/2)^k * A080253(k) * (x-1/2)^(n-k)
and
(10)... R_n(x) = sum {k = 0..n} binomial(n,k) * (-1)^k * A007047(k) * (x+1)^(n-k).
SUMS OF POWERS OF INTEGERS
The row polynomials satisfy the difference equation
(11)... 2*R_n(x+1) - R_n(x) = x^n,
and so may be used to evaluate the weighted sum of powers of integers
(12)... sum {k = 0..n-1} 2^k * k^m = 2^n*R_m(n) - R_m(0).
For example, m = 3 gives
(13)... sum {k = 0..n-1} 2^k * k^3 = 2^n*(n^3-6*n^2+18*n-26) + 26.
More generally we have
(14)... sum {k = 0..n-1} 2^k * (x + k)^m = 2^n * R_m(x + n) - R_m(x).
RELATIONS WITH OTHER SEQUENCES
(15)... Row sums [1,-1,3,-13,75,...] = (-1)^n*A000670(n).
(16)... Alt. row sums [1,-3,11,-51,299,...] = (-1)^n * A007047(n).
(17)... Column 0: (-1)^n * A000629(n).
(18)... (-2)^n * R_n(1/2) = A080253(n).
(19)... R_n(1-x) = (-1)^n * P_n(x),
where P_n(x) are the row generating polynomials of A154921.
This provides the connection between (12) and the result
(20)... sum {k = 0..n-1} (1/2)^k * k^m = 2*P_m(0) - (1/2)^(n-1) * P_m(n).

Extensions

Typo corrected by Peter Bala, Nov 05 2010

A154926 Signed version of Pascal's triangle. Diagonal positive, rest negative.

Original entry on oeis.org

1, -1, 1, -1, -2, 1, -1, -3, -3, 1, -1, -4, -6, -4, 1, -1, -5, -10, -10, -5, 1, -1, -6, -15, -20, -15, -6, 1, -1, -7, -21, -35, -35, -21, -7, 1, -1, -8, -28, -56, -70, -56, -28, -8, 1
Offset: 1

Views

Author

Mats Granvik, Jan 17 2009

Keywords

Comments

Matrix inverse of this lower triangular array is A154921. Signs in columns as in sequence A153881.
Exponential Riordan array [2-exp(x),x]. - Paul Barry, Apr 06 2011

Examples

			From _Mats Granvik_, Jul 24 2009: (Start)
Table begins:
.1
-1....1
-1...-2....1
-1...-3...-3....1
-1...-4...-6...-4....1
-1...-5..-10..-10...-5....1
-1...-6..-15..-20..-15...-6....1
-1...-7..-21..-35..-35..-21...-7....1
-1...-8..-28..-56..-70..-56..-28...-8....1
-1...-9..-36..-84.-126.-126..-84..-36...-9....1
-1..-10..-45.-120.-210.-252.-210.-120..-45..-10....1
-1..-11..-55.-165.-330.-462.-462.-330.-165..-55..-11....1
(End)
		

Crossrefs

A162313 Triangular array P*(2*I - P^2)^-1, where P is Pascal's triangle A007318 and I is the identity matrix.

Original entry on oeis.org

1, 3, 1, 17, 6, 1, 147, 51, 9, 1, 1697, 588, 102, 12, 1, 24483, 8485, 1470, 170, 15, 1, 423857, 146898, 25455, 2940, 255, 18, 1, 8560947, 2966999, 514143, 59395, 5145, 357, 21, 1, 197613377, 68487576, 11867996, 1371048, 118790, 8232, 476, 24, 1
Offset: 0

Views

Author

Peter Bala, Jul 01 2009

Keywords

Comments

Unsigned inverse of A162315.
The row generating polynomials of this triangle may be used to evaluate the weighted sums of powers of odd numbers
(1)... 1^m + 2*3^m + 4*5^m + ... + 2^n*(2*n+1)^m
and also the sums
(2)... 1^m + (1/2)*3^m + (1/4)*5^m + ... + (1/2)^n*(2*n+1)^m.
See the Formula section below.
We make a few remarks about the general array M(a) := a*P*(I-a*P^2)^-1, where a <> 1, and its connection with weighted sums of powers of odd numbers. The present case corresponds to a = 1/2. Compare with the remarks in A162312.
The array M(a) begins
/
| a/(1-a)
| (a^2+a)/(1-a)^2 ................. a/(1-a)
| (a^3+6*a^2+a)/(1-a)^3 ........... 2*(a^2+a)/(1-a)^2 ... a/(1-a)
(a^4+23*a^3+23*a^2+a)/(1-a)^4 ...
| .
\ .
In the first column we recognize the numerator polynomials as the Eulerian polynomials of type B. See A060187.
The e.g.f. for this array is
(3)... a*exp((x+1)*t)/(1-a*exp(2*t)) = a/(1-a) +[(a^2+a)/(1-a)^2 + a/(1-a)*x]*t + [(a^3+6*a^2+a)/(1-a)^3 + 2*(a^2+a)*x/(1-a)^2 + a/(1-a)*x^2]*t^2/2! + ....
The row polynomials P_m(x), which depend on a, satisfy the difference equation
(4)... P_m(x) - a*P_m(x+2) = a*(x+1)^m.
for m >= 1.
The first few values are
P_0(x) = a/(1-a), P_1(x) = a*x/(1-a) + (a^2+a)/(1-a)^2 and
P_2(x) = a*x^2/(1-a) + 2*(a^2+a)*x/(1-a)^2 + (a^3+6*a^2+a)/(1-a)^3.
Using (4) leads to the evaluations of the weighted sums of powers of even and odd positive integers:
(5)... Sum_{k = 1..n} a^k*(2*k)^m = P_m(1) - a^n*P_m(2*n+1)
and
(6)... Sum_{k = 1..n} a^k*(2*k-1)^m = P_m(0) - a^n*P_m(2*n),
with m = 0,1,2,... and a <> 1.
In the remaining case a = 1 the sums are evaluated in terms of the Bernoulli polynomials.

Examples

			Triangle begins
  n\k|.......0.......1......2......3......4......5......6
  =======================================================
  0..|.......1
  1..|.......3.......1
  2..|......17.......6......1
  3..|.....147......51......9......1
  4..|....1697.....588....102.....12......1
  5..|...24483....8485...1470....170.....15......1
  6..|..423857..146898..25455...2940....255.....18......1
  ...
		

Crossrefs

A000629, A007318, A060187, A080253 (column 0), A154921, A162312, A162314 (row sums), A162315 (unsigned inverse).

Programs

  • Mathematica
    m = 8;
    P = Table[Binomial[n, k], {n, 0, m}, {k, 0, m}];
    T = P.Inverse[2 IdentityMatrix[m+1] - P.P];
    Table[T[[n+1, k+1]], {n, 0, m}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 02 2019 *)

Formula

TABLE ENTRIES
(1)... T(n,k) = binomial(n,k)*A080253(n-k).
GENERATING FUNCTION
(2)... exp((x+1)*t)/(2-exp(2*t)) = 1 + (3+x)*t + (17+6*x+x^2)*t^2/2!
+ ....
The e.g.f. can also be written as
(3)... exp(x*t)*G(t), where G(t) = exp(t)/(2-exp(2*t)) is the e.g.f. for A080253.
ROW GENERATING POLYNOMIALS
The row generating polynomials R_n(x) form an Appell sequence. The first few values are R_0(x) = 1, R_1(x) = 3 + x, R_2(x) = 17 + 6*x + x^2 and R_3(x) = 147 + 51*x + 9*x^2 + x^3.
The row polynomials may be recursively computed by means of
(4)... R_n(x) = (x+1)^n + Sum_{k=0..n-1} 2^(n-k)*binomial(n,k)*R_k(x).
An explicit formula is
(5)... R_n(x) = Sum_{j = 0..n} Sum_{k = 0..j} (-1)^(j-k)*binomial(j,k)*(x+2*k+1)^n.
There is also a representation as an infinite series
(6)... R_n(x) = (1/2)*Sum_{k >= 0} (1/2)^k*(x+2*k+1)^n.
SUMS OF POWERS OF INTEGERS
The row polynomials satisfy the difference equation
(7)... 2*R_n(x) - R_n(x+2) = (x+1)^n,
and hence may be used to evaluate the weighted sums of powers of odd integers
(8)... Sum_{k=0..n-1} (1/2)^k*(2*k+1)^m = 2*R_m(0)-1/2^(n-1)*R_m(2*n)
as well as
(9)... Sum_{k=0..n-1} 2^k*(2*k+1)^m = (-1)^m*(2^n*R_m(-2*n)-R_m(0)).
For example, m = 2 gives
(10)... Sum_{k=0..n-1} (1/2)^k*(2*k+1)^2 = 34-2^(1-n)*(4*n^2+12*n+17)
and
(11)... Sum_{k = 0..n-1} 2^k*(2*k+1)^2 = 2^n*(4*n^2 - 12*n + 17)-17.
RELATIONS WITH OTHER SEQUENCES
(12)... Row sums = [1,4,24,208,2400,...] = 2^n*A000629(n) = A162314(n).
(13)... Column 0 = [1,3,17,147,1697,...] = A080253.
The identity
(14)... R_n(2*x-1) = 2^n*P_n(x),
where P_n(x) are the row generating polynomials of A154921, provides a surprising connection between (6) and the result
(15)... Sum_{k = 0..n-1} (1/2)^k*k^m = 2*P_m(0) - (1/2)^(n-1)*P_m(n).

A099880 Number of preferential arrangements (or simple hierarchies) of 2*n labeled elements with two kinds of elements (where each kind has n elements).

Original entry on oeis.org

1, 2, 18, 260, 5250, 136332, 4327092, 162309576, 7024896450, 344582629820, 18890850749628, 1144656941236536, 75963981061424820, 5479642938171428600, 426894499408073653800, 35720957482170932284560, 3195135789350678836128450, 304234032845362459798904220
Offset: 0

Views

Author

Thomas Wieder, Nov 02 2004

Keywords

Comments

The unlabeled case seems to be given by A003480, which can be generated by the following combstruct command: SeqUnionU := [S, {S=Sequence(Set(U,card>=1), card>=1), U=Union(a,b), a=Atom, b=Atom},unlabeled]; [seq(count(SeqUnionU, size=n), n=0..20)]; .

Examples

			Let a[1], a[2],...,a[n] and b[1],b[2],...,b[n] denote two kinds "a" and "b" of labeled elements where each kind as n elements in total.
Let ":" denote a level, e.g., if the elements a[1] and a[2] are on level L=1 and the element b[1] is on level L=2 then a[1]a[2]:b[1] is a preferrential arrangement (a simple hierarchy) with two levels.
Then for n=2 we have a(2) = 18 arrangements: a[1]a[2]; a[1]:a[2]; a[2]:a[1]; a[1]b[1]; a[1]:b[1]; b[1]:a[1]; a[1]b[2]; a[1]:b[2]; b[2]:a[1]; a[2]b[1]; a[2]:b[1]; b[1]:a[2]; a[2]b[2]; a[2]:b[2]; b[2]:a[2]; b[1]b[2]; b[1]:b[2]; b[2]:b[1].
		

Crossrefs

Programs

  • Maple
    a:=n-> add(binomial(2*n, n)*(Stirling2(n, k))*k!, k=0..n): seq(a(n), n=0..16); # Zerinvary Lajos, Oct 19 2006
    # second Maple program:
    b:= proc(n) b(n):= `if`(n=0, 1, add(b(n-j)/j!, j=1..n)) end:
    a:= n-> b(n)*(2*n)!/n!:
    seq(a(n), n=0..20);  # Alois P. Heinz, Feb 03 2019
  • Mathematica
    f[n_] := Sum[l! StirlingS2[n, l] Binomial[2n, n], {l, n}]; Table[ f[n], {n, 0, 16}] (* Robert G. Wilson v, Nov 04 2004 *)

Formula

a(n) = binomial(2*n, n) * Sum_{k=0..n} k! * Stirling2(n, k).
a(n) = binomial(2*n, n) * A000670(n).
a(n) = A154921(2n,n). - Mats Granvik, Feb 07 2009

Extensions

More terms from Robert G. Wilson v, Nov 04 2004
a(0) corrected and edited by Alois P. Heinz, Feb 03 2019

A363732 Triangle read by rows. The triangle algorithm applied to (-1)^n/n!.

Original entry on oeis.org

1, -2, 1, 5, -4, 1, -15, 15, -6, 1, 52, -60, 30, -8, 1, -203, 260, -150, 50, -10, 1, 877, -1218, 780, -300, 75, -12, 1, -4140, 6139, -4263, 1820, -525, 105, -14, 1, 21147, -33120, 24556, -11368, 3640, -840, 140, -16, 1, -115975, 190323, -149040, 73668, -25578, 6552, -1260, 180, -18, 1
Offset: 0

Views

Author

Peter Luschny, Jun 18 2023

Keywords

Comments

The triangle algorithm, as understood here, is a transformation that maps a sequence of integers (a(n) : n >= 0) to a polynomial sequence. A polynomial sequence is a sequence of polynomials (P(n,x) : n >= 0) with degree(P(n, x)) = n for all n >= 0.
The polynomials P(n, x) are recursively defined by P(n, x) = p(n, 0, x), where the initial sequence is p(0, m, x) = a(m), and for n > 0 is given by
p(n, m, x) = (m + 1)*p(n - 1, m + 1, x) - (m + 1 - x)*p(n - 1, m, x).
Here we identify the polynomial sequence with the infinite lower triangular array of its coefficients, T(n, k) = [x^k] P(n, x). We call the mapping (a(n) : n >= 0) -> (T(n, k) : 0 <= k <= n) the 'triangle algorithm', following the lead of Kawasaki and Ohno.
Evaluating P(n, x) at different values of x gives rise to a multitude of other sequences; in particular, the transformation a(n) -> b(n) = P(n, 1) will be called the Akiyama-Tanigawa transform of a.
The triangle algorithm was studied by Akiyama and Tanigawa, Chen, Imatomi, Arakawa and Kaneko, Kawasaki and Ohno, and others, at first in connection with the Bernoulli and Poly-Bernoulli numbers.
.
The paradigmatic examples are:
a(n) = 1 -> x^n, the standard base of polynomials, A023531.
a(n) = n + 1 -> binomial(n, k), Pascal triangle, A007318.
a(n) = n + 1 -> P(n, 1) powers of 2, A000079.
a(n) = n + 1 -> P(n, 0) the all 1's sequence A000012.
a(n) = 2^n -> [x^k] P(n, x), A154921.
a(n) = 2^n -> P(n, 0) Fubini numbers, A000670.
a(n) = 2^n -> P(n, 1) ordered set partitions of subsets of [n], A000629.
a(n) = 2^n -> P(n,-1) osp. of [n] with even number of blocks, A052841.
a(n) = 1 / (n + 1) -> [x^k] B(n, x), Bernoulli polynomials, A196838/A196839.
a(n) = 1 / (n + 1) -> B(n, 1), the Bernoulli numbers, A164555/A027642.
a(n) = Chen(n) -> skp(n, x), Swiss-Knife polynomials, A153641.
a(n) = Chen(n) -> P(n, 0), 2^n*Euler(n, 1/2) = Euler(n), A122045.
a(n) = Chen(n) -> P(n, 1), 2^n*Euler(n, 1), A155585.
a(n) = (-1)^n/n! -> [x^k] P(n, x) this "Bell" triangle.
a(n) = (-1)^n/n! -> (-1)^n*P(n, 1) = Bell(n), A000110.
a(n) = (-1)^n/n! -> (-1)^n*P(n,-1) = 2-Bell(n), A005493.
a(n) = 1/n! -> (-1)^n*P(n, 1) = complementary Bell(n), A000587.
a(n) = 1/n! -> (-1)^n*P(n,-1) = complementary 2-Bell(n), A074051.
(For Chen's sequence see A363524.)
.
The present sequence deals with the case of the Bell numbers. In contrast to Aitken's array A011971 and its variants A123346 and A011972, the Bell numbers do not appear as a column of the triangle but as row sums (times (-1)^n), i.e., as values of the associated polynomials at x = 1. Comparing this with a similar situation with the Bernoulli numbers/polynomials, our triangle could be viewed as a more organic generalization of the Bell numbers. Indeed, the names 'Bell triangle' and 'Bell polynomials' would be justified here; but these are already assigned to other concepts.

Examples

			The triangle T(n, k) starts:
  [0]     1;
  [1]    -2,      1;
  [2]     5,     -4,     1;
  [3]   -15,     15,    -6,      1;
  [4]    52,    -60,    30,     -8,    1;
  [5]  -203,    260,  -150,     50,  -10,    1;
  [6]   877,  -1218,   780,   -300,   75,  -12,   1;
  [7] -4140,   6139, -4263,   1820, -525,  105, -14,   1;
  [8] 21147, -33120, 24556, -11368, 3640, -840, 140, -16, 1;
		

Crossrefs

Cf. A293037 (row sums), A000110 (row sums, unsigned), A005493 (alternating row sums, signed).

Programs

  • Maple
    TA := proc(a, n, m, x) option remember; if n = 0 then a(m) else
    normal((m + 1)*TA(a, n - 1, m + 1, x) - (m + 1 - x)*TA(a, n - 1, m, x)) fi end:
    seq(seq(coeff(TA(n -> (-1)^n/n!, n, 0, x), x, k), k = 0..n), n = 0..10);
  • Mathematica
    (* rows[0..n], n[0..oo] *)
    (* row[n]= *)
    n=9;r={};For[a=n+1,a>0,a--,AppendTo[r,(-1)^(a+1)*Sum[StirlingS2[a,k],{k,0,a}]*Product[(2*(a+j))/(2*j+2),{j,0,n-a}]]];r
    (* columns[1..n], n[0..oo] *)
    (* column[n]= *)
    n=0;c={};For[a=1,a<15,a++,AppendTo[c,(-1)^(a+1)*Sum[StirlingS2[a,k],{k,0,a}]*Product[(2*(a+j-1))/(2*j),{j,1,n}]]];c
    (* sequence *)
    s={};For[n=0,n<15,n++,For[a=n+1,a>0,a--,AppendTo[s,(-1)^(a+1)*Sum[StirlingS2[a,k],{k,0,a}]*Product[(2*(a+j))/(2*j+2),{j,0,n-a}]]]];s
    (* Detlef Meya, Jun 22 2023 *)
  • SageMath
    def a(n): return (-1)^n / factorial(n)
    @cached_function
    def p(n, m):
        R = PolynomialRing(QQ, "x")
        if n == 0: return R(a(m))
        return R((m + 1)*p(n - 1, m + 1) - (m + 1 - x)*p(n - 1, m))
    for n in range(10): print(p(n, 0).list())
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