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)

A002162 Decimal expansion of the natural logarithm of 2.

Original entry on oeis.org

6, 9, 3, 1, 4, 7, 1, 8, 0, 5, 5, 9, 9, 4, 5, 3, 0, 9, 4, 1, 7, 2, 3, 2, 1, 2, 1, 4, 5, 8, 1, 7, 6, 5, 6, 8, 0, 7, 5, 5, 0, 0, 1, 3, 4, 3, 6, 0, 2, 5, 5, 2, 5, 4, 1, 2, 0, 6, 8, 0, 0, 0, 9, 4, 9, 3, 3, 9, 3, 6, 2, 1, 9, 6, 9, 6, 9, 4, 7, 1, 5, 6, 0, 5, 8, 6, 3, 3, 2, 6, 9, 9, 6, 4, 1, 8, 6, 8, 7
Offset: 0

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Newton calculated the first 16 terms of this sequence.
Area bounded by y = tan x, y = cot x, y = 0. - Clark Kimberling, Jun 26 2020
Choose four values independently and uniformly at random from the unit interval [0,1]. Sort them, and label them a,b,c,d from least to greatest (so that a b^2+c^2. - Akiva Weinberger, Dec 02 2024
Define the trihyperboloid to be the intersection of the three solid hyperboloids x^2+y^2-z^2<1, x^2-y^2+z^2<1, and -x^2+y^2+z^2<1. This fits perfectly within the cube [-1,1]^3. Then this is the ratio of the volume of the trihyperboloid to its bounding cube. - Akiva Weinberger, Dec 02 2024

Examples

			0.693147180559945309417232121458176568075500134360255254120680009493393...
		

References

  • G. Boros and V. H. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004.
  • Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 227.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 250.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3, 2.21, 6.2, and 7.2.
  • 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).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 25 and appendix A, equations 25:14:3 and A:7:3 at pages 232, 670.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 29.

Crossrefs

Cf. A016730 (continued fraction), A002939, A008288, A142979, A142992.

Programs

  • Mathematica
    RealDigits[N[Log[2],200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)
    RealDigits[Log[2],10,120][[1]] (* Harvey P. Dale, Jan 25 2024 *)
  • PARI
    { default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } \\ Harry J. Smith, Apr 21 2009

Formula

log(2) = Sum_{k>=1} 1/(k*2^k) = Sum_{j>=1} (-1)^(j+1)/j.
log(2) = Integral_{t=0..1} dt/(1+t).
log(2) = (2/3) * (1 + Sum_{k>=1} 2/((4*k)^3-4*k)) (Ramanujan).
log(2) = 4*Sum_{k>=0} (3-2*sqrt(2))^(2*k+1)/(2*k+1) (Y. Luke). - R. J. Mathar, Jul 13 2006
log(2) = 1 - (1/2)*Sum_{k>=1} 1/(k*(2*k+1)). - Jaume Oliver Lafont, Jan 06 2009, Jan 08 2009
log(2) = 4*Sum_{k>=0} 1/((4*k+1)*(4*k+2)*(4*k+3)). - Jaume Oliver Lafont, Jan 08 2009
log(2) = 7/12 + 24*Sum_{k>=1} 1/(A052787(k+4)*A000079(k)). - R. J. Mathar, Jan 23 2009
From Alexander R. Povolotsky, Jul 04 2009: (Start)
log(2) = (1/4)*(3 - Sum_{n>=1} 1/(n*(n+1)*(2*n+1))).
log(2) = (230166911/9240 - Sum_{k>=1} (1/2)^k*(11/k + 10/(k+1) + 9/(k+2) + 8/(k+3) + 7/(k+4) + 6/(k+5) - 6/(k+7) - 7/(k+8) - 8/(k+9) - 9/(k+10) - 10/(k+11)))/35917. (End)
log(2) = A052882/A000670. - Mats Granvik, Aug 10 2009
From log(1-x-x^2) at x=1/2, log(2) = (1/2)*Sum_{k>=1} L(k)/(k*2^k), where L(n) is the n-th Lucas number (A000032). - Jaume Oliver Lafont, Oct 24 2009
log(2) = Sum_{k>=1} 1/(cos(k*Pi/3)*k*2^k) (cf. A176900). - Jaume Oliver Lafont, Apr 29 2010
log(2) = (Sum_{n>=1} 1/(n^2*(n+1)^2*(2*n+1)) + 11)/16. - Alexander R. Povolotsky, Jan 13 2011
log(2) = ((Sum_{n>=1} (2*n+1)/(Sum_{k=1..n} k^2)^2)+396)/576. - Alexander R. Povolotsky, Jan 14 2011
From Alexander R. Povolotsky, Dec 16 2008: (Start)
log(2) = 105*(319/44100 - Sum_{n>=1} 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7))).
log(2) = 319/420 - (3/2)*Sum_{n>=1} 1/(6*n^2+39*n+63). (End)
log(2) = Sum_{k>=1} A191907(2,k)/k. - Mats Granvik, Jun 19 2011
log(2) = Integral_{x=0..oo} 1/(1 + e^x) dx. - Jean-François Alcover, Mar 21 2013
log(2) = lim_{s->1} zeta(s)*(1-1/2^(s-1)). - Mats Granvik, Jun 18 2013
From Peter Bala, Dec 10 2013: (Start)
log(2) = 2*Sum_{n>=1} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
log(2) = (1/3)*Sum_{n >= 0} (5*n+4)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(1/2)^n = (1/12)*Sum_{n >= 0} (28*n+17)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(-1/4)^n.
log(2) = (3/16)*Sum_{n >= 0} (14*n+11)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(1/4)^n = (1/12)*Sum_{n >= 0} (34*n+25)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(-1/18)^n. For more series of this type see the Bala link.
See A142979 for series acceleration formulas for log(2) obtained from the Mercator series log(2) = Sum_{n >= 1} (-1)^(n+1)/n. See A142992 for series for log(2) related to the root lattice C_n. (End)
log(2) = lim_{n->oo} Sum_{k=2^n..2^(n+1)-1} 1/k. - Richard R. Forberg, Aug 16 2014
From Peter Bala, Feb 03 2015: (Start)
log(2) = (2/3)*Sum_{k >= 0} 1/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*9^k). Both satisfy the same second-order recurrence equation u(n) = (40*n + 16)*u(n-1) - 36*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(2) = (2/3)*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))). Cf. A002391, A073000 and A105531 for similar expansions. (End)
log(2) = Sum_{n>=1} (Zeta(2*n)-1)/n. - Vaclav Kotesovec, Dec 11 2015
From Peter Bala, Oct 30 2016: (Start)
Asymptotic expansions:
for N even, log(2) - Sum_{k = 1..N/2} (-1)^(k-1)/k ~ (-1)^(N/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1. See Borwein et al., Theorem 1 (b);
for N odd, log(2) - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/k ~ (-1)^((N-1)/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), by Borwein et al., Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then set x = (N - 1)/2, where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. (End)
log(2) = lim_{n->oo} Sum_{k=1..n} sin(1/(n+k)). See Mathematical Reflections link. - Michel Marcus, Jan 07 2017
log(2) = Sum_{n>=1} A006519(n) / ((1 + 2^A006519(n)) * A000265(n) * (1 + A000265(n))). - Nicolas Nagel, Mar 19 2018
From Amiram Eldar, Jul 02 2020: (Start)
Equals Sum_{k>=2} zeta(k)/2^k.
Equals -Sum_{k>=2} log(1 - 1/k^2).
Equals Sum_{k>=1} 1/A002939(k).
Equals Integral_{x=0..Pi/3} tan(x) dx. (End)
log(2) = Integral_{x=0..Pi/2} (sec(x) - tan(x)) dx. - Clark Kimberling, Jul 08 2020
From Peter Bala, Nov 14 2020: (Start)
log(2) = Integral_{x = 0..1} (x - 1)/log(x) dx (Boros and Moll, p. 97).
log(2) = (1/2)*Integral_{x = 0..1} (x + 2)*(x - 1)^2/log(x)^2 dx.
log(2) = (1/4)*Integral_{x = 0..1} (x^2 + 3*x + 4)*(x - 1)^3/log(x)^3 dx. (End)
log(2) = 2*arcsinh(sqrt(2)/4) = 2*sqrt(2)*Sum_{n >= 0} (-1)^n*C(2*n,n)/ ((8*n+4)*32^n) = 3*Sum_{n >= 0} (-1)^n/((8*n+4)*(2^n)*C(2*n,n)). - Peter Bala, Jan 14 2022
log(2) = Integral_{x=0..oo} ( e^(-x) * (1-e^(-2x)) * (1-e^(-4x)) * (1-e^(-6x)) ) / ( x * (1-e^(-14x)) ) dx (see Crux Mathematicorum link). - Bernard Schott, Jul 11 2022
From Peter Bala, Oct 22 2023: (Start)
log(2) = 23/32 + 2!^3/16 * Sum_{n >= 1} (-1)^n * (n + 1)/(n*(n + 1)*(n + 2))^2 = 707/1024 - 4!^3/(16^2 * 2!^2) * Sum_{n >= 1} (-1)^n * (n + 2)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^2 = 42611/61440 + 6!^3/(16^3 * 3!^2) * Sum_{n >= 1} (-1)^n * (n + 3)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6))^2.
More generally, it appears that for k >= 0, log(2) = c(k) + (2*k)!^3/(16^k * k!^2) * Sum_{n >= 1} (-1)^(n+k+1) * (n + k)/(n*(n + 1)*...*(n + 2*k))^2 , where c(k) is a rational approximation to log(2). The first few values of c(k) are [0, 23/32, 707/1024, 42611/61440, 38154331/55050240, 76317139/110100480, 26863086823/38755368960, ...].
Let P(n,k) = n*(n + 1)*...*(n + k).
Conjecture: for k >= 0 and r even with r - 1 <= k, the series Sum_{n >= 1} (-1)^n * (d/dn)^r (P(n,k)) / (P(n,k)^2 = A(r,k)*log(2) + B(r,k), where A(r,k) and B(r,k) are both rational numbers. (End)
From Peter Bala, Nov 13 2023: (Start)
log(2) = 5/8 + (1/8)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)^2 / ( k*(k + 1) )^4
= 257/384 + (3!^5/2^9)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)^2*(2*k + 5) / ( k*(k + 1)*(k + 2)*(k + 3) )^4
= 267515/393216 + (5!^5/2^19)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)*(2*k + 5)^2*(2*k + 7)*(2*k + 9) / ( k*(k + 1)*(k + 2)*(k + 3)*(k + 4)*(k + 5) )^4
log(2) = 3/4 - 1/128 * Sum_{k >= 0} (-1/16)^k * (10*k + 12)*binomial(2*k+2,k+1)/ ((k + 1)*(2*k + 3)). The terms of the series are O(1/(k^(3/2)*4^n)). (End)
log(2) = eta(1) is a period, where eta(x) is the Dirichlet eta function. - Andrea Pinos, Mar 19 2024
log(2) = K_{n>=0} (n^2 + [n=0])/1, where K is the Gauss notation for an infinite continued fraction. In the expanded form, log(2) = 1/(1 + 1/(1 + 4/(1 + 9/1 + 16/(1 + 25/(1 + ... (see Clawson at p. 227). - Stefano Spezia, Jul 01 2024
log(2) = lim_{n->oo} Sum_{k=1..n} 1/(n + k) = lim_{x->0} (2^x - 1)/x = lim_{x->0} (2^x - 2^(-x))/(2*x) (see Finch). - Stefano Spezia, Oct 19 2024
From Colin Linzer, Nov 08 2024: (Start)
log(2) = Integral_{t=0...oo} (1 - tanh(t)) dt.
log(2) = Integral_{t=0...1} arctanh(t) dt.
log(2) = (1/2) * Integral_{t=-1...1} |arctanh(t)| dt. (End)
log(2) = 1 + Sum_{n >= 1} (-1)^n/(n*(4*n^2 - 1)) = 1/2 + (1/2)*Sum_{n >= 1} 1/(n*(4*n^2 - 1)). - Peter Bala, Jan 07 2025
log(2) = Integral_{x=0..1} Integral_{y=0..1} 1/((1 - x*y)*(1 + x)*(1 + y)) dy dx. - Kritsada Moomuang, May 22 2025

A024167 a(n) = n!*(1 - 1/2 + 1/3 - ... + c/n), where c = (-1)^(n+1).

Original entry on oeis.org

1, 1, 5, 14, 94, 444, 3828, 25584, 270576, 2342880, 29400480, 312888960, 4546558080, 57424792320, 948550176000, 13869128448000, 256697973504000, 4264876094976000, 87435019510272000, 1627055289796608000, 36601063093905408000, 754132445894209536000
Offset: 1

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Stirling transform of (-1)^n*a(n-1) = [0, 1, -1, 5, -14, 94, ...] is A000629(n-2) = [0, 1, 2, 6, 26, ...]. - Michael Somos, Mar 04 2004
Stirling transform of a(n) = [1, 1, 5, 14, 94, ...] is A052882(n) = [1, 2, 9, 52, 375, ...]. - Michael Somos, Mar 04 2004
a(n) is the number of n-permutations that have a cycle with length greater than n/2. - Geoffrey Critzer, May 28 2009
From Jens Voß, May 07 2010: (Start)
a(4n) is divisible by 6*n + 1 for all n >= 1; the quotient of a(4*n) and 6*n+1 is A177188(n).
a(4*n+3) is divisible by 6*n + 5 for all n >= 0; the quotient of a(4*n+3) and 6*n + 5 is A177174(n). (End)

Examples

			G.f. = x + x^2 + 5*x^3 + 14*x^4 + 94*x^5 + 444*x^6 + 3828*x^7 + 25584*x^8 + ...
		

Crossrefs

Programs

  • Maple
    a := n -> n!*(log(2) - (-1)^n*LerchPhi(-1, 1, n+1));
    seq(simplify(a(n)), n=1..20); # Peter Luschny, Dec 27 2018
  • Mathematica
    f[k_] := k (-1)^(k + 1)
    t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 18}]    (* A024167 signed *)
    (* Clark Kimberling, Dec 30 2011 *)
    a[ n_] := If[ n < 0, 0, n! Sum[ -(-1)^k / k, {k, n}]]; (* Michael Somos, Nov 28 2013 *)
    a[ n_] := If[ n < 0, 0, n! (PolyGamma[n + 1] - PolyGamma[(n + Mod[n, 2, 1]) / 2])]; (* Michael Somos, Nov 28 2013 *)
    a[ n_] := If[ n < 1, 0, (-1)^Quotient[n, 2] SymmetricPolynomial[ n - 1, Table[ -(-1)^k k, {k, n}]]]; (* Michael Somos, Nov 28 2013 *)
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( log(1 + x + x * O(x^n)) / (1 - x), n))}; /* Michael Somos, Mar 02 2004 */
    
  • PARI
    x='x+O('x^33); Vec(serlaplace(log(1+x)/(1-x))) \\ Joerg Arndt, Dec 27 2018
    
  • Python
    def A():
        a, b, n = 1, 1, 2
        yield(a)
        while True:
            yield(a)
            b, a = a, a + b * n * n
            n += 1
    a = A(); print([next(a) for  in range(20)]) # _Peter Luschny, May 19 2020

Formula

E.g.f.: log(1 + x)/(1 - x). - Vladeta Jovovic, Aug 25 2002
a(n) = a(n-1) + a(n-2) * (n-1)^2, n > 1. - Michael Somos, Oct 29 2002
b(n) = n! satisfies the above recurrence with b(1) = 1, b(2) = 2. This gives the finite continued fraction expansion a(n)/n! = 1/(1 + 1^2/(1 + 2^2/(1 + 3^2/(1 + ... + (n-1)^2/1)))). Cf. A142979. - Peter Bala, Jul 17 2008
a(n) = A081358(n) - A092691(n). - Gary Detlefs, Jul 09 2010
E.g.f.: (x/(x-1))/G(0) where G(k) = -1 + (x-1)*k + x*(k+1)^2/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 18 2012
a(n) ~ log(2)*n!. - Daniel Suteu, Dec 03 2016
a(n) = (1/2)*n!*((-1)^n*(digamma((n+1)/2) - digamma((n+2)/2)) + log(4)). - Daniel Suteu, Dec 03 2016
a(n) = n!*(log(2) - (-1)^n*LerchPhi(-1, 1, n+1)). - Peter Luschny, Dec 27 2018
a(n) = A054651(n,n-1). - Pontus von Brömssen, Oct 25 2020
a(n) = Sum_{k=0..n} (-1)^k*k!*A094587(n, k+1). - Mélika Tebni, Jun 20 2022
a(n) = n * a(n-1) - (-1)^n * (n-1)! for n > 1. - Werner Schulte, Oct 20 2024

Extensions

More terms from Benoit Cloitre, Jan 27 2002
a(21)-a(22) from Pontus von Brömssen, Oct 25 2020

A154921 Triangle read by rows, T(n, k) = binomial(n, k) * Sum_{j=0..n-k} E(n-k, j)*2^j, where E(n, k) are the Eulerian numbers A173018(n, k), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 13, 9, 3, 1, 75, 52, 18, 4, 1, 541, 375, 130, 30, 5, 1, 4683, 3246, 1125, 260, 45, 6, 1, 47293, 32781, 11361, 2625, 455, 63, 7, 1, 545835, 378344, 131124, 30296, 5250, 728, 84, 8, 1, 7087261, 4912515, 1702548, 393372, 68166, 9450, 1092, 108, 9, 1
Offset: 0

Views

Author

Mats Granvik, Jan 17 2009

Keywords

Comments

Previous name: Matrix inverse of A154926.
A000670 appears in the first column. A052882 appears in the second column. A000027 and A045943 appear as diagonals. An alternative to calculating the matrix inverse of A154926 is to move the term in the lower right corner to a position in the same column and calculate the determinant instead, which yields the same answer.
Matrix inverse of (2*I - P), where P is Pascal's triangle and I the identity matrix. See A162312 for the matrix inverse of (2*P - I) and some general remarks about arrays of the form M(a) := (I - a*P)^-1 and their connection with weighted sums of powers of integers. The present array equals (1/2)*M(1/2). - Peter Bala, Jul 01 2009
From Mats Granvik, Aug 11 2009: (Start)
The values in this triangle can be seen as permanents of the Pascal triangle analogous to the method in the Redheffer matrix. The elements satisfy (T(n,k)/T(n,k-1))*k = (T(n-1,k)/T(n,k))*n which converges to log(2) as n->oo and k->0. More generally to calculate log(x) multiply the negative values in A154926 by 1/(x-1) and calculate the matrix inverse. Then (T(n,k)/T(n,k-1))*k and (T(n-1,k)/T(n,k))*n in the resulting triangle converge to log(x).
This method for calculating log(x) converges faster than the Taylor series when x is greater than 5 or so. See chapter on Taylor series in Spiegel for comparison. (End)
Exponential Riordan array [1/(2-exp(x)),x]. - Paul Barry, Apr 06 2011
T(n,k) is the number of ordered set partitions of {1,2,...,n} such that the first block contains k elements. For k=0 the first block contains arbitrarily many elements. - Geoffrey Critzer, Jul 22 2013
A natural (signed) refinement of these polynomials is given by the Appell sequence e.g.f. e^(xt)/ f(t) = exp[tP.(x)] with the formal Taylor series f(x) = 1 + x[1] x + x[2] x^2/2! + ... and with raising operator R = x - d[log(f(D)]/dD (cf. A263634). - Tom Copeland, Nov 06 2015

Examples

			From _Peter Bala_, Jul 01 2009: (Start)
Triangle T(n, k) begins:
n\k|     0     1     2     3     4     5     6
==============================================
0  |     1
1  |     1     1
2  |     3     2     1
3  |    13     9     3     1
4  |    75    52    18     4     1
5  |   541   375   130    30     5     1
6  |  4683  3246  1125   260    45     6     1
...
(End)
From _Mats Granvik_, Aug 11 2009: (Start)
Row 4 equals 75,52,18,4,1 because permanents of:
  1,0,0,0,1  1,0,0,0,0  1,0,0,0,0  1,0,0,0,0  1,0,0,0,0
  1,1,0,0,0  1,1,0,0,1  1,1,0,0,0  1,1,0,0,0  1,1,0,0,0
  1,2,1,0,0  1,2,1,0,0  1,2,1,0,1  1,2,1,0,0  1,2,1,0,0
  1,3,3,1,0  1,3,3,1,0  1,3,3,1,0  1,3,3,1,1  1,3,3,1,0
  1,4,6,4,0  1,4,6,4,0  1,4,6,4,0  1,4,6,4,0  1,4,6,4,1
are:
     75         52         18          4          1
(End)
		

References

  • Murray R. Spiegel, Mathematical handbook, Schaum's Outlines, p. 111.

Crossrefs

Cf. A000629 (row sums), A000670, A007047, A052822 (column 1), A052841 (alt. row sums), A080253, A162312, A162313.
Cf. A263634, A099880 (T(2n,n)).

Programs

  • Maple
    A154921_row := proc(n) local i,p; p := proc(n,x) option remember; local k;
    if n = 0 then 1 else add(p(k,0)*binomial(n,k)*(1+x^(n-k)),k=0..n-1) fi end:
    seq(coeff(p(n,x),x,i),i=0..n) end: for n from 0 to 5 do A154921_row(n) od;
    # Peter Luschny, Jul 15 2012
    T := (n,k) -> binomial(n,k)*add(combinat:-eulerian1(n-k,j)*2^j, j=0..n-k):
    seq(print(seq(T(n,k), k=0..n)),n=0..6); # Peter Luschny, Feb 07 2015
    # third Maple program:
    b:= proc(n) b(n):= `if`(n=0, 1, add(b(n-j)/j!, j=1..n)) end:
    T:= (n, k)-> n!/k! *b(n-k):
    seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Feb 03 2019
    # fourth Maple program:
    p := proc(n, m) option remember; if n = 0 then 1 else
        (m + x)*p(n - 1, m) + (m + 1)*p(n - 1, m + 1) fi end:
    row := n -> local k; seq(coeff(p(n, 0), x, k), k = 0..n):
    for n from 0 to 6 do row(n) od;  # Peter Luschny, Jun 23 2023
  • Mathematica
    nn = 8; a = Exp[x] - 1;
    Map[Select[#, # > 0 &] &,
      Transpose[
       Table[Range[0, nn]! CoefficientList[
    Series[x^n/n!/(1 - a), {x, 0, nn}], x], {n, 0, nn}]]] // Grid (* Geoffrey Critzer, Jul 22 2013 *)
    E1[n_ /; n >= 0, 0] = 1; E1[n_, k_] /; k < 0 || k > n = 0; E1[n_, k_] := E1[n, k] = (n - k) E1[n - 1, k - 1] + (k + 1) E1[n - 1, k];
    T[n_, k_] := Binomial[n, k] Sum[E1[n - k, j] 2^j, {j, 0, n - k}];
    Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 30 2018, after Peter Luschny *)
  • Sage
    @CachedFunction
    def Poly(n, x):
        return 1 if n == 0 else add(Poly(k,0)*binomial(n,k)*(x^(n-k)+1) for k in range(n))
    R = PolynomialRing(ZZ, 'x')
    for n in (0..6): print(R(Poly(n,x)).list()) # Peter Luschny, Jul 15 2012

Formula

From Peter Bala, Jul 01 2009: (Start)
TABLE ENTRIES
(1) T(n,k) = binomial(n,k)*A000670(n-k).
GENERATING FUNCTION
(2) exp(x*t)/(2-exp(t)) = 1 + (1+x)*t + (3+2*x+x^2)*t^2/2! + ....
PROPERTIES OF THE ROW POLYNOMIALS
The row generating polynomials R_n(x) form an Appell sequence. They appear in the study of the poset of power sets [Nelsen and Schmidt].
The first few values are R_0(x) = 1, R_1(x) = 1+x, R_2(x) = 3+2*x+x^2 and R_3(x) = 13+9*x+3*x^2+x^3.
The row polynomials may be recursively computed by means of
(3) R_n(x) = x^n + Sum_{k = 0..n-1} binomial(n,k)*R_k(x).
Explicit formulas include
(4) R_n(x) = (1/2)*Sum_{k >= 0} (1/2)^k*(x+k)^n,
(5) R_n(x) = Sum_{j = 0..n} Sum_{k = 0..j} (-1)^(j-k)*binomial(j,k) *(x+k)^n,
and
(6) R_n(x) = Sum_{j = 0..n} Sum_{k = j..n} k!*Stirling2(n,k) *binomial(x,k-j).
SUMS OF POWERS OF INTEGERS
The row polynomials satisfy the difference equation
(7) 2*R_m(x) - R_m(x+1) = x^m,
which easily leads to the evaluation of the weighted sums of powers of integers
(8) Sum_{k = 1..n-1} (1/2)^k*k^m = 2*R_m(0) - (1/2)^(n-1)*R_m(n).
For example, m = 2 gives
(9) Sum_{k = 1..n-1} (1/2)^k*k^2 = 6 - (1/2)^(n-1)*(n^2+2*n+3).
More generally we have
(10) Sum_{k=0..n-1} (1/2)^k*(x+k)^m = 2*R_m(x) - (1/2)^(n-1)*R_m(x+n).
RELATIONS WITH OTHER SEQUENCES
Sequences in the database given by particular values of the row polynomials are
(11) A000670(n) = R_n(0)
(12) A052841(n) = R_n(-1)
(13) A000629(n) = R_n(1)
(14) A007047(n) = R_n(2)
(15) A080253(n) = 2^n*R_n(1/2).
This last result is the particular case (x = 0) of the result that the polynomials 2^n*R_n(1/2+x/2) are the row generating polynomials for A162313.
The above formulas should be compared with those for A162312. (End)
From Peter Luschny, Jul 15 2012: (Start)
(16) A151919(n) = R_n(1/3)*3^n*(-1)^n
(17) A052882(n) = [x^1] R_n(x)
(18) A045943(n) = [x^(n-1)] R_n+1(x)
(19) A099880(n) = [x^n] R_2n(x). (End)
The coefficients in ascending order of x^i of the polynomials p{0}(x) = 1 and p{n}(x) = Sum_{k=0..n-1} binomial(n,k)*p{k}(0)*(1+x^(n-k)). - Peter Luschny, Jul 15 2012

Extensions

New name by Peter Luschny, Feb 07 2015

A007525 Decimal expansion of log_2 e.

Original entry on oeis.org

1, 4, 4, 2, 6, 9, 5, 0, 4, 0, 8, 8, 8, 9, 6, 3, 4, 0, 7, 3, 5, 9, 9, 2, 4, 6, 8, 1, 0, 0, 1, 8, 9, 2, 1, 3, 7, 4, 2, 6, 6, 4, 5, 9, 5, 4, 1, 5, 2, 9, 8, 5, 9, 3, 4, 1, 3, 5, 4, 4, 9, 4, 0, 6, 9, 3, 1, 1, 0, 9, 2, 1, 9, 1, 8, 1, 1, 8, 5, 0, 7, 9, 8, 8, 5, 5, 2, 6, 6, 2, 2, 8, 9, 3, 5, 0, 6, 3, 4, 4, 4, 9, 6, 9, 9
Offset: 1

Views

Author

Keywords

Comments

Around 1670, James Gregory discovered by inversion of 1 - 1/2 + 1/3 - 1/4 + 1/5 - ... = log(2) that 1 + 1/2 - 1/12 + 1/24 - 19/720 + (27/1440 = 3/160) - 863/60480 + ... = 1/log(2). See formula with A002206 and A002207. See also A141417 signed /A091137; case i = 0 in A165313. First row in array p. 36 of the reference. - Paul Curtz, Sep 12 2011
This constant 1/log(2) is also related to the asymptotic evaluation of the maximum number of subtraction steps required to compute gcd(m, n) by the binary Euclidean algorithm, m and n being odd and chosen at random. - Jean-François Alcover, Jun 23 2014, after Steven Finch

Examples

			1.442695040888963407359924681...
		

References

  • Paul Curtz, Intégration numérique des systèmes différentiels .. , note n° 12, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 2.18 Porter-Hensley constants, p. 159.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 25, equation 25:14:3 at page 232.

Crossrefs

Programs

Formula

Equals lim_{n->infinity} A000670(n)/A052882(n). - Mats Granvik, Aug 10 2009
Equals Sum_{k>=-1} A002206(k)/A002207(k). - Paul Curtz, Sep 12 2011
Also equals integral_{x>=2} 1/(x*log(x)^2). - Jean-François Alcover, May 24 2013
1/log(2) = Sum_{n = -infinity..infinity} (2^n / (1 + 2^2^n)). - Nicolas Nagel, Mar 16 2018
More generally: 1/log(2) = Sum_{n = -infinity..infinity} (2^(n+x) / (1 + 2^2^(n+x))) for all real x. - Nicolas Nagel, Jul 02 2019
From Amiram Eldar, Jun 04 2023: (Start)
Equals 1 + Sum_{k>=1} 1/(2^k * (1 + 2^(1/2^k))).
Equals Product_{k>=1} ((1 + 2^(1/2^k))/2). (End)

A053525 Expansion of e.g.f.: (1-x)/(2-exp(x)).

Original entry on oeis.org

1, 0, 1, 4, 23, 166, 1437, 14512, 167491, 2174746, 31374953, 497909380, 8619976719, 161667969646, 3265326093109, 70663046421208, 1631123626335707, 40004637435452866, 1038860856732399105, 28476428717448349996, 821656049857815980455
Offset: 0

Views

Author

N. J. A. Sloane, Jan 15 2000

Keywords

Comments

The number of connected labeled threshold graphs on n vertices. - Sam Spiro, Sep 22 2019
Also the number of 2-interval parking functions of size n. - Sam Spiro, Sep 24 2019

Examples

			G.f. = 1 + x^2 + 4*x^3 + 23*x^4 + 166*x^5 + 1437*x^6 + 14512*x^7 + ...
		

References

  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.4(a).

Crossrefs

Programs

  • Magma
    m:=25; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( (1-x)/(2-Exp(x)) )); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Mar 15 2019
    
  • Maple
    A053525 := proc(n) option remember;
    `if`(n < 2, 1 - n, add(binomial(n, k) * A053525(k), k = 0..n-1)) end:
    seq(A053525(n), n = 0..20); # Peter Luschny, Oct 24 2021
  • Mathematica
    With[{nn=20},CoefficientList[Series[(1-x)/(2-Exp[x]),{x,0,nn}], x] Range[0,nn]!] (* Harvey P. Dale, May 17 2012 *)
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( (1 - x) / (2 - exp(x + x*O(x^n))), n))}; /* Michael Somos, Aug 01 2016 */
    
  • Sage
    m = 25; T = taylor((1-x)/(2-exp(x)), x, 0, m); [factorial(n)*T.coefficient(x, n) for n in (0..m)] # G. C. Greubel, Mar 15 2019

Formula

a(n) = c(n) - n*c(n-1) where c() = A000670.
a(n) ~ n!/2 * (1-log(2))/(log(2))^(n+1). - Vaclav Kotesovec, Dec 08 2012
Binomial transform is A005840. - Michael Somos, Aug 01 2016
a(n) = Sum_{k=0..n-1} binomial(n, k) * a(k), n>1. - Michael Somos, Aug 01 2016
a(n) = A005840(n) / 2, n>1. - Michael Somos, Aug 01 2016
E.g.f. A(x) satisfies (1 - x) * A'(x) = A(x) * (x - 2 + 2*A(x)). - Michael Somos, Aug 01 2016
a(n) = Sum_{k=1..n-1} (n-k)*A008292(n-1,k-1)*2^(k-1), valid for n>=2. - Sam Spiro, Sep 22 2019

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

A282507 Triangular array read by rows. T(n,k) is the number of chain topologies on an n-set with exactly k open sets where one of the open sets is a single point set, n >= 2, 3 <= k <= n+1.

Original entry on oeis.org

2, 3, 6, 4, 24, 24, 5, 70, 180, 120, 6, 180, 900, 1440, 720, 7, 434, 3780, 10920, 12600, 5040, 8, 1008, 14448, 67200, 134400, 120960, 40320, 9, 2286, 52164, 367416, 1134000, 1723680, 1270080, 362880, 10, 5100, 181500, 1864800, 8341200, 19051200, 23284800, 14515200, 3628800
Offset: 2

Views

Author

Geoffrey Critzer, Feb 16 2017

Keywords

Comments

A chain topology is a topology that can be totally ordered by inclusion.

Examples

			Triangle begins:
  2;
  3,   6;
  4,  24,  24;
  5,  70, 180,  120;
  6, 180, 900, 1440, 720;
  ...
		

Crossrefs

Cf. A119741 where the topologies are further restricted.
Row sums = A052882.
Cf. A019538.

Programs

  • Mathematica
    nn = 10; Map[Select[#, # > 0 &] &, Drop[Range[0, nn]! CoefficientList[Series[x/(1 - y (Exp[x] - 1)), {x, 0, nn}], {x, y}], 2]] // Grid

Formula

E.g.f.: y^2*x/(1 - y*(exp(x) - 1)). Generally for chain topologies where the smallest nonempty open set has size m: (x^m/m!) * y^2/(1 - y*(exp(x) - 1)).
A conjecture I made to Loic Foissy, who replied: sequence T(n,k) counts surjective maps [n]->> [k] such that k is obtained exactly once, whereas sequence A019538 b(n,k) counts surjective maps [n]->> [k]. To construct an element for T(n,k), you may choose the element of [n] giving k (n choices), then a surjection from the n-1 remaining elements to [k-1] (b(n-1,k-1) choices). This gives T(n,k) = n * b(n-1,k-1), if k,n>1. - Tom Copeland, Nov 10 2023 [So it is now a theorem, not a conjecture, right? - N. J. A. Sloane, Dec 23 2023]

A092692 Expansion of e.g.f. -log(1-x)/(1-x^2).

Original entry on oeis.org

0, 1, 1, 8, 18, 184, 660, 8448, 42000, 648576, 4142880, 74972160, 586776960, 12174658560, 113020427520, 2643856588800, 28432576972800, 740051782041600, 9056055981772800, 259500083163955200, 3562946373482496000
Offset: 0

Views

Author

Michael Somos, Mar 04 2004

Keywords

Comments

Stirling transform of (-1)^n*a(n-1)=[0,1,-1,8,-18,...] is A052882(n-1)=[0,1,2,9,52,...].

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[-Log[1-x]/(1-x^2), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Jul 01 2018 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(-log(1-x+x*O(x^n))/(1-x^2),n))
    
  • PARI
    {a(n)=if(n<0, 0, n!*sum(k=1, n, ((n-k+1)%2)/k))} /* Michael Somos, Sep 19 2006 */

Formula

E.g.f.: -log(1-x)/(1-x^2).
a(n) = n!*Sum_{k=1..n} (-1)^(n-k)*Harmonic(k). - Vladeta Jovovic, Aug 14 2005
a(n) ~ n! * (log(n) + gamma - (-1)^n * log(2)) / 2, where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jul 01 2018

A355720 Expansion of e.g.f. exp( x/(2 - exp(x)) ).

Original entry on oeis.org

1, 1, 3, 16, 113, 986, 10237, 123096, 1680737, 25668766, 433329461, 8009178596, 160802065393, 3483842906610, 80992799730221, 2010720004254856, 53081510001375041, 1484613248976841958, 43846812123456425221, 1363477059263944382604
Offset: 0

Views

Author

Seiichi Manyama, Jul 15 2022

Keywords

Crossrefs

Programs

  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(exp(x/(2-exp(x)))))
    
  • PARI
    a000670(n) = sum(k=0, n, k!*stirling(n, k, 2));
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=1, i, j*a000670(j-1)*binomial(i-1, j-1)*v[i-j+1])); v;

Formula

a(0) = 1; a(n) = Sum_{k=1..n} A052882(k) * binomial(n-1,k-1) * a(n-k).
a(n) ~ n^(n - 1/4) * exp(sqrt(2*n) - 1/4 - n) / (sqrt(2) * log(2)^n). - Vaclav Kotesovec, Jul 15 2022
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