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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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)

A001813 Quadruple factorial numbers: a(n) = (2n)!/n!.

Original entry on oeis.org

1, 2, 12, 120, 1680, 30240, 665280, 17297280, 518918400, 17643225600, 670442572800, 28158588057600, 1295295050649600, 64764752532480000, 3497296636753920000, 202843204931727360000, 12576278705767096320000, 830034394580628357120000, 58102407620643984998400000
Offset: 0

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Author

Keywords

Comments

Counts binary rooted trees (with out-degree <= 2), embedded in plane, with n labeled end nodes of degree 1. Unlabeled version gives Catalan numbers A000108.
Define a "downgrade" to be the permutation which places the items of a permutation in descending order. We are concerned with permutations that are identical to their downgrades. Only permutations of order 4n and 4n+1 can have this property; the number of permutations of length 4n having this property are equinumerous with those of length 4n+1. If a permutation p has this property then the reversal of this permutation also has it. a(n) = number of permutations of length 4n and 4n+1 that are identical to their downgrades. - Eugene McDonnell (eemcd(AT)mac.com), Oct 26 2003
Number of broadcast schemes in the complete graph on n+1 vertices, K_{n+1}. - Calin D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
Hankel transform is A137565. - Paul Barry, Nov 25 2009
The e.g.f. of 1/a(n) = n!/(2*n)! is (exp(sqrt(x)) + exp(-sqrt(x)) )/2. - Wolfdieter Lang, Jan 09 2012
From Tom Copeland, Nov 15 2014: (Start)
Aerated with intervening zeros (1,0,2,0,12,0,120,...) = a(n) (cf. A123023 and A001147), the e.g.f. is e^(t^2), so this is the base for the Appell sequence with e.g.f. e^(t^2) e^(x*t) = exp(P(.,x),t) (reverse A059344, cf. A099174, A066325 also). P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for e^(-t^2)e^(x*t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), e.g., (P(.,t))^n = P(n,t).
Equals A000407*2 with leading 1 added. (End)
a(n) is also the number of square roots of any permutation in S_{4*n} whose disjoint cycle decomposition consists of 2*n transpositions. - Luis Manuel Rivera Martínez, Mar 04 2015
Self-convolution gives A076729. - Vladimir Reshetnikov, Oct 11 2016
For n > 1, it follows from the formula dated Aug 07 2013 that a(n) is a Zumkeller number (A083207). - Ivan N. Ianakiev, Feb 28 2017
For n divisible by 4, a(n/4) is the number of ways to place n points on an n X n grid with pairwise distinct abscissae, pairwise distinct ordinates, and 90-degree rotational symmetry. For n == 1 (mod 4), the number of ways is a((n-1)/4) because the center point can be considered "fixed". For 180-degree rotational symmetry see A006882, for mirror symmetry see A000085, A135401, and A297708. - Manfred Scheucher, Dec 29 2017

Examples

			The following permutations of order 8 and their reversals have this property:
  1 7 3 5 2 4 0 6
  1 7 4 2 5 3 0 6
  2 3 7 6 1 0 4 5
  2 4 7 1 6 0 3 5
  3 2 6 7 0 1 5 4
  3 5 1 7 0 6 2 4
		

References

  • D. E. Knuth, The Art of Computer Programming, Vol. 4, Section 7.2.1.6, Eq. 32.
  • L. C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181.
  • Eugene McDonnell, "Magic Squares and Permutations" APL Quote-Quad 7.3 (Fall, 1976)
  • R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    List([0..20],n->Factorial(2*n)/Factorial(n)); # Muniru A Asiru, Nov 01 2018
    
  • Magma
    [Factorial(2*n)/Factorial(n): n in [0..20]]; // Vincenzo Librandi, Oct 09 2018
    
  • Maple
    A001813 := n->(2*n)!/n!;
    A001813 := n -> mul(k, k = select(k-> k mod 4 = 2,[$1 .. 4*n])):
    seq(A001813(n), n=0..16);  # Peter Luschny, Jun 23 2011
  • Mathematica
    Table[(2n)!/n!, {n,0,20}] (* Harvey P. Dale, May 02 2011 *)
  • Maxima
    makelist(binomial(n+n, n)*n!,n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n)=binomial(n+n,n)*n! \\ Charles R Greathouse IV, Jun 15 2011
    
  • PARI
    first(n) = x='x+O('x^n); Vec(serlaplace((1 - 4*x)^(-1/2))) \\ Iain Fox, Jan 01 2018 (corrected by Iain Fox, Jan 11 2018)
    
  • Python
    from math import factorial
    def A001813(n): return factorial(n<<1)//factorial(n) # Chai Wah Wu, Feb 14 2023
  • Sage
    [binomial(2*n,n)*factorial(n) for n in range(0, 17)] # Zerinvary Lajos, Dec 03 2009
    

Formula

E.g.f.: (1-4*x)^(-1/2).
a(n) = (2*n)!/n! = Product_{k=0..n-1} (4*k + 2) = A081125(2*n).
Integral representation as n-th moment of a positive function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/4)/(sqrt(x)*2*sqrt(Pi)) dx, n >= 0. This representation is unique. - Karol A. Penson, Sep 18 2001
Define a'(1)=1, a'(n) = Sum_{k=1..n-1} a'(n-k)*a'(k)*C(n, k); then a(n)=a'(n+1). - Benoit Cloitre, Apr 27 2003
With interpolated zeros (1, 0, 2, 0, 12, ...) this has e.g.f. exp(x^2). - Paul Barry, May 09 2003
a(n) = A000680(n)/A000142(n)*A000079(n) = Product_{i=0..n-1} (4*i + 2) = 4^n*Pochhammer(1/2, n) = 4^n*GAMMA(n+1/2)/sqrt(Pi). - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
For asymptotics, see the Robinson paper.
a(k) = (2*k)!/k! = Sum_{i=1..k+1} |A008275(i,k+1)| * k^(i-1). - André F. Labossière, Jun 21 2007
a(n) = 12*A051618(a) n >= 2. - Zerinvary Lajos, Feb 15 2008
a(n) = A000984(n)*A000142(n). - Zerinvary Lajos, Mar 25 2008
a(n) = A016825(n-1)*a(n-1). - Roger L. Bagula, Sep 17 2008
a(n) = (-1)^n*A097388(n). - D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
From Paul Barry, Jan 15 2009: (Start)
G.f.: 1/(1-2x/(1-4x/(1-6x/(1-8x/(1-10x/(1-... (continued fraction);
a(n) = (n+1)!*A000108(n). (End)
a(n) = Sum_{k=0..n} A132393(n,k)*2^(2n-k). - Philippe Deléham, Feb 10 2009
G.f.: 1/(1-2x-8x^2/(1-10x-48x^2/(1-18x-120x^2/(1-26x-224x^2/(1-34x-360x^2/(1-42x-528x^2/(1-... (continued fraction). - Paul Barry, Nov 25 2009
a(n) = A173333(2*n,n) for n>0; cf. A006963, A001761. - Reinhard Zumkeller, Feb 19 2010
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n, M = an infinite square production matrix as follows:
2, 2, 0, 0, 0, 0, ...
4, 4, 4, 0, 0, 0, ...
6, 6, 6, 6, 0, 0, ...
8, 8, 8, 8, 8, 0, ...
...
(End)
a(n) = (-2)^n*Sum_{k=0..n} 2^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/Q(0), where Q(k) = 1 + x*(4*k+2) - x*(4*k+4)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 18 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(8*k+4)/(x*(8*k+4) - 1 + 8*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x/(2*x + 1/(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
D-finite with recurrence: a(n) = (4*n-6)*a(n-2) + (4*n-3)*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 07 2013
Sum_{n>=0} 1/a(n) = (exp(1/4)*sqrt(Pi)*erf(1/2) + 2)/2 = 1 + A214869, where erf(x) is the error function. - Ilya Gutkovskiy, Nov 10 2016
Sum_{n>=0} (-1)^n/a(n) = 1 - sqrt(Pi)*erfi(1/2)/(2*exp(1/4)), where erfi(x) is the imaginary error function. - Amiram Eldar, Feb 20 2021
a(n) = 1/([x^n] hypergeom([1], [1/2], x/4)). - Peter Luschny, Sep 13 2024
a(n) = 2^n*n!*JacobiP(n, -1/2, -n, 3). - Peter Luschny, Jan 22 2025
G.f.: 2F0(1,1/2;;4x). - R. J. Mathar, Jun 07 2025

Extensions

More terms from James Sellers, May 01 2000

A046899 Triangle in which n-th row is {binomial(n+k,k), k=0..n}, n >= 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 10, 20, 1, 5, 15, 35, 70, 1, 6, 21, 56, 126, 252, 1, 7, 28, 84, 210, 462, 924, 1, 8, 36, 120, 330, 792, 1716, 3432, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 1, 11, 66, 286, 1001
Offset: 0

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Author

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Comments

C(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Row sums are A001700.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) of waist k (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
If T(r,c), r=0,1,2,..., c=1,2,...,(r+1), are the triangle elements, then for r > 0, T(r,c) = binomial(r+c-1,c-1) = M(r,c) is the number of monotonic mappings from an ordered set of r elements into an ordered set of c elements. For example, there are 15 monotonic mappings from an ordered set of 4 elements into an ordered set of 3 elements. For c > r+1, use the identity M(r,c) = M(c-1,r+1) = T(c-1,r+1). For example, there are 210 monotonic mappings from an ordered set of 4 elements into an ordered set of 7 elements, because M(4,7) = T(6,5) = 210. Number of monotonic endomorphisms in a set of r elements, M(r,r), therefore appear on the second diagonal of the triangle which coincides with A001700. - Stanislav Sykora, May 26 2012
Start at the origin. Flip a fair coin to determine steps of (1,0) or (0,1). Stop when you are a (perpendicular) distance of n steps from the x axis or the y axis. For k = 0,1,...,n-1, C(n-1,k)/2^(n+k) is the probability that you will stop on the point (n,k). This is equal to the probability that you will stop on the point (k,n). Hence, Sum_{k=0..n} C(n,k)/2^(n+k) = 1. - Geoffrey Critzer, May 13 2017

Examples

			The triangle is the lower triangular part of the square array:
  1|  1,  1,   1,   1,    1,    1,     1,     1,     1, ...
  1,  2|  3,   4,   5,    6,    7,     8,     9,    10, ...
  1,  3,  6|  10,  15,   21,   28,    36,    45,    55, ...
  1,  4, 10,  20|  35,   56,   84,   120,   165,   220, ...
  1,  5, 15,  35,  70|  126,  210,   330,   495,   715, ...
  1,  6, 21,  56, 126,  252|  462,   792,  1287,  2002, ...
  1,  7, 28,  84, 210,  462,  924|  1716,  3003,  5005, ...
  1,  8, 36, 120, 330,  792, 1716,  3432|  6435, 11440, ...
  1,  9, 45, 165, 495, 1287, 3003,  6435, 12870| 24310, ...
  1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620| ...
The array read by antidiagonals gives the binomial triangle.
From _Reinhard Zumkeller_, Jul 27 2012: (Start)
Take the first n elements of the n-th diagonal (NW to SE) of left half of Pascal's triangle and write it as n-th row on the triangle on the right side, see above
  0:                 1                    1
  1:               1   _                  1  2
  2:             1   2  __                1  3  6
  3:           1   3  __  __              1  4 10 20
  4:         1   4   6  __  __            1  5 15 35 70
  5:       1   5  10  __  __  __          1  6 21 56 .. ..
  6:     1   6  15  20  __  __  __        1  7 28 .. .. .. ..
  7:   1   7  21  35  __  __  __  __      1  8 .. .. .. .. .. ..
  8: 1   8  28  56  70  __  __  __  __    1 .. .. .. .. .. .. .. .. (End)
		

References

  • H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83.

Crossrefs

Programs

  • Haskell
    import Data.List (transpose)
    a046899 n k = a046899_tabl !! n !! k
    a046899_row n = a046899_tabl !! n
    a046899_tabl = zipWith take [1..] $ transpose a007318_tabl
    -- Reinhard Zumkeller, Jul 27 2012
    
  • Magma
    /* As triangle */ [[Binomial(n+k, n): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 18 2015
    
  • Maple
    for n from 0 to 10 do seq( binomial(n+m,n), m = 0 .. n) od; # Zerinvary Lajos, Dec 09 2007
  • Mathematica
    t[n_, k_] := Binomial[n + k, n]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 12 2013 *)
  • PARI
    /* same as in A092566 but use */
    steps=[[1, 0], [1, 0] ];
    /* Joerg Arndt, Jul 01 2011 */
    
  • SageMath
    for n in (0..9):
        print([multinomial(n, k) for k in (0..n)]) # Peter Luschny, Dec 24 2020

Formula

T(n,k) = A092392(n,n-k), k = 0..n. - Reinhard Zumkeller, Jul 27 2012
T(n,k) = A178300(n,k), n>0, k = 1..n. - L. Edson Jeffery, Jul 23 2014
T(n,k) = (n + 1)*hypergeom([-n, 1 - k], [2], 1). - Peter Luschny, Jan 09 2022
T(n,k) = hypergeom([-n, -k], [1], 1). - Peter Luschny, Mar 21 2024
G.f.: 1/((1-2x*y*C(x*y))*(1-x*C(x*y))), where C(x) is the g.f. for A000108, the Catalan numbers. - Michael D. Weiner, Jul 31 2024

Extensions

More terms from James Sellers

A007685 a(n) = Product_{k=1..n} binomial(2*k,k).

Original entry on oeis.org

1, 2, 12, 240, 16800, 4233600, 3911846400, 13425456844800, 172785629592576000, 8400837310791045120000, 1552105098192510332190720000, 1094904603628138948657963991040000, 2960792853328653706847125274154762240000, 30794022150329995743434211126374020153344000000
Offset: 0

Views

Author

Keywords

References

  • H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    [seq(mul(binomial(2*k,k),k=1..n),n=0..16)];
  • Mathematica
    Table[Product[Binomial[2*k, k], {k, 1, n}], {n, 0, 50}] (* G. C. Greubel, Feb 02 2017 *)
  • PARI
    a(n) = prod(k=1,n, binomial(2*k, k)); \\ Michel Marcus, Sep 18 2015

Formula

a(0) = 1, a(n) = (2^(2*n)*a(n - 1)*Gamma(n + 1/2))/(sqrt(Pi)*Gamma(n + 1)). - Ilya Gutkovskiy, Sep 18 2015
a(n) = (2^(n^2 + n - 1/24)*A^(3/2)*Pi^(-n/2 - 1/4)*BarnesG(n + 3/2))/(e^(1/8)*BarnesG(n + 2)), where A is the Glaisher-Kinkelin constant (A074962), BarnesG is the Barnes G-function. - Ilya Gutkovskiy, Sep 18 2015
a(n) ~ A^(3/2) * 2^(n^2 + n - 7/24) * exp(n/2 - 1/8) / (Pi^((n+1)/2) * n^(n/2 + 3/8)), where A = A074962 is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Apr 16 2016
For n>0, a(n) = 2^((n+1)/2) * sqrt(BarnesG(2*n)) * Gamma(2*n) / (n * BarnesG(n)^2 * Gamma(n)^(7/2)). - Vaclav Kotesovec, Apr 20 2024
Product_{1 <= j <= i <= n} (i + j)/(i - j + 1). - Peter Bala, Oct 25 2024

A103718 Triangle of coefficients of certain polynomials used with prime numbers as variables in the computation of the array A103728.

Original entry on oeis.org

1, 2, -1, 5, -4, 1, 17, -17, 7, -1, 74, -85, 45, -11, 1, 394, -499, 310, -100, 16, -1, 2484, -3388, 2359, -910, 196, -22, 1, 18108, -26200, 19901, -8729, 2282, -350, 29, -1, 149904, -227708, 185408, -89733, 26985, -5082, 582, -37, 1, 1389456, -2199276, 1896380, -993005, 332598, -72723, 10320, -915, 46, -1
Offset: 0

Views

Author

Wolfdieter Lang, Feb 24 2005

Keywords

Comments

The g.f. for the sequence {b(N,p)}, with b(N,p) the number of cyclically inequivalent two-color, N bead necklaces with p beads of one color and N-p beads of the other color is, for prime numbers p, G(p(n),x):=P(p(n)-1,x)/((1-x)^(p(n)-1)*(1-x^p(n))), with the numerator polynomial P(p(n)-1,x):= sum(r(n,k)*x^k,k=0..p(n)-1) and the row polynomials of this triangle r(n,k):=sum(a(k,m)*p(n)^m,m=0..k). p(n)=A000040(n) (prime numbers).
Row sums (signed) give A000142(k)=k!. Row sums (unsigned) coincide with A007680(k)=(2*k+1)*k!, k>=0.
The (unsigned) column sequences are, for m=0..10: A000774, A081052, A103719-A103727.

Examples

			Triangle begins:
    1;
    2,   -1;
    5,   -4,    1;
   17,  -17,    7,   -1;
   74,  -85,   45,  -11,    1;
  394, -499,  310, -100,   16,   -1;
  ...
		

Crossrefs

Cf. A008275.

Programs

  • Mathematica
    a[0, 0] = 1; a[k_, 0] := (k - 1)! + k*a[k - 1, 0]; a[k_, m_]:= If[kIndranil Ghosh, Mar 11 2017 *)
  • PARI
    a(k, m) = if(m==0, if(k==0, 1, (k - 1)! + k*a(k - 1, 0)) , if(kIndranil Ghosh, Mar 11 2017

Formula

a(k, m) = ((-1)^m)*(|S1(k+1, m+1)| + |S1(k+1, m+2)|) = ((-1)^m)*(|S1(k+2, m+2)|-k*|S1(k+1, m+2)|), with the (signed) Stirling number triangle S1(n, m) = A048994(n, m), n >= m >= 0.
a(0, 0)=1, a(k, 0) = (k-1)! + k*a(k-1, 0); a(k, m) = -a(k-1, m-1) + k*a(k-1, m), m > 0 and a(k, m)=0 if k < m.
Let B = (n+1)-st row of Stirling cycle numbers (unsigned, A008275); say a,b,c,d,.... Then n-th row of present triangle = ((a+b), (b+c), (c+d), ..., (d)). E.g., 4th row of the Stirling cycle numbers = (6, 11, 6, 1). Then third row of A103718 = ((6+11), (11+6), (6+1), (1)) = (17, 17, 7, 1). - Gary W. Adamson, May 07 2006

Extensions

More terms from Indranil Ghosh, Mar 11 2017

A167546 The ED1 array read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 2, 4, 1, 6, 12, 7, 1, 24, 48, 32, 10, 1, 120, 240, 160, 62, 13, 1, 720, 1440, 960, 384, 102, 16, 1, 5040, 10080, 6720, 2688, 762, 152, 19, 1, 40320, 80640, 53760, 21504, 6144, 1336, 212, 22, 1
Offset: 1

Views

Author

Johannes W. Meijer, Nov 10 2009

Keywords

Comments

The coefficients in the upper right triangle of the ED1 array (m > n) were found with the a(n,m) formula while the coefficients in the lower left triangle of the ED1 array (m <= n) were found with the recurrence relation, see below. We use for the array rows the letter n (>= 1) and for the array columns the letter m (>= 1).
Our procedure for finding the coefficients in the lower left triangle can be compared with the procedure that De Smit and Lenstra used to fill in the hole in the middle of 'The Print Gallery' by M. C. Escher, see the links. In this lithograph Escher made use of the so-called Droste effect, hence we propose to call this square array of numbers the ED1 array.
For the ED2, ED3 and ED4 arrays see A167560, A167572 and A167584.

Examples

			The ED1 array begins with:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1
1, 4, 7, 10, 13, 16, 19, 22, 25, 28
2, 12, 32, 62, 102, 152, 212, 282, 362, 452
6, 48, 160, 384, 762, 1336, 2148, 3240, 4654, 6432
24, 240, 960, 2688, 6144, 12264, 22200, 37320, 59208, 89664
120, 1440, 6720, 21504, 55296, 122880, 245640, 452880, 783144, 1285536
		

Crossrefs

A000012, A016777, 2*A005891, A167547, A167548 and A167549 equal the first sixth rows of the array.
A000142 equals the first column of the array.
A167550 equals the a(n, n+1) diagonal of the array.
A047053 equals the a(n, n) diagonal of the array.
A167558 equals the a(n+1, n) diagonal of the array.
A167551 equals the row sums of the ED1 array read by antidiagonals.
A167552 is a triangle related to the a(n) formulas of rows of the ED1 array.
A167556 is a triangle related to the GF(z) formulas of the rows of the ED1 array.
A167557 is the lower left triangle of the ED1 array.
Cf. A068424 (the (m-1)!/(m-n-1)! factor), A007680 (the (2*n-1)*(n-1)! factor).
Cf. A167560 (ED2 array), A167572 (ED3 array), A167584 (ED4 array).

Programs

  • Maple
    nmax:=10; mmax:=10; for n from 1 to nmax do for m from 1 to n do a(n,m) := 4^(m-1)*(m-1)!*(n-1+m-1)!/(2*m-2)! od; for m from n+1 to mmax do a(n,m):= (2*n-1)*(n-1)! + sum((-1)^(k-1)*binomial(n-1,k)*a(n,m-k),k=1..n-1) od; od: for n from 1 to nmax do for m from 1 to n do d(n,m):=a(n-m+1,m) od: od: T:=1: for n from 1 to nmax do for m from 1 to n do a(T):= d(n,m): T:=T+1: od: od: seq(a(n),n=1..T-1);
  • Mathematica
    nmax = 10; mmax = 10; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, a[n, m] = 4^(m - 1)*(m - 1)!*((n - 1 + m - 1)!/(2*m - 2)!)]; For[m = n + 1, m <= mmax, m++, a[n, m] = (2*n - 1)*(n - 1)! + Sum[(-1)^(k - 1)*Binomial[n - 1, k]*a[n, m - k], {k, 1, n - 1}]]; ]; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, d[n, m] = a[n - m + 1, m]]; ]; t = 1; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, a[t] = d[n, m]; t = t + 1]]; Table[a[n], {n, 1, t - 1}] (* Jean-François Alcover, Dec 20 2011, translated from Maple *)

Formula

a(n,m) = (2*(m-1)!/(m-n-1)!)*Integral_{y>=0} sinh(y*(2*n-1))/cosh(y)^(2*m-1) for m > n.
The (n-1)-differences of the n-th array row lead to the recurrence relation
Sum_{k=0..n-1} (-1)^k*binomial(n-1,k)*a(n,m-k) = (2*n-1)*(n-1)!
which in its turn leads to, see also A167557,
a(n,m) = 4^(m-1)*(m-1)!*(n+m-2)!/(2*m-2)! for m <= n.

A130152 Triangle read by rows: T(n,k) = number of permutations p of [n] such that max(|p(i)-i|)=k (n>=1, 0<=k<=n-1).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 4, 9, 10, 1, 7, 23, 47, 42, 1, 12, 60, 157, 274, 216, 1, 20, 151, 503, 1227, 1818, 1320, 1, 33, 366, 1669, 4833, 10402, 13656, 9360, 1, 54, 877, 5472, 18827, 50879, 96090, 115080, 75600, 1, 88, 2088, 17531, 75693, 234061, 569602, 966456, 1077840, 685440, 1, 143, 4937, 55135, 304900, 1076807, 3111243, 6791994, 10553640, 11123280, 6894720
Offset: 1

Views

Author

Emeric Deutsch, May 27 2007

Keywords

Comments

Row sums are the factorials. T(n,n) = (n-2)!*(2n-3) = A007680(n-2) (for n>=2). T(n,1) = Fibonacci(n+1)-1 = A000071(n+1). Sum_{k=0..n-1} k*T(n,k) = A130153(n). For the statistic max(p(i)-i) see A056151.

Examples

			T(4,1) = 4 because we have 1243, 1324, 2134 and 2143.
Triangle starts:
  1;
  1,  1;
  1,  2,  3;
  1,  4,  9,  10;
  1,  7, 23,  47,  42;
  1, 12, 60, 157, 274, 216;
  ...
		

Crossrefs

Row sums give A000142.
T(n,floor(n/2)) gives A323807.

Programs

  • Maple
    with(combinat): for n from 1 to 7 do P:=permute(n): for i from 0 to n-1 do ct[i]:=0 od: for j from 1 to n! do if max(seq(abs(P[j][i]-i),i=1..n))=0 then ct[0]:=ct[0]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=1 then ct[1]:=ct[1]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=2 then ct[2]:=ct[2]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=3 then ct[3]:=ct[3]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=4 then ct[4]:=ct[4]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=5 then ct[5]:=ct[5]+1 elif max(seq(abs(P[j][i]-i),i=1..n))=6 then ct[6]:=ct[6]+1 else fi od: a[n]:=seq(ct[i],i=0..n-1): od: for n from 1 to 7 do a[n] od; # a cumbersome program to obtain, by straightforward counting, the first 7 rows of the triangle
    n := 8: st := proc (p) max(seq(abs(p[j]-j), j = 1 .. nops(p))) end proc: with(combinat): P := permute(n): f := sort(add(t^st(P[i]), i = 1 .. factorial(n))); # program gives the row generating polynomial for the specified n - Emeric Deutsch, Aug 13 2009
    # second Maple program:
    b:= proc(s) option remember; (n-> `if`(n=0, 1, add((p-> add(
          coeff(p, x, i)*x^max(i, abs(n-j)), i=0..degree(p)))(
            b(s minus {j})), j=s)))(nops(s))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..n-1))(b({$1..n})):
    seq(T(n), n=1..10);  # Alois P. Heinz, Jan 21 2019
    # third Maple program:
    A:= proc(n, k) option remember; LinearAlgebra[Permanent](
          Matrix(n, (i, j)-> `if`(abs(i-j)<=k, 1, 0)))
        end:
    T:= (n, k)-> A(n, k)-A(n, k-1):
    seq(seq(T(n, k), k=0..n-1), n=1..10);  # Alois P. Heinz, Jan 22 2019
  • Mathematica
    (* from second Maple program: *)
    b[s_List] := b[s] = Function[n, If[n == 0, 1, Sum[Function[p, Sum[ Coefficient[p, x, i]*x^Max[i, Abs[n - j]], {i, 0, Exponent[p, x]}]][b[s ~Complement~ {j}]], {j, s}]]][Length[s]];
    T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, n-1}]][b[Range[n]] ];
    Table[T[n], {n, 1, 11}] // Flatten
    (* from third Maple program: *)
    A[n_, k_] := A[n, k] = Permanent[Table[If[Abs[i-j] <= k, 1, 0], {i, 1, n}, {j, 1, n}]];
    T[n_, k_] := A[n, k] - A[n, k - 1];
    Table[Table[T[n, k], {k, 0, n - 1}], {n, 1, 11}] // Flatten (* Jean-François Alcover, Dec 06 2019, after Alois P. Heinz *)

Formula

T(n,k) = A306209(n,k) - A306209(n,k-1) for k > 0, T(n,0) = 1. - Alois P. Heinz, Jan 29 2019

Extensions

More terms from R. J. Mathar, Oct 15 2007

A007019 a(n) = (2n+1)! / 2^n.

Original entry on oeis.org

1, 3, 30, 630, 22680, 1247400, 97297200, 10216206000, 1389404016000, 237588086736000, 49893498214560000, 12623055048283680000, 3786916514485104000000, 1329207696584271504000000, 539658324813214230624000000, 250941121038144617240160000000, 132496911908140357902804480000000
Offset: 0

Views

Author

Keywords

Comments

Denominators of coefficients of the Taylor series of sinh(sqrt(2*x))/(sqrt(2*x)). - J. Zurita (jrzurita(AT)inaoep.mx), Dec 01 2007

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Numerators: A002067, erf(x): A007680.

Programs

  • Magma
    [Factorial(2*n+1)/2^n: n in [0..25]]; // Vincenzo Librandi, May 14 2011
    
  • Maple
    a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]*(2*n-1)*(n-1) od: seq(a[n], n=1..14); # Zerinvary Lajos, Mar 08 2008
  • Mathematica
    Table[(2n+1)!/2^n,{n,0,20}] (* Harvey P. Dale, May 13 2011 *)
  • PARI
    a(n) = (2*n+1)!/2^n; \\ Altug Alkan, Aug 27 2018

Formula

sin(x)*cosh(x) = Sum_{n>=0} (-1)^floor(n/2)*x^(2n+1)/a(n). - Benoit Cloitre, Feb 02 2002
a(n) = Product_{k=0..n-1} (A000217(n+1) - A000217(k)). - Anton Zakharov, Sep 14 2016
a(n) ~ sqrt(Pi)*2^(n+2)*n^(2*n+3/2)/exp(2*n). - Ilya Gutkovskiy, Sep 14 2016
a(n) = Product_{j=1..n} T(2j) (where T(k) is the k-th triangular number). For example: a(3) = T(2)*T(4)*T(6) (that is, 630 = 3*10*21). - Rigoberto Florez, Aug 26 2018
From Amiram Eldar, Jun 25 2020: (Start)
Sum_{n>=0} 1/a(n) = sinh(sqrt(2))/sqrt(2).
Sum_{n>=0} (-1)^n/a(n) = sin(sqrt(2))/sqrt(2). (End)

A060524 Triangle read by rows: T(n,k) = number of degree-n permutations with k odd cycles, k=0..n, n >= 0.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 5, 0, 1, 9, 0, 14, 0, 1, 0, 89, 0, 30, 0, 1, 225, 0, 439, 0, 55, 0, 1, 0, 3429, 0, 1519, 0, 91, 0, 1, 11025, 0, 24940, 0, 4214, 0, 140, 0, 1, 0, 230481, 0, 122156, 0, 10038, 0, 204, 0, 1, 893025, 0, 2250621, 0, 463490, 0, 21378, 0, 285, 0, 1, 0
Offset: 0

Views

Author

Vladeta Jovovic, Apr 01 2001

Keywords

Comments

The row polynomials t(n,x):=Sum_{k=0..n} T(n,k)*x^k satisfy the recurrence relation t(n,x) = x*t(n-1,x) + ((n-1)^2)*t(n-2,x); t(-1,x)=0, t(0,x)=1. - Wolfdieter Lang, see above.
This is an example of a Sheffer triangle (coefficient triangle for Sheffer polynomials). In the umbral calculus (see the Roman reference given under A048854) s(n,x) := Sum_{k=0..n} T(n,k)*x^k would be called Sheffer polynomials for (1/cosh(t),tanh(t)), which translates to the e.g.f. for column number k>=0 given by (1/sqrt(1-x^2))*((arctanh(x))^k)/k!. The e.g.f. given below is rewritten in this Sheffer context as (1/sqrt(1-x^2))*exp(y*log(sqrt((1+x)/(1-x))))= (1/sqrt(1-x^2))*exp(y*arctanh(x)). The rows of the Jabotinsky type triangle |A049218| provide the coefficients of the associated polynomials. - Wolfdieter Lang, Feb 24 2005
The solution of the differential-difference relation f(n+1,x)= (d/dx)f(n,x) + (n^2)*f(n-1,x), n >= 1, with inputs f(0,x) and f(1,x) = (d/dx)f(0,x) is f(n,x) = t(n,d_x)*f(0,x), with the differential operator d_x:=d/dx and the row polynomials t(n,x) defined above. This problem appears in a computation of thermo field dynamics where f(0,x)=1/cosh(x). See the triangle A060081. - Wolfdieter Lang, Feb 24 2005
The inverse of the Sheffer matrix T with elements T(n,k) is the Sheffer matrix A060081. - Wolfdieter Lang, Jul 22 2005
T(n,k)=0 if n-k= 1(mod 2), else T(n,k) = sum of M2(n,p), p from {1,...,A000041(n)} restricted to partitions with exactly k odd parts and any nonnegative number of even parts. For the M2-multinomial numbers in A-St order see A036039(n,p). - Wolfdieter Lang, Aug 07 2007

Examples

			Triangle begins:
  [1],
  [0, 1],
  [1, 0, 1],
  [0, 5, 0, 1],
  [9, 0, 14, 0, 1],
  [0, 89, 0, 30, 0, 1],
  [225, 0, 439, 0, 55, 0, 1],
  [0, 3429, 0, 1519, 0, 91, 0, 1],
  [11025, 0, 24940, 0, 4214, 0, 140, 0, 1],
  [0, 230481, 0, 122156, 0, 10038, 0, 204, 0, 1],
  [893025, 0, 2250621, 0, 463490, 0, 21378, 0, 285, 0, 1],
  [0, 23941125, 0, 14466221, 0, 1467290, 0, 41778, 0, 385, 0, 1],
  ...
Signed version begins:
  [1],
  [0, 1],
  [-1, 0, 1],
  [0, -5, 0, 1],
  [9, 0, -14, 0, 1],
  [0, 89, 0, -30, 0, 1],
  [-225, 0, 439, 0, -55, 0, 1],
  [0, -3429, 0, 1519, 0, -91, 0, 1],
  ...
From _Peter Bala_, Feb 23 2024: (Start)
Maple can verify the following series for Pi:
Row 1 polynomial R(1, x) = x:
Pi = 3 + 4*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(1, 2*n)*R(1, 2*n+2)).
Row 3 polynomial R(3, x) = 5*x + x^3:
(3/2)^2 * Pi = 7 + 4*(3^4)*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(3, 2*n)*R(3, 2*n+2)).
Row 5 polynomial R(5, x) = 89*x + 30*x^3 + x^5:
((3*5)/(2*4))^2 * Pi = 11 + 4*(3*5)^4*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(5, 2*n)*R(5, 2*n+2)). (End)
		

Crossrefs

Cf. A060338, A060523, A094368, A028353 (col 1), A103916 (col 2), A103917 (col 3), A103918 (col 4).
Cf. A111594 (associated Sheffer polynomials), A142979, A142983.

Programs

  • Maple
    with(combinat):
    b:= proc(n, i) option remember; expand(`if`(n=0, 1, `if`(i<1, 0,
          add(multinomial(n, n-i*j, i$j)*(i-1)!^j/j!*b(n-i*j, i-1)*
          `if`(irem(i, 2)=1, x^j, 1), j=0..n/i))))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2)):
    seq(T(n), n=0..12);  # Alois P. Heinz, Mar 09 2015
    # alternative
    A060524 := proc(n,k)
        option remember;
        if nR. J. Mathar, Jul 06 2023
  • Mathematica
    nn = 6; Range[0, nn]! CoefficientList[
       Series[(1 - x^2)^(-1/2) ((1 + x)/(1 - x))^(y/2), {x, 0, nn}], {x, y}] // Grid  (* Geoffrey Critzer, Aug 28 2012 *)

Formula

E.g.f.: (1+x)^((y-1)/2)/(1-x)^((y+1)/2).
T(n, k) = T(n-1, k-1) + ((n-1)^2)*T(n-2, k); T(-1, k):=0, T(n, -1):=0, T(0, 0)=1, T(n, k)=0 if nWolfdieter Lang, see above.
The Meixner polynomials defined by S_0(x)=1, S_1(x) = x; S_{n+1}(x) = x*S_n(x) - n^2*S_{n-1}(x) give a signed version of this triangle (cf. A060338). - N. J. A. Sloane, May 30 2013
From Peter Bala, Apr 10 2024: (Start)
The n-th row polynomial R(n, x) satisfies
(4*n + 2)*R(n, x) = (x + 1)*R(n, x+2) - (x - 1)*R(n, x-2).
Series for Pi involving the row polynomials R(n, x): for n >= 0 there holds
((2*n + 1)!!/(2^n*n!))^2 * Pi = (4*n + 3) + 4*((2*n + 1)!!^4) * Sum_{k >= 1} (-1)^(k+1)/((2*k + 1)*R(2*n+1, 2*k)*R(2*n+1, 2*k+2)). Cf. A142979 and A142983.
R(2*n, 0) = A001147(n)^2 = A001818(n); R(2*n+1, 0) = 0.
R(n, 1) = n! = A000142(n).
R(2*n, 2) = (4*n + 1)*A001147(n)^2 = (4*n + 1)*((2*n)!/(2^n*n!))^2;
R(2*n+1, 2) = 2*A001447(n+1)^2 = 2*(2*n + 1)!^2/(n!^2*4^n).
R(n, 3) = (2*n + 1)*n! = A007680(n). (End)

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

Original entry on oeis.org

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

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Author

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

Keywords

Comments

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

Examples

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

Crossrefs

Programs

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

Formula

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

Extensions

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