A284005 -id:A284005 - cp's OEIS frontend

A341392 a(n) = A284005(n) / (1 + A000120(n))!.

1, 1, 2, 1, 4, 2, 3, 1, 8, 4, 6, 2, 9, 3, 4, 1, 16, 8, 12, 4, 18, 6, 8, 2, 27, 9, 12, 3, 16, 4, 5, 1, 32, 16, 24, 8, 36, 12, 16, 4, 54, 18, 24, 6, 32, 8, 10, 2, 81, 27, 36, 9, 48, 12, 15, 3, 64, 16, 20, 4, 25, 5, 6, 1, 64, 32, 48, 16, 72, 24, 32, 8, 108, 36, 48, 12, 64, 16, 20, 4, 162, 54, 72, 18, 96, 24, 30, 6, 128

Offset: 0

Mikhail Kurkov, Feb 10 2021 [verification needed]

From Antti Karttunen, Feb 10 2021: (Start)
This sequence can be represented as a binary tree. Each child to the left is obtained by multiplying its parent with (1+{binary weight of its breadth-first-wise index in the tree}), while each child to the right is just a clone of its parent:
                                      1
                                      |
                   ...................1...................
                  2                                       1
        4......../ \........2                   3......../ \........1
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
    8       4           6       2           9       3           4       1
  16 8    12 4        18 6     8 2        27 9    12 3        16 4     5 1
etc.
(End)
This sequence and A243499 have the same set of values on intervals from 2^m to 2^(m+1) - 1 for m >= 0. - Mikhail Kurkov, Jun 18 2021 [verification needed]
FindStat provides a sequence of mappings between this sequence and A000110 starting from collection [Set partitions] (see Links section for illustration). - Mikhail Kurkov, May 20 2023 [verification needed]

Alois P. Heinz, Table of n, a(n) for n = 0..16384
Michael De Vlieger, This sequence as a binary tree showing rows 0 <= r <= 5.
Michael De Vlieger, This sequence as a binary tree showing rows 0 <= r <= 8.
FindStat, Illustration of the sequence of mappings

Cf. A000120, A000142, A007814, A036987, A053645, A243499, A284005, A329369 (similar recurrence).

a:= proc(n) option remember; `if`(n=0, 1,
      a(iquo(n, 2, 'd'))*`if`(d=1, 1, add(i, i=Bits[Split](n+1))))
    end:
seq(a(n), n=0..120);  # Alois P. Heinz, Jun 23 2021

Array[DivisorSigma[0, Apply[Times, Map[#1^#2 & @@ # &, FactorInteger[#1] /. {p_, e_} /; e == 1 :> {Times @@ Prime@ Range@ PrimePi@ p, e}]]]/#2 & @@ {Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ #, (1 + Count[#, 1])!} &@ IntegerDigits[#, 2] &, 89, 0] (* Michael De Vlieger, Feb 24 2021 *)

A284005(n) = { my(k=if(n, logint(n, 2)), s=1); prod(i=0, k, s+=bittest(n, k-i)); }; \\ From A284005
A341392(n) = (A284005(n)/((1 + hammingweight(n))!)); \\ Antti Karttunen, Feb 10 2021

A341392(n) = if(!n,1,if(n%2, A341392((n-1)/2), (1+hammingweight(n))*A341392(n/2))); \\ Antti Karttunen, Feb 10 2021

a(n) = A284005(n) / (1 + A000120(n))! = A284005(n) / A000142(1 + A000120(n)).
a(2n+1) = a(n) for n >= 0.
a(2n) = (1 + A000120(n))*a(n) = A243499(2*A059894(n)) = a(n) + a(2n - 2^A007814(n)) for n > 0 with a(0) = 1.
[2*a(n) - 1 = A329369(n)] = A036987(A053645(n)).
From Mikhail Kurkov, Apr 24 2023: (Start)
a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m, k)*a(2^k*n) for m >= 0, n >= 0 with a(0) = 1.
a(n) = a(f(n)) + Sum_{k=0..floor(log_2(n))-1} (1 - T(n, k))*a(f(n) + 2^k*(1 - T(n, k))) for n > 1 with a(0) = 1, a(1) = 1, where f(n) = A053645(n) and where T(n, k) = floor(n/2^k) mod 2. (End) [verification needed]

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

N. J. A. Sloane

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

			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)

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Index entries for "core" sequences
Index entries for related partition-counting sequences

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

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

R:=PowerSeriesRing(Rationals(), 40);
Coefficients(R!(Laplace( 1/(2-Exp(x)) ))); // G. C. Greubel, Jun 11 2024

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

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 *)

makelist(sum(stirling2(n,k)*k!,k,0,n),n,0,12); /* Emanuele Munarini, Jul 07 2011 */

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 */

{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 */

Vec(serlaplace(1/(2-exp('x+O('x^66))))) /* Joerg Arndt, Jul 10 2011 */

{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 */

{a(n) = if( n<1, n==0, sum(k=1, n, binomial(n, k) * a(n-k)))}; /* Michael Somos, Jul 16 2017 */

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

@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

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)

A283477 If 2n = 2^e1 + 2^e2 + ... + 2^ek [e1 .. ek distinct], then a(n) = A002110(e1) * A002110(e2) * ... * A002110(ek).

Original entry on oeis.org

1, 2, 6, 12, 30, 60, 180, 360, 210, 420, 1260, 2520, 6300, 12600, 37800, 75600, 2310, 4620, 13860, 27720, 69300, 138600, 415800, 831600, 485100, 970200, 2910600, 5821200, 14553000, 29106000, 87318000, 174636000, 30030, 60060, 180180, 360360, 900900, 1801800, 5405400, 10810800, 6306300, 12612600, 37837800, 75675600

Offset: 0

Antti Karttunen, Mar 16 2017

nonn

a(n) = Product of distinct primorials larger than one, obtained as Product_{i} A002110(1+i), where i ranges over the zero-based positions of the 1-bits present in the binary representation of n.
This sequence can be represented as a binary tree. Each child to the left is obtained as A283980(k), and each child to the right is obtained as 2*A283980(k), when their parent contains k:
                                      1
                                      |
                   ...................2....................
                  6                                       12
       30......../ \........60                 180......../ \......360
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
  210       420      1260       2520     6300       12600   37800       75600
etc.

Cf. A129912 (same terms, but sorted into ascending order).
Cf. A000120, A001221, A001222, A002110, A005187, A005361, A006068, A007814, A007913, A019565, A029931, A030308, A046660, A048675, A051903, A056169, A065120, A070939, A072411, A090880, A097248, A108951, A124757, A248663, A276075, A276085, A283475, A283483, A283980, A283984, A283985, A284001, A284002, A284003, A284005, A324287, A324289, A324341, A324342, A324343.
Cf. A005940, A052330, A322827, A323505 for other similar trees.
Cf. also A260443.

Table[Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e == 1 :> {Times @@ Prime@ Range@ PrimePi@ p, e}] &[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2]], {n, 0, 43}] (* Michael De Vlieger, Mar 18 2017 *)

A283477(n) = prod(i=0,exponent(n),if(bittest(n,i),vecprod(primes(1+i)),1)) \\ Edited by M. F. Hasler, Nov 11 2019

from sympy import prime, primerange, factorint
from operator import mul
from functools import reduce
def P(n): return reduce(mul, [i for i in primerange(2, n + 1)])
def a108951(n):
    f = factorint(n)
    return 1 if n==1 else reduce(mul, [P(i)**f[i] for i in f])
def a019565(n): return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1 # after Chai Wah Wu
def a(n): return a108951(a019565(n))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 22 2017

from sympy import primorial
from math import prod
def A283477(n): return prod(primorial(i) for i, b in enumerate(bin(n)[:1:-1],1) if b =='1') # Chai Wah Wu, Dec 08 2022

(define (A283477 n) (A108951 (A019565 n)))
;; Recursive "binary tree" implementation, using memoization-macro definec:
(definec (A283477 n) (cond ((zero? n) 1) ((even? n) (A283980 (A283477 (/ n 2)))) (else (* 2 (A283980 (A283477 (/ (- n 1) 2)))))))

a(0) = 1; a(2n) = A283980(a(n)), a(2n+1) = 2*A283980(a(n)).
Other identities. For all n >= 0 (or for n >= 1):
a(2n+1) = 2*a(2n).
a(n) = A108951(A019565(n)).
A097248(a(n)) = A283475(n).
A007814(a(n)) = A051903(a(n)) = A000120(n).
A001221(a(n)) = A070939(n).
A001222(a(n)) = A029931(n).
A048675(a(n)) = A005187(n).
A248663(a(n)) = A006068(n).
A090880(a(n)) = A283483(n).
A276075(a(n)) = A283984(n).
A276085(a(n)) = A283985(n).
A046660(a(n)) = A124757(n).
A056169(a(n)) = A065120(n). [seems to be]
A005361(a(n)) = A284001(n).
A072411(a(n)) = A284002(n).
A007913(a(n)) = A284003(n).
A000005(a(n)) = A284005(n).
A324286(a(n)) = A324287(n).
A276086(a(n)) = A324289(n).
A267263(a(n)) = A324341(n).
A276150(a(n)) = A324342(n). [subsequences in the latter are converging towards this sequence]
G.f.: Product_{k>=0} (1 + prime(k + 1)# * x^(2^k)), where prime()# = A002110. - Ilya Gutkovskiy, Aug 19 2019

More formulas and the binary tree illustration added by Antti Karttunen, Mar 19 2017

Four more linking formulas added by Antti Karttunen, Feb 25 2019

A329369 Number of permutations of {1,2,...,m} with excedance set constructed by taking m-i (0 < i < m) if b(i-1) = 1 where b(k)b(k-1)...b(1)b(0) (0 <= k < m-1) is the binary expansion of n.

Original entry on oeis.org

1, 1, 3, 1, 7, 3, 7, 1, 15, 7, 17, 3, 31, 7, 15, 1, 31, 15, 37, 7, 69, 17, 37, 3, 115, 31, 69, 7, 115, 15, 31, 1, 63, 31, 77, 15, 145, 37, 81, 7, 245, 69, 155, 17, 261, 37, 77, 3, 391, 115, 261, 31, 445, 69, 145, 7, 675, 115, 245, 15, 391, 31, 63, 1, 127, 63

Offset: 0

Mikhail Kurkov, Nov 12 2019

Another version of A152884.
The excedance set of a permutation p of {1,2,...,m} is the set of indices i such that p(i) > i; it is a subset of {1,2,...,m-1}.
Great work on this subject was done by R. Ehrenborg and E. Steingrimsson, so most of the formulas given below are just their results translated into the language of the sequences which are related to the binary expansion of n.
Conjecture 1: equivalently, number of open tours by a biased rook on a specific f(n) X 1 board, which ends on a white cell, where f(n) = A070941(n) = floor(log_2(2n)) + 1 and cells are colored white or black according to the binary representation of 2n. A cell is colored white if the binary digit is 0 and a cell is colored black if the binary digit is 1. A biased rook on a white cell moves only to the left and otherwise moves only to the right. - Mikhail Kurkov, May 18 2021
Conjecture 2: this sequence is an inverse modulo 2 binomial transform of A284005. - Mikhail Kurkov, Dec 15 2021

			a(1) = 1 because the 1st excedance set is {m-1} and the permutations of {1,2,...,m} with such excedance set are 21, 132, 1243, 12354 and so on, i.e., for a given m we always have 1 permutation.
a(2) = 3 because the 2nd excedance set is {m-2} and the permutations of {1,2,...,m} with such excedance set are 213, 312, 321, 1324, 1423, 1432, 12435, 12534, 12543 and so on, i.e., for a given m we always have 3 permutations.
a(3) = 1 because the 3rd excedance set is {m-2, m-1} and the permutations of {1,2,...,m} with such excedance set are 231, 1342, 12453 and so on, i.e., for a given m we always have 1 permutation.

Mikhail Kurkov, Table of n, a(n) for n = 0..8191
Max Alekseyev, Recursion for the sum with Stirling numbers of both kinds, answer to question on MathOverflow (2024).
R. Ehrenborg and E. Steingrimsson, The excedance set of a permutation, Advances in Appl. Math., 24, 284-299, 2000.
Peter J. Taylor, Correctness of the algorithm for the A329369, A347205 and related sequences, answer to question on MathOverflow (2024).

Cf. A000120, A000523, A001511, A007814, A008292, A023416, A036987, A053645, A062253, A062254, A062255, A063250, A080791, A091344, A091347, A091348, A110501, A136126, A152884 (lexicographic order), A173018, A329718, A344902, A344947, A357990, A358612, A373183, A379817.
Similar recurrences: A006014, A124758, A217924, A243499, A284005, A290615, A341392, A347204, A347205.
Cf. A360288.

g:= proc(n) option remember;  2^padic[ordp](n, 2) end:
a:= proc(n) option remember; `if`(n=0, 1, (h-> a(h)+
     `if`(n::odd, 0, (t-> a(h-t)+a(n-t))(g(h))))(iquo(n, 2)))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Jan 30 2023

a[n_] := a[n] = Which[n == 0, 1, OddQ[n], a[(n-1)/2], True, a[n/2] + a[n/2 - 2^IntegerExponent[n/2, 2]] + a[n - 2^IntegerExponent[n/2, 2]]];
a /@ Range[0, 65] (* Jean-François Alcover, Feb 13 2020 *)

upto(n) = my(A, v1); v1 = vector(n+1, i, 0); v1[1] = 1; for(i=1, n, v1[i+1] = v1[i\2+1] + if(i%2, 0, A = 1 << valuation(i/2, 2); v1[i/2-A+1] + v1[i-A+1])); v1 \\ Mikhail Kurkov, Jun 06 2024

a(2n+1) = a(n) for n >= 0.
a(2n) = a(n) + a(n - 2^f(n)) + a(2n - 2^f(n)) for n > 0 with a(0) = 1 where f(n) = A007814(n) (equivalent to proposition 2.1 at the page 286, see R. Ehrenborg and E. Steingrimsson link).
a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n) = a(2^m*n) + a(2^(m-1)*(2n+1)) + a(2^(m-1)*(4n+1)) for m > 0, n >= 0 (equivalent to proposition 2.5 at the page 287, see R. Ehrenborg and E. Steingrimsson link).
a(2n) = a(2*g(n)) + a(2n - 2^h(n)) + a(2*g(n) + 2^h(n)) for n > 0 with a(0) = 1 where g(n) = A053645(n), h(n) = A063250(n) (equivalent to proposition 2.1 at the page 286, see R. Ehrenborg and E. Steingrimsson link).
a(2n) = 2*a(n + g(n)) + a(2*g(n)) for n > 0, floor(n/3) < 2^(floor(log_2(n))-1) (in other words, for 2^m + k where 0 <= k < 2^(m-1), m > 0) with a(0) = 1 (just a special case of the previous formula, because for 2^m + k where 0 <= k < 2^(m-1), m > 0 we have 2^h(n) = n - g(n)).
a(2n) = a(f(n,-1)) + a(f(n,0)) + a(f(n,1)) for n > 0 with a(0) = 1 where f(n,k) = 2*(f(floor(n/2),k) + n mod 2) + k*A036987(n) for n > 1 with f(1,k) = abs(k) (equivalent to a(2n) = a(2*g(n)) + a(2n - 2^h(n)) + a(2*g(n) + 2^h(n))).
a(n) = Sum_{j=0..2^wt(n) - 1} (-1)^(wt(n) - wt(j)) Product_{k=0..wt(n) - 1} (1 + wt(floor(j/2^k)))^T(n,k) for n > 0 with a(0) = 1 where wt(n) = A000120(n), T(n,k) = T(floor(n/2), k - n mod 2) for k > 0 with T(n,0) = A001511(n) (equivalent to theorem 6.3 at page 296, see R. Ehrenborg and E. Steingrimsson link). Here T(n, k) - 1 for k > 0 is the length of the run of zeros between k-th pair of ones from the right side in the binary expansion of n. Conjecture 1: this formula is equivalent to inverse modulo 2 binomial transform of A284005.
Sum_{k=0..2^n-1} a(k) = (n+1)! for n >= 0.
a((4^n-1)/3) = A110501(n+1) for n >= 0.
a(2^2*(2^n-1)) = A091344(n+1),
a(2^3*(2^n-1)) = A091347(n+1),
a(2^4*(2^n-1)) = A091348(n+1).
More generally, a(2^m*(2^n-1)) = a(2^n*(2^m-1)) = S(n+1,m) for n >= 0, m >= 0 where S(n,m) = Sum_{k=1..n} k!*k^m*Stirling2(n,k)*(-1)^(n-k) (equivalent to proposition 6.5 at the page 297, see R. Ehrenborg and E. Steingrimsson link).
Conjecture 2: a(n) = (1 + A023416(n))*a(g(n)) + Sum_{k=0..floor(log_2(n))-1} (1-R(n,k))*a(g(n) + 2^k*(1 - R(n,k))) for n > 1 with a(0) = 1, a(1) = 1, where g(n) = A053645(n) and where R(n,k) = floor(n/2^k) mod 2 (at this moment this is the only formula here, which is not related to R. Ehrenborg's and E. Steingrimsson's work and arises from another definition given above, exactly conjectured definition with a biased rook). Here R(n,k) is the (k+1)-th bit from the right side in the binary expansion of n. - Mikhail Kurkov, Jun 23 2021
From Mikhail Kurkov, Jan 23 2023: (Start)
The formulas below are not related to R. Ehrenborg's and E. Steingrimsson's work.
Conjecture 3: a(n) = A357990(n, 1) for n >= 0.
Conjecture 4: a(2^m*(2k+1)) = Sum_{i=1..wt(k) + 2} i!*i^m*A358612(k, i)*(-1)^(wt(k) - i) for m >= 0, k >= 0 where wt(n) = A000120(n).
Conjecture 5: a(2^m*(2^n - 2^p - 1)) = Sum_{i=1..n} i!*i^m*(-1)^(n - i)*((i - p + 1)*Stirling2(n, i) - Stirling2(n - p, i - p) + Sum_{j=0..p-2} (p - j - 1)*Stirling2(n - p, i - j)/j! Sum_{k=0..j} (i - k)^p*binomial(j, k)*(-1)^k) for n > 2, m >= 0, 0 < p < n - 1. Here we consider that Stirling2(n, k) = 0 for n >= 0, k < 0. (End)
Conjecture 6: a(2^m*n + q) = Sum_{i=A001511(n+1)..A000120(n)+1} A373183(n, i)*a(2^m*(2^(i-1)-1) + q) for n >= 0, m >= 0, q >= 0. Note that this formula is recursive for n != 2^k - 1. Also, it is not related to R. Ehrenborg's and E. Steingrimsson's work. - Mikhail Kurkov, Jun 05 2024
From Mikhail Kurkov, Jul 10 2024: (Start)
a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=1..m+1} a(2^i*k)*(-1)^(m-i+1)*Sum_{j=i..m+1} j^n*Stirling1(j, i)*Stirling2(m+1, j) for m >= 0, n >= 0, k >= 0 with a(0) = 1.
Proof: start with a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n) given above and rewrite it as a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=0..m} binomial(m+1, i) a(2^i*(2^(n-1)*(2k+1) - 1)).
Then conjecture that a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=1..m+1} a(2^i*k)*f(n, m, i). From that it is obvious that f(0, m, i) = [i = (m+1)].
After that use a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=0..m} binomial(m+1, i) Sum_{j=1..i+1} a(2^j*k)*f(n-1, i, j) = Sum_{i=1..m+1} a(2^i*k) Sum_{j=i-1..m} binomial(m+1, j)*f(n-1, j, i). From that it is obvious that f(n, m, i) = Sum_{j=i-1..m} binomial(m+1, j)*f(n-1, j, i).
Finally, all we need is to show that basic conditions and recurrence for f(n, m, i) gives f(n, m, i) = (-1)^(m-i+1)*Sum_{j=i..m+1} j^n*Stirling1(j, i)*Stirling2(m+1, j) (see Max Alekseyev link).
a(2^m*(2k+1)) = a(2^(m-1)*k) + (m+1)*a(2^m*k) + Sum_{i=1..m-1} a(2^m*k + 2^i) for m > 0, k >= 0.
Proof: start with a(2^(m+1)*(2k+1)) = a(2^m*k) + (m+2)*a(2^(m+1)*k) + Sum_{i=1..m} a(2^(m+1)*k + 2^i).
Then use a(2^m*(4k+1)) = a(2^m*k) + (m+1)*a(2^(m+1)*k) + Sum_{i=1..m-1} a(2^(m+1)*k + 2^i).
From that we get a(2^(m+1)*(2k+1)) - a(2^m*k) - (m+2)*a(2^(m+1)*k) - a(2^(m+1)*k + 2^m) = a(2^m*(4k+1)) - a(2^m*k) - (m+1)*a(2^(m+1)*k).
Finally, a(2^(m+1)*(2k+1)) = a(2^(m+1)*k) + a(2^m*(2*k+1)) + a(2^m*(4k+1)) which agrees with the a(2^m*(2n+1)) = a(2^m*n) + a(2^(m-1)*(2n+1)) + a(2^(m-1)*(4n+1)) given above.
This formula can be considered as an alternative to a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n). There are algorithms for both these formulas that allow you to calculate them without recursion. However, even though it is necessary to calculate binomial coefficients in the mentioned formula, the triple-looped algorithm for it still works faster (see Peter J. Taylor link).
It looks like you can also change v2 in the mentioned algorithm to vector with elements a(2^m*(2^(i+A007814(n+1)-1)-1) + q) to get a(2^m*n + q) instead of a(n). This may have common causes with formula that uses A373183 given above. (End)
From Mikhail Kurkov, Jan 27 2025: (Start)
The formulas below are not related to R. Ehrenborg's and E. Steingrimsson's work.
Conjecture 7: A008292(n+1,k+1) = Sum_{i=0..2^n-1} [A000120(i) = k]*a(i) for n >= 0, k >= 0.
Conjecture 8: a(2^m*(2^n*(2k+1)-1)) = Sum_{i=0..m} Sum_{j=0..m-i} Sum_{q=0..i} binomial(m-i,j)*(m-j+1)^n*a(2^(q+1)*k)*L(m,i,q)*(-1)^j for m >= 0, n > 0, k >= 0 where L(n,k,m) = W(n-m,k-m,m+1) for n > 0, 0 <= k < n, 0 <= m <= k and where W(n,k,m) = (k+m)*W(n-1,k,m) + (n-k)*W(n-1,k-1,m) + [m > 1]*W(n,k,m-1) for 0 <= k < n, m > 0 with W(0,0,m) = 1, W(n,k,m) = 0 for n < 0 or k < 0.
In particular, W(n, k, 1) = A173018(n, k), W(n, k, 2) = A062253(n, k), W(n, k, 3) = A062254(n, k) and W(n, k, 4) = A062255(n, k).
Conjecture 9: a(n) = b(n,wt(n)) for n >= 0 where b(2n+1,k) = b(n,k) + (wt(n)-k+2)*b(n,k-1), b(2n,k) = (wt(n)-k+1)*b(2n+1,k) for n > 0, k > 0 with b(n,0) = A341392(n) for n >= 0, b(0,k) = 0 for k > 0 and where wt(n) = A000120(n) (see A379817).
More generally, a(2^m*(2k+1)) = ((m+1)!)^2*b(k,wt(k)-m) - Sum_{j=1..m} Stirling1(m+2,j+1)*a(2^(j-1)*(2k+1)) for m >= 0, k >= 0. Here we also consider that b(n,k) = 0 for k < 0. (End)
Conjecture 10: if we change b(n,0) = A341392(n) given above to b(n,0) = A341392(n)*x^n, then nonzero terms of the resulting polynomials for b(n,wt(n)) form c(n,k) such that a(n) = Sum_{k=0..A080791(n)} c(n,k) for n >= 0 where c(n,k) = (Product_{i=0..k-1} (1 + 1/A000120(floor(n/2^(A000523(n)-i))))) * Sum_{j=max{0,k-A080791(n)+A080791(A053645(n))}..A080791(A053645(n))} c(A053645(n),j) for n > 0, k >= 0 with c(0,0) = 1, c(0,k) = 0 for k > 0. - Mikhail Kurkov, Jun 19 2025

A111528 Square table, read by antidiagonals, where the g.f. for row n+1 is generated by: xR_{n+1}(x) = (1+nx - 1/R_n(x))/(n+1) with R_0(x) = Sum_{n>=0} n!*x^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 1, 4, 13, 24, 1, 1, 5, 22, 71, 120, 1, 1, 6, 33, 148, 461, 720, 1, 1, 7, 46, 261, 1156, 3447, 5040, 1, 1, 8, 61, 416, 2361, 10192, 29093, 40320, 1, 1, 9, 78, 619, 4256, 23805, 99688, 273343, 362880, 1, 1, 10, 97, 876, 7045, 48096, 263313

Offset: 0

Paul D. Hanna, Aug 06 2005

			Table begins:
  1, 1,  2,   6,   24,   120,    720,    5040,     40320, ...
  1, 1,  3,  13,   71,   461,   3447,   29093,    273343, ...
  1, 1,  4,  22,  148,  1156,  10192,   99688,   1069168, ...
  1, 1,  5,  33,  261,  2361,  23805,  263313,   3161781, ...
  1, 1,  6,  46,  416,  4256,  48096,  591536,   7840576, ...
  1, 1,  7,  61,  619,  7045,  87955, 1187845,  17192275, ...
  1, 1,  8,  78,  876, 10956, 149472, 2195208,  34398288, ...
  1, 1,  9,  97, 1193, 16241, 240057, 3804353,  64092553, ...
  1, 1, 10, 118, 1576, 23176, 368560, 6262768, 112784896, ...
Rows are generated by logarithms of factorial series:
log(1 + x + 2*x^2 + 6*x^3 + 24*x^4 + ... n!*x^n + ...) = x + (3/2)*x^2 + (13/3)*x^3 + (71/4)*x^4 + (461/5)*x^5 + ...
(1/2)*log(1 + 2*x + 6*x^2 + ... + ((n+1)!/1!)*x^n + ...) = x + (4/2)*x^2 + (22/3)*x^3 + (148/4)*x^4 + (1156/5)*x^5 + ...
(1/3)*log(1 + 3*x + 12*x^2 + 60*x^3 + ... + ((n+2)!/2!)*x^n + ...) = x + (5/2)*x^2 + (33/3)*x^3 + (261/4)*x^4 + (2361/5)*x^5 +...
G.f. of row n may be expressed by the continued fraction:
R_n(x) = 1/(1+n*x - (n+1)*x/(1+(n+1)*x - (n+2)*x/(1+(n+2)*x -...
or recursively by: R_n(x) = 1/(1+n*x - (n+1)*x*R_{n+1}(x)).

A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf: A003319 (row 1), A111529 (row 2), A111530 (row 3), A111531 (row 4), A111532 (row 5), A111533 (row 6), A111534 (diagonal).

Similar recurrences: A124758, A243499, A284005, A329369, A341392.

T := (n, k) -> coeff(series(hypergeom([n+1, 1], [], x)/hypergeom([n, 1], [], x), x, 21), x, k):
#display as a sequence
seq(seq(T(n-k, k), k = 0..n), n = 0..10);
# display as a square array
seq(print(seq(T(n, k), k = 0..10)), n = 0..10); # Peter Bala, Jul 16 2022

T[n_, k_] := T[n, k] = Which[n < 0 || k < 0, 0, k == 0 || k == 1, 1, n == 0, k!, True, (T[n - 1, k + 1] - T[n - 1, k])/n - Sum[T[n, j]*T[n - 1, k - j], {j, 1, k - 1}]]; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2018 *)

{T(n,k)=if(n<0||k<0,0,if(k==0||k==1,1,if(n==0,k!, (T(n-1,k+1)-T(n-1,k))/n-sum(j=1,k-1,T(n,j)*T(n-1,k-j)))))}
for(n=0,10,for(k=0,10,print1(T(n,k),", ")); print(""))

{T(n,k)=if(n<0||k<0,0,if(k==0,1,if(n==0,k!, k/n*polcoeff(log(sum(m=0,k,(n-1+m)!/(n-1)!*x^m)),k))))}
for(n=0,10,for(k=0,10,print1(T(n,k),", ")); print(""))

T(n, 0) = 1, T(0, k) = k!, otherwise for n>=1 and k>=1:
T(n, k) = (T(n-1, k+1) - T(n-1, k))/n - Sum_{j=1..k-1} T(n, j)*T(n-1, k-j).
T(n, k) = (k/n)*[x^k] log(Sum_{m=0..k} (n-1+m)!/(n-1)!*x^m).
T(n, k) = Sum_{j = 0..k} A089949(k, j)*n^(k-j). - Philippe Deléham, Aug 08 2005
R_n(x) = -((n-1)!/n)/Sum_{i>=1} (i+n-2)!*x^i, n > 0. - Vladeta Jovovic, May 06 2006
G.f. of row R may be expressed by the continued fraction: W(0), where W(k) = 1 - x*(k+1)/( x*(k+1) - 1/(1 - x*(k+1+R)/( x*(k+1+R) - 1/W(k+1) ))). - Sergei N. Gladkovskii, Aug 26 2013
Conjecture: T(n, k) = b(2^(k-1) - 1, n) for k > 0 with T(n, 0) = 1 where b(n, m) = b(floor(n/2), m) + b(floor((2n - 2^A007814(n))/2), m) + m*b(A025480(n-1), m) for n > 0 with b(0, m) = 1. - Mikhail Kurkov, Dec 16 2021
From Peter Bala, Jul 11 2022: (Start)
O.g.f. for row n, n >= 1: R(n,x) = ( Sum_{k >= 0} (n+k)!/n!*x^k )/( Sum_{k >= 0} (n-1+k)!/(n-1)!*x^k ).
R(n,x)/(1 - n*x*R(n,x)) = Sum_{k >= 0} (n+k)!/n!*x^k.
For n >= 0, R(n,x) satisfies the Riccati equation x^2*d/dx(R(n,x)) + n*x*R(n,x)^2 - (1 + (n-1)*x)*R(n,x) + 1 = 0 with R(n,0) = 1.
Apply Stokes 1982 to find that for n >= 0, R(n,x) = 1/(1 - x/(1 - (n+1)*x/(1 - 2*x/(1 - (n+2)*x/(1 - 3*x/(1 - (n+3)*x/(1 - 4*x/(1 - (n+4)*x/(1 - ...))))))))), a continued fraction of Stieltjes type. (End)

A011965 Second differences of Bell numbers.

Original entry on oeis.org

1, 2, 7, 27, 114, 523, 2589, 13744, 77821, 467767, 2972432, 19895813, 139824045, 1028804338, 7905124379, 63287544055, 526827208698, 4551453462543, 40740750631417, 377254241891064, 3608700264369193, 35613444194346451, 362161573323083920, 3790824599495473121

Offset: 0

N. J. A. Sloane

Number of partitions of n+3 with at least one singleton and with the smallest element in a singleton equal to 3. Alternatively, number of partitions of n+3 with at least one singleton and with the largest element in a singleton equal to n+1. - Olivier GERARD, Oct 29 2007

Out of the A005493(n) set partitions with a specific two elements clustered separately, number that have a different set of two elements clustered separately. - Andrey Goder (andy.goder(AT)gmail.com), Dec 17 2007

Olivier Gérard and Karol A. Penson, A budget of set partition statistics, in preparation, unpublished as of Sep 22 2011.

Chai Wah Wu, Table of n, a(n) for n = 0..570(terms 0..250 from Alois P. Heinz).
Martin Cohn, Shimon Even, Karl Menger, Jr. and Philip K. Hooper, On the Number of Partitionings of a Set of n Distinct Objects, Amer. Math. Monthly 69 (1962), no. 8, 782--785. MR1531841.
Martin Cohn, Shimon Even, Karl Menger, Jr. and Philip K. Hooper, On the Number of Partitionings of a Set of n Distinct Objects, Amer. Math. Monthly 69 (1962), no. 8, 782--785. MR1531841. [Annotated scanned copy]
Adam M. Goyt and Lara K. Pudwell, Avoiding colored partitions of two elements in the pattern sense, arXiv preprint arXiv:1203.3786 [math.CO], 2012.
Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.

Cf. A000110, A005493, A106436.

Similar recurrences: A006014, A090365, A124758, A217924, A243499, A284005, A290615, A329369, A341392, A347204, A347205, A350309.

[Bell(n+2) -2*Bell(n+1) + Bell(n): n in [0..40]]; // G. C. Greubel, Jan 07 2025

a:= n-> add((-1)^k*binomial(2,k)*combinat['bell'](n+k), k=0..2): seq(a(n), n=0..20);  # Alois P. Heinz, Sep 05 2008

Differences[BellB[Range[0, 30]], 2] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)

# requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
from itertools import accumulate
A011965_list, blist, b = [1], [1, 2], 2
for _ in range(1000):
    blist = list(accumulate([b]+blist))
    b = blist[-1]
    A011965_list.append(blist[-3])
# Chai Wah Wu, Sep 02 2014

# or Sagemath
b=bell_number
print([b(n+2) -2*b(n+1) +b(n) for n in range(41)]) # G. C. Greubel, Jan 07 2025

a(n) = A005493(n) - A005493(n-1).
E.g.f.: exp(exp(x)-1)*(exp(2*x)-exp(x)+1). - Vladeta Jovovic, Feb 11 2003
a(n) = A000110(n) - 2*A000110(n-1) + A000110(n-2). - Andrey Goder (andy.goder(AT)gmail.com), Dec 17 2007
G.f.: G(0) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k+2*x-1) - x*(2*k+1)*(2*k+3)*(2*x*k+2*x-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+3*x-1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 19 2012
G.f.: 1 - G(0) where G(k) = 1 - 1/(1-k*x-2*x)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 17 2013
G.f.: 1 - 1/x + (1-x)^2/x/(G(0)-x) where G(k) = 1 - x*(k+1)/(1 - x/G(k+1) ); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 26 2013
G.f.: G(0)*(1-1/x) where G(k) = 1 - 1/(1-x*(k+1))/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 07 2013
a(n) ~ n^2 * Bell(n) / LambertW(n)^2 * (1 - 2*LambertW(n)/n). - Vaclav Kotesovec, Jul 28 2021
Conjecture: a(n) = Sum_{k=0..2^n - 1} b(k) for n >= 0 where b(2n+1) = b(n) + b(A025480(n-1)), b(2n) = b(n - 2^f(n)) + b(2n - 2^f(n)) + b(A025480(n-1)) for n > 0 with b(0) = b(1) = 1 and where f(n) = A007814(n). Also b((4^n - 1)/3) = A141154(n+1). - Mikhail Kurkov, Jan 27 2022

A217924 a(n) = n! * [x^n] exp(2*exp(x) - x - 2). Row sums of triangle A217537.

Original entry on oeis.org

1, 1, 3, 9, 35, 153, 755, 4105, 24323, 155513, 1064851, 7760745, 59895203, 487397849, 4166564147, 37298443977, 348667014723, 3395240969785, 34365336725715, 360837080222761, 3923531021460707, 44108832866004121, 511948390801374835, 6126363766802713481

Offset: 0

Peter Luschny, Oct 15 2012

nonn

The inverse binomial transform of a(n) is A194689.
A087981(n) = Sum_{k=0..n} (-1)^k*s(n+1,k+1)*a(k);
|A000023(n)| = |Sum_{k=0..n} (-1)^(n-k)*s(n,k)*a(k)|
where s(n,k) are the unsigned Stirling numbers of first kind.
a(n) is the number of inequivalent set partitions of {1,2,...,n} where two blocks are considered equivalent when one can be obtained from the other by an alternating (even) permutation. - Geoffrey Critzer, Mar 17 2013

			a(3)=9 because we have: {1,2,3}; {1,3,2}; {1}{2,3}; {1}{3,2}; {2}{1,3}; {2}{3,1}; {3}{1,2}; {3}{2,1}; {1}{2}{3}. - _Geoffrey Critzer_, Mar 17 2013

Vaclav Kotesovec, Table of n, a(n) for n = 0..556

Similar recurrences: A124758, A243499, A284005, A329369, A341392, A372205.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!(Laplace( Exp(2*Exp(x) -x-2) ))); // G. C. Greubel, Jan 09 2025

egf := exp(2*exp(x) - x - 2): ser := series(egf, x, 25):
seq(n!*coeff(ser, x, n), n = 0..23);  # Peter Luschny, Apr 22 2024

nn=23;Range[0,nn]!CoefficientList[Series[Exp[2 Exp[x]-x-2],{x,0,nn}],x]  (* Geoffrey Critzer, Mar 17 2013 *)
nmax = 25; CoefficientList[Series[1/(1 - x + ContinuedFractionK[-2*k*x^2 , 1 - (k + 1)*x, {k, 1, nmax}]), {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 25 2017 *)

a(n):=sum(sum(binomial(n,k-j)*2^j*(-1)^(k-j)*stirling2(n-k+j,j),j,0,k),k,0,n); /* Vladimir Kruchinin, Feb 28 2015 */

def A217924_list(n):
    T = A217537_triangle(n)
    return [add(T.row(n)) for n in range(n)]
A217924_list(24)

def A217924_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( exp(2*exp(x)-x-2) ).egf_to_ogf().list()
print(A217924_list(40)) # G. C. Greubel, Jan 09 2025

G.f.: 1/Q(0) where Q(k) = 1 + x*k - x/(1 - 2*x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 06 2013
E.g.f.: exp(2*exp(x) - x - 2). -  Geoffrey Critzer, Mar 17 2013
G.f.: 1/Q(0), where Q(k) = 1 - (k+1)*x - 2*(k+1)*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)/( 2*x^2*(k+1) - (1-x-x*k)*(1-2*x-x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 19 2013
a(n) = Sum_{k=0..n} Sum_{j=0..k} binomial(n,k-j)*2^j*(-1)^(k-j)*Stirling2(n-k+j,j). - Vladimir Kruchinin, Feb 28 2015
a(n) = exp(-2) * Sum_{k>=0} 2^k * (k - 1)^n / k!. - Ilya Gutkovskiy, Jun 27 2020
Conjecture: a(n) = Sum_{k=0..2^n-1} A372205(k). - Mikhail Kurkov, Nov 21 2021 [Rewritten by Peter Luschny, Apr 22 2024]
a(n) ~ 2 * n^(n-1) * exp(n/LambertW(n/2) - n - 2) / (sqrt(1 + LambertW(n/2)) * LambertW(n/2)^(n-1)). - Vaclav Kotesovec, Jun 26 2022

Name extended by a formula of Geoffrey Critzer by Peter Luschny, Apr 22 2024

A284001 a(n) = A005361(A283477(n)).

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 4, 6, 1, 2, 4, 6, 8, 12, 18, 24, 1, 2, 4, 6, 8, 12, 18, 24, 16, 24, 36, 48, 54, 72, 96, 120, 1, 2, 4, 6, 8, 12, 18, 24, 16, 24, 36, 48, 54, 72, 96, 120, 32, 48, 72, 96, 108, 144, 192, 240, 162, 216, 288, 360, 384, 480, 600, 720, 1, 2, 4, 6, 8, 12, 18, 24, 16, 24, 36, 48, 54, 72, 96, 120, 32, 48, 72, 96, 108, 144, 192, 240, 162, 216, 288, 360, 384, 480

Offset: 0

Antti Karttunen, Mar 18 2017

a(n) is the product of elements of the multiset that covers an initial interval of positive integers with multiplicities equal to the parts of the n-th composition in standard order (graded reverse-lexicographic, A066099). This composition is obtained by taking the set of positions of 1's in the reversed binary expansion of n, prepending 0, taking first differences, and reversing again. For example, the 13th composition is (1,2,1) giving the multiset {1,2,2,3} with product 12, so a(13) = 12. - Gus Wiseman, Apr 26 2020

Antti Karttunen, Table of n, a(n) for n = 0..8191
Index entries for sequences related to binary expansion of n

Cf. A005361, A283477, A284002, A284005.
Row products of A095684.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Weighted sum is A029931.
- Necklaces are A065609.
- Sum is A070939.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon words are A275692.
- Distinct parts are counted by A334028.
Cf. A003963, A034691, A053645, A057335, A305936, A318284, A333764, A333940, A333942, A334030.

Table[Times @@ FactorInteger[#][[All, -1]] &[Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e == 1 :> {Times @@ Prime@ Range@ PrimePi@ p, e}] &[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2]]], {n, 0, 93}] (* Michael De Vlieger, Mar 18 2017 *)

A005361(n) = factorback(factor(n)[, 2]); \\ From A005361
A034386(n) = prod(i=1, primepi(n), prime(i));
A108951(n) = { my(f=factor(n)); prod(i=1, #f~, A034386(f[i, 1])^f[i, 2]) };  \\ From A108951
A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ From A019565
A283477(n) = A108951(A019565(n));
A284001(n) = A005361(A283477(n));

(define (A284001 n) (A005361 (A283477 n)))

a(n) = A005361(A283477(n)).
a(n) = A003963(A057335(n)). - Gus Wiseman, Apr 26 2020
a(n) = A284005(A053645(n)) for n > 0 with a(0) = 1. - Mikhail Kurkov, Jun 05 2021 [verification needed]

A347205 a(2n+1) = a(n) for n >= 0, a(2n) = a(n) + a(n - 2^A007814(n)) for n > 0 with a(0) = 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 6, 3, 4, 1, 5, 4, 7, 3, 9, 5, 7, 2, 10, 6, 9, 3, 10, 4, 5, 1, 6, 5, 9, 4, 12, 7, 10, 3, 14, 9, 14, 5, 16, 7, 9, 2, 15, 10, 16, 6, 19, 9, 12, 3, 20, 10, 14, 4, 15, 5, 6, 1, 7, 6, 11, 5, 15, 9, 13, 4, 18, 12, 19, 7, 22, 10, 13

Offset: 0

Mikhail Kurkov, Aug 23 2021

Scatter plot might be called "Cypress forest on a windy day". - Antti Karttunen, Nov 30 2021

Antti Karttunen, Table of n, a(n) for n = 0..16384
J. Abate and W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011) # 11.2.6.
Index entries for sequences with interesting graphs or plots

Cf. A000108, A006318, A007814, A059894, A108524, A115197, A116867.

Similar recurrences: A124758, A243499, A284005, A329369, A341392.

a[0] = 1; a[n_] := a[n] = If[OddQ[n], a[(n - 1)/2], a[n/2] + a[n/2 - 2^IntegerExponent[n/2, 2]]]; Array[a, 100, 0] (* Amiram Eldar, Sep 06 2021 *)

a(n) = if (n==0, 1, if (n%2, a(n\2), a(n/2) + a(n/2 - 2^valuation(n/2, 2)))); \\ Michel Marcus, Sep 09 2021

a(2n+1) = a(n) for n >= 0.
a(2n) = a(n) + a(n - 2^A007814(n)) = a(2*A059894(n)) for n > 0 with a(0) = 1.
Sum_{k=0..2^n - 1} a(k) = A000108(n+1) for n >= 0.
a((4^n - 1)/3) = A000108(n) for n >= 0.
a(2^m*(2^n - 1)) = binomial(n + m, n) for n >= 0, m >= 0.
Generalization:
b(2n+1, p, q) = b(n, p, q) for n >= 0.
b(2n, p, q) = p*b(n, p, q) + q*b(n - 2^A007814(n), p, q) = for n > 0 with b(0, p, q) = 1.
Conjectured formulas: (Start)
Sum_{k=0..2^n - 1} b(k, 2, 1) = A006318(n) for n >= 0.
Sum_{k=0..2^n - 1} b(k, 2, 2) = A115197(n) for n >= 0.
Sum_{k=0..2^n - 1} b(k, 3, 1) = A108524(n+1) for n >= 0.
Sum_{k=0..2^n - 1} b(k, 3, 3) = A116867(n) for n >= 0.
b((4^n - 1)/3, p, q) is generalized Catalan number C(p, q; n). (End)
Conjecture: a(n) = T(n, wt(n)+1), a(2n) = Sum_{k=1..wt(n)+1} T(n, k) where T(2n+1, k) = T(n, k) for 1 <= k <= wt(n)+1, T(2n+1, wt(n)+2) = T(n, wt(n)+1), T(2n, k) = Sum_{i=1..k} T(n, i) for 1 <= k <= wt(n)+1 with T(0, 1) = 1. - Mikhail Kurkov, Dec 13 2024

A090365 Shifts 1 place left under the INVERT transform of the BINOMIAL transform of this sequence.

Original entry on oeis.org

1, 1, 3, 11, 47, 225, 1177, 6625, 39723, 251939, 1681535, 11764185, 86002177, 655305697, 5193232611, 42726002123, 364338045647, 3215471252769, 29331858429241, 276224445794785, 2682395337435723, 26832698102762435, 276221586866499839, 2923468922184615897

Offset: 0

Paul D. Hanna, Nov 26 2003

nonn

The Hankel transform of this sequence is A000178(n+1); example: det([1,1,3; 1,3,11; 3,11,47]) = 12. - Philippe Deléham, Mar 02 2005
a(n) appears to be the number of indecomposable permutations (A003319) of [n+1] that avoid both of the dashed patterns 32-41 and 41-32. - David Callan, Aug 27 2014
This is true: A nonempty permutation avoids 32-41 and 41-32 if and only if all its components do so. So if A(x) denotes the g.f. for indecomposable {32-41,41-32}-avoiders, then F(x):=1/(1-A(x)) is the g.f. for all {32-41,41-32}-avoiders. From A074664, F(x)=1/x(1-1/B(x)) where B(x) is the o.g.f. for the Bell numbers. Solve for A(x). - David Callan, Jul 21 2017
The Hankel transform of this sequence without the a(0)=1 term is also A000178(n+1). - Michael Somos, Oct 02 2024

Alois P. Heinz, Table of n, a(n) for n = 0..574

Cf. A090366, A090367.
Cf. A074664.
Similar recurrences: A006014, A124758, A217924, A243499, A284005, A290615, A329369, A341392, A347204, A347205.

bintr:= proc(p) proc(n) add(p(k) *binomial(n,k), k=0..n) end end:
invtr:= proc(p) local b;
           b:= proc(n) option remember; local i;
                `if`(n<1, 1, add(b(n-i) *p(i-1), i=1..n+1))
               end;
        end:
b:= invtr(bintr(a)):
a:= n-> `if`(n<0, 0, b(n-1)):
seq(a(n), n=0..30);  # Alois P. Heinz, Jun 28 2012

a[n_] := Module[{A, B}, A = 1+x; For[k=1, k <= n, k++, B = (A /. x -> x/(1 - x))/(1-x) + O[x]^n // Normal; A = 1 + x*A*B]; SeriesCoefficient[A, {x, 0, n}]]; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Oct 23 2016, adapted from PARI *)

{a(n)=local(A); if(n<0,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A,x, x/(1-x))/(1-x)+x*O(x^n); A=1+x*A*B);polcoeff(A,n,x))}

G.f.: A(x) satisfies A(x) = 1/(1 - A(x/(1-x))*x/(1-x) ).
a(n) = Sum_{k = 0..n} A085838(n, k). - Philippe Deléham, Jun 04 2004
G.f.: 1/x-1-1/(B(x)-1) where B(x) = g.f. for A000110 the Bell numbers. - Vladeta Jovovic, Aug 08 2004
a(n) = Sum_{k=0..n} A094456(n,k). - Philippe Deléham, Nov 07 2007
G.f.: 1/(1-x/(1-2x/(1-x/(1-3x/(1-x/(1-4x/(1-x/(1-5x/(1-... (continued fraction). - Paul Barry, Feb 25 2010
From Sergei N. Gladkovskii, Jan 06 2012  - May 12 2013: (Start)
Continued fractions:
G.f.: 1 - x/(G(0)+x); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1).
G.f.: 1/x - 1/2 + (x^2-4)/(4*U(0)-2*x^2+8) where U(k) = k*(2*k+3)*x^2 + x - 2 - (2-x+2*k*x)*(2+3*x+2*k*x)*(k+1)*x^2/U(k+1).
G.f.: 1/x+1/(U(0)-1) where U(k) = -x*k + 1 - x - x^2*(k+1)/U(k+1).
G.f.: (1 - U(0))/x - 1 where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1).
G.f.: (1 - U(0))/x where U(k) = 1 - x*(k+1)/(1-x/U(k+1)).
G.f.: 1/x + 1/( G(0)-1) where G(k) = 1 - x/(1 - x*(2*k+1)/(1 - x/(1 - x*(2*k+2)/ G(k+1) ))).
G.f.:1/x + 1/( G(0) - 1 ) where G(k) = 1 - x/(1 - x*(k+1)/G(k+1) ).
G.f.: (1 - Q(0))/x where Q(k) = 1 + x/(x*k - 1 )/Q(k+1).
G.f.: 1/x - 1/x/Q(0), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))).
(End)
Conjecture: a(n) = b(2^(n-1) - 1) for n > 0 with a(0) = 1 where b(n) = b((n - 2^f(n))/2) + b(floor((2n - 2^f(n))/2)) + b(A025480(n-1)) for n > 0 with b(0) = 1 and where f(n) = A007814(n). - Mikhail Kurkov, Jan 11 2022

cp's OEIS Frontend

A341392 a(n) = A284005(n) / (1 + A000120(n))!.

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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].

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A283477 If 2n = 2^e1 + 2^e2 + ... + 2^ek [e1 .. ek distinct], then a(n) = A002110(e1) * A002110(e2) * ... * A002110(ek).

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A329369 Number of permutations of {1,2,...,m} with excedance set constructed by taking m-i (0 < i < m) if b(i-1) = 1 where b(k)b(k-1)...b(1)b(0) (0 <= k < m-1) is the binary expansion of n.

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A111528 Square table, read by antidiagonals, where the g.f. for row n+1 is generated by: x*R_{n+1}(x) = (1+n*x - 1/R_n(x))/(n+1) with R_0(x) = Sum_{n>=0} n!*x^n.

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A011965 Second differences of Bell numbers.

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A111528 Square table, read by antidiagonals, where the g.f. for row n+1 is generated by: xR_{n+1}(x) = (1+nx - 1/R_n(x))/(n+1) with R_0(x) = Sum_{n>=0} n!*x^n.