cp's OEIS Frontend

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

Showing 1-10 of 59 results. Next

A256595 Triangle A074909(n) with 0's as second column.

Original entry on oeis.org

1, 1, 0, 1, 0, 3, 1, 0, 6, 4, 1, 0, 10, 10, 5, 1, 0, 15, 20, 15, 6, 1, 0, 21, 35, 35, 21, 7, 1, 0, 28, 56, 70, 56, 28, 8, 1, 0, 36, 84, 126, 126, 84, 36, 9, 1, 0, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0, 55, 165, 330, 462, 462, 330, 165, 55, 11
Offset: 0

Views

Author

Paul Curtz, Apr 03 2015

Keywords

Comments

For Bernoulli numbers, B(1) excluded.
B(n) is calculated via
B(0) = 1;
B(0) + 0 = 1;
B(0) + 0 + 3*B(2) = 3/2;
B(0) + 0 + 6*B(2) + 4*B(3) = 2;
etc.
The diagonal is A026741(n+1)/A040001(n).
Row sums: 1, 1, 4, 11, 26, 57, ..., essentially Euler numbers A000295. See A130103, A008292 and A173018.
There is an infinitude of Bernoulli number sequences. They are of the form
B(n,q) = 1, q, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, ... .
Chronologically, the first, and the most regular, is, for q=1/2, A164555(n)/A027642(n), from Jacob Bernoulli (1654-1705), published in Ars Conjectandi in 1713 and(?) Seko Kowa (1642-1708) in 1712. See A159688. The second is, for q=-1/2, B(n,-1/2) = A027641(n)/A027642(n), from B(n,1/2) via Pascal's triangle. We could choose Be(n,q) instead of B(n,q) to avoid confusion with Sloane's B(n,p) for A027641(n)/A027642(n) (p=-1), A164555(n)/A027642(n) (p=1), A164558(n)/A027642(n) (p=2), A157809(n)/A027642(n) (p=3), ..., successive binomial transforms of the previous sequence.
This motivates the proposal of the (independent of q) sequence Bernoulli(n+2):
B(n+2) = 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, ... and its inverse binomial transform. See A190339.

Examples

			1,
1, 0,
1, 0,  3,
1, 0,  6, 4,
1, 0, 10, 10,  5,
1, 0, 15, 20, 15, 6,
1, 0, 21, 35, 35, 21, 7,
etc.
		

References

  • Jacob Bernoulli, Ars Conjectandi (1713).

Crossrefs

Programs

A193630 Augmentation of the triangle A074909. See Comments.

Original entry on oeis.org

1, 1, 2, 1, 5, 7, 1, 9, 28, 33, 1, 14, 74, 181, 191, 1, 20, 159, 637, 1333, 1297, 1, 27, 300, 1767, 5906, 11029, 10063, 1, 35, 517, 4190, 20256, 59324, 101351, 87669, 1, 44, 833, 8873, 58339, 244125, 645146, 1024949, 847015, 1, 54, 1274, 17241, 147680
Offset: 0

Views

Author

Clark Kimberling, Aug 01 2011

Keywords

Comments

For an introduction to the unary operation "augmentation" as applied to triangular arrays or sequences of polynomials, see A193091.
Regarding A193630, writing the general term as w(n,k),
w(n,n): A104981
w(n,n-1): A156629

Examples

			First five rows of A193607:
1
1...2
1...5....7
1...9....28...33
1...14...74...181...191
		

Crossrefs

Programs

  • Mathematica
    p[n_, k_] := Binomial[n + 1, k];
    Table[p[n, k], {n, 0, 7}, {k, 0, n}]  (* A074909 *)
    m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}]
    TableForm[m[4]]
    w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1];
    v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]};
    v[n_] := v[n - 1].m[n]
    TableForm[Table[v[n], {n, 0, 6}]] (* A193630 *)
    Flatten[Table[v[n], {n, 0, 8}]]

A210381 Triangle by rows, derived from the beheaded Pascal's triangle, A074909.

Original entry on oeis.org

1, 0, 2, 0, 1, 3, 0, 1, 3, 4, 0, 1, 4, 6, 5, 0, 1, 5, 10, 10, 6, 0, 1, 6, 15, 20, 15, 7, 0, 1, 7, 21, 35, 35, 21, 8, 0, 1, 8, 28, 56, 70, 56, 28, 9, 0, 1, 9, 36, 84, 126, 126, 84, 36, 10, 0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 11
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Mar 20 2012

Keywords

Comments

Row sums of the triangle = 2^n.
Let the triangle = an infinite lower triangular matrix, M. Then M * The Bernoulli numbers, A027641/A027642 as a vector V = [1, -1, 0, 0, 0,...]. M * the Bernoulli sequence variant starting [1, 1/2, 1/6,...] = [1, 1, 1,...]. M * 2^n: [1, 2, 4, 8,...] = A027649. M * 3^n = A255463; while M * [1, 2, 3,...] = A047859, and M * A027649 = A027650.
Row sums of powers of the triangle generate the Poly-Bernoulli number sequences shown in the array of A099594. - Gary W. Adamson, Mar 21 2012
Triangle T(n,k) given by (0, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 25 2012

Examples

			{1},
{0, 2},
{0, 1, 3},
{0, 1, 3, 4},
{0, 1, 4, 6, 5},
{0, 1, 5, 10, 10, 6},
{0, 1, 6, 15, 20, 15, 7},
{0, 1, 7, 21, 35, 35, 21, 8},
{0, 1, 8, 28, 56, 70, 56, 28, 9},
{0, 1, 9, 36, 84, 126, 126, 84, 36, 10},
{0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 11}
...
		

References

  • Konrad Knopp, Elements of the Theory of Functions, Dover, 1952,pp 117-118.

Crossrefs

Programs

  • Mathematica
    t2[n_, m_] = If[m - 1 <= n, Binomial[n, m - 1], 0];
    O2 = Table[Table[If[n == m, t2[n, m] + 1, t2[n, m]], {m, 0, n}], {n, 0, 10}];
    Flatten[O2]

Formula

Partial differences of the beheaded Pascal's triangle A074909 starting from the top, by columns.
G.f.: (1-x)/(1-x-2*y*x+y*x^2+y^2*x^2). - Philippe Deléham, Mar 25 2012
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0) = T(2,1) = 1, T(1,0) = T(2,0) = 0, T(1,1) = 2, T(2,2) = 3 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 25 2012

A212362 Triangle by rows, binomial transform of the beheaded Pascal's triangle A074909.

Original entry on oeis.org

1, 2, 2, 4, 7, 3, 8, 19, 15, 4, 16, 47, 52, 26, 5, 32, 111, 155, 110, 40, 6, 64, 255, 426, 385, 200, 57, 7, 128, 575, 1113, 1211, 805, 329, 77, 8, 256, 1279, 2808, 3556, 2856, 1498, 504, 100, 9, 512, 2815, 6903, 9948, 9324, 5922, 2562, 732, 126, 10
Offset: 0

Views

Author

Gary W. Adamson, Jun 29 2012

Keywords

Comments

Row sums of the triangle inverse = A027641/A027642, the Bernoulli numbers; (1, -1/2, 1/6, 0, -1/30,...)

Examples

			First few rows of the triangle are:
    1;
    2,    2;
    4,    7,    3;
    8,   19,   15,    4
   16,   47,   52,   26,    5;
   32,  111,  155,  110,   40,    6;
   64,  255,  426,  385,  200,   57,   7;
  128,  575, 1113, 1211,  805,  329,  77,   8;
  256, 1279, 2808, 3556, 2856, 1498, 504, 100, 9;
  ...
		

Crossrefs

Cf. A074909, A027641/A027642, A027649 (row sums), A006589 (2nd column), A106515.

Programs

  • Magma
    A074909:= func< n,k | k lt 0 or k gt n select 0 else Binomial(n+1, k) >;
    A212362:= func< n,k | (&+[ Binomial(n,j)*A074909(j, k) : j in [0..n]]) >;
    [A212362(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 05 2021
    
  • Maple
    A212362 := proc(n,k)
            add( binomial(n,i)*A074909(i,k),i=0..n) ;
    end proc: # R. J. Mathar, Aug 03 2015
  • Mathematica
    T[n_, k_]= 2^(n-k)*Binomial[n+1, k] + (2^(n-k) -1)*Binomial[n, k-1];
    Table[T[n, k] , {n,0,12}, {k,0,n}] //Flatten (* G. C. Greubel, Aug 05 2021 *)
  • Sage
    def T(n, k): return 2^(n-k)*binomial(n+1, k) + (2^(n-k) - 1)*binomial(n, k-1)
    flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Aug 05 2021

Formula

Binomial transform of the beheaded Pascal's triangle (A074909) as a matrix. (The beheaded Pascal matrix deletes the rightmost border of 1's.)
From G. C. Greubel, Aug 05 2021: (Start)
T(n, k) = Sum_{j=0..n} binomial(n, j)*binomial(j+1, k) - binomial(n, k-1), with T(n, 0) = 2^n.
T(n, k) = 2^(n-k)*binomial(n+1, k) + (2^(n-k) - 1)*binomial(n, k-1).
Sum_{k=0..n} T(n, k) = A027649(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A106515(n). (End)

Extensions

a(22) corrected by G. C. Greubel, Aug 05 2021

A227396 Triangle A074909(n) with the first column equal to 1 followed by -A000027(n) instead of A000012.

Original entry on oeis.org

1, -1, 2, -2, 3, 3, -3, 4, 6, 4, -4, 5, 10, 10, 5, -5, 6, 15, 20, 15, 6, -6, 7, 21, 35, 35, 21, 7, -7, 8, 28, 56, 70, 56, 28, 8, -8, 9, 36, 84, 126, 126, 84, 36, 9, -9, 10, 45, 120, 210, 252, 210, 120, 45, 10, -10
Offset: 0

Views

Author

Paul Curtz, Sep 20 2013

Keywords

Comments

Triangle leading to A164555(n)/A027642(n).
Starting from B(0)=1, the Bernoulli numbers B(n) with B(1)=1/2 are such that
1*B(0) = 1
-1*B(0) +2*B(1)= 0 --> B(1)=1/2
-2*B(0) +3*B(1) +3*B(2) = 0 --> B(2)=1/6
-3*B(0) +4*B(1) +6*B(2) +4*B(3) = 0 --> B(3)=0
-4*B(0) +5*B(1) +10*B(2) +10*B(3) +5*B(4) = 0 --> B(4)=-1/30 etc.
Row sum of A: A130103(n+1).
Row sum's absolute values of A: A145071(n).

Examples

			a(n) triangle is A:
1
-1 2
-2 3  3
-3 4  6  4
-4 5 10 10  5
-5 6 15 20 15  6
-6 7 21 35 35 21 7  etc.
		

Formula

T(n,k) = A074909(n,k) for n>0 and k>0, T(0,0)=1, T(n,0)=-n for n>0.

A000027 The positive integers. Also called the natural numbers, the whole numbers or the counting numbers, but these terms are ambiguous.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77
Offset: 1

Views

Author

Keywords

Comments

For some authors, the terms "natural numbers" and "counting numbers" include 0, i.e., refer to the nonnegative integers A001477; the term "whole numbers" frequently also designates the whole set of (signed) integers A001057.
a(n) is smallest positive integer which is consistent with sequence being monotonically increasing and satisfying a(a(n)) = n (cf. A007378).
Inverse Euler transform of A000219.
The rectangular array having A000027 as antidiagonals is the dispersion of the complement of the triangular numbers, A000217 (which triangularly form column 1 of this array). The array is also the transpose of A038722. - Clark Kimberling, Apr 05 2003
For nonzero x, define f(n) = floor(nx) - floor(n/x). Then f=A000027 if and only if x=tau or x=-tau. - Clark Kimberling, Jan 09 2005
Numbers of form (2^i)*k for odd k (i.e., n = A006519(n)*A000265(n)); thus n corresponds uniquely to an ordered pair (i,k) where i=A007814, k=A000265 (with A007814(2n)=A001511(n), A007814(2n+1)=0). - Lekraj Beedassy, Apr 22 2006
If the offset were changed to 0, we would have the following pattern: a(n)=binomial(n,0) + binomial(n,1) for the present sequence (number of regions in 1-space defined by n points), A000124 (number of regions in 2-space defined by n straight lines), A000125 (number of regions in 3-space defined by n planes), A000127 (number of regions in 4-space defined by n hyperplanes), A006261, A008859, A008860, A008861, A008862 and A008863, where the last six sequences are interpreted analogously and in each "... by n ..." clause an offset of 0 has been assumed, resulting in a(0)=1 for all of them, which corresponds to the case of not cutting with a hyperplane at all and therefore having one region. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
Define a number of points on a straight line to be in general arrangement when no two points coincide. Then these are the numbers of regions defined by n points in general arrangement on a straight line, when an offset of 0 is assumed. For instance, a(0)=1, since using no point at all leaves one region. The sequence satisfies the recursion a(n) = a(n-1) + 1. This has the following geometrical interpretation: Suppose there are already n-1 points in general arrangement, thus defining the maximal number of regions on a straight line obtainable by n-1 points, and now one more point is added in general arrangement. Then it will coincide with no other point and act as a dividing wall thereby creating one new region in addition to the a(n-1)=(n-1)+1=n regions already there, hence a(n)=a(n-1)+1. Cf. the comments on A000124 for an analogous interpretation. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
The sequence a(n)=n (for n=1,2,3) and a(n)=n+1 (for n=4,5,...) gives to the rank (minimal cardinality of a generating set) for the semigroup I_n\S_n, where I_n and S_n denote the symmetric inverse semigroup and symmetric group on [n]. - James East, May 03 2007
The sequence a(n)=n (for n=1,2), a(n)=n+1 (for n=3) and a(n)=n+2 (for n=4,5,...) gives the rank (minimal cardinality of a generating set) for the semigroup PT_n\T_n, where PT_n and T_n denote the partial transformation semigroup and transformation semigroup on [n]. - James East, May 03 2007
"God made the integers; all else is the work of man." This famous quotation is a translation of "Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk," spoken by Leopold Kronecker in a lecture at the Berliner Naturforscher-Versammlung in 1886. Possibly the first publication of the statement is in Heinrich Weber's "Leopold Kronecker," Jahresberichte D.M.V. 2 (1893) 5-31. - Clark Kimberling, Jul 07 2007
Binomial transform of A019590, inverse binomial transform of A001792. - Philippe Deléham, Oct 24 2007
Writing A000027 as N, perhaps the simplest one-to-one correspondence between N X N and N is this: f(m,n) = ((m+n)^2 - m - 3n + 2)/2. Its inverse is given by I(k)=(g,h), where g = k - J(J-1)/2, h = J + 1 - g, J = floor((1 + sqrt(8k - 7))/2). Thus I(1)=(1,1), I(2)=(1,2), I(3)=(2,1) and so on; the mapping I fills the first-quadrant lattice by successive antidiagonals. - Clark Kimberling, Sep 11 2008
a(n) is also the mean of the first n odd integers. - Ian Kent, Dec 23 2008
Equals INVERTi transform of A001906, the even-indexed Fibonacci numbers starting (1, 3, 8, 21, 55, ...). - Gary W. Adamson, Jun 05 2009
These are also the 2-rough numbers: positive integers that have no prime factors less than 2. - Michael B. Porter, Oct 08 2009
Totally multiplicative sequence with a(p) = p for prime p. Totally multiplicative sequence with a(p) = a(p-1) + 1 for prime p. - Jaroslav Krizek, Oct 18 2009
Triangle T(k,j) of natural numbers, read by rows, with T(k,j) = binomial(k,2) + j = (k^2-k)/2 + j where 1 <= j <= k. In other words, a(n) = n = binomial(k,2) + j where k is the largest integer such that binomial(k,2) < n and j = n - binomial(k,2). For example, T(4,1)=7, T(4,2)=8, T(4,3)=9, and T(4,4)=10. Note that T(n,n)=A000217(n), the n-th triangular number. - Dennis P. Walsh, Nov 19 2009
Hofstadter-Conway-like sequence (see A004001): a(n) = a(a(n-1)) + a(n-a(n-1)) with a(1) = 1, a(2) = 2. - Jaroslav Krizek, Dec 11 2009
a(n) is also the dimension of the irreducible representations of the Lie algebra sl(2). - Leonid Bedratyuk, Jan 04 2010
Floyd's triangle read by rows. - Paul Muljadi, Jan 25 2010
Number of numbers between k and 2k where k is an integer. - Giovanni Teofilatto, Mar 26 2010
Generated from a(2n) = r*a(n), a(2n+1) = a(n) + a(n+1), r = 2; in an infinite set, row 2 of the array shown in A178568. - Gary W. Adamson, May 29 2010
1/n = continued fraction [n]. Let barover[n] = [n,n,n,...] = 1/k. Then k - 1/k = n. Example: [2,2,2,...] = (sqrt(2) - 1) = 1/k, with k = (sqrt(2) + 1). Then 2 = k - 1/k. - Gary W. Adamson, Jul 15 2010
Number of n-digit numbers the binary expansion of which contains one run of 1's. - Vladimir Shevelev, Jul 30 2010
From Clark Kimberling, Jan 29 2011: (Start)
Let T denote the "natural number array A000027":
1 2 4 7 ...
3 5 8 12 ...
6 9 13 18 ...
10 14 19 25 ...
T(n,k) = n+(n+k-2)*(n+k-1)/2. See A185787 for a list of sequences based on T, such as rows, columns, diagonals, and sub-arrays. (End)
The Stern polynomial B(n,x) evaluated at x=2. See A125184. - T. D. Noe, Feb 28 2011
The denominator in the Maclaurin series of log(2), which is 1 - 1/2 + 1/3 - 1/4 + .... - Mohammad K. Azarian, Oct 13 2011
As a function of Bernoulli numbers B_n (cf. A027641: (1, -1/2, 1/6, 0, -1/30, 0, 1/42, ...)): let V = a variant of B_n changing the (-1/2) to (1/2). Then triangle A074909 (the beheaded Pascal's triangle) * [1, 1/2, 1/6, 0, -1/30, ...] = the vector [1, 2, 3, 4, 5, ...]. - Gary W. Adamson, Mar 05 2012
Number of partitions of 2n+1 into exactly two parts. - Wesley Ivan Hurt, Jul 15 2013
Integers n dividing u(n) = 2u(n-1) - u(n-2); u(0)=0, u(1)=1 (Lucas sequence A001477). - Thomas M. Bridge, Nov 03 2013
For this sequence, the generalized continued fraction a(1)+a(1)/(a(2)+a(2)/(a(3)+a(3)/(a(4)+...))), evaluates to 1/(e-2) = A194807. - Stanislav Sykora, Jan 20 2014
Engel expansion of e-1 (A091131 = 1.71828...). - Jaroslav Krizek, Jan 23 2014
a(n) is the number of permutations of length n simultaneously avoiding 213, 231 and 321 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
a(n) is also the number of permutations simultaneously avoiding 213, 231 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014
a(n) = least k such that 2*Pi - Sum_{h=1..k} 1/(h^2 - h + 3/16) < 1/n. - Clark Kimberling, Sep 28 2014
a(n) = least k such that Pi^2/6 - Sum_{h=1..k} 1/h^2 < 1/n. - Clark Kimberling, Oct 02 2014
Determinants of the spiral knots S(2,k,(1)). a(k) = det(S(2,k,(1))). These knots are also the torus knots T(2,k). - Ryan Stees, Dec 15 2014
As a function, the restriction of the identity map on the nonnegative integers {0,1,2,3...}, A001477, to the positive integers {1,2,3,...}. - M. F. Hasler, Jan 18 2015
See also A131685(k) = smallest positive number m such that c(i) = m (i^1 + 1) (i^2 + 2) ... (i^k+ k) / k! takes integral values for all i>=0: For k=1, A131685(k)=1, which implies that this is a well defined integer sequence. - Alexander R. Povolotsky, Apr 24 2015
a(n) is the number of compositions of n+2 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
Does not satisfy Benford's law [Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017
Parametrization for the finite multisubsets of the positive integers, where, for p_j the j-th prime, n = Product_{j} p_j^(e_j) corresponds to the multiset containing e_j copies of j ('Heinz encoding' -- see A056239, A003963, A289506, A289507, A289508, A289509). - Christopher J. Smyth, Jul 31 2017
The arithmetic function v_1(n,1) as defined in A289197. - Robert Price, Aug 22 2017
For n >= 3, a(n)=n is the least area that can be obtained for an irregular octagon drawn in a square of n units side, whose sides are parallel to the axes, with 4 vertices that coincide with the 4 vertices of the square, and the 4 remaining vertices having integer coordinates. See Affaire de Logique link. - Michel Marcus, Apr 28 2018
a(n+1) is the order of rowmotion on a poset defined by a disjoint union of chains of length n. - Nick Mayers, Jun 08 2018
Number of 1's in n-th generation of 1-D Cellular Automata using Rules 50, 58, 114, 122, 178, 186, 206, 220, 238, 242, 250 or 252 in the Wolfram numbering scheme, started with a single 1. - Frank Hollstein, Mar 25 2019
(1, 2, 3, 4, 5, ...) is the fourth INVERT transform of (1, -2, 3, -4, 5, ...). - Gary W. Adamson, Jul 15 2019

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 1.
  • T. M. Apostol, Modular Functions and Dirichlet Series in Number Theory, Springer-Verlag, 1990, page 25.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 22.
  • W. Fulton and J. Harris, Representation theory: a first course, (1991), page 149. [From Leonid Bedratyuk, Jan 04 2010]
  • I. S. Gradstein and I. M. Ryshik, Tables of series, products, and integrals, Volume 1, Verlag Harri Deutsch, 1981.
  • R. E. Schwartz, You Can Count on Monsters: The First 100 numbers and Their Characters, A. K. Peters and MAA, 2010.
  • 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

A001477 = nonnegative numbers.
Partial sums of A000012.
Cf. A026081 = integers in reverse alphabetical order in U.S. English, A107322 = English name for number and its reverse have the same number of letters, A119796 = zero through ten in alphabetical order of English reverse spelling, A005589, etc. Cf. A185787 (includes a list of sequences based on the natural number array A000027).
Cf. Boustrophedon transforms: A000737, A231179;
Cf. A038722 (mirrored when seen as triangle), A056011 (boustrophedon).
Cf. A048993, A048994, A000110 (see the Feb 03 2015 formula).

Programs

Formula

a(2k+1) = A005408(k), k >= 0, a(2k) = A005843(k), k >= 1.
Multiplicative with a(p^e) = p^e. - David W. Wilson, Aug 01 2001
Another g.f.: Sum_{n>0} phi(n)*x^n/(1-x^n) (Apostol).
When seen as an array: T(k, n) = n+1 + (k+n)*(k+n+1)/2. Main diagonal is 2n*(n+1)+1 (A001844), antidiagonal sums are n*(n^2+1)/2 (A006003). - Ralf Stephan, Oct 17 2004
Dirichlet generating function: zeta(s-1). - Franklin T. Adams-Watters, Sep 11 2005
G.f.: x/(1-x)^2. E.g.f.: x*exp(x). a(n)=n. a(-n)=-a(n).
Series reversion of g.f. A(x) is x*C(-x)^2 where C(x) is the g.f. of A000108. - Michael Somos, Sep 04 2006
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = u^2 - v - 4*u*v. - Michael Somos, Oct 03 2006
Convolution of A000012 (the all-ones sequence) with itself. - Tanya Khovanova, Jun 22 2007
a(n) = 2*a(n-1)-a(n-2); a(1)=1, a(2)=2. a(n) = 1+a(n-1). - Philippe Deléham, Nov 03 2008
a(n) = A000720(A000040(n)). - Juri-Stepan Gerasimov, Nov 29 2009
a(n+1) = Sum_{k=0..n} A101950(n,k). - Philippe Deléham, Feb 10 2012
a(n) = Sum_{d | n} phi(d) = Sum_{d | n} A000010(d). - Jaroslav Krizek, Apr 20 2012
G.f.: x * Product_{j>=0} (1+x^(2^j))^2 = x * (1+2*x+x^2) * (1+2*x^2+x^4) * (1+2*x^4+x^8) * ... = x + 2x^2 + 3x^3 + ... . - Gary W. Adamson, Jun 26 2012
a(n) = det(binomial(i+1,j), 1 <= i,j <= n). - Mircea Merca, Apr 06 2013
E.g.f.: x*E(0), where E(k) = 1 + 1/(x - x^3/(x^2 + (k+1)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 03 2013
From Wolfdieter Lang, Oct 09 2013: (Start)
a(n) = Product_{k=1..n-1} 2*sin(Pi*k/n), n > 1.
a(n) = Product_{k=1..n-1} (2*sin(Pi*k/(2*n)))^2, n > 1.
These identities are used in the calculation of products of ratios of lengths of certain lines in a regular n-gon. For the first identity see the Gradstein-Ryshik reference, p. 62, 1.392 1., bringing the first factor there to the left hand side and taking the limit x -> 0 (L'Hôpital). The second line follows from the first one. Thanks to Seppo Mustonen who led me to consider n-gon lengths products. (End)
a(n) = Sum_{j=0..k} (-1)^(j-1)*j*binomial(n,j)*binomial(n-1+k-j,k-j), k>=0. - Mircea Merca, Jan 25 2014
a(n) = A052410(n)^A052409(n). - Reinhard Zumkeller, Apr 06 2014
a(n) = Sum_{k=1..n^2+2*n} 1/(sqrt(k)+sqrt(k+1)). - Pierre CAMI, Apr 25 2014
a(n) = floor(1/sin(1/n)) = floor(cot(1/(n+1))) = ceiling(cot(1/n)). - Clark Kimberling, Oct 08 2014
a(n) = floor(1/(log(n+1)-log(n))). - Thomas Ordowski, Oct 10 2014
a(k) = det(S(2,k,1)). - Ryan Stees, Dec 15 2014
a(n) = 1/(1/(n+1) + 1/(n+1)^2 + 1/(n+1)^3 + ...). - Pierre CAMI, Jan 22 2015
a(n) = Sum_{m=0..n-1} Stirling1(n-1,m)*Bell(m+1), for n >= 1. This corresponds to Bell(m+1) = Sum_{k=0..m} Stirling2(m, k)*(k+1), for m >= 0, from the fact that Stirling2*Stirling1 = identity matrix. See A048993, A048994 and A000110. - Wolfdieter Lang, Feb 03 2015
a(n) = Sum_{k=1..2n-1}(-1)^(k+1)*k*(2n-k). In addition, surprisingly, a(n) = Sum_{k=1..2n-1}(-1)^(k+1)*k^2*(2n-k)^2. - Charlie Marion, Jan 05 2016
G.f.: x/(1-x)^2 = (x * r(x) *r(x^3) * r(x^9) * r(x^27) * ...), where r(x) = (1 + x + x^2)^2 = (1 + 2x + 3x^2 + 2x^3 + x^4). - Gary W. Adamson, Jan 11 2017
a(n) = floor(1/(Pi/2-arctan(n))). - Clark Kimberling, Mar 11 2020
a(n) = Sum_{d|n} mu(n/d)*sigma(d). - Ridouane Oudra, Oct 03 2020
a(n) = Sum_{k=1..n} phi(gcd(n,k))/phi(n/gcd(n,k)). - Richard L. Ollerton, May 09 2021
a(n) = S(n-1, 2), with the Chebyshev S-polynomials A049310. - Wolfdieter Lang, Mar 09 2023
From Peter Bala, Nov 02 2024: (Start)
For positive integer m, a(n) = (1/m)* Sum_{k = 1..2*m*n-1} (-1)^(k+1) * k * (2*m*n - k) = (1/m) * Sum_{k = 1..2*m*n-1} (-1)^(k+1) * k^2 * (2*m*n - k)^2 (the case m = 1 is given above).
a(n) = Sum_{k = 0..3*n} (-1)^(n+k+1) * k * binomial(3*n+k, 2*k). (End)

Extensions

Links edited by Daniel Forgues, Oct 07 2009.

A000007 The characteristic function of {0}: a(n) = 0^n.

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0

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Comments

Changing the offset to 1 gives the arithmetical function a(1) = 1, a(n) = 0 for n > 1, the identity function for Dirichlet multiplication (see Apostol). - N. J. A. Sloane
Changing the offset to 1 makes this the decimal expansion of 1. - N. J. A. Sloane, Nov 13 2014
Hankel transform (see A001906 for definition) of A000007 (powers of 0), A000012 (powers of 1), A000079 (powers of 2), A000244 (powers of 3), A000302 (powers of 4), A000351 (powers of 5), A000400 (powers of 6), A000420 (powers of 7), A001018 (powers of 8), A001019 (powers of 9), A011557 (powers of 10), A001020 (powers of 11), etc. - Philippe Deléham, Jul 07 2005
This is the identity sequence with respect to convolution. - David W. Wilson, Oct 30 2006
a(A000004(n)) = 1; a(A000027(n)) = 0. - Reinhard Zumkeller, Oct 12 2008
The alternating sum of the n-th row of Pascal's triangle gives the characteristic function of 0, a(n) = 0^n. - Daniel Forgues, May 25 2010
The number of maximal self-avoiding walks from the NW to SW corners of a 1 X n grid. - Sean A. Irvine, Nov 19 2010
Historically there has been some disagreement as to whether 0^0 = 1. Graphing x^0 seems to support that conclusion, but graphing 0^x instead suggests that 0^0 = 0. Euler and Knuth have argued in favor of 0^0 = 1. For some calculators, 0^0 triggers an error, while in Mathematica, 0^0 is Indeterminate. - Alonso del Arte, Nov 15 2011
Another consequence of changing the offset to 1 is that then this sequence can be described as the sum of Moebius mu(d) for the divisors d of n. - Alonso del Arte, Nov 28 2011
With the convention 0^0 = 1, 0^n = 0 for n > 0, the sequence a(n) = 0^|n-k|, which equals 1 when n = k and is 0 for n >= 0, has g.f. x^k. A000007 is the case k = 0. - George F. Johnson, Mar 08 2013
A fixed point of the run length transform. - Chai Wah Wu, Oct 21 2016

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 30.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Crossrefs

Characteristic function of {g}: this sequence (g = 0), A063524 (g = 1), A185012 (g = 2), A185013 (g = 3), A185014 (g = 4), A185015 (g = 5), A185016 (g = 6), A185017 (g = 7). - Jason Kimberley, Oct 14 2011
Characteristic function of multiples of g: this sequence (g = 0), A000012 (g = 1), A059841 (g = 2), A079978 (g = 3), A121262 (g = 4), A079998 (g = 5), A079979 (g = 6), A082784 (g = 7). - Jason Kimberley, Oct 14 2011

Programs

  • Haskell
    a000007 = (0 ^)
    a000007_list = 1 : repeat 0
    -- Reinhard Zumkeller, May 07 2012, Mar 27 2012
    
  • Magma
    [1] cat [0:n in [1..100]]; // Sergei Haller, Dec 21 2006
    
  • Maple
    A000007 := proc(n) if n = 0 then 1 else 0 fi end: seq(A000007(n), n=0..20);
    spec := [A, {A=Z} ]: seq(combstruct[count](spec, size=n+1), n=0..20);
  • Mathematica
    Table[If[n == 0, 1, 0], {n, 0, 99}]
    Table[Boole[n == 0], {n, 0, 99}] (* Michael Somos, Aug 25 2012 *)
    Join[{1},LinearRecurrence[{1},{0},102]] (* Ray Chandler, Jul 30 2015 *)
    PadRight[{1},120,0] (* Harvey P. Dale, Jul 18 2024 *)
  • PARI
    {a(n) = !n};
    
  • Python
    def A000007(n): return int(n==0) # Chai Wah Wu, Feb 04 2022

Formula

Multiplicative with a(p^e) = 0. - David W. Wilson, Sep 01 2001
a(n) = floor(1/(n + 1)). - Franz Vrabec, Aug 24 2005
As a function of Bernoulli numbers (cf. A027641: (1, -1/2, 1/6, 0, -1/30, ...)), triangle A074909 (the beheaded Pascal's triangle) * B_n as a vector = [1, 0, 0, 0, 0, ...]. - Gary W. Adamson, Mar 05 2012
a(n) = Sum_{k = 0..n} exp(2*Pi*i*k/(n+1)) is the sum of the (n+1)th roots of unity. - Franz Vrabec, Nov 09 2012
a(n) = (1-(-1)^(2^n))/2. - Luce ETIENNE, May 05 2015
a(n) = 1 - A057427(n). - Alois P. Heinz, Jan 20 2016
From Ilya Gutkovskiy, Sep 02 2016: (Start)
Binomial transform of A033999.
Inverse binomial transform of A000012. (End)

A000670 Fubini numbers: number of preferential arrangements of n labeled elements; or number of weak orders on n labeled elements; or number of ordered partitions of [n].

Original entry on oeis.org

1, 1, 3, 13, 75, 541, 4683, 47293, 545835, 7087261, 102247563, 1622632573, 28091567595, 526858348381, 10641342970443, 230283190977853, 5315654681981355, 130370767029135901, 3385534663256845323, 92801587319328411133, 2677687796244384203115, 81124824998504073881821
Offset: 0

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Comments

Number of ways n competitors can rank in a competition, allowing for the possibility of ties.
Also number of asymmetric generalized weak orders on n points.
Also called the ordered Bell numbers.
A weak order is a relation that is transitive and complete.
Called Fubini numbers by Comtet: counts formulas in Fubini theorem when switching the order of summation in multiple sums. - Olivier Gérard, Sep 30 2002 [Named after the Italian mathematician Guido Fubini (1879-1943). - Amiram Eldar, Jun 17 2021]
If the points are unlabeled then the answer is a(0) = 1, a(n) = 2^(n-1) (cf. A011782).
For n>0, a(n) is the number of elements in the Coxeter complex of type A_{n-1}. The corresponding sequence for type B is A080253 and there one can find a worked example as well as a geometric interpretation. - Tim Honeywill and Paul Boddington, Feb 10 2003
Also number of labeled (1+2)-free posets. - Detlef Pauly, May 25 2003
Also the number of chains of subsets starting with the empty set and ending with a set of n distinct objects. - Andrew Niedermaier, Feb 20 2004
From Michael Somos, Mar 04 2004: (Start)
Stirling transform of A007680(n) = [3,10,42,216,...] gives [3,13,75,541,...].
Stirling transform of a(n) = [1,3,13,75,...] is A083355(n) = [1,4,23,175,...].
Stirling transform of A000142(n) = [1,2,6,24,120,...] is a(n) = [1,3,13,75,...].
Stirling transform of A005359(n-1) = [1,0,2,0,24,0,...] is a(n-1) = [1,1,3,13,75,...].
Stirling transform of A005212(n-1) = [0,1,0,6,0,120,0,...] is a(n-1) = [0,1,3,13,75,...].
(End)
Unreduced denominators in convergent to log(2) = lim_{n->infinity} n*a(n-1)/a(n).
a(n) is congruent to a(n+(p-1)p^(h-1)) (mod p^h) for n >= h (see Barsky).
Stirling-Bernoulli transform of 1/(1-x^2). - Paul Barry, Apr 20 2005
This is the sequence of moments of the probability distribution of the number of tails before the first head in a sequence of fair coin tosses. The sequence of cumulants of the same probability distribution is A000629. That sequence is twice the result of deletion of the first term of this sequence. - Michael Hardy (hardy(AT)math.umn.edu), May 01 2005
With p(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j,i) = the j-th part of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i=1..p(n)} (n!/(Product_{j=1..p(i)} p(i,j)!)) * (p(i)!/(Product_{j=1..d(i)} m(i,j)!)). - Thomas Wieder, May 18 2005
The number of chains among subsets of [n]. The summed term in the new formula is the number of such chains of length k. - Micha Hofri (hofri(AT)wpi.edu), Jul 01 2006
Occurs also as first column of a matrix-inversion occurring in a sum-of-like-powers problem. Consider the problem for any fixed natural number m>2 of finding solutions to the equation Sum_{k=1..n} k^m = (k+1)^m. Erdős conjectured that there are no solutions for n, m > 2. Let D be the matrix of differences of D[m,n] := Sum_{k=1..n} k^m - (k+1)^m. Then the generating functions for the rows of this matrix D constitute a set of polynomials in n (for varying n along columns) and the m-th polynomial defining the m-th row. Let GF_D be the matrix of the coefficients of this set of polynomials. Then the present sequence is the (unsigned) first column of GF_D^-1. - Gottfried Helms, Apr 01 2007
Assuming A = log(2), D is d/dx and f(x) = x/(exp(x)-1), we have a(n) = (n!/2*A^(n+1)) Sum_{k=0..n} (A^k/k!) D^n f(-A) which gives Wilf's asymptotic value when n tends to infinity. Equivalently, D^n f(-a) = 2*( A*a(n) - 2*a(n-1) ). - Martin Kochanski (mjk(AT)cardbox.com), May 10 2007
List partition transform (see A133314) of (1,-1,-1,-1,...). - Tom Copeland, Oct 24 2007
First column of A154921. - Mats Granvik, Jan 17 2009
A slightly more transparent interpretation of a(n) is as the number of 'factor sequences' of N for the case in which N is a product of n distinct primes. A factor sequence of N of length k is of the form 1 = x(1), x(2), ..., x(k) = N, where {x(i)} is an increasing sequence such that x(i) divides x(i+1), i=1,2,...,k-1. For example, N=70 has the 13 factor sequences {1,70}, {1,2,70}, {1,5,70}, {1,7,70}, {1,10,70}, {1,14,70}, {1,35,70}, {1,2,10,70}, {1,2,14,70}, {1,5,10,70}, {1,5,35,70}, {1,7,14,70}, {1,7,35,70}. - Martin Griffiths, Mar 25 2009
Starting (1, 3, 13, 75, ...) = row sums of triangle A163204. - Gary W. Adamson, Jul 23 2009
Equals double inverse binomial transform of A007047: (1, 3, 11, 51, ...). - Gary W. Adamson, Aug 04 2009
If f(x) = Sum_{n>=0} c(n)*x^n converges for every x, then Sum_{n>=0} f(n*x)/2^(n+1) = Sum_{n>=0} c(n)*a(n)*x^n. Example: Sum_{n>=0} exp(n*x)/2^(n+1) = Sum_{n>=0} a(n)*x^n/n! = 1/(2-exp(x)) = e.g.f. - Miklos Kristof, Nov 02 2009
Hankel transform is A091804. - Paul Barry, Mar 30 2010
It appears that the prime numbers greater than 3 in this sequence (13, 541, 47293, ...) are of the form 4n+1. - Paul Muljadi, Jan 28 2011
The Fi1 and Fi2 triangle sums of A028246 are given by the terms of this sequence. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
The modified generating function A(x) = 1/(2-exp(x))-1 = x + 3*x^2/2! + 13*x^3/3! + ... satisfies the autonomous differential equation A' = 1 + 3*A + 2*A^2 with initial condition A(0) = 0. Applying [Bergeron et al., Theorem 1] leads to two combinatorial interpretations for this sequence: (A) a(n) gives the number of plane-increasing 0-1-2 trees on n vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 2 colors. (B) a(n) gives the number of non-plane-increasing 0-1-2 trees on n vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 4 colors. Examples are given below. - Peter Bala, Aug 31 2011
Starting with offset 1 = the eigensequence of A074909 (the beheaded Pascal's triangle), and row sums of triangle A208744. - Gary W. Adamson, Mar 05 2012
a(n) = number of words of length n on the alphabet of positive integers for which the letters appearing in the word form an initial segment of the positive integers. Example: a(2) = 3 counts 11, 12, 21. The map "record position of block containing i, 1<=i<=n" is a bijection from lists of sets on [n] to these words. (The lists of sets on [2] are 12, 1/2, 2/1.) - David Callan, Jun 24 2013
This sequence was the subject of one of the earliest uses of the database. Don Knuth, who had a computer printout of the database prior to the publication of the 1973 Handbook, wrote to N. J. A. Sloane on May 18, 1970, saying: "I have just had my first real 'success' using your index of sequences, finding a sequence treated by Cayley that turns out to be identical to another (a priori quite different) sequence that came up in connection with computer sorting." A000670 is discussed in Exercise 3 of Section 5.3.1 of The Art of Computer Programming, Vol. 3, 1973. - N. J. A. Sloane, Aug 21 2014
Ramanujan gives a method of finding a continued fraction of the solution x of an equation 1 = x + a2*x^2 + ... and uses log(2) as the solution of 1 = x + x^2/2 + x^3/6 + ... as an example giving the sequence of simplified convergents as 0/1, 1/1, 2/3, 9/13, 52/75, 375/541, ... of which the sequence of denominators is this sequence, while A052882 is the numerators. - Michael Somos, Jun 19 2015
For n>=1, a(n) is the number of Dyck paths (A000108) with (i) n+1 peaks (UD's), (ii) no UUDD's, and (iii) at least one valley vertex at every nonnegative height less than the height of the path. For example, a(2)=3 counts UDUDUD (of height 1 with 2 valley vertices at height 0), UDUUDUDD, UUDUDDUD. These paths correspond, under the "glove" or "accordion" bijection, to the ordered trees counted by Cayley in the 1859 reference, after a harmless pruning of the "long branches to a leaf" in Cayley's trees. (Cayley left the reader to infer the trees he was talking about from examples for small n and perhaps from his proof.) - David Callan, Jun 23 2015
From David L. Harden, Apr 09 2017: (Start)
Fix a set X and define two distance functions d,D on X to be metrically equivalent when d(x_1,y_1) <= d(x_2,y_2) iff D(x_1,y_1) <= D(x_2,y_2) for all x_1, y_1, x_2, y_2 in X.
Now suppose that we fix a function f from unordered pairs of distinct elements of X to {1,...,n}. Then choose positive real numbers d_1 <= ... <= d_n such that d(x,y) = d_{f(x,y)}; the set of all possible choices of the d_i's makes this an n-parameter family of distance functions on X. (The simplest example of such a family occurs when n is a triangular number: When that happens, write n = (k 2). Then the set of all distance functions on X, when |X| = k, is such a family.) The number of such distance functions, up to metric equivalence, is a(n).
It is easy to see that an equivalence class of distance functions gives rise to a well-defined weak order on {d_1, ..., d_n}. To see that any weak order is realizable, choose distances from the set of integers {n-1, ..., 2n-2} so that the triangle inequality is automatically satisfied. (End)
a(n) is the number of rooted labeled forests on n nodes that avoid the patterns 213, 312, and 321. - Kassie Archer, Aug 30 2018
From A.H.M. Smeets, Nov 17 2018: (Start)
Also the number of semantic different assignments to n variables (x_1, ..., x_n) including simultaneous assignments. From the example given by Joerg Arndt (Mar 18 2014), this is easily seen by replacing
"{i}" by "x_i := expression_i(x_1, ..., x_n)",
"{i, j}" by "x_i, x_j := expression_i(x_1, .., x_n), expression_j(x_1, ..., x_n)", i.e., simultaneous assignment to two different variables (i <> j),
similar for simultaneous assignments to more variables, and
"<" by ";", i.e., the sequential constructor. These examples are directly related to "Number of ways n competitors can rank in a competition, allowing for the possibility of ties." in the first comment.
From this also the number of different mean definitions as obtained by iteration of n different mean functions on n initial values. Examples:
the AGM(x1,x2) = AGM(x2,x1) is represented by {arithmetic mean, geometric mean}, i.e., simultaneous assignment in any iteration step;
Archimedes's scheme (for Pi) is represented by {geometric mean} < {harmonic mean}, i.e., sequential assignment in any iteration step;
the geometric mean of two values can also be observed by {arithmetic mean, harmonic mean};
the AGHM (as defined in A319215) is represented by {arithmetic mean, geometric mean, harmonic mean}, i.e., simultaneous assignment, but there are 12 other semantic different ways to assign the values in an AGHM scheme.
By applying power means (also called Holder means) this can be extended to any value of n. (End)
Total number of faces of all dimensions in the permutohedron of order n. For example, the permutohedron of order 3 (a hexagon) has 6 vertices + 6 edges + 1 2-face = 13 faces, and the permutohedron of order 4 (a truncated octahedron) has 24 vertices + 36 edges + 14 2-faces + 1 3-face = 75 faces. A001003 is the analogous sequence for the associahedron. - Noam Zeilberger, Dec 08 2019
Number of odd multinomial coefficients N!/(a_1!*a_2!*...*a_k!). Here each a_i is positive, and Sum_{i} a_i = N (so 2^{N-1} multinomial coefficients in all), where N is any positive integer whose binary expansion has n 1's. - Richard Stanley, Apr 05 2022 (edited Oct 19 2022)
From Peter Bala, Jul 08 2022: (Start)
Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 16 we obtain the sequence [1, 1, 3, 13, 11, 13, 11, 13, 11, 13, ...], with an apparent period of 2 beginning at a(4). Cf. A354242.
More generally, we conjecture that the same property holds for integer sequences having an e.g.f. of the form G(exp(x) - 1), where G(x) is an integral power series. (End)
a(n) is the number of ways to form a permutation of [n] and then choose a subset of its descent set. - Geoffrey Critzer, Apr 29 2023
This is the Akiyama-Tanigawa transform of A000079, the powers of two. - Shel Kaphan, May 02 2024

Examples

			Let the points be labeled 1,2,3,...
a(2) = 3: 1<2, 2<1, 1=2.
a(3) = 13 from the 13 arrangements: 1<2<3, 1<3<2, 2<1<3, 2<3<1, 3<1<2, 3<2<1, 1=2<3 1=3<2, 2=3<1, 1<2=3, 2<1=3, 3<1=2, 1=2=3.
Three competitors can finish in 13 ways: 1,2,3; 1,3,2; 2,1,3; 2,3,1; 3,1,2; 3,2,1; 1,1,3; 2,2,1; 1,3,1; 2,1,2; 3,1,1; 1,2,2; 1,1,1.
a(3) = 13. The 13 plane increasing 0-1-2 trees on 3 vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 2 colors, are:
........................................................
........1 (x3 colors).....1(x2 colors)....1(x2 colors)..
........|................/.\............./.\............
........2 (x3 colors)...2...3...........3...2...........
........|...............................................
........3...............................................
......====..............====............====............
.Totals 9......+..........2....+..........2....=..13....
........................................................
a(4) = 75. The 75 non-plane increasing 0-1-2 trees on 4 vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 4 colors, are:
...............................................................
.....1 (x3).....1(x4).......1(x4).....1(x4)........1(x3).......
.....|........./.\........./.\......./.\...........|...........
.....2 (x3)...2...3.(x3)..3...2(x3).4...2(x3)......2(x4).......
.....|.............\...........\.........\......../.\..........
.....3.(x3).........4...........4.........3......3...4.........
.....|.........................................................
.....4.........................................................
....====......=====........====......====.........====.........
Tots 27....+....12......+...12....+...12.......+...12...=...75.
From _Joerg Arndt_, Mar 18 2014: (Start)
The a(3) = 13 strings on the alphabet {1,2,3} containing all letters up to the maximal value appearing and the corresponding ordered set partitions are:
01:  [ 1 1 1 ]     { 1, 2, 3 }
02:  [ 1 1 2 ]     { 1, 2 } < { 3 }
03:  [ 1 2 1 ]     { 1, 3 } < { 2 }
04:  [ 2 1 1 ]     { 2, 3 } < { 1 }
05:  [ 1 2 2 ]     { 1 } < { 2, 3 }
06:  [ 2 1 2 ]     { 2 } < { 1, 3 }
07:  [ 2 2 1 ]     { 3 } < { 1, 2 }
08:  [ 1 2 3 ]     { 1 } < { 2 } < { 3 }
09:  [ 1 3 2 ]     { 1 } < { 3 } < { 2 }
00:  [ 2 1 3 ]     { 2 } < { 1 } < { 3 }
11:  [ 2 3 1 ]     { 3 } < { 1 } < { 2 }
12:  [ 3 1 2 ]     { 2 } < { 3 } < { 1 }
13:  [ 3 2 1 ]     { 3 } < { 2 } < { 1 }
(End)
		

References

  • Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.
  • Norman Biggs, E. Keith Lloyd and Robin J. Wilson, Graph Theory 1736-1936, Oxford, 1976, p. 44 (P(x)).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 183 (see R_n).
  • Kenneth S. Brown, Buildings, Springer-Verlag, 1988.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 228.
  • Jean-Marie De Koninck, Ces nombres qui nous fascinent, Entry 13, pp 4, Ellipses, Paris 2008.
  • P. J. Freyd, On the size of Heyting semi-lattices, preprint, 2002.
  • Ian P. Goulden and David M. Jackson, Combinatorial Enumeration, John Wiley and Sons, N.Y., 1983.
  • Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd Ed., 1994, exercise 7.44 (pp. 378, 571).
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • Donald E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, 1973, Section 5.3.1, Problem 3.
  • M. Muresan, Generalized Fubini numbers, Stud. Cerc. Mat., Vol. 37, No. 1 (1985), pp. 70-76.
  • Paul Peart, Hankel determinants via Stieltjes matrices. Proceedings of the Thirty-first Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 2000). Congr. Numer. 144 (2000), 153-159.
  • S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay 1957 Vol. 1, see page 19.
  • Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nuernberg, Jul 27 1994.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Richard P. Stanley, Enumerative Combinatorics, Wadsworth, Vol. 1, 1986; see Example 3.15.10, p. 146.
  • Jack van der Elsen, Black and White Transformations, Shaker Publishing, Maastricht, 2005, p. 18.

Crossrefs

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

Programs

  • Haskell
    a000670 n = a000670_list !! n
    a000670_list = 1 : f [1] (map tail $ tail a007318_tabl) where
       f xs (bs:bss) = y : f (y : xs) bss where y = sum $ zipWith (*) xs bs
    -- Reinhard Zumkeller, Jul 26 2014
    
  • Magma
    R:=PowerSeriesRing(Rationals(), 40);
    Coefficients(R!(Laplace( 1/(2-Exp(x)) ))); // G. C. Greubel, Jun 11 2024
  • Maple
    A000670 := proc(n) option remember; local k; if n <=1 then 1 else add(binomial(n,k)*A000670(n-k),k=1..n); fi; end;
    with(combstruct); SeqSetL := [S, {S=Sequence(U), U=Set(Z,card >= 1)},labeled]; seq(count(SeqSetL,size=j),j=1..12);
    with(combinat): a:=n->add(add((-1)^(k-i)*binomial(k, i)*i^n, i=0..n), k=0..n): seq(a(n), n=0..18); # Zerinvary Lajos, Jun 03 2007
    a := n -> add(combinat:-eulerian1(n,k)*2^k,k=0..n): # Peter Luschny, Jan 02 2015
    a := n -> (polylog(-n, 1/2)+`if`(n=0,1,0))/2: seq(round(evalf(a(n),32)), n=0..20); # Peter Luschny, Nov 03 2015
    # next Maple program:
    b:= proc(n, k) option remember;
         `if`(n=0, k!, k*b(n-1, k)+b(n-1, k+1))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..20);  # Alois P. Heinz, Aug 04 2021
  • Mathematica
    Table[(PolyLog[-z, 1/2] + KroneckerDelta[z])/2, {z, 0, 20}] (* Wouter Meeussen *)
    a[0] = 1; a[n_]:= a[n]= Sum[Binomial[n, k]*a[n-k], {k, 1, n}]; Table[a[n], {n, 0, 30}] (* Roger L. Bagula and Gary W. Adamson, Sep 13 2008 *)
    t = 30; Range[0, t]! CoefficientList[Series[1/(2 - Exp[x]), {x, 0, t}], x] (* Vincenzo Librandi, Mar 16 2014 *)
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1 / (2 - Exp@x), {x, 0, n}]]; (* Michael Somos, Jun 19 2015 *)
    Table[Sum[k^n/2^(k+1),{k,0,Infinity}],{n,0,20}] (* Vaclav Kotesovec, Jun 26 2015 *)
    Table[HurwitzLerchPhi[1/2, -n, 0]/2, {n, 0, 20}] (* Jean-François Alcover, Jan 31 2016 *)
    Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*((i+r)^(n-r)/(i!*(k-i-r)!)), {i, 0, k-r}], {k, r, n}]; Fubini[0, 1] = 1; Table[Fubini[n, 1], {n, 0, 20}] (* Jean-François Alcover, Mar 31 2016 *)
    Eulerian1[0, 0] = 1; Eulerian1[n_, k_] := Sum[(-1)^j (k-j+1)^n Binomial[n+1, j], {j, 0, k+1}]; Table[Sum[Eulerian1[n, k] 2^k, {k, 0, n}], {n, 0, 20}] (* Jean-François Alcover, Jul 13 2019, after Peter Luschny *)
    Prepend[Table[-(-1)^k HurwitzLerchPhi[2, -k, 0]/2, {k, 1, 50}], 1] (* Federico Provvedi,Sep 05 2020 *)
    Table[Sum[k!*StirlingS2[n,k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 22 2020 *)
  • Maxima
    makelist(sum(stirling2(n,k)*k!,k,0,n),n,0,12); /* Emanuele Munarini, Jul 07 2011 */
    
  • Maxima
    a[0]:1$ a[n]:=sum(binomial(n,k)*a[n-k],k,1,n)$ A000670(n):=a[n]$ makelist(A000670(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( subst( 1 / (1 - y), y, exp(x + x*O(x^n)) - 1), n))}; /* Michael Somos, Mar 04 2004 */
    
  • PARI
    Vec(serlaplace(1/(2-exp('x+O('x^66))))) /* Joerg Arndt, Jul 10 2011 */
    
  • PARI
    {a(n)=polcoeff(sum(m=0,n,m!*x^m/prod(k=1,m,1-k*x+x*O(x^n))),n)} /* Paul D. Hanna, Jul 20 2011 */
    
  • PARI
    {a(n) = if( n<1, n==0, sum(k=1, n, binomial(n, k) * a(n-k)))}; /* Michael Somos, Jul 16 2017 */
    
  • Python
    from math import factorial
    from sympy.functions.combinatorial.numbers import stirling
    def A000670(n): return sum(factorial(k)*stirling(n,k) for k in range(n+1)) # Chai Wah Wu, Nov 08 2022
    
  • Sage
    @CachedFunction
    def A000670(n) : return 1 if n == 0 else add(A000670(k)*binomial(n,k) for k in range(n))
    [A000670(n) for n in (0..20)] # Peter Luschny, Jul 14 2012
    

Formula

a(n) = Sum_{k=0..n} k! * StirlingS2(n,k) (whereas the Bell numbers A000110(n) = Sum_{k=0..n} StirlingS2(n,k)).
E.g.f.: 1/(2-exp(x)).
a(n) = Sum_{k=1..n} binomial(n, k)*a(n-k), a(0) = 1.
The e.g.f. y(x) satisfies y' = 2*y^2 - y.
a(n) = A052856(n) - 1, if n>0.
a(n) = A052882(n)/n, if n>0.
a(n) = A076726(n)/2.
a(n) is asymptotic to (1/2)*n!*log_2(e)^(n+1), where log_2(e) = 1.442695... [Barthelemy80, Wilf90].
For n >= 1, a(n) = (n!/2) * Sum_{k=-infinity..infinity} of (log(2) + 2 Pi i k)^(-n-1). - Dean Hickerson
a(n) = ((x*d/dx)^n)(1/(2-x)) evaluated at x=1. - Karol A. Penson, Sep 24 2001
For n>=1, a(n) = Sum_{k>=1} (k-1)^n/2^k = A000629(n)/2. - Benoit Cloitre, Sep 08 2002
Value of the n-th Eulerian polynomial (cf. A008292) at x=2. - Vladeta Jovovic, Sep 26 2003
First Eulerian transform of the powers of 2 [A000079]. See A000142 for definition of FET. - Ross La Haye, Feb 14 2005
a(n) = Sum_{k=0..n} (-1)^k*k!*Stirling2(n+1, k+1)*(1+(-1)^k)/2. - Paul Barry, Apr 20 2005
a(n) + a(n+1) = 2*A005649(n). - Philippe Deléham, May 16 2005 - Thomas Wieder, May 18 2005
Equals inverse binomial transform of A000629. - Gary W. Adamson, May 30 2005
a(n) = Sum_{k=0..n} k!*( Stirling2(n+2, k+2) - Stirling2(n+1, k+2) ). - Micha Hofri (hofri(AT)wpi.edu), Jul 01 2006
Recurrence: 2*a(n) = (a+1)^n where superscripts are converted to subscripts after binomial expansion - reminiscent of Bernoulli numbers' B_n = (B+1)^n. - Martin Kochanski (mjk(AT)cardbox.com), May 10 2007
a(n) = (-1)^n * n! * Laguerre(n,P((.),2)), umbrally, where P(j,t) are the polynomials in A131758. - Tom Copeland, Sep 27 2007
Formula in terms of the hypergeometric function, in Maple notation: a(n) = hypergeom([2,2...2],[1,1...1],1/2)/4, n=1,2..., where in the hypergeometric function there are n upper parameters all equal to 2 and n-1 lower parameters all equal to 1 and the argument is equal to 1/2. Example: a(4) = evalf(hypergeom([2,2,2,2],[1,1,1],1/2)/4) = 75. - Karol A. Penson, Oct 04 2007
a(n) = Sum_{k=0..n} A131689(n,k). - Philippe Deléham, Nov 03 2008
From Peter Bala, Jul 01 2009: (Start)
Analogy with the Bernoulli numbers.
We enlarge upon the above comment of M. Kochanski.
The Bernoulli polynomials B_n(x), n = 0,1,..., are given by the formula
(1)... B_n(x) := Sum_{k=0..n} binomial(n,k)*B(k)*x^(n-k),
where B(n) denotes the sequence of Bernoulli numbers B(0) = 1,
B(1) = -1/2, B(2) = 1/6, B(3) = 0, ....
By analogy, we associate with the present sequence an Appell sequence of polynomials {P_n(x)} n >= 0 defined by
(2)... P_n(x) := Sum_{k=0..n} binomial(n,k)*a(k)*x^(n-k).
These polynomials have similar properties to the Bernoulli polynomials.
The first few values are P_0(x) = 1, P_1(x) = x + 1,
P_2(x) = x^2 + 2*x + 3, P_3(x) = x^3 + 3*x^2 + 9*x + 13 and
P_4(x) = x^4 + 4*x^3 + 18*x^2 + 52*x + 75. See A154921 for the triangle of coefficients of these polynomials.
The e.g.f. for this polynomial sequence is
(3)... exp(x*t)/(2 - exp(t)) = 1 + (x + 1)*t + (x^2 + 2*x + 3)*t^2/2! + ....
The polynomials satisfy the difference equation
(4)... 2*P_n(x - 1) - P_n(x) = (x - 1)^n,
and so may be used to evaluate the weighted sums of powers of integers
(1/2)*1^m + (1/2)^2*2^m + (1/2)^3*3^m + ... + (1/2)^(n-1)*(n-1)^m
via the formula
(5)... Sum_{k=1..n-1} (1/2)^k*k^m = 2*P_m(0) - (1/2)^(n-1)*P_m(n),
analogous to the evaluation of the sums 1^m + 2^m + ... + (n-1)^m in terms of Bernoulli polynomials.
This last result can be generalized to
(6)... Sum_{k=1..n-1} (1/2)^k*(k+x)^m = 2*P_m(x)-(1/2)^(n-1)*P_m(x+n).
For more properties of the polynomials P_n(x), refer to A154921.
For further information on weighted sums of powers of integers and the associated polynomial sequences, see A162312.
The present sequence also occurs in the evaluation of another sum of powers of integers. Define
(7)... S_m(n) := Sum_{k=1..n-1} (1/2)^k*((n-k)*k)^m, m = 1,2,....
Then
(8)... S_m(n) = (-1)^m *[2*Q_m(-n) - (1/2)^(n-1)*Q_m(n)],
where Q_m(x) are polynomials in x given by
(9)... Q_m(x) = Sum_{k=0..m} a(m+k)*binomial(m,k)*x^(m-k).
The first few values are Q_1(x) = x + 3, Q_2(x) = 3*x^2 + 26*x + 75
and Q_3(x) = 13*x^3 + 225*x^2 + 1623*x + 4683.
For example, m = 2 gives
(10)... S_2(n) := Sum_{k=1..n-1} (1/2)^k*((n-k)*k)^2
= 2*(3*n^2 - 26*n + 75) - (1/2)^(n-1)*(3*n^2 + 26*n + 75).
(End)
G.f.: 1/(1-x/(1-2*x/(1-2*x/(1-4*x/(1-3*x/(1-6*x/(1-4*x/(1-8*x/(1-5*x/(1-10*x/(1-6*x/(1-... (continued fraction); coefficients of continued fraction are given by floor((n+2)/2)*(3-(-1)^n)/2 (A029578(n+2)). - Paul Barry, Mar 30 2010
G.f.: 1/(1-x-2*x^2/(1-4*x-8*x^2/(1-7*x-18*x^2/(1-10*x-32*x^2/(1../(1-(3*n+1)*x-2*(n+1)^2*x^2/(1-... (continued fraction). - Paul Barry, Jun 17 2010
G.f.: A(x) = Sum_{n>=0} n!*x^n / Product_{k=1..n} (1-k*x). - Paul D. Hanna, Jul 20 2011
a(n) = A074206(q_1*q_2*...*q_n), where {q_i} are distinct primes. - Vladimir Shevelev, Aug 05 2011
The adjusted e.g.f. A(x) := 1/(2-exp(x))-1, has inverse function A(x)^-1 = Integral_{t=0..x} 1/((1+t)*(1+2*t)). Applying [Dominici, Theorem 4.1] to invert the integral yields a formula for a(n): Let f(x) = (1+x)*(1+2*x). Let D be the operator f(x)*d/dx. Then a(n) = D^(n-1)(f(x)) evaluated at x = 0. Compare with A050351. - Peter Bala, Aug 31 2011
a(n) = D^n*(1/(1-x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A052801. - Peter Bala, Nov 25 2011
From Sergei N. Gladkovskii, from Oct 2011 to Oct 2013: (Start)
Continued fractions:
G.f.: 1+x/(1-x+2*x*(x-1)/(1+3*x*(2*x-1)/(1+4*x*(3*x-1)/(1+5*x*(4*x-1)/(1+... or 1+x/(U(0)-x), U(k) = 1+(k+2)*(k*x+x-1)/U(k+1).
E.g.f.: 1 + x/(G(0)-2*x) where G(k) = x + k + 1 - x*(k+1)/G(k+1).
E.g.f. (2 - 2*x)*(1 - 2*x^3/(8*x^2 - 4*x + (x^2 - 4*x + 2)*G(0)))/(x^2 - 4*x + 2) where G(k) = k^2 + k*(x+4) + 2*x + 3 - x*(k+1)*(k+3)^2 /G(k+1).
G.f.: 1 + x/G(0) where G(k) = 1 - 3*x*(k+1) - 2*x^2*(k+1)*(k+2)/G(k+1).
G.f.: 1/G(0) where G(k) = 1 - x*(k+1)/( 1 - 2*x*(k+1)/G(k+1) ).
G.f.: 1 + x/Q(0), where Q(k) = 1 - 3*x*(2*k+1) - 2*x^2*(2*k+1)*(2*k+2)/( 1 - 3*x*(2*k+2) - 2*x^2*(2*k+2)*(2*k+3)/Q(k+1) ).
G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)^2/( 2*x^2*(k+1)^2 - (1-x-3*x*k)*(1-4*x-3*x*k)/T(k+1) ). (End)
a(n) is always odd. For odd prime p and n >= 1, a((p-1)*n) = 0 (mod p). - Peter Bala, Sep 18 2013
a(n) = log(2)* Integral_{x>=0} floor(x)^n * 2^(-x) dx. - Peter Bala, Feb 06 2015
For n > 0, a(n) = Re(polygamma(n, i*log(2)/(2*Pi))/(2*Pi*i)^(n+1)) - n!/(2*log(2)^(n+1)). - Vladimir Reshetnikov, Oct 15 2015
a(n) = Sum_{k=1..n} (k*b2(k-1)*(k)!*Stirling2(n, k)), n>0, a(0)=1, where b2(n) is the n-th Bernoulli number of the second kind. - Vladimir Kruchinin, Nov 21 2016
Conjecture: a(n) = Sum_{k=0..2^(n-1)-1} A284005(k) for n > 0 with a(0) = 1. - Mikhail Kurkov, Jul 08 2018
a(n) = A074206(k) for squarefree k with n prime factors. In particular a(n) = A074206(A002110(n)). - Amiram Eldar, May 13 2019
For n > 0, a(n) = -(-1)^n / 2 * PHI(2, -n, 0), where PHI(z, s, a) is the Lerch zeta function. - Federico Provvedi, Sep 05 2020
a(n) = Sum_{s in S_n} Product_{i=1..n} binomial(i,s(i)-1), where s ranges over the set S_n of permutations of [n]. - Jose A. Rodriguez, Feb 02 2021
Sum_{n>=0} 1/a(n) = 2.425674839121428857970063350500499393706641093287018840857857170864211946122664... - Vaclav Kotesovec, Jun 17 2021
From Jacob Sprittulla, Oct 05 2021: (Start)
The following identities hold for sums over Stirling numbers of the second kind with even or odd second argument:
a(n) = 2 * Sum_{k=0..floor(n/2)} ((2k)! * Stirling2(n,2*k) ) - (-1)^n = 2*A052841-(-1)^n
a(n) = 2 * Sum_{k=0..floor(n/2)} ((2k+1)!* Stirling2(n,2*k+1))+ (-1)^n = 2*A089677+(-1)^n
a(n) = Sum_{k=1..floor((n+1)/2)} ((2k-1)!* Stirling2(n+1,2*k))
a(n) = Sum_{k=0..floor((n+1)/2)} ((2k)! * Stirling2(n+1,2*k+1)). (End)

A027641 Numerator of Bernoulli number B_n.

Original entry on oeis.org

1, -1, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 7, 0, -3617, 0, 43867, 0, -174611, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -7709321041217, 0, 2577687858367, 0, -26315271553053477373, 0, 2929993913841559, 0, -261082718496449122051
Offset: 0

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Author

Keywords

Comments

a(n)/A027642(n) (Bernoulli numbers) provide the a-sequence for the Sheffer matrix A094816 (coefficients of orthogonal Poisson-Charlier polynomials). See the W. Lang link under A006232 for a- and z-sequences for Sheffer matrices. The corresponding z-sequence is given by the rationals A130189(n)/A130190(n).
Harvey (2008) describes a new algorithm for computing Bernoulli numbers. His method is to compute B(k) modulo p for many small primes p and then reconstruct B(k) via the Chinese Remainder Theorem. The time complexity is O(k^2 log(k)^(2+eps)). The algorithm is especially well-suited to parallelization. - Jonathan Vos Post, Jul 09 2008
Regard the Bernoulli numbers as forming a vector = B_n, and the variant starting (1, 1/2, 1/6, 0, -1/30, ...), (i.e., the first 1/2 has sign +) as forming a vector Bv_n. The relationship between the Pascal triangle matrix, B_n, and Bv_n is as follows: The binomial transform of B_n = Bv_n. B_n is unchanged when multiplied by the Pascal matrix with rows signed (+-+-, ...), i.e., (1; -1,-1; 1,2,1; ...). Bv_n is unchanged when multiplied by the Pascal matrix with columns signed (+-+-, ...), i.e., (1; 1,-1; 1,-2,1; 1,-3,3,-1; ...). - Gary W. Adamson, Jun 29 2012
The sequence of the Bernoulli numbers B_n = a(n)/A027642(n) is the inverse binomial transform of the sequence {A164555(n)/A027642(n)}, illustrated by the fact that they appear as top row and left column in A190339. - Paul Curtz, May 13 2016
Named by de Moivre (1773; "the numbers of Mr. James Bernoulli") after the Swiss mathematician Jacob Bernoulli (1655-1705). - Amiram Eldar, Oct 02 2023

Examples

			B_n sequence begins 1, -1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, -691/2730, 0, 7/6, 0, -3617/510, ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 810.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
  • Harold T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 230.
  • Harold M. Edwards, Riemann's Zeta Function, Academic Press, NY, 1974; see p. 11.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.6.1.
  • Herman H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.6.
  • L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 137.
  • Hans Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1.

Crossrefs

This is the main entry for the Bernoulli numbers and has all the references, links and formulas. Sequences A027642 (the denominators of B_n) and A000367/A002445 = B_{2n} are also important!
A refinement is A194587.

Programs

  • Magma
    [Numerator(Bernoulli(n)): n in [0..40]]; // Vincenzo Librandi, Mar 17 2014
    
  • Maple
    B := n -> add((-1)^m*m!*Stirling2(n, m)/(m+1), m=0..n);
    B := n -> bernoulli(n);
    seq(numer(bernoulli(n)), n=0..40); # Zerinvary Lajos, Apr 08 2009
  • Mathematica
    Table[ Numerator[ BernoulliB[ n]], {n, 0, 40}] (* Robert G. Wilson v, Oct 11 2004 *)
    Numerator[ Range[0, 40]! CoefficientList[ Series[x/(E^x - 1), {x, 0, 40}], x]]
    Numerator[CoefficientList[Series[PolyGamma[1, 1/x]/x - x, {x, 0, 40}, Assumptions -> x > 0], x]] (* Vladimir Reshetnikov, Apr 24 2013 *)
  • Maxima
    B(n):=(-1)^((n))*sum((stirling1(n,k)*stirling2(n+k,n))/binomial(n+k,k),k,0,n);
    makelist(num(B(n)),n,0,20); /* Vladimir Kruchinin, Mar 16 2013 */
    
  • PARI
    a(n)=numerator(bernfrac(n))
    
  • Python
    from sympy import bernoulli
    from fractions import Fraction
    [bernoulli(i).as_numer_denom()[0] for i in range(51)]  # Indranil Ghosh, Mar 18 2017
    
  • Python
    from sympy import bernoulli
    def A027641(n): return bernoulli(n).p
    print([A027641(n) for n in range(80)])  # M. F. Hasler, Jun 11 2019
  • SageMath
    [bernoulli(n).numerator() for n in range(41)]  # Peter Luschny, Feb 19 2016
    
  • SageMath
    # Alternatively:
    def A027641_list(len):
        f, R, C = 1, [1], [1]+[0]*(len-1)
        for n in (1..len-1):
            f *= n
            for k in range(n, 0, -1):
                C[k] = C[k-1] / (k+1)
            C[0] = -sum(C[k] for k in (1..n))
            R.append((C[0]*f).numerator())
        return R
    A027641_list(41)  # Peter Luschny, Feb 20 2016
    

Formula

E.g.f: x/(exp(x) - 1); take numerators.
Recurrence: B^n = (1+B)^n, n >= 2 (interpreting B^j as B_j).
B_{2n}/(2n)! = 2*(-1)^(n-1)*(2*Pi)^(-2n) Sum_{k>=1} 1/k^(2n) (gives asymptotics) - Rademacher, p. 16, Eq. (9.1). In particular, B_{2*n} ~ (-1)^(n-1)*2*(2*n)!/(2*Pi)^(2*n).
Sum_{i=1..n-1} i^k = ((n+B)^(k+1)-B^(k+1))/(k+1) (interpreting B^j as B_j).
B_{n-1} = - Sum_{r=1..n} (-1)^r binomial(n, r) r^(-1) Sum_{k=1..r} k^(n-1). More concisely, B_n = 1 - (1-C)^(n+1), where C^r is replaced by the arithmetic mean of the first r n-th powers of natural numbers in the expansion of the right-hand side. [Bergmann]
Sum_{i>=1} 1/i^(2k) = zeta(2k) = (2*Pi)^(2k)*|B_{2k}|/(2*(2k)!).
B_{2n} = (-1)^(m-1)/2^(2m+1) * Integral{-inf..inf, [d^(m-1)/dx^(m-1) sech(x)^2 ]^2 dx} (see Grosset/Veselov).
Let B(s,z) = -2^(1-s)(i/Pi)^s s! PolyLog(s,exp(-2*i*Pi/z)). Then B(2n,1) = B_{2n} for n >= 1. Similarly the numbers B(2n+1,1), which might be called Co-Bernoulli numbers, can be considered, and it is remarkable that Leonhard Euler in 1755 already calculated B(3,1) and B(5,1) (Opera Omnia, Ser. 1, Vol. 10, p. 351). (Cf. the Luschny reference for a discussion.) - Peter Luschny, May 02 2009
The B_n sequence is the left column of the inverse of triangle A074909, the "beheaded" Pascal's triangle. - Gary W. Adamson, Mar 05 2012
From Sergei N. Gladkovskii, Dec 04 2012: (Start)
E.g.f. E(x)= 2 - x/(tan(x) + sec(x) - 1)= Sum_{n>=0} a(n)*x^n/n!, a(n)=|B(n)|, where B(n) is Bernoulli number B_n.
E(x)= 2 + x - B(0), where B(k)= 4*k+1 + x/(2 + x/(4*k+3 - x/(2 - x/B(k+1)))); (continued fraction, 4-step). (End)
E.g.f.: x/(exp(x)-1)= U(0); U(k)= 2*k+1 - x(2*k+1)/(x + (2*k+2)/(1 + x/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 05 2012
E.g.f.: 2*(x-1)/(x*Q(0)-2) where Q(k) = 1 + 2*x*(k+1)/((2*k+1)*(2*k+3) - x*(2*k+1)*(2*k+3)^2/(x*(2*k+3) + 4*(k+1)*(k+2)/Q(k+1))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 26 2013
a(n) = numerator(B(n)), B(n) = (-1)^n*Sum_{k=0..n} Stirling1(n,k) * Stirling2(n+k,n) / binomial(n+k,k). - Vladimir Kruchinin, Mar 16 2013
E.g.f.: x/(exp(x)-1) = E(0) where E(k) = 2*k+1 - x/(2 + x/E(k+1)); (continued fraction). - Sergei N. Gladkovskii, Mar 16 2013
G.f. for Bernoulli(n) = a(n)/A027642(n): psi_1(1/x)/x - x, where psi_n(z) is the polygamma function, psi_n(z) = (d/dz)^(n+1) log(Gamma(z)). - Vladimir Reshetnikov, Apr 24 2013
E.g.f.: 2*E(0) - 2*x, where E(k)= x + (k+1)/(1 + 1/(1 - x/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 10 2013
B_n = Sum_{m=0..n} (-1)^m *A131689(n, m)/(m + 1), n >= 0. See one of the Maple programs. - Wolfdieter Lang, May 05 2017
a(n) = numerator((-1)^n*A155585(n-1)*n/(4^n-2^n)), for n>=1. - Mats Granvik, Nov 26 2017
From Artur Jasinski, Dec 30 2020: (Start)
a(n) = numerator(-2*cos(Pi*n/2)*Gamma(n+1)*zeta(n)/(2*Pi)^n), for n=0 and n>1.
a(n) = numerator(-n*zeta(1-n)), for n=0 and n>1. (End)
a(n) = numerator(Sum_{k=0..n-1} (-1)^(k-1)*k!*Stirling2(n-1,k) / ((k+1)*(k+2))), for n>0 (see Jha link). - Bill McEachen, Jul 17 2025

A019538 Triangle of numbers T(n,k) = k!*Stirling2(n,k) read by rows (n >= 1, 1 <= k <= n).

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 14, 36, 24, 1, 30, 150, 240, 120, 1, 62, 540, 1560, 1800, 720, 1, 126, 1806, 8400, 16800, 15120, 5040, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880, 1, 1022, 55980, 818520, 5103000, 16435440, 29635200, 30240000, 16329600, 3628800
Offset: 1

Views

Author

N. J. A. Sloane and Manfred Goebel (goebel(AT)informatik.uni-tuebingen.de), Dec 11 1996

Keywords

Comments

Number of ways n labeled objects can be distributed into k nonempty parcels. Also number of special terms in n variables with maximal degree k.
In older terminology these are called differences of 0. - Michael Somos, Oct 08 2003
Number of surjections (onto functions) from an n-element set to a k-element set.
Also coefficients (in ascending order) of so-called ordered Bell polynomials.
(k-1)!*Stirling2(n,k-1) is the number of chain topologies on an n-set having k open sets [Stephen].
Number of set compositions (ordered set partitions) of n items into k parts. Number of k dimensional 'faces' of the n dimensional permutohedron (see Simion, p. 162). - Mitch Harris, Jan 16 2007
Correction of comment before: Number of (n-k)-dimensional 'faces' of the permutohedron of order n (an (n-1)-dimensional polytope). - Tilman Piesk, Oct 29 2014
This array is related to the reciprocal of an e.g.f. as sketched in A133314. For example, the coefficient of the fourth-order term in the Taylor series expansion of 1/(a(0) + a(1) x + a(2) x^2/2! + a(3) x^3/3! + ...) is a(0)^(-5) * {24 a(1)^4 - 36 a(1)^2 a(2) a(0) + [8 a(1) a(3) + 6 a(2)^2] a(0)^2 - a(4) a(0)^3}. The unsigned coefficients characterize the P3 permutohedron depicted on page 10 in the Loday link with 24 vertices (0-D faces), 36 edges (1-D faces), 6 squares (2-D faces), 8 hexagons (2-D faces) and 1 3-D permutohedron. Summing coefficients over like dimensions gives A019538 and A090582. Compare to A133437 for the associahedron. - Tom Copeland, Sep 29 2008, Oct 07 2008
Further to the comments of Tom Copeland above, the permutohedron of type A_3 can be taken as the truncated octahedron. Its dual is the tetrakis hexahedron, a simplicial polyhedron, with f-vector (1,14,36,24) giving the fourth row of this triangle. See the Wikipedia entry and [Fomin and Reading p. 21]. The corresponding h-vectors of permutohedra of type A give the rows of the triangle of Eulerian numbers A008292. See A145901 and A145902 for the array of f-vectors for type B and type D permutohedra respectively. - Peter Bala, Oct 26 2008
Subtriangle of triangle in A131689. - Philippe Deléham, Nov 03 2008
Since T(n,k) counts surjective functions and surjective functions are "consistent", T(n,k) satisfies a binomial identity, namely, T(n,x+y) = Sum_{j=0..n} C(n,j)*T(j,x)*T(n-j,y). For definition of consistent functions and a generalized binomial identity, see "Toy stories and combinatorial identities" in the link section below. - Dennis P. Walsh, Feb 24 2012
T(n,k) is the number of labeled forests on n+k vertices satisfying the following two conditions: (i) each forest consists of exactly k rooted trees with roots labeled 1, 2, ..., k; (ii) every root has at least one child vertex. - Dennis P. Walsh, Feb 24 2012
The triangle is the inverse binomial transform of triangle A028246, deleting the left column and shifting up one row. - Gary W. Adamson, Mar 05 2012
See A074909 for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
E.g.f. for the shifted signed polynomials is G(x,t) = (e^t-1)/[1+(1+x)(e^t-1)] = 1-(1+x)(e^t-1) + (1+x)^2(e^t-1)^2 - ... (see also A008292 and A074909), which has the infinitesimal generator g(x,u)d/du = [(1-x*u)(1-(1+x)u)]d/du, i.e., exp[t*g(x,u)d/du]u eval. at u=0 gives G(x,t), and dG(x,t)/dt = g(x,G(x,t)). The compositional inverse is log((1-xt)/(1-(1+x)t)). G(x,t) is a generating series associated to the generalized Hirzebruch genera. See the G. Rzadowski link for the relation of the derivatives of g(x,u) to solutions of the Riccatt differential equation, soliton solns. to the KdV equation, and the Eulerian and Bernoulli numbers. In addition A145271 connects products of derivatives of g(x,u) and the refined Eulerian numbers to the inverse of G(x,t), which gives the normalized, reverse face polynomials of the simplices (A135278, divided by n+1). See A028246 for the generator g(x,u)d/dx. - Tom Copeland, Nov 21 2014
For connections to toric varieties and Eulerian polynomials, see the Dolgachev and Lunts and the Stembridge links. - Tom Copeland, Dec 31 2015
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra (this entry) and stellahedra. - Tom Copeland, Nov 14 2016
T(n, k) appears in a Worpitzky identity relating monomials to binomials: x^n = Sum_{k=1..n} T(n, k)*binomial(x,k), n >= 1. See eq. (11.) of the Worpitzky link on p. 209. The relation to the Eulerian numbers is given there in eqs. (14.) and (15.). See the formula below relating to A008292. See also Graham et al. eq. (6.10) (relating monomials to falling factorials) on p. 248 (2nd ed. p. 262). The Worpitzky identity given in the Graham et al. reference as eq. (6.37) (2nd ed. p. 269) is eq. (5.), p. 207, of Worpitzky. - Wolfdieter Lang, Mar 10 2017
T(n, m) is also the number of minimum clique coverings and minimum matchings in the complete bipartite graph K_{m,n}. - Eric W. Weisstein, Apr 26 2017
From the Hasan and Franco and Hasan papers: The m-permutohedra for m=1,2,3,4 are the line segment, hexagon, truncated octahedron and omnitruncated 5-cell. The first three are well-known from the study of elliptic models, brane tilings and brane brick models. The m+1 torus can be tiled by a single (m+2)-permutohedron. Relations to toric Calabi-Yau Kahler manifolds are also discussed. - Tom Copeland, May 14 2020
From Manfred Boergens, Jul 25 2021: (Start)
Number of n X k binary matrices with row sums = 1 and no zero columns. These matrices are a subset of the matrices defining A183109.
The distribution into parcels in the leading comment can be regarded as a covering of [n] by tuples (A_1,...,A_k) in P([n])^k with nonempty and disjoint A_j, with P(.) denoting the power set (corrected for clarity by Manfred Boergens, May 26 2024). For the non-disjoint case see A183109 and A218695.
For tuples with "nonempty" dropped see A089072. For tuples with "nonempty and disjoint" dropped see A092477 and A329943 (amendment by Manfred Boergens, Jun 24 2024). (End)

Examples

			The triangle T(n, k) begins:
  n\k 1    2     3      4       5        6        7        8        9      10
  1:  1
  2:  1    2
  3:  1    6     6
  4:  1   14    36     24
  5:  1   30   150    240     120
  6:  1   62   540   1560    1800      720
  7:  1  126  1806   8400   16800    15120     5040
  8:  1  254  5796  40824  126000   191520   141120    40320
  9:  1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... Reformatted and extended - _Wolfdieter Lang_, Oct 04 2014
---------------------------------------------------------------------------
T(4,1) = 1: {1234}. T(4,2) = 14: {1}{234} (4 ways), {12}{34} (6 ways), {123}{4} (4 ways). T(4,3) = 36: {12}{3}{4} (12 ways), {1}{23}{4} (12 ways), {1}{2}{34} (12 ways). T(4,4) = 1: {1}{2}{3}{4} (1 way).
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 89, ex. 1; also p. 210.
  • Miklos Bona, Combinatorics of Permutations, Chapman and Hall,2004, p.12.
  • G. Boole, A Treatise On The Calculus of Finite Differences, Dover Publications, 1960, p. 20.
  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 212.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, 1989, p. 155. Also eqs.(6.10) and (6.37).
  • Kiran S. Kedlaya and Andrew V. Sutherland, Computing L -Series of Hyperelliptic Curves in Algorithmic Number Theory Lecture Notes in Computer Science Volume 5011/2008.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 5.6.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
  • J. F. Steffensen, Interpolation, 2nd ed., Chelsea, NY, 1950, see p. 54.
  • A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.
  • E. Whittaker and G. Robinson, The Calculus of Observations, Blackie, London, 4th ed., 1949; p. 7.

Crossrefs

Row sums give A000670. Maximal terms in rows give A002869. Central terms T(2k-1,k) give A233734.
Diagonal is n! (A000142). 2nd diagonal is A001286. 3rd diagonal is A037960.
Reflected version of A090582. A371568 is another version.
See also the two closely related triangles: A008277(n, k) = T(n, k)/k! (Stirling numbers of second kind) and A028246(n, k) = T(n, k)/k.
Cf. A033282 'faces' of the associahedron.
Cf. A008292, A047969, A145901, A145902. - Peter Bala, Oct 26 2008
Visible in the 3-D array in A249042.
See also A000182.

Programs

  • Haskell
    a019538 n k = a019538_tabl !! (n-1) !! (k-1)
    a019538_row n = a019538_tabl !! (n-1)
    a019538_tabl = iterate f [1] where
       f xs = zipWith (*) [1..] $ zipWith (+) ([0] ++ xs) (xs ++ [0])
    -- Reinhard Zumkeller, Dec 15 2013
    
  • Maple
    with(combinat): A019538 := (n,k)->k!*stirling2(n,k);
  • Mathematica
    Table[k! StirlingS2[n, k], {n, 9}, {k, n}] // Flatten
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, sum(i=0, k, (-1)^i * binomial(k, i) * (k-i)^n))}; /* Michael Somos, Oct 08 2003 */
    
  • Sage
    def T(n, k): return factorial(k)*stirling_number2(n,k) # Danny Rorabaugh, Oct 10 2015

Formula

T(n, k) = k*(T(n-1, k-1)+T(n-1, k)) with T(0, 0) = 1 [or T(1, 1) = 1]. - Henry Bottomley, Mar 02 2001
E.g.f.: (y*(exp(x)-1) - exp(x))/(y*(exp(x)-1) - 1). - Vladeta Jovovic, Jan 30 2003
Equals [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deléham's operator defined in A084938.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*j^n*binomial(k, j). - Mario Catalani (mario.catalani(AT)unito.it), Nov 28 2003. See Graham et al., eq. (6.19), p. 251. For a proof see Bert Seghers, Jun 29 2013.
Sum_{k=0..n} T(n, k)(-1)^(n-k) = 1, Sum_{k=0..n} T(n, k)(-1)^k = (-1)^n. - Mario Catalani (mario.catalani(AT)unito.it), Dec 11 2003
O.g.f. for n-th row: polylog(-n, x/(1+x))/(x+x^2). - Vladeta Jovovic, Jan 30 2005
E.g.f.: 1 / (1 + t*(1-exp(x))). - Tom Copeland, Oct 13 2008
From Peter Bala, Oct 26 2008: (Start)
O.g.f. as a continued fraction: 1/(1 - x*t/(1 - (x + 1)*t/(1 - 2*x*t/(1 - 2*(x + 1)*t/(1 - ...))))) = 1 + x*t + (x + 2*x^2)*t^2 + (x + 6*x^2 + 6*x^3)*t^3 + ... .
The row polynomials R(n,x), which begin R(1,x) = x, R(2,x) = x + 2*x^2, R(3,x) = x + 6*x^2 + 6*x^3, satisfy the recurrence x*d/dx ((x + 1)*R(n,x)) = R(n+1,x). It follows that the zeros of R(n,x) are real and negative (apply Corollary 1.2 of [Liu and Wang]).
Since this is the triangle of f-vectors of the (simplicial complexes dual to the) type A permutohedra, whose h-vectors form the Eulerian number triangle A008292, the coefficients of the polynomial (x-1)^n*R(n,1/(x-1)) give the n-th row of A008292. For example, from row 3 we have x^2 + 6*x + 6 = 1 + 4*y + y^2, where y = x + 1, producing [1,4,1] as the third row of A008292. The matrix product A008292 * A007318 gives the mirror image of this triangle (see A090582).
For n,k >= 0, T(n+1,k+1) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*[(j+1)^(n+1) - j^(n+1)]. The matrix product of Pascal's triangle A007318 with the current array gives (essentially) A047969. This triangle is also related to triangle A047969 by means of the S-transform of [Hetyei], a linear transformation of polynomials whose value on the basis monomials x^k is given by S(x^k) = binomial(x,k). The S-transform of the shifted n-th row polynomial Q(n,x) := R(n,x)/x is S(Q(n,x)) = (x+1)^n - x^n. For example, from row 3 we obtain S(1 + 6*x + 6*x^2) = 1 + 6*x + 6*x*(x-1)/2 = 1 + 3*x + 3*x^2 = (x+1)^3 - x^3. For fixed k, the values S(Q(n,k)) give the nonzero entries in column (k-1) of the triangle A047969 (the Hilbert transform of the Eulerian numbers). (End)
E.g.f.: (exp(x)-1)^k = sum T(n,k)x^n/n!. - Vladimir Kruchinin, Aug 10 2010
T(n,k) = Sum_{i=1..k} A(n,i)*Binomial(n-i,k-i) where A(n,i) is the number of n-permutations that have i ascending runs, A008292.
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f. A(x,t) = -1 + 1/(1+t*(1-exp(x))), the comp. inverse in x is B(x,t) = log(((1+t)/t) - 1/(t(1+x))).
With h(x,t) = 1/(dB/dx)= (1+x)((1+t)(1+x)-1), the row polynomial P(n,t) is given by (h(x,t)*d/dx)^n x, eval. at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t), with P(0,t)=0.
(A factor of -1/n! was removed by Copeland on Aug 25 2016.) (End)
The term linear in x of [x*h(d/dx,t)]^n 1 gives the n-th row polynomial. (See A134685.) - Tom Copeland, Nov 07 2011
Row polynomials are given by D^n(1/(1-x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. - Peter Bala, Nov 25 2011
T(n,x+y) = Sum_{j=0..n} binomial(n,j)*T(j,x)*T(n-j,y). - Dennis P. Walsh, Feb 24 2012
Let P be a Rota-Baxter operator of weight 1 satisfying the identity P(x)*P(y) = P(P(x)*y) + P(x*P(y)) + P(x*y). Then P(1)^2 = P(1) + 2*P^2(1). More generally, Guo shows that P(1)^n = Sum_{k=1..n} T(n,k)*P^k(1). - Peter Bala, Jun 08 2012
Sum_{i=1..n} (-1)^i*T(n,i)/i = 0, for n > 1. - Leonid Bedratyuk, Aug 09 2012
T(n, k) = Sum_{j=0..k} (-1)^j*binomial(k, j)*(k-j)^n. [M. Catalani's re-indexed formula from Nov 28 2003] Proof: count the surjections of [n] onto [k] with the inclusion-exclusion principle, as an alternating sum of the number of functions from [n] to [k-j]. - Bert Seghers, Jun 29 2013
n-th row polynomial = 1/(1 + x)*( Sum_{k>=0} k^n*(x/(1 + x))^k ), valid for x in the open interval (-1/2, inf). See Tanny link. Cf. A145901. - Peter Bala, Jul 22 2014
T(n,k) = k * A141618(n,k-1) / binomial(n,k-1). - Tom Copeland, Oct 25 2014
Sum_{n>=0} n^k*a^n = Sum_{i=1..k} (a / (1 - a))^i * T(k, i)/(1-a) for |a| < 1. - David A. Corneth, Mar 09 2015
From Peter Bala, May 26 2015: (Start)
The row polynomials R(n,x) satisfy (1 + x)*R(n,x) = (-1)^n*x*R(n,-(1 + x)).
For a fixed integer k, the expansion of the function A(k,z) := exp( Sum_{n >= 1} R(n,k)*z^n/n ) has integer coefficients and satisfies the functional equation A(k,z)^(k + 1) = BINOMIAL(A(k,z))^k, where BINOMIAL(F(z))= 1/(1 - z)*F(z/(1 - z)) denotes the binomial transform of the o.g.f. F(z). Cf. A145901. For cases see A084784 (k = 1), A090352 (k = 2), A090355 (k = 3), A090357 (k = 4), A090362 (k = 5) and A084785 (k = -2 with z -> -z).
A(k,z)^(k + 1) = A(-(k + 1),-z)^k and hence BINOMIAL(A(k,z)) = A(-(k + 1),-z). (End)
From Tom Copeland, Oct 19 2016: (Start)
Let a(1) = 1 + x + B(1) = x + 1/2 and a(n) = B(n) = (B.)^n, where B(n) are the Bernoulli numbers defined by e^(B.t) = t / (e^t-1), then t / e^(a.t) = t / [(x + 1) * t + exp(B.t)] = (e^t - 1) /[ 1 + (x + 1) (e^t - 1)] = exp(p.(x)t), where (p.(x))^n = p_n(x) are the shifted, signed row polynomials of this array: p_0(x) = 0, p_1(x) = 1, p_2(x) = -(1 + 2 x), p_3(x) = 1 + 6 x + 6 x^2, ... and p_n(x) = n * b(n-1), where b(n) are the partition polynomials of A133314 evaluated with these a(n).
Sum_{n > 0} R(n,-1/2) x^n/n! = 2 * tanh(x/2), where R(n,x) = Sum_{k = 1..n} T(n,k) x^(k-1) are the shifted row polynomials of this entry, so R(n,-1/2) = 4 * (2^(n+1)-1) B(n+1)/(n+1). (Cf. A000182.)
(End)
Also the Bernoulli numbers are given by B(n) = Sum_{k =1..n} (-1)^k T(n,k) / (k+1). - Tom Copeland, Nov 06 2016
G.f. for column k: k! x^k / Product_{i=1..k} (1-i*x). - Robert A. Russell, Sep 25 2018
a(j) <= A183109(j). - Manfred Boergens, Jul 25 2021
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