cp's OEIS Frontend

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

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A008292 Triangle of Eulerian numbers T(n,k) (n >= 1, 1 <= k <= n) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 11, 11, 1, 1, 26, 66, 26, 1, 1, 57, 302, 302, 57, 1, 1, 120, 1191, 2416, 1191, 120, 1, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1
Offset: 1

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Author

N. J. A. Sloane, Mar 15 1996

Keywords

Comments

The indexing used here follows that given in the classic books by Riordan and Comtet. For two other versions see A173018 and A123125. - N. J. A. Sloane, Nov 21 2010
Coefficients of Eulerian polynomials. Number of permutations of n objects with k-1 rises. Number of increasing rooted trees with n+1 nodes and k leaves.
T(n,k) = number of permutations of [n] with k runs. T(n,k) = number of permutations of [n] requiring k readings (see the Knuth reference). T(n,k) = number of permutations of [n] having k distinct entries in its inversion table. - Emeric Deutsch, Jun 09 2004
T(n,k) = number of ways to write the Coxeter element s_{e1}s_{e1-e2}s_{e2-e3}s_{e3-e4}...s_{e_{n-1}-e_n} of the reflection group of type B_n, using s_{e_k} and as few reflections of the form s_{e_i+e_j}, where i = 1, 2, ..., n and j is not equal to i, as possible. - Pramook Khungurn (pramook(AT)mit.edu), Jul 07 2004
Subtriangle for k>=1 and n>=1 of triangle A123125. - Philippe Deléham, Oct 22 2006
T(n,k)/n! also represents the n-dimensional volume of the portion of the n-dimensional hypercube cut by the (n-1)-dimensional hyperplanes x_1 + x_2 + ... x_n = k, x_1 + x_2 + ... x_n = k-1; or, equivalently, it represents the probability that the sum of n independent random variables with uniform distribution between 0 and 1 is between k-1 and k. - Stefano Zunino, Oct 25 2006
[E(.,t)/(1-t)]^n = n!*Lag[n,-P(.,t)/(1-t)] and [-P(.,t)/(1-t)]^n = n!*Lag[n, E(.,t)/(1-t)] umbrally comprise a combinatorial Laguerre transform pair, where E(n,t) are the Eulerian polynomials and P(n,t) are the polynomials in A131758. - Tom Copeland, Sep 30 2007
From Tom Copeland, Oct 07 2008: (Start)
G(x,t) = 1/(1 + (1-exp(x*t))/t) = 1 + 1*x + (2+t)*x^2/2! + (6+6*t+t^2)*x^3/3! + ... gives row polynomials for A090582, the reverse f-polynomials for the permutohedra (see A019538).
G(x,t-1) = 1 + 1*x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... gives row polynomials for A008292, the h-polynomials for permutohedra (Postnikov et al.).
G((t+1)*x, -1/(t+1)) = 1 + (1+t)*x + (1+3*t+2*t^2)*x^2/2! + ... gives row polynomials for A028246.
(End)
A subexceedant function f on [n] is a map f:[n] -> [n] such that 1 <= f(i) <= i for all i, 1 <= i <= n. T(n,k) equals the number of subexceedant functions f of [n] such that the image of f has cardinality k [Mantaci & Rakotondrajao]. Example T(3,2) = 4: if we identify a subexceedant function f with the word f(1)f(2)...f(n) then the subexceedant functions on [3] are 111, 112, 113, 121, 122 and 123 and four of these functions have an image set of cardinality 2. - Peter Bala, Oct 21 2008
Further to the comments of Tom Copeland above, the n-th row of this triangle is the h-vector of the simplicial complex dual to a permutohedron of type A_(n-1). The corresponding f-vectors are the rows of A019538. For example, 1 + 4*x + x^2 = y^2 + 6*y + 6 and 1 + 11*x + 11*x^2 + x^3 = y^3 + 14*y^2 + 36*y + 24, where x = y + 1, give [1,6,6] and [1,14,36,24] as the third and fourth rows of A019538. The Hilbert transform of this triangle (see A145905 for the definition) is A047969. See A060187 for the triangle of Eulerian numbers of type B (the h-vectors of the simplicial complexes dual to permutohedra of type B). See A066094 for the array of h-vectors of type D. For tables of restricted Eulerian numbers see A144696 - A144699. - Peter Bala, Oct 26 2008
For a natural refinement of A008292 with connections to compositional inversion and iterated derivatives, see A145271. - Tom Copeland, Nov 06 2008
The polynomials E(z,n) = numerator(Sum_{k>=1} (-1)^(n+1)*k^n*z^(k-1)) for n >=1 lead directly to the triangle of Eulerian numbers. - Johannes W. Meijer, May 24 2009
From Walther Janous (walther.janous(AT)tirol.com), Nov 01 2009: (Start)
The (Eulerian) polynomials e(n,x) = Sum_{k=0..n-1} T(n,k+1)*x^k turn out to be also the numerators of the closed-form expressions of the infinite sums:
S(p,x) = Sum_{j>=0} (j+1)^p*x^j, that is
S(p,x) = e(p,x)/(1-x)^(p+1), whenever |x| < 1 and p is a positive integer.
(Note the inconsistent use of T(n,k) in the section listing the formula section. I adhere tacitly to the first one.) (End)
If n is an odd prime, then all numbers of the (n-2)-th and (n-1)-th rows are in the progression k*n+1. - Vladimir Shevelev, Jul 01 2011
The Eulerian triangle is an element of the formula for the r-th successive summation of Sum_{k=1..n} k^j which appears to be Sum_{k=1..n} T(j,k-1) * binomial(j-k+n+r, j+r). - Gary Detlefs, Nov 11 2011
Li and Wong show that T(n,k) counts the combinatorially inequivalent star polygons with n+1 vertices and sum of angles (2*k-n-1)*Pi. An equivalent formulation is: define the total sign change S(p) of a permutation p in the symmetric group S_n to be equal to Sum_{i=1..n} sign(p(i)-p(i+1)), where we take p(n+1) = p(1). T(n,k) gives the number of permutations q in S_(n+1) with q(1) = 1 and S(q) = 2*k-n-1. For example, T(3,2) = 4 since in S_4 the permutations (1243), (1324), (1342) and (1423) have total sign change 0. - Peter Bala, Dec 27 2011
Xiong, Hall and Tsao refer to Riordan and mention that a traditional Eulerian number A(n,k) is the number of permutations of (1,2...n) with k weak exceedances. - Susanne Wienand, Aug 25 2014
Connections to algebraic geometry/topology and characteristic classes are discussed in the Buchstaber and Bunkova, the Copeland, the Hirzebruch, the Lenart and Zainoulline, the Losev and Manin, and the Sheppeard links; to the Grassmannian, in the Copeland, the Farber and Postnikov, the Sheppeard, and the Williams links; and to compositional inversion and differential operators, in the Copeland and the Parker links. - Tom Copeland, Oct 20 2015
The bivariate e.g.f. noted in the formulas is related to multiplying edges in certain graphs discussed in the Aluffi-Marcolli link. See p. 42. - Tom Copeland, Dec 18 2016
Distribution of left children in treeshelves is given by a shift of the Eulerian numbers. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. See A278677, A278678 or A278679 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016
The row polynomial P(n, x) = Sum_{k=1..n} T(n, k)*x^k appears in the numerator of the o.g.f. G(n, x) = Sum_{m>=0} S(n, m)*x^m with S(n, m) = Sum_{j=0..m} j^n for n >= 1 as G(n, x) = Sum_{k=1..n} P(n, x)/(1 - x)^(n+2) for n >= 0 (with 0^0=1). See also triangle A131689 with a Mar 31 2017 comment for a rewritten form. For the e.g.f see A028246 with a Mar 13 2017 comment. - Wolfdieter Lang, Mar 31 2017
For relations to Ehrhart polynomials, volumes of polytopes, polylogarithms, the Todd operator, and other special functions, polynomials, and sequences, see A131758 and the references therein. - Tom Copeland, Jun 20 2017
For relations to values of the Riemann zeta function at integral arguments, see A131758 and the Dupont reference. - Tom Copeland, Mar 19 2018
Normalized volumes of the hypersimplices, attributed to Laplace. (Cf. the De Loera et al. reference, p. 327.) - Tom Copeland, Jun 25 2018

Examples

			The triangle T(n, k) begins:
n\k 1    2     3      4       5       6      7     8    9 10 ...
1:  1
2:  1    1
3:  1    4     1
4:  1   11    11      1
5:  1   26    66     26       1
6:  1   57   302    302      57       1
7:  1  120  1191   2416    1191     120      1
8:  1  247  4293  15619   15619    4293    247     1
9:  1  502 14608  88234  156190   88234  14608   502    1
10: 1 1013 47840 455192 1310354 1310354 455192 47840 1013  1
... Reformatted. - _Wolfdieter Lang_, Feb 14 2015
-----------------------------------------------------------------
E.g.f. = (y) * x^1 / 1! + (y + y^2) * x^2 / 2! + (y + 4*y^2 + y^3) * x^3 / 3! + ... - _Michael Somos_, Mar 17 2011
Let n=7. Then the following 2*7+1=15 consecutive terms are 1(mod 7): a(15+i), i=0..14. - _Vladimir Shevelev_, Jul 01 2011
Row 3: The plane increasing 0-1-2 trees on 3 vertices (with the number of colored vertices shown to the right of a vertex) are
.
.   1o (1+t)         1o t         1o t
.   |                / \          / \
.   |               /   \        /   \
.   2o (1+t)      2o     3o    3o    2o
.   |
.   |
.   3o
.
The total number of trees is (1+t)^2 + t + t = 1 + 4*t + t^2.
		

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.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 106.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 243.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 260.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 254; 2nd. ed., p. 268.[Worpitzky's identity (6.37)]
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1998, Vol. 3, p. 47 (exercise 5.1.4 Nr. 20) and p. 605 (solution).
  • Meng Li and Ron Goldman. "Limits of sums for binomial and Eulerian numbers and their associated distributions." Discrete Mathematics 343.7 (2020): 111870.
  • Anthony Mendes and Jeffrey Remmel, Generating functions from symmetric functions, Preliminary version of book, available from Jeffrey Remmel's home page http://math.ucsd.edu/~remmel/
  • K. Mittelstaedt, A stochastic approach to Eulerian numbers, Amer. Math. Mnthly, 127:7 (2020), 618-628.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
  • R. Sedgewick and P. Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Figure M3416, Academic Press, 1995.
  • H. S. Wall, Analytic Theory of Continued Fractions, Chelsea, 1973, see p. 208.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 101.

Crossrefs

Programs

  • GAP
    Flat(List([1..10],n->List([1..n],k->Sum([0..k],j->(-1)^j*(k-j)^n*Binomial(n+1,j))))); # Muniru A Asiru, Jun 29 2018
    
  • Haskell
    import Data.List (genericLength)
    a008292 n k = a008292_tabl !! (n-1) !! (k-1)
    a008292_row n = a008292_tabl !! (n-1)
    a008292_tabl = iterate f [1] where
       f xs = zipWith (+)
         (zipWith (*) ([0] ++ xs) (reverse ks)) (zipWith (*) (xs ++ [0]) ks)
         where ks = [1 .. 1 + genericLength xs]
    -- Reinhard Zumkeller, May 07 2013
    
  • Magma
    Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >; [[Eulerian(n,k): k in [0..n-1]]: n in [1..10]]; // G. C. Greubel, Apr 15 2019
  • Maple
    A008292 := proc(n,k) option remember; if k < 1 or k > n then 0; elif k = 1 or k = n then 1; else k*procname(n-1,k)+(n-k+1)*procname(n-1,k-1) ; end if; end proc:
  • Mathematica
    t[n_, k_] = Sum[(-1)^j*(k-j)^n*Binomial[n+1, j], {j, 0, k}];
    Flatten[Table[t[n, k], {n, 1, 10}, {k, 1, n}]] (* Jean-François Alcover, May 31 2011, after Michael Somos *)
    Flatten[Table[CoefficientList[(1-x)^(k+1)*PolyLog[-k, x]/x, x], {k, 1, 10}]] (* Vaclav Kotesovec, Aug 27 2015 *)
    Table[Tally[
       Count[#, x_ /; x > 0] & /@ (Differences /@
          Permutations[Range[n]])][[;; , 2]], {n, 10}] (* Li Han, Oct 11 2020 *)
  • PARI
    {T(n, k) = if( k<1 || k>n, 0, if( n==1, 1, k * T(n-1, k) + (n-k+1) * T(n-1, k-1)))}; /* Michael Somos, Jul 19 1999 */
    
  • PARI
    {T(n, k) = sum( j=0, k, (-1)^j * (k-j)^n * binomial( n+1, j))}; /* Michael Somos, Jul 19 1999 */
    
  • PARI
    {A(n,c)=c^(n+c-1)+sum(i=1,c-1,(-1)^i/i!*(c-i)^(n+c-1)*prod(j=1,i,n+c+1-j))}
    
  • Python
    from sympy import binomial
    def T(n, k): return sum([(-1)**j*(k - j)**n*binomial(n + 1, j) for j in range(k + 1)])
    for n in range(1, 11): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, Nov 08 2017
    
  • R
    T <- function(n, k) {
      S <- numeric()
      for (j in 0:k) S <- c(S, (-1)^j*(k-j)^n*choose(n+1, j))
      return(sum(S))
    }
    for (n in 1:10){
      for (k in 1:n) print(T(n,k))
    } # Indranil Ghosh, Nov 08 2017
    
  • Sage
    [[sum((-1)^j*binomial(n+1, j)*(k-j)^n for j in (0..k)) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Feb 23 2019
    

Formula

T(n, k) = k * T(n-1, k) + (n-k+1) * T(n-1, k-1), T(1, 1) = 1.
T(n, k) = Sum_{j=0..k} (-1)^j * (k-j)^n * binomial(n+1, j).
Row sums = n! = A000142(n) unless n=0. - Michael Somos, Mar 17 2011
E.g.f. A(x, q) = Sum_{n>0} (Sum_{k=1..n} T(n, k) * q^k) * x^n / n! = q * ( e^(q*x) - e^x ) / ( q*e^x - e^(q*x) ) satisfies dA / dx = (A + 1) * (A + q). - Michael Somos, Mar 17 2011
For a column listing, n-th term: T(c, n) = c^(n+c-1) + Sum_{i=1..c-1} (-1)^i/i!*(c-i)^(n+c-1)*Product_{j=1..i} (n+c+1-j). - Randall L Rathbun, Jan 23 2002
From John Robertson (jpr2718(AT)aol.com), Sep 02 2002: (Start)
Four characterizations of Eulerian numbers T(i, n):
1. T(0, n)=1 for n>=1, T(i, 1)=0 for i>=1, T(i, n) = (n-i)T(i-1, n-1) + (i+1)T(i, n-1).
2. T(i, n) = Sum_{j=0..i} (-1)^j*binomial(n+1,j)*(i-j+1)^n for n>=1, i>=0.
3. Let C_n be the unit cube in R^n with vertices (e_1, e_2, ..., e_n) where each e_i is 0 or 1 and all 2^n combinations are used. Then T(i, n)/n! is the volume of C_n between the hyperplanes x_1 + x_2 + ... + x_n = i and x_1 + x_2 + ... + x_n = i+1. Hence T(i, n)/n! is the probability that i <= X_1 + X_2 + ... + X_n < i+1 where the X_j are independent uniform [0, 1] distributions. - See Ehrenborg & Readdy reference.
4. Let f(i, n) = T(i, n)/n!. The f(i, n) are the unique coefficients so that (1/(r-1)^(n+1)) Sum_{i=0..n-1} f(i, n) r^{i+1} = Sum_{j>=0} (j^n)/(r^j) whenever n>=1 and abs(r)>1. (End)
O.g.f. for n-th row: (1-x)^(n+1)*polylog(-n, x)/x. - Vladeta Jovovic, Sep 02 2002
Triangle T(n, k), n>0 and k>0, read by rows; given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] (positive integers interspersed with 0's) where DELTA is Deléham's operator defined in A084938.
Sum_{k=1..n} T(n, k)*2^k = A000629(n). - Philippe Deléham, Jun 05 2004
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n} E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
From the relations between the h- and f-polynomials of permutohedra and reciprocals of e.g.f.s described in A049019: (t-1)((t-1)d/dx)^n 1/(t-exp(x)) evaluated at x=0 gives the n-th Eulerian row polynomial in t and the n-th row polynomial in (t-1) of A019538 and A090582. From the Comtet and Copeland references in A139605: ((t+exp(x)-1)d/dx)^(n+1) x gives pairs of the Eulerian polynomials in t as the coefficients of x^0 and x^1 in its Taylor series expansion in x. - Tom Copeland, Oct 05 2008
G.f: 1/(1-x/(1-x*y/1-2*x/(1-2*x*y/(1-3*x/(1-3*x*y/(1-... (continued fraction). - Paul Barry, Mar 24 2010
If n is odd prime, then the following consecutive 2*n+1 terms are 1 modulo n: a((n-1)*(n-2)/2+i), i=0..2*n. This chain of terms is maximal in the sense that neither the previous term nor the following one are 1 modulo n. - _Vladimir Shevelev, Jul 01 2011
From Peter Bala, Sep 29 2011: (Start)
For k = 0,1,2,... put G(k,x,t) := x -(1+2^k*t)*x^2/2 +(1+2^k*t+3^k*t^2)*x^3/3-(1+2^k*t+3^k*t^2+4^k*t^3)*x^4/4+.... Then the series reversion of G(k,x,t) with respect to x gives an e.g.f. for the present table when k = 0 and for A008517 when k = 1.
The e.g.f. B(x,t) := compositional inverse with respect to x of G(0,x,t) = (exp(x)-exp(x*t))/(exp(x*t)-t*exp(x)) = x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... satisfies the autonomous differential equation dB/dx = (1+B)*(1+t*B) = 1 + (1+t)*B + t*B^2.
Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the Eulerian polynomials: A(n,t) counts plane increasing trees on n vertices where each vertex has outdegree <= 2, the vertices of outdegree 1 come in 1+t colors and the vertices of outdegree 2 come in t colors. An example is given below. Cf. A008517. Applying [Dominici, Theorem 4.1] gives the following method for calculating the Eulerian polynomials: Let f(x,t) = (1+x)*(1+t*x) and let D be the operator f(x,t)*d/dx. Then A(n+1,t) = D^n(f(x,t)) evaluated at x = 0.
(End)
With e.g.f. A(x,t) = G[x,(t-1)]-1 in Copeland's 2008 comment, the compositional inverse is Ainv(x,t) = log(t-(t-1)/(1+x))/(t-1). - Tom Copeland, Oct 11 2011
T(2*n+1,n+1) = (2*n+2)*T(2*n,n). (E.g., 66 = 6*11, 2416 = 8*302, ...) - Gary Detlefs, Nov 11 2011
E.g.f.: (1-y) / (1 - y*exp( (1-y)*x )). - Geoffrey Critzer, Nov 10 2012
From Peter Bala, Mar 12 2013: (Start)
Let {A(n,x)} n>=1 denote the sequence of Eulerian polynomials beginning [1, 1 + x, 1 + 4*x + x^2, ...]. Given two complex numbers a and b, the polynomial sequence defined by R(n,x) := (x+b)^n*A(n+1,(x+a)/(x+b)), n >= 0, satisfies the recurrence equation R(n+1,x) = d/dx((x+a)*(x+b)*R(n,x)). These polynomials give the row generating polynomials for several triangles in the database including A019538 (a = 0, b = 1), A156992 (a = 1, b = 1), A185421 (a = (1+i)/2, b = (1-i)/2), A185423 (a = exp(i*Pi/3), b = exp(-i*Pi/3)) and A185896 (a = i, b = -i).
(End)
E.g.f.: 1 + x/(T(0) - x*y), where T(k) = 1 + x*(y-1)/(1 + (k+1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 07 2013
From Tom Copeland, Sep 18 2014: (Start)
A) Bivariate e.g.f. A(x,a,b)= (e^(ax)-e^(bx))/(a*e^(bx)-b*e^(ax)) = x + (a+b)*x^2/2! + (a^2+4ab+b^2)*x^3/3! + (a^3+11a^2b+11ab^2+b^3)x^4/4! + ...
B) B(x,a,b)= log((1+ax)/(1+bx))/(a-b) = x - (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 - (a^3+a^2b+ab^2+b^3)x^4/4 + ... = log(1+u.*x), with (u.)^n = u_n = h_(n-1)(a,b) a complete homogeneous polynomial, is the compositional inverse of A(x,a,b) in x (see Drake, p. 56).
C) A(x) satisfies dA/dx = (1+a*A)(1+b*A) and can be written in terms of a Weierstrass elliptic function (see Buchstaber & Bunkova).
D) The bivariate Eulerian row polynomials are generated by the iterated derivative ((1+ax)(1+bx)d/dx)^n x evaluated at x=0 (see A145271).
E) A(x,a,b)= -(e^(-ax)-e^(-bx))/(a*e^(-ax)-b*e^(-bx)), A(x,-1,-1) = x/(1+x), and B(x,-1,-1) = x/(1-x).
F) FGL(x,y) = A(B(x,a,b) + B(y,a,b),a,b) = (x+y+(a+b)xy)/(1-ab*xy) is called the hyperbolic formal group law and related to a generalized cohomology theory by Lenart and Zainoulline. (End)
For x > 1, the n-th Eulerian polynomial A(n,x) = (x - 1)^n * log(x) * Integral_{u>=0} (ceiling(u))^n * x^(-u) du. - Peter Bala, Feb 06 2015
Sum_{j>=0} j^n/e^j, for n>=0, equals Sum_{k=1..n} T(n,k)e^k/(e-1)^(n+1), a rational function in the variable "e" which evaluates, approximately, to n! when e = A001113 = 2.71828... - Richard R. Forberg, Feb 15 2015
For a fixed k, T(n,k) ~ k^n, proved by induction. - Ran Pan, Oct 12 2015
From A145271, multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by g_n = (d/dx)^n (1+a*x)*(1+b*x) evaluated at x= 0, i.e., g_0 = 1, g_1 = (a+b), g_2 = 2ab, and g_n = 0 otherwise, to obtain the tridiagonal matrix VP with VP(n,k) = binomial(n,k) g_(n-k). Then the m-th bivariate row polynomial of this entry is P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^(m-1) (1, a+b, 2ab, 0, ...)^T, where S is the shift matrix A129185, representing differentiation in the divided powers basis x^n/n!. Also, P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^m (0, 1, 0, ...)^T. - Tom Copeland, Aug 02 2016
Cumulatively summing a row generates the n starting terms of the n-th differences of the n-th powers. Applying the finite difference method to x^n, these terms correspond to those before constant n! in the lowest difference row. E.g., T(4,k) is summed as 0+1=1, 1+11=12, 12+11=23, 23+1=4!. See A101101, A101104, A101100, A179457. - Andy Nicol, May 25 2024

Extensions

Thanks to Michael Somos for additional comments.
Further comments from Christian G. Bower, May 12 2000

A000828 E.g.f. cos(x)/(cos(x) - sin(x)).

Original entry on oeis.org

1, 1, 2, 8, 40, 256, 1952, 17408, 177280, 2031616, 25866752, 362283008, 5535262720, 91620376576, 1633165156352, 31191159799808, 635421069967360, 13753735117275136, 315212388819402752, 7625476699018231808
Offset: 0

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Comments

For a refinement of these numbers see A185896.
A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...|x_n|} = {1,2...,n}. Let x_1,...,x_n be a signed permutation. Then we say 0,x_1,...,x_n,0 is a snake of type S(n;0,0) when 0 < x_1 > x_2 < ... 0. For example, 0 4 -3 -1 -2 0 is a snake of type S(4;0,0). Then a(n) equals the cardinality of S(n;0,0) [Verges]. An example is given below. - Peter Bala, Sep 02 2011
Original name was: E.g.f. cos(x)*(cos(x)+sin(x)) /cos(2*x). - Arkadiusz Wesolowski, Jul 25 2012
Number of plane (that is, ordered) increasing 0-1-2 trees on n vertices where the vertices of outdegree 1 or 2 come in two colors. An example is given below. - Peter Bala, Oct 10 2012

Examples

			a(3) = 8: The eight snakes of type S(3;0,0) are
0 1 -2 3 0, 0 1 -3 2 0, 0 2 1 3 0, 0 2 -1 3 0, 0 2 -3 1 0,
0 3 1 2 0, 0 3 -1 2 0, 0 3 -2 1 0.
1 + x + 2*x^2 + 8*x^3 + 40*x^4 + 256*x^5 + 1952*x^6 + 17408*x^7 + ...
a(3) = 8: The eight increasing 0-1-2 trees on 3 vertices are
..1o (x2 colors)......1o (x2 colors)......1o (x2 colors).....
...|................./.\................./.\.................
..2o (x2 colors)...2o...o3.............3o...o2...............
...|
..3o
Totals.......................................................
...4......+...........2.........+.........2....=...8.........
		

Crossrefs

Programs

  • Maple
    A000828 := n -> (-1)^((n-1)*n/2)*4^n*(Euler(n,1/2)+Euler(n,1))/2: # Peter Luschny, Nov 25 2010
  • Mathematica
    a[n_] := (-1)^((n-1)*n/2)*4^n*(EulerE[n, 1/2] + EulerE[n, 1])/2; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Nov 22 2012, after Peter Luschny *)
  • Maxima
    a(n):=sum(if evenp(n+k) then (-1)^((n+k)/2)*sum(j!/n!*stirling2(n,j)*2^(n-j)*(-1)^(n+j-k)*binomial(j-1,k-1),j,k,n),k,1,n); /* Vladimir Kruchinin, Aug 18 2010 */
    
  • PARI
    my(x='x + O('x^30)); Vec(serlaplace(cos(x)/(cos(x)-sin(x)))) \\ Michel Marcus, Nov 21 2020

Formula

E.g.f.: 1/(1- tan(x)). - Emeric Deutsch, Sep 10 2001
a(n) = A000831(n)/2 for n>0. - Peter Luschny, Nov 25 2010
a(n) = Sum_{k=1..n, n+k is even} (-1)^((n+k)/2)*Sum_{j=k..n} j!/n!*Stirling2(n,j)*2^(n-j)*(-1)^(n+j-k)*binomial(j-1,k-1), n>0. - Vladimir Kruchinin, Aug 18 2010
a(n) = (-1)^((n^2-n)/2)*4^n*(E_{n}(1/2)+E_{n}(1))/2 for n >= 0, where E_{n}(x) is an Euler polynomial. - Peter Luschny, Nov 25 2010
From Peter Bala, Sep 02 2011: (Start)
a(n) = (2*i)^(n-1)*Sum_{k = 1..n} (-1)^(n-k)*k!* Stirling2(n,k) * ((1-i)/2)^(k-1), where i = sqrt(-1).
a(n) = 2^(n-1)*A000111(n) for n >= 1.
Let f(x) = 1+x^2 and define the effect of the operator D on a function g(x) by D(g(x)) = d/dx(f(x)*g(x)). Then for n >= 0, a(n+1) = D^n(1) evaluated at x = 1. (End)
From Sergei N. Gladkovskii, Dec 09 2011 - Dec 23 2013: (Start) Continued fractions:
E.g.f.: 1 + x/(G(0)-x); G(k) = 2*k + 1 - (x^2)/G(k+1).
E.g.f.: 1 + x/(U(0)-2*x) where U(k) = 4*k+1 + x/(1+x/(4*k+3 - x/(1- x/U(k+1)))).
E.g.f.: 1 + x/(U(0)-x) where U(k) = 2*k+1 - x^2/U(k+1).
G.f.: 1 + x/G(0) where G(k) = 1 - x*(2*k+2) - 2*x^2*(k+1)*(k+2)/G(k+1).
E.g.f.: 1 + x/T(0) where T(k) = 4*k+1 - x/(1 - x/(4*k+3 + x/(1 + x/T(k+1)))).
G.f.: 1 + x/Q(0) where Q(k) = 1 - 2*x*(2*k+1) - 2*x^2*(2*k+1)*(2*k+2)/(1 - 2*x*(2*k+2) - 2*x^2*(2*k+2)*(2*k+3)/Q(k+1)).
G.f.: 1 + x/(1-2*x)*T(0) where T(k) = 1 - 2*x^2*(k+1)*(k+2)/( 2*x^2*(k+1)*(k+2) - (1 - 2*x*(k+1))*(1 - 2*x*(k+2))/T(k+1)).
E.g.f.: T(0) where T(k) = 1 + x/(4*k+1 - x/(1 - x/( 4*k+3 + x/T(k+1)))). (End)
G.f.: 1 /(1 - 1*x /(1 - 1*x /(1 - 4*x /(1 - 2*6*x^2 /(1 - 6*x /(1 - 4*x /(1 - 4*x /(1 - 10*x /(1 - 5*12*x^2 /(1 - 12*x / ...)))))))))). - Michael Somos, May 12 2012
a(n) ~ n! * 2^(2*n+1)/Pi^(n+1). - Vaclav Kotesovec, Jun 21 2013
a(0) = a(1) = 1; a(n) = 2 * Sum_{k=1..n-1} binomial(n-1,k) * a(k) * a(n-k-1). - Ilya Gutkovskiy, Nov 21 2020
From Peter Bala, Dec 04 2021: (Start)
F(x) = exp(2*x)*(exp(2*x) - 1)/(exp(4*x) + 1) = x + 2*x^2/2! - 8*x^3/3! - 40*x^4/4! + 256*x^5/5! + 1952*x^6/6! - - + + ... is the e.g.f. for the sequence [1, 2, -8, -40, 256, 1952, ...], a signed version of this sequence without the first term.
Let G(x) = x + 2*x^2 - 8*x^3 - 40*x^4 + 256*x^5 + 1952*x^6 - - + + ... be the corresponding o.g.f. We have the continued fraction representation G(x) = x/(1 - 2*x + 12*x^2/(1 + 20*x^2/(1 - 2*x + 56*x^2/(1 + 72*x^2/(1 - 2*x + ... + 4*n*(4*n-1)*x^2/(1 + 4*n*(4*n+1)*x^2/(1 - 2*x + ... ))))))).
The inverse binomial transform 1/(1 + x)*G(x/(1 + x)) = x - 11*x^3 + 361*x^5 - 24611*x^7 + - ... is a g.f. for a signed and aerated version of A000464. (End)

Extensions

Name changed by Arkadiusz Wesolowski, Jul 25 2012

A008303 Triangle read by rows: T(n,k) (n >= 1, 0 <= k <= ceiling(n/2)-1) = number of permutations of [n] with k peaks.

Original entry on oeis.org

1, 2, 4, 2, 8, 16, 16, 88, 16, 32, 416, 272, 64, 1824, 2880, 272, 128, 7680, 24576, 7936, 256, 31616, 185856, 137216, 7936, 512, 128512, 1304832, 1841152, 353792, 1024, 518656, 8728576, 21253376, 9061376, 353792, 2048, 2084864, 56520704, 222398464, 175627264, 22368256
Offset: 1

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Comments

From Petros Hadjicostas, Aug 06 2019: (Start)
André (1895) first defined these numbers. In his notation, T(n, k) = Q(n+1, 2*(k+1)) for n >= 1 and 0 <= k <= ceiling(n/2)-1.
His triangle is as follows (p. 148):
Q_{2,2}
Q_{3,2}
Q_{4,2} Q_{4,4}
Q_{5,2} Q_{5,4}
Q_{6,2} Q_{6,4} Q_{6,6}
Q_{7,2} Q_{7,4} Q_{7,6}
...
He has Q(n, s) = 0 when either s is odd, or n <= 1, or s > n. Also, Q_{n,2} = 2^(n-2) for n >= 2.
His recurrence is Q(n, s) = s * Q(n-1, s) + (n - s + 1) * Q(n-1, s-2) for n >= 3 and s >= 2. (Obviously, for s odd, we get Q(n, s) = 0 + 0 = 0.)
In terms of the current array, André's (1895) recurrence becomes T(n, k) = (2*k + 2) * T(n-1, k) + (n - 2*k) * T(n-1, k-1) for n >= 2 and 1 <= k <= n with T(n, 0) = 2^(n-1) for n >= 1. In this recurrence, we assume T(n, k) = 0 for k >= ceiling(n/2) or k < 0. (End)
From Petros Hadjicostas, Aug 07 2019: (Start)
We clarify further the quantity Q(n, s) defined by André (1895). In his paper, André considers circular permutations of [n] and deals with maxima, minima, and so-called "séquences" in a permutation.
The term "séquence" in a permutation, as used by André in several of his papers in the 19th century, means a list of consecutive numbers in the permutation that go from a maximum to a minimum, or vice versa, and do not contain any interior minima or maxima. This terminology is also repeated in Ex. 13 (pp. 260-261) by Comtet (even though he refers to the corresponding indices rather than the numbers in the permutation itself).
Some authors call these so-called "séquences" (defined by André and Comtet) "alternate runs" (or just "runs"). Here we are actually dealing with "circular runs" if we read these so-called "séquences" in ascending order in one of the two directions on a circle.
Q(n, s) is the number of circular permutations of [n] (out of the (n-1)! in total) that have exactly s of these so-called "séquences" ("alternate runs").
André (1895) proves that, in a circular permutation of [n], the number of maxima equals the number of minima and that the number of his so-called "séquences" ("alternate runs") is always even (i.e., Q(n, s) = 0 for s odd).
He also shows that, if v = floor(n/2), then the only possible values for the length of a so-called "séquence" ("alternate run") in a circular permutation of [n] are 2, 4, ..., 2*v. That is why Q(n, s) = 0 when either s is odd, or n <= 1, or s > n.
Note that Sum_{t = 1..floor(n/2)} Q_{n, 2*t} = Sum_{t = 1..floor(n/2)} T(n-1, t-1) = (n-1)! = total number of circular permutations of [n].
Since T(n, k) = Q(n+1, 2*(k+1)) for n >= 1 and 0 <= k <= ceiling(n/2)-1, we conclude that the number of (linear) permutations of [n] with k peaks equals the number of circular permutations of [n+1] with exactly 2*(k+1) of these so-called "séquences" ("alternate runs"). (End)
From Petros Hadjicostas, Aug 08 2019: (Start)
The author of this array indirectly assumes that a "peak" of a (linear) permutation of [n] is an interior maximum of the permutation; i.e., we ignore maxima at the endpoints of a permutation.
Similarly, a "valley" of a (linear) permutation of [n] is an interior minimum of the permutation; i.e., we ignore minima at the endpoints of the permutation.
Since the complement of a permutation a_1 a_2 ... a_n (using one-line notation, not cycle notation) is (n+1-a_1) (n+1-a_2) ... (n+1-a_n), it follows that, for n >= 2 and 0 <= k <= ceiling(n/2) - 1, T(n, k) is also the number of (linear) permutations of [n] with exactly k valleys. (End)

Examples

			Triangle T(n,k) (with rows n >= 1 and columns k >= 0) starts as follows:
  [ 1]    1;
  [ 2]    2;
  [ 3]    4,       2;
  [ 4]    8,      16;
  [ 5]   16,      88,      16;
  [ 6]   32,     416,     272;
  [ 7]   64,    1824,    2880,     272;
  [ 8]  128,    7680,   24576,    7936;
  [ 9]  256,   31616,  185856,  137216,    7936;
  [10]  512,  128512, 1304832, 1841152,  353792;
    A000079, A000431, A000487, A000517, A179708, ...
T(3,1) = 2 because we have 132 and 231.
From _Petros Hadjicostas_, Aug 07 2019: (Start)
In terms of André's (1895) notation (see the comments above), we have Q(4, 2) = T(3, 0) = 4 and Q(4, 4) = T(3, 1) = 2.
Out of the (4-1)! = 6 circular permutations of [4], each of the permutations 1324 and 1423 has exactly 4 so-called "séquences" ("alternate runs"), while each of the rest (1234, 1243, 1342, and 1432) has exactly 2 so-called "séquences" ("alternate runs").
In detail, we list the so-called "séquences" ("alternate runs") of the above circular permutations:
  1234 --> 1234 and 41 (maximum 4 and minimum 1).
  1243 --> 124 and 431 (maximum 4 and minimum 1).
  1324 --> 13, 32, 24, and 41 (maxima 3, 4, and minima 1, 2).
  1342 --> 134 and 421 (maximum 4 and minimum 1).
  1423 --> 14, 42, 23, and 31 (maxima 3, 4 and minima 1, 2),
  1432 --> 14 and 4321 (maximum 4 and minimum 1).
(End)
		

References

  • Florence Nightingale David and D. E. Barton, Combinatorial Chance, Charles Griffin, 1962; see Table 10.6, p. 163. [They use the notation T_{N,t^*}^{**}, where N is the length of the permutation and t^* is the number of peaks in the permutation. They also give André's recurrence. So, here n = N and k = t^*. - Petros Hadjicostas, Aug 09 2019]
  • Florence Nightingale David, Maurice George Kendall, and D. E. Barton, Symmetric Functions and Allied Tables, Cambridge, 1966, p. 261, Table 7.3.
  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, John Wiley and Sons, N.Y., 1983, Ex. 3.3.46. - Emeric Deutsch, Jul 26 2009
  • T. K. Petersen, Eulerian Numbers, Birkhäuser, 2015, Chapter 4.

Crossrefs

From Emeric Deutsch, Jul 26 2009: (Start)
Sum of entries in row n is n! = A000142(n).
T(n,0) = 2^(n-1) = A000079(n-1).
T(n,1) = A000431(n).
T(n,2) = A000487(n).
T(n,3) = A000517(n).
T(2n, n-1) = T(2n+1, n) = A000182(n+1) (the tangent numbers). (End)
Columns k = 0-6 give: A011782, A000431, A000487, A000517, A179708, A179709, A179710.

Programs

  • Maple
    # The Maple program yields (by straightforward counting) the generating polynomial of the row n specified in the program.
    n := 8: with(combinat): P := permute(n): st := proc (p) local ct, j: ct := 0: for j from 2 to nops(p)-1 do if p[j-1] < p[j] and p[j+1] < p[j] then ct := ct+1 else end if end do: ct end proc: sort(add(t^st(P[j]), j = 1 .. factorial(n))); # Emeric Deutsch, Jul 26 2009
    # Second Maple program:
    a := 1+sqrt(1-t): b := 1-sqrt(1-t): G := (exp(b*z)-exp(a*z))/(b*exp(a*z)-a*exp(b*z)): Gser := simplify(series(G, z = 0, 15)): for n to 12 do P[n] := sort(expand(factorial(n)*coeff(Gser, z, n))) end do: for n to 12 do seq(coeff(P[n], t, j), j = 0 .. ceil((1/2)*n)-1) end do; # yields sequence in triangular form - Emeric Deutsch, Jul 26 2009
    # Third Maple program:
    b:= proc(u, o, t) option remember; expand(`if`(u+o=0, 1,
          add(b(u-j, o+j-1, 0)*x^t, j=1..u)+
          add(b(u+j-1, o-j, 1), j=1..o)))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
    seq(T(n), n=1..15);  # Alois P. Heinz, May 22 2014
    # Recurrence of D. André (1895).
    T := proc(n, k) option remember;
    if n < 1 or 2*k > (n-1) then return 0 fi;
    if k = 0 then return 2^(n-1) fi;
    (2*k + 2)*T(n-1, k) + (n - 2*k)*T(n-1, k-1) end:
    seq(seq(T(n, k), k=0..(n-1)/2), n=1..12); # Peter Luschny, Aug 06 2019
  • Mathematica
    From Luc Roy, Jul 08 2010: (Start)
    It appears that one-half of the sequence A008303 can be obtained with this Mathematica program:
    Expand[CoefficientList[Simplify[InverseSeries[Integrate[
    Series[(1 + m Sinh[x]^2)^(-1), {x, 0, 15}, {m, 0, 15}], x]]], x]
    Denominator[CoefficientList[Series[Exp[x], {x, 0, 15}], x]]]
    (* Mathematica Output of Luc Roy's program *)
    {0, 1, 0, 2 m, 0, 8 m + 16 m^2, 0, 32 m + 416 m^2 + 272 m^3, 0, 128 m + 7680 m^2 + 24576 m^3 + 7936 m^4, 0, 512 m + 128512 m^2 + 1304832 m^3 + 1841152 m^4 + 353792 m^5, 0, 2048 m + 2084864 m^2 + 56520704 m^3 + 222398464 m^4 + 175627264 m^5 + 22368256 m^6, 0, 8192 m + 33497088 m^2 + 2230947840 m^3 + 20261765120 m^4 + 41731645440 m^5 + 21016670208 m^6 + 1903757312 m^7}
    (End)
    (* Another Mathematica program *)
    m = 14; a = 1 + Sqrt[1 - t]; b = 1 - Sqrt[1 - t];
    g[z_] = (E^(b*z) - E^(a*z))/(b*E^(a*z) - a*E^(b*z));
    gser = Series[g[z], {z, 0, m}];
    Do[p[n]=n!*Coefficient[gser, z, n]//Simplify, {n, 0, m}];
    Flatten[ Table[ Coefficient[p[n], t, j], {n, 0, m}, {j, 0, Ceiling[n/2] - 1}]]
    (* Jean-François Alcover, May 27 2011, after Emeric Deutsch *)
    (* To get the triangle from Jean-François Alcover's Mathematica program *)
    FormTable[Table[Coefficient[p[n], t, j], {n, 0, m}, {j, 0, Ceiling[n/2] - 1}]]
    (* Petros Hadjicostas, Aug 06 2019 *)
    gf := Sqrt[x - 1] Cot[y Sqrt[x - 1]] - 1; ser := Series[1/gf, {y, 0, 16}];
    cy[n_] := n! Coefficient[ser, y, n]; row[n_] := CoefficientList[cy[n], x];
    Table[row[n], {n, 1, 12}] // Flatten (* Peter Luschny, Aug 06 2019 *)
  • PARI
    {T(n, k) = if(n<1, 0, my(z = sqrt(1 - y + y*O(y^(n\2)))); n!*polcoef(polcoef(z/(z - tanh(x*z)), n, x), k))}; /* Michael Somos, May 24 2023 */

Formula

From Emeric Deutsch, Jul 26 2009: (Start)
E.g.f.: G(t,z)=[exp(bz)-exp(az)]/[b*exp(az)-a*exp(bz)], where a+b=2 and ab=t, i.e., a=1+sqrt(1-t), b=1-sqrt(1-t) (see the Goulden-Jackson reference). -
Sum_{k>=0} k*T(n,k) = n!*(n-2)/3 = A090672(n-1).
Row n has ceiling(n/2) terms. (End)
E.g.f.: tan(t*sqrt(x-1))/(sqrt(x-1)-tan(t*sqrt(x-1))) = Sum_{n>=0} P(n,x)*t^n/n! = t + 2*t^2/2! + (4+2*x)*t^3/3! + (8+16*x)*t^4/4! + .... The row generating polynomials P(n,x) satisfy x^(n-1)*P(n,1+1/x^2) = R(n-1,x), where R(n,x) are the row polynomials of A185896. A000670(n) = (3/2)^(n-1)*P(n,8/9). - Peter Bala, Oct 14 2011
From Jinyuan Wang, Dec 28 2020: (Start)
T(n, k) = (n - 2*k + 2)*T(n-1, k-1) + 2*k*T(n-1, k) for n > 1 and k > 1; T(n, 1) = 2^(n - 1); T(1, k) = 0 for k > 1.
T(2*k-1, k) = A000182(k). (End)
From Ammar Khatab, Aug 17 2024: (Start)
T(2*n,k) = 4^(n-k+1)* Sum_{p=0..k} (-1)^p * (2*p+2*n-2*k-1)/(p+2*n-2*k-1) binomial(p+2*n-2*k-1,p) (A008292(2*n,k-p+1)+A008292(2*n,2*n+p-k) ) for n>0.
T(2*n+1,k) = 4^(n-k)* Sum_{p=0..k} (-1)^p * (p+n-k)/(p+2*n-2*k) binomial(p+2*n-2*k,p) (A008292(2*n+1,k-p+1)+A008292(2*n,2*n+p-k+1) ) for k<>n. (End)

Extensions

Additional comments from Emeric Deutsch, May 08 2004
More terms from R. J. Mathar and Vladeta Jovovic, Jun 26 2007
Corrected by Emeric Deutsch, Jul 26 2009
Edited definition - N. J. A. Sloane, May 25 2023

A104035 Triangle T(n,k), 0 <= k <= n, read by rows, defined by T(0,0) = 1; T(0,k) = 0 if k>0 or if k<0; T(n,k) = k*T(n-1,k-1) + (k+1)*T(n-1,k+1).

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 0, 5, 0, 6, 5, 0, 28, 0, 24, 0, 61, 0, 180, 0, 120, 61, 0, 662, 0, 1320, 0, 720, 0, 1385, 0, 7266, 0, 10920, 0, 5040, 1385, 0, 24568, 0, 83664, 0, 100800, 0, 40320, 0, 50521, 0, 408360, 0, 1023120, 0, 1028160, 0, 362880, 50521, 0, 1326122, 0, 6749040
Offset: 0

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Author

Philippe Deléham, Apr 06 2005

Keywords

Comments

Or, triangle of coefficients (with exponents in increasing order) in polynomials Q_n(u) defined by d^n sec x / dx^n = Q_n(tan x)*sec x.
Interpolates between factorials and Euler (or secant) numbers. Related to Springer numbers.
Companion triangles are A155100 (derivative polynomials of tangent function) and A185896 (derivative polynomials of squared secant function).
A combinatorial interpretation for the polynomial Q_n(u) as the generating function for a sign change statistic on certain types of signed permutation can be found in [Verges]. A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...,|x_n|} = {1,2,...,n}. They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube.
Let x_1,...,x_n be a signed permutation. Adjoin x_0 = 0 to the front of the permutation and x_(n+1) = (-1)^n*(n+1) to the end to form x_0,x_1,...,x_n,x_(n+1). Then x_0,x_1,...,x_n,x_(n+1) is a snake of type S(n;0) when x_0 < x_1 > x_2 < ... x_(n+1). For example, 0 3 -1 2 -4 is a snake of type S(3;0).
Let sc be the number of sign changes through a snake ... sc = #{i, 0 <= i <= n, x_i*x_(i+1) < 0}. For example, the snake 0 3 -1 2 -4 has sc = 3. The polynomial Q_n(u) is the generating function for the sign change statistic on snakes of type S(n;0): ... Q_n(u) = sum {snakes in S(n;0)} u^sc. See the example section below for the cases n = 2 and n = 3.
PRODUCTION MATRIX
Let D = subdiag(1,2,3,...) be the array with the indicated sequence on the first subdiagonal and zeros elsewhere and let C = transpose(D). The production matrix for this triangle is C+D: the first row of (C+D)^n is the n-th row of this triangle. D represents the derivative operator d/dx and C represents the operator p(x) -> x*d/dx(x*p(x)) acting on the basis monomials {x^n}n>=0. See Formula (1) below.

Examples

			The polynomials Q_0(u) through Q_6(u) (with exponents in decreasing order) are:
  1
  u
  2*u^2 + 1
  6*u^3 + 5*u
  24*u^4 + 28*u^2 + 5
  120*u^5 + 180*u^3 + 61*u
  720*u^6 + 1320*u^4 + 662*u^2 + 61
Triangle begins:
  1
  0 1
  1 0 2
  0 5 0 6
  5 0 28 0 24
  0 61 0 180 0 120
  61 0 662 0 1320 0 720
  0 1385 0 7266 0 10920 0 5040
  1385 0 24568 0 83664 0 100800 0 40320
  0 50521 0 408360 0 1023120 0 1028160 0 362880
  50521 0 1326122 0 6749040 0 13335840 0 11491200 0 3628800
  0 2702765 0 30974526 0 113760240 0 185280480 0 139708800 0 39916800
  2702765 0 98329108 0 692699304 0 1979524800 0 2739623040 0 1836172800 0 479001600
Examples of sign change statistic sc on snakes of type S(n;0)
= = = = = = = = = = = = = = = = = = = = = =
.....Snakes....# sign changes sc.......u^sc
= = = = = = = = = = = = = = = = = = = = = =
n=2
...0 1 -2 3...........2.................u^2
...0 2  1 3...........0.................1
...0 2 -1 3...........2.................u^2
yields Q_2(u) = 2*u^2 + 1.
n=3
...0 1 -2  3 -4.......3.................u^3
...0 1 -3  2 -4.......3.................u^3
...0 1 -3 -2 -4.......1.................u
...0 2  1  3 -4.......1.................u
...0 2 -1  3 -4.......3.................u^3
...0 2 -3  1 -4.......3.................u^3
...0 2 -3 -2 -4.......1.................u
...0 3  1  2 -4.......1.................u
...0 3 -1  2 -4.......3.................u^3
...0 3 -2  1 -4.......3.................u^3
...0 3 -2 -1 -4.......1.................u
yields Q_3(u) = 6*u^3 + 5*u.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, p. 287.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see pp. 445 and 469.

Crossrefs

See A008294 for another version of this triangle.
Setting u=0,1,2,3,4 gives A000364, A001586, A156129, A156131, A156132.
Setting u=sqrt(2) gives A156134 and A156138; u=sqrt(3) gives A002437 and A002439.

Programs

Formula

T(n, n) = n!; T(n, 0) = 0 if n = 2m + 1; T(n, 0) = A000364(m) if n = 2m.
Sum_{k>=0} T(m, k)*T(n, k) = T(m+n, 0).
Sum_{k>=0} T(n, k) = A001586(n): Springer numbers.
G.f.: Sum_{n >= 0} Q_n(u)*t^n/n! = 1/(cos t - u sin t).
From Peter Bala: (Start)
RECURRENCE RELATION
For n>=0,
(1)... Q_(n+1)(u) = d/du Q_n(u) + u*d/du(u*Q_n(u))
... = (1+u^2)*d/du Q_n(u) + u*Q_n(u),
with starting condition Q_0(u) = 1. Compare with Formula (4) of A186492.
RELATION WITH TYPE B EULERIAN NUMBERS
(2)... Q_n(u) = ((u+i)/2)^n*B(n,(u-i)/(u+i)), where i = sqrt(-1) and
[B(n,u)]n>=0 = [1,1+u,1+6*u+u^2,1+23*u+23*u^2+u^3,...] is the sequence of type B Eulerian polynomials (with a factor of u removed) - see A060187.
(End)
T(n,0) = abs(A122045(n)). - Reinhard Zumkeller, Apr 27 2014

Extensions

Entry revised by N. J. A. Sloane, Nov 06 2009

A155100 Triangle read by rows: coefficients in polynomials P_n(u) arising from the expansion of D^(n-1) (tan x) in increasing powers of tan x for n>=1 and 1 for n=0.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 2, 2, 0, 8, 0, 6, 0, 16, 0, 40, 0, 24, 16, 0, 136, 0, 240, 0, 120, 0, 272, 0, 1232, 0, 1680, 0, 720, 272, 0, 3968, 0, 12096, 0, 13440, 0, 5040, 0, 7936, 0, 56320, 0, 129024, 0, 120960, 0, 40320, 7936, 0, 176896, 0, 814080, 0, 1491840
Offset: 0

Views

Author

N. J. A. Sloane, Nov 05 2009

Keywords

Comments

The definition is d^(n-1) tan x / dx^n = P_n(tan x) for n>=1 and 1 for n=0.
Interpolates between factorials and tangent numbers.
From Peter Bala, Mar 02 2011: (Start)
Companion triangles are A104035 and A185896.
A combinatorial interpretation for the polynomial P_n(t) as the generating function for a sign change statistic on certain types of signed permutation can be found in [Verges].
A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...,|x_n|} = {1,2,...,n}.
They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube.
Let x_1,...,x_n be a signed permutation and put x_0 = -(n+1) and x_(n+1) = (-1)^n*(n+1). Then x_0,x_1,...,x_n,x_(n+1) is a snake of type S(n) when x_0 < x_1 > x_2 < ... x_(n+1). For example, -5 4 -3 -1 -2 5 is a snake of type S(4).
Let sc be the number of sign changes through a snake sc = #{i, 0 <= i <= n, x_i*x_(i+1) < 0}. For example, the snake -5 4 -3 -1 -2 5 has sc = 3.
The polynomial P_(n+1)(t) is the generating function for the sign change statistic on snakes of type S(n): P_(n+1)(t) = sum {snakes in S(n)} t^sc.
See the example section below for the cases n=1 and n=2.
(End)
Equals A107729 when the first column is removed. - Georg Fischer, Jul 26 2023

Examples

			The polynomials P_{-1}(u) through P_6(u) with exponents in decreasing order:
      1
      u
      u^2 +    1
    2*u^3 +    2*u
    6*u^4 +    8*u^2 +    2
   24*u^5 +   40*u^3 +   16*u
  120*u^6 +  240*u^4 +  136*u^2 +  16
  720*u^7 + 1680*u^5 + 1232*u^3 + 272*u
  ...
Triangle begins:
  1
  0, 1
  1, 0, 1
  0, 2, 0, 2
  2, 0, 8, 0, 6
  0, 16, 0, 40, 0, 24
  16, 0, 136, 0, 240, 0, 120
  0, 272, 0, 1232, 0, 1680, 0, 720
  272, 0, 3968, 0, 12096, 0, 13440, 0, 5040
  0, 7936, 0, 56320, 0, 129024, 0, 120960, 0, 40320
  7936, 0, 176896, 0, 814080, 0, 1491840, 0, 1209600, 0, 362880
  0, 353792, 0, 3610112, 0, 12207360, 0, 18627840, 0, 13305600, 0, 3628800
  ...
From _Peter Bala_, Feb 07 2011: (Start)
Examples of sign change statistic sc on snakes of type S(n):
    Snakes     # sign changes sc  t^sc
  ===========  =================  ====
n=1:
  -2  1 -2 ........... 2 ........ t^2
  -2 -1 -2 ........... 0 ........ 1
                  yields P_2(t) = 1 + t^2;
n=2:
  -3  1 -2  3 ........ 3 ........ t^3
  -3  2  1  3 ........ 1 ........ t
  -3  2 -1  3 ........ 3 ........ t^3
  -3 -1 -2  3 ........ 1 ........ t
                  yields P_3(t) = 2*t + 2*t^3. (End)
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, p. 287.

Crossrefs

For other versions of this triangle see A008293, A101343.
A104035 is a companion triangle.
Highest order coefficients give factorials A000142. Constant terms give tangent numbers A000182. Other coefficients: A002301.
Setting u=1 in P_n gives A000831, u=2 gives A156073, u=3 gives A156075, u=4 gives A156076, u=1/2 gives A156102.
Setting u=sqrt(2) in P_n gives A156108 and A156122; setting u=sqrt(3) gives A156103 and A000436.

Programs

  • Maple
    P:=proc(n) option remember;
    if n=-1 then RETURN(1); elif n=0 then RETURN(u); else RETURN(expand((u^2+1)*diff(P(n-1),u))); fi;
    end;
    for n from -1 to 12 do t1:=series(P(n),u,20); lprint(seriestolist(t1)); od:
    # Alternatively:
    with(PolynomialTools): seq(print(CoefficientList(`if`(i=0,1,D@@(i-1))(tan),tan)), i=0..7); # Peter Luschny, May 19 2015
  • Mathematica
    p[n_, u_] := D[Tan[x], {x, n}] /. Tan[x] -> u /. Sec[x] -> Sqrt[1 + u^2] // Expand; p[-1, u_] = 1; Flatten[ Table[ CoefficientList[ p[n, u], u], {n, -1, 9}]] (* Jean-François Alcover, Jun 28 2012 *)
    T[ n_, k_] := Which[n<0, Boole[n==-1 && k==0], n==0, Boole[k==1], True, (k-1)*T[n-1, k-1] + (k+1)*T[n-1, k+1]]; (* Michael Somos, Jul 09 2024 *)
  • PARI
    {T(n, k) = if(n<0, n==-1 && k==0, n==0, k==1, (k-1)*T(n-1, k-1) + (k+1)*T(n-1, k+1))}; /* Michael Somos, Jul 09 2024 */

Formula

If the polynomials are denoted by P_n(u), we have the recurrence P_{-1}=1, P_0 = u, P_n = (u^2+1)*dP_{n-1}/du.
G.f.: Sum_{n >= 0} P_n(u) t^n/n! = (sin t + u*cos t)/(cos t - u sin t). [Hoffman]
From Peter Bala, Feb 07 2011: (Start)
RELATION WITH BERNOULLI NUMBERS A000367 AND A002445
Put T(n,t) = P_n(i*t), where i = sqrt(-1). We have the definite integral evaluation, valid when both m and n are >=1 and m+n >= 4:
int( T(m,t)*T(n,t)/(1-t^2), t = -1..1) = (-1)^((m-n)/2)*2^(m+n-1)*Bernoulli(m+n-2).
The case m = n is equivalent to the result of [Grosset and Veselov]. The methods used there extend to the general case.
RELATION WITH OTHER ROW POLYNOMIALS
The following three identities hold for n >= 1:
P_(n+1)(t) = (1+t^2)*R(n-1,t) where R(n,t) is the n-th row polynomial of A185896.
P_(n+1)(t) = (-2*i)^n*(t-i)*R(n,-1/2+1/2*i*t), where i = sqrt(-1) and R(n,x) is an ordered Bell polynomial, that is, the n-th row polynomial of A019538.
P_(n+1)(t) = (t-i)*(t+i)^n*A(n,(t-i)/(t+i)), where {A(n,t)}n>=1 = [1,1+t,1+4*t+t^2,1+11*t+11*t^2+t^3,...] is the sequence of Eulerian polynomials - see A008292. (End)
T(n,k) = cos((n+k)*Pi/2) * Sum_{p=0..n-1} A008292(n-1,p+1) Sum_{j=0..k}(-1)^(p+j+1) * binomial(p+1,k-j) *binomial(n-p-1,j) for n>1. - Ammar Khatab, Aug 15 2024

Extensions

Name clarified by Peter Luschny, May 25 2015

A214406 Triangle of second-order Eulerian numbers of type B.

Original entry on oeis.org

1, 1, 1, 1, 8, 3, 1, 33, 71, 15, 1, 112, 718, 744, 105, 1, 353, 5270, 14542, 9129, 945, 1, 1080, 33057, 191384, 300291, 129072, 10395, 1, 3265, 190125, 2033885, 6338915, 6524739, 2071215, 135135, 1, 9824, 1038780, 18990320, 103829590, 204889344, 150895836, 37237680, 2027025
Offset: 0

Views

Author

Peter Bala, Jul 17 2012

Keywords

Comments

The second-order Eulerian numbers A008517 count Stirling permutations by ascents. A Stirling permutation of order n is a permutation of the multiset {1,1,2,2,...,n,n} such that for each i, 1 <= i <= n, the elements lying between the two occurrences of i are larger than i.
We define a signed Stirling permutation of order n to be a vector (x_0, x_1, ..., x_(2*n)) such that x_0 = 0 and (|x_1|, ... ,|x_(2*n)|) is a Stirling permutation of order n. We say that a signed Stirling permutation (x_0, x_1, ... , x_(2*n)) has an ascent at position j, 0 <= j <= 2*n-1, if |x_j| < |x_(j+1)|. We define T(n,k), the second-order Eulerian numbers of type B, as the number of signed Stirling permutations of order n having k ascents. An example is given below.

Examples

			Row 2: [1,8,3]:
Signed Stirling permutations of order 2
= = = = = = = = = = = = = = = = = = = =
..............ascents...................ascents
(0 2 2 1 1)......1.......(0 -2 -2 1 1).....1
(0 1 2 2 1)......2.......(0 1 -2 -2 1).....2
(0 1 1 2 2)......2.......(0 1 1 -2 -2).....1
(0 2 2 -1 -1)....1.......(0 -2 -2 -1 -1)...1
(0 -1 2 2 -1)....1.......(0 -1 -2 -2 -1)...1
(0 -1 -1 2 2)....1.......(0 -1 -1 -2 -2)...0
............................................
Triangle begins
.n\k.|..0.....1......2.......3......4........5......6
= = = = = = = = = = = = = = = = = = = = = = = = = = =
..0..|..1
..1..|..1.....1
..2..|..1.....8......3
..3..|..1....33.....71......15
..4..|..1...112....718.....744....105
..5..|..1...353...5270...14542...9129......945
..6..|..1..1080..33057..191384..300291..129072..10395
...
Recurrence example: T(4,2) = 11*T(3,1) + 5*T(3,2) = 11*33 + 5*71 = 718.
		

Crossrefs

Cf. A001813 (row sums), A008517, A039755, A185896, A034940.

Programs

  • Mathematica
    T[n_, k_] /; 0 < k <= n := T[n, k] = (4n-2k-1) T[n-1, k-1] + (2k+1) T[n-1, k]; T[, 0] = 1; T[, _] = 0;
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 11 2019 *)

Formula

T(n,k) = Sum_{i = 0..k} (-1)^(i-k)*binomial(2*n+1,k-i)*S(n+i,i), where S(n,k) = 1/(2^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n = A039755(n,k).
It appears that Sum_{k = 0..n} (-1)^(k+1)*T(n,k)/((2*n-k)*binomial(2*n,k)) = (-1)^n *(2^n-2)*Bernoulli(n)/n.
Recurrence equation: T(n,k) = (4*n-2*k-1)*T(n-1,k-1) + (2*k+1)*T(n-1,k), for n,k >= 0.
The row polynomials R(n,x) may be calculated by means of the recurrence equation R(0,x) = 1 and for n >= 0, R(n,x^2) = (1 - x^2)^(2*n)*d/dx( x/(1-x^2)^(2*n-1)*R(n-1,x^2) ). Equivalently, x*R(n,x^2)/(1 - x^2)^(2*n+1)) = D^n(x), where D is the differential operator x/(1 - x^2)*d/dx.
Another recurrence is R(n+1,x) = 2*x*(1 - x)*d/dx(R(n,x)) + (1 + (4*n+1)*x)*R(n,x). It follows that the row polynomials R(n,x) have only real zeros (apply Liu and Wang, Corollary 1.2 with f(x) = R(n,x) and g(x) = R'(n,x)).
For n >= 0, the rational functions Q(n,x) := R(n,x)/(1 - x)^(2*n+1) are the o.g.f.'s for the diagonals of the type B Stirling numbers of the second kind A039755. They appear to satisfy the semi-orthogonality property Integral_{x = 0..oo} (1 - x)*Q(n,x)*Q(m,x) dx = (-1)^n*(2^(n+m) - 2)*Bernoulli(n+m)/(n+m), for n, m >= 0 but excluding the case (n,m) = (0,0). A similar result holds for the row polynomials of A185896.
Row sums are A001813.
Define functions F(n,z) := Sum_{k >= 0} (2*k+1)^(k+n)*z^k/k!, n = 0,1,2,.... Then exp(-x/2)*F(n,x/2*exp(-x)) = R(n,x)/(1 - x)^(2*n+1). - Peter Bala, Jul 26 2012

Extensions

Missing 1 in data inserted by Jean-François Alcover, Nov 11 2019

A080795 Number of minimax trees on n nodes.

Original entry on oeis.org

1, 1, 4, 20, 128, 1024, 9856, 110720, 1421312, 20525056, 329334784, 5812797440, 111923560448, 2334639652864, 52444850814976, 1262260748288000, 32405895451246592, 883950436237705216, 25530268718794276864
Offset: 0

Views

Author

Emanuele Munarini, Mar 14 2003

Keywords

Comments

A minimax tree is (i) rooted, (ii) binary (i.e., each node has at most two sons), (iii) topological (i.e., the left son is different from the right son), (iv) labeled (i.e., there is a bijection between the nodes and a finite totally ordered set). Moreover it has the following property: (v) the label of each node x is the minimum or the maximum of all the labels of the nodes of the subtree whose root is x.

Crossrefs

Programs

  • Maple
    w := (sqrt(2) - 1)/2:
    seq(simplify((2*sqrt(2))^(n-1)*add(k!*Stirling2(n, k)*w^(k-1), k = 1..n)), n = 1..20); # Peter Bala, Oct 31 2024
  • Mathematica
    Range[0, 18]! CoefficientList[ Series[ Tanh[ ArcTanh[ Sqrt[2]] - Sqrt[2] x]/Sqrt[2], {x, 0, 18}], x] (* Robert G. Wilson v *)
  • PARI
    {Stirling2(n,k)=(1/k!)*sum(j=0,k,(-1)^j*binomial(k,j)*(k-j)^n)}
    /* Finite sum given by Peter Bala: */
    {a(n)=local(w=(sqrt(2)-1)/2);if(n==0,1,round((2*sqrt(2))^(n-1)*sum(k=1,n,k!*Stirling2(n,k)*w^(k-1))))}

Formula

E.g.f.: ( tanh(arctanh(sqrt(2)) - sqrt(2)*x) )/sqrt(2) = sqrt(2)/2* (1 + (3-2*sqrt(2))* exp(2*sqrt(2)*x) )/( 1 - (3-2*sqrt(2))* exp(2*sqrt(2)*x) ).
Recurrence: a(n+1) = 2*(Sum_{k=0..n} binomial(n,k)*a(k)*a(n-k)) - 0^n.
a(2*n) = 2^n * A006154(2*n), n>0 (conjectured). - Ralf Stephan, Apr 29 2004
For n>0, a(n) = sqrt(2)^(3*n+1)*Sum_{k>=0} k^n/(1+sqrt(2))^(2*k). - Benoit Cloitre, Jan 12 2005
From Peter Bala, Jan 30 2011: (Start)
A finite sum equivalent to the previous formula of Benoit Cloitre is
a(n) = (2*sqrt(2))^(n-1)*Sum_{k = 1..n} k!*Stirling2(n,k)*w^(k-1), for n >= 1, with w = (sqrt(2) - 1)/2.
This formula can be used to prove congruences for a(n). For example, a(p) == (-1)^((p^2-1)/8) (mod p) for odd prime p.
For similar formulas for labeled plane and non-plane unary-binary trees see A080635 and A000111 respectively.
For a sequence of related polynomials see A185419. For a recursive table to calculate a(n) see A185420.
The e.g.f. A(x) satisfies the autonomous differential equation d/dx (A(x)) = 2*A(x)^2 - 1. (End)
From Peter Bala, Aug 26 2011: (Start)
The inverse function A(x)^(-1) of the generating function A(x) satisfies A(x)^(-1) = Integral_{t = 1..x} 1/(2*t^2 - 1) dt.
Let f(x) = 2*x^2 - 1. Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0 (see A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x)). Then by [Dominici, Theorem 4.1] we have a(n+1) = D^n[f](1).
For n >= 1 we have a(n) = (2 + sqrt(2))^(n-1)*A(n, 3 - 2*sqrt(2)), where {A(n, x)}n>=1 = [1, 1 + x, 1 + 4*x + x^2, 1 + 11*x + 11*x^2 + x^3, ...] denotes the sequence of Eulerian polynomials (see A008292).
a(n+1) = (-1)^n*(sqrt(-2))^n * R(n, sqrt(-2)) where R(n, x) are the polynomials defined in A185896 (derivative polynomials associated with the function sec^2(x)). (End)
G.f.: 1 + x/G(0) where G(k) = 1 - 4*x*(k+1) - 2*x^2*(k+1)*(k+2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 11 2013
G.f.: 1 + x/(G(0) -x), where G(k) = 1 - x*(k+1) - 2*x*(k+1)/(1 - x*(k+2)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
E.g.f.: sqrt(2)*( -1/2 + (3+2*sqrt(2))/(4 + 2*sqrt(2)- E(0) )), where E(k) = 2 + 2*sqrt(2)*x/( 2*k+1 - 2*sqrt(2)*x/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 27 2013
a(n) ~ n! * 2^((3*n+1)/2) / (log(3+2*sqrt(2)))^(n+1). - Vaclav Kotesovec, Feb 25 2014

A186695 A Galton triangle: T(n,k) = (2k-1)*(T(n-1,k) + T(n-1,k-1)): a type B analog of the ordered Bell numbers A019538.

Original entry on oeis.org

1, 1, 3, 1, 12, 15, 1, 39, 135, 105, 1, 120, 870, 1680, 945, 1, 363, 4950, 17850, 23625, 10395, 1, 1092, 26565, 159600, 373275, 374220, 135135, 1, 3279, 138285, 1303155, 4795875, 8222445, 6621615, 2027025
Offset: 1

Views

Author

Peter Bala, Mar 26 2011

Keywords

Comments

The row polynomials R(n,x) of A019538 satisfy the recurrence relation R(n+1,x) = x*d/dx((1+x)*R(n,x)), and have the expansion R(n,x) = Sum_{k = 1..n} k!*Stirling2(n,k)*x^k.
Here we consider a sequence of polynomials P(n,x) (n>=1) defined by means of the similar recursion P(n+1,x) = x*d/dx((1+x^2)*P(n,x)), with starting value P(1,x) = x.
The first few polynomials are P(1,x) = x, P(2,x) = x + 3*x^3, P(3,x) = x + 12*x^3 + 15*x^5, and P(4,x) = x + 39*x^3 + 135*x^5 + 105*x^7.
Clearly, the P(n,x) are odd polynomials of the form P(n,x) = Sum_{k = 1..n} T(n,k)*x^(2*k-1).
This triangle lists the coefficients T(n,k). They are related to A039755, the type B Stirling numbers of Suter, by T(n,k) = (2*k-1)!!*A039755(n-1,k-1).

Examples

			Triangle begins
  n\k.|..1.....2.....3......4......5......6
  =========================================
  ..1.|..1
  ..2.|..1.....3
  ..3.|..1....12....15
  ..4.|..1....39...135....105
  ..5.|..1...120...870...1680....945
  ..6.|..1...363..4950..17850..23625..10395
  ..
Examples of recurrence relation
  T(4,3) = 5*(T(3,3)+T(3,2)) = 5*(15+12) = 135;
  T(6,4) = 7*(T(5,4)+T(5,3)) = 7*(1680+870) = 17850.
		

Crossrefs

Programs

  • Maple
    A186695 := proc(n, k) option remember; if k < 1 or k > n then 0; elif k = 1 then 1; else (2*k-1)*(procname(n-1, k) + procname(n-1, k-1)) ; end if; end proc: seq(seq(A186695(n,k),k = 1..n),n = 1..10);
  • Mathematica
    T[n_, k_] := (2k-1)! Sum[(-1)^(k-j-1) (2j+1)^(n-1) Binomial[k-1, j], {j, 0, k-1}] / (2^(k-1) (k-1)!)^2;
    Table[T[n, k], {n, 1, 8}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 02 2019 *)

Formula

T(n+1,k+1) = ((2*k+1)!/(2^k*k!)^2)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.
Recurrence relation: T(n,k) = (2k-1)*(T(n-1,k) + T(n-1,k-1)) with boundary conditions T(n,1) = 1, T(1,k) = 0 for k >= 2.
E.g.f.: F(x,t) = (x/(1+x))*(exp(t)/sqrt((1+x) - x*exp(2*t)) - 1) = Sum_{n>=1} R(n,x)*t^n/n! = x*t + (x + 3*x^2)*t^2/2! + (x + 12*x^2 + 15*x^3)*t^3/3! + ....
Compare with the e.g.f. for A019538, which is (x/(1+x))*(exp(t)/((1+x) - x*exp(t))-1).
The row polynomials R(n,x) are related to the polynomials P(n,x) of the comments section by P(n,x) = 1/x*R(n,x^2).
The generating function F(x,t) satisfies the partial differential equation d/dt(F) = 2*x*(1+x)*d/dx(F) + (x-1)*F + x.
It follows that the polynomials P(n,x) := Sum_{k = 1..n} T(n,k)*x^(2*k-1) satisfy the recurrence P(n+1,x) = x*d/dx((1+x^2)*P(n,x)), with P(1,x) = x. (Cf. the recurrence relation for row polynomials of A185896.)
The recurrence relation for T(n,k) given above now follows.
The row polynomials R(n,x) = Sum_{k = 1..n} T(n,k)*x^k satisfy R(n,-x-1) = (-1)^n*((1+x)/x)*S(n,x), where S(n,x) is the n-th row polynomial of A187075.
In addition, R(n,1/(x-1)) = (1/(x-1)^n)*Q(n-1,x), where Q(n,x) is the n-th row polynomial of A156919.
Row sums are [1,4,28,280,3616...] = 1/2*A124212(n) for n >= 1.
Main diagonal is [1,3,15,105,...] = A001147(k) for k >= 1.
Put S(n) = sum {k = 1..n} (-1)^k*T(n,k)/(k+1). Then for m>=2, S(2*m-1) = S(2*m) = (4^m-1)*Bernoulli(2*m)/m.
From Peter Bala, Aug 30 2016: (Start)
n-th row polynomial R(n,x) = 1/(1 + x)^(3/2) * Sum_{k >= 0} (1/4)^k*(x/(1 + x))^k*binomial(2*k,k)*(2*k + 1)^n.
R(n,x) = (1/(1 + x))*Sum_{k = 0..n} binomial(2*k,k)*A145901(n,k)*(x/4)^k. (End)

A189392 E.g.f. tan(x)/(1-tan(x)-tan(x)^2).

Original entry on oeis.org

0, 1, 2, 14, 88, 856, 8912, 115184, 1644928, 26916736, 484527872, 9646440704, 208891091968, 4908327894016, 124094242629632, 3363100573005824, 97195058375262208, 2984961531977629696, 97056253813754888192, 3331252527903082348544
Offset: 0

Views

Author

Vladimir Kruchinin, Apr 21 2011

Keywords

Examples

			G.f. = x + 2*x^2 + 14*x^3 + 88*x^4 + 856*x^5 + 8912*x^6 + 115184*x^7 + ...
		

Crossrefs

Cf. A185896.

Programs

  • Mathematica
    nn = 20; Range[0, nn]! CoefficientList[Series[Tan[x]/(1 - Tan[x] - Tan[x]^2), {x, 0, nn}], x] (* T. D. Noe, Mar 14 2013 *)
  • Maxima
    a(n):=sum((sum(binomial(i,k-i-1),i,ceiling((k-1)/2),k-1))*((-1)^(n-k)+1)*sum(binomial(j-1,k-1)*j!*2^(n-j-1)*(-1)^((n+k)/2+j)*stirling2(n,j),j,k,n),k,1,n);
    
  • PARI
    x='x+O('x^66); /* that many terms */
    egf=tan(x)/(1-tan(x)-tan(x)^2); /* = x + x^2 + 7/3*x^3 + 11/3*x^4 +... */
    Vec(serlaplace(egf)) /* Joerg Arndt, Apr 22 2011 */

Formula

E.g.f.: A(tan(x)) where A(x) = x/(1-x-x^2) is the g.f. of A000045.
a(n)=sum(k=1..n, (sum(i=ceiling((k-1)/2)..k-1, binomial(i,k-i-1)))*((-1)^(n-k)+1)*sum(j=k..n, binomial(j-1,k-1)*j!*2^(n-j-1)*(-1)^((n+k)/2+j)*stirling2(n,j)));
From Peter Bala, Mar 12 2013: (Start)
With an offset of -1 the e.g.f is 1/(cos(2*x) - 1/2*sin(2*x))^2 = 1 + 2*x + 14*x^2/2! + 88*x^3/3! + .... This relates this sequence to A185896. Define a sequence of polynomials R(n,x) by the recurrence R(n+1,x) = d/dx{(x^2 + 4)*R(n,x)} with R(1,x) = 1. Then a(n) = R(n,1). Let M be the array with the sequence [2,3,4,...] on the main superdiagonal and the sequence [4,8,12,...] on the main subdiagonal and 0's everywhere else. Then a(n+1) equals the sum of the entries in row 1 of M^n.
(End)
a(n) ~ n! / (5*(arctan(sqrt(5)/2-1/2))^(n+1)). - Vaclav Kotesovec, Jun 02 2013
G.f.: x/Q(0), where Q(k) = 1 - 2*x*(2*k+1) - 5*x^2*(2*k+1)*(2*k+2)/( 1 - 2*x*(2*k+2) - 5*x^2*(2*k+2)*(2*k+3)/Q(k+1) ) ; (continued fraction). - Sergei N. Gladkovskii, Sep 23 2013
E.g.f.: W(0) -1, where W(k) = 1 + x/( 4*k+1 - x/( 1 - 4*x/( 4*k+3 + 4*x/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 27 2014
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