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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A001586 Generalized Euler numbers, or Springer numbers.

Original entry on oeis.org

1, 1, 3, 11, 57, 361, 2763, 24611, 250737, 2873041, 36581523, 512343611, 7828053417, 129570724921, 2309644635483, 44110959165011, 898621108880097, 19450718635716001, 445777636063460643, 10784052561125704811, 274613643571568682777, 7342627959965776406281
Offset: 0

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From Peter Bala, Feb 02 2011: (Start)
The Springer numbers were originally considered by Glaisher (see references). They are a type B analog of the zigzag numbers A000111 for the group of signed permutations.
COMBINATORIAL INTERPRETATIONS
Several combinatorial interpretations of the Springer numbers are known:
1) a(n) gives the number of Weyl chambers in the principal Springer cone of the Coxeter group B_n of symmetries of an n dimensional cube. An example can be found in [Arnold - The Calculus of snakes...].
2) Arnold found an alternative combinatorial interpretation of the Springer numbers in terms of snakes. Snakes are a generalization of alternating permutations to the group of signed permutations. 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. A snake of type B_n is a signed permutation (x_1,x_2,...,x_n) such that 0 < x_1 > x_2 < ... x_n. For example, (3,-4,-2,-5,1,-6) is a snake of type B_6. a(n) gives the number of snakes of type B_n [Arnold]. The cases n=2 and n=3 are given in the Example section below.
3) The Springer numbers also arise in the study of the critical points of functions; they count the topological types of odd functions with 2*n critical values [Arnold, Theorem 35].
4) Let F_n be the set of plane rooted forests satisfying the following conditions:
... each root has exactly one child, and each of the other internal nodes has exactly two (ordered) children,
... there are n nodes labeled by integers from 1 to n, but some leaves can be non-labeled (these are called empty leaves), and labels are increasing from each root down to the leaves. Then a(n) equals the cardinality of F_n. An example and proof are given in [Verges, Theorem 4.5].
OTHER APPEARANCES OF THE SPRINGER NUMBERS
1) Hoffman has given a connection between Springer numbers, snakes and the successive derivatives of the secant and tangent functions.
2) For integer N the quarter Gauss sums Q(N) are defined by ... Q(N) := Sum_{r = 0..floor(N/4)} exp(2*Pi*I*r^2/N). In the cases N = 1 (mod 4) and N = 3 (mod 4) an asymptotic series for Q(N) as N -> inf that involves the Springer numbers has been given by Evans et al., see 1.32 and 1.33.
For a sequence of polynomials related to the Springer numbers see A185417. For a table to recursively compute the Springer numbers see A185418.
(End)
Similar to the way in which the signed Euler numbers A122045 are 2^n times the value of the Euler polynomials at 1/2, the generalized signed Euler numbers A188458 can be seen as 2^n times the value of generalized Euler polynomials at 1/2. These are the Swiss-Knife polynomials A153641. A recursive definition of these polynomials is given in A081658. - Peter Luschny, Jul 19 2012
a(n) is the number of reverse-complementary updown permutations of [2n]. For example, the updown permutation 241635 is reverse-complementary because its complement is 536142, which is the same as its reverse, and a(2)=3 counts 1324, 2413, 3412. - David Callan, Nov 29 2012
a(n) = |2^n G(n,1/2;-1)|, a specialization of the Appell sequence of polynomials umbrally formed by G(n,x;t) = (G(.,0;t) + x)^n from the Grassmann polynomials G(n,0;t) of A046802 enumerating the cells of the positive Grassmannians. - Tom Copeland, Oct 14 2015
Named "Springer numbers" after the Dutch mathematician Tonny Albert Springer (1926-2011). - Amiram Eldar, Jun 13 2021

Examples

			a(2) = 3: The three snakes of type B_2 are
  (1,-2), (2,1), (2,-1).
a(3) = 11: The 11 snakes of type B_3 are
  (1,-2,3), (1,-3,2), (1,-3,-2),
  (2,1,3), (2,-1,3), (2,-3,1), (2,-3,-1),
  (3,1,2), (3,-1,2), (3,-2,1), (3,-2,-1).
		

References

  • V. I. Arnold, Springer numbers and Morsification spaces. J. Algebraic Geom., Vol. 1, No. 2 (1992), pp. 197-214.
  • J. W. L. Glaisher, "On the coefficients in the expansions of cos x/cos 2x and sin x/cos 2x", Quart. J. Pure and Applied Math., Vol. 45 (1914), pp. 187-222.
  • J. W. L. Glaisher, On the Bernoullian function, Q. J. Pure Appl. Math., Vol. 29 (1898), pp. 1-168.
  • Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nürnberg, 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).
  • Tonny Albert Springer, Remarks on a combinatorial problem, Nieuw Arch. Wisk., Vol. 19, No. 3 (1971), pp. 30-36.

Crossrefs

Row 2 of A349271.
Bisections are A000281 and A000464. Overview in A349264.
Related polynomials are given in A098432, A081658 and A153641.
Cf. A046802.

Programs

  • Maple
    a := proc(n) local k; (-1)^iquo(n,2)*add(2^k*binomial(n,k)*euler(k),k=0..n) end; # Peter Luschny, Jul 08 2009
    a := n -> (-1)^(n+iquo(n,2))*2^(3*n+1)*(Zeta(0,-n,1/8) - Zeta(0,-n,5/8)):
    seq(a(n),n=0..21); # Peter Luschny, Mar 11 2015
  • Mathematica
    n=21; CoefficientList[Series[1/(Cos[x]-Sin[x]), {x, 0, n}], x] * Table[k!, {k, 0, n}] (* Jean-François Alcover, May 18 2011 *)
    Table[Abs[Numerator[EulerE[n,1/4]]],{n,0,35}] (* Harvey P. Dale, May 18 2011 *)
  • PARI
    {a(n) = if(n<0, 0, n! * polcoeff( 1 / (cos(x + x * O(x^n)) - sin(x + x * O(x^n))), n))}; /* Michael Somos, Feb 03 2004 */
    
  • PARI
    {a(n) = my(an); if(n<2, n>=0, an = vector(n+1, m, 1); for(m=2, n, an[m+1] = 2*an[m] + an[m-1] + sum(k=0, m-3, binomial(m-2, k) * (an[k+1] * an[m-1-k] + 2*an[k+2] * an[m-k] - an[k+3] * an[m-1-k]))); an[n+1])}; /* Michael Somos, Feb 03 2004 */
    
  • PARI
    /* Explicit formula by Peter Bala: */
    {a(n)=((1+I)/2)^n*sum(k=0,n,((1-I)/(1+I))^k*sum(j=0,k,(-1)^(k-j)*binomial(n+1,k-j)*(2*j+1)^n))}
    
  • Sage
    @CachedFunction
    def p(n,x) :
        if n == 0 : return 1
        w = -1 if n%2 == 0 else 0
        v =  1 if n%2 == 0 else -1
        return v*add(p(k,0)*binomial(n,k)*(x^(n-k)+w) for k in range(n)[::2])
    def A001586(n) : return abs(2^n*p(n, 1/2))
    [A001586(n) for n in (0..21)] # Peter Luschny, Jul 19 2012

Formula

E.g.f.: 1/(cos(x) - sin(x)).
Values at 1 of polynomials Q_n() defined in A104035. - N. J. A. Sloane, Nov 06 2009
a(n) = numerator of abs(Euler(n,1/4)). - N. J. A. Sloane, Nov 07 2009
Let B_n(x) = Sum_{k=0.. n*(n-1)/2} b(n,k)*x^k, where b(n,k) is number of n-node acyclic digraphs with k arcs, cf. A081064; then a(n) = |B_n(-2)|. - Vladeta Jovovic, Jan 25 2005
G.f. A(x) = y satisfies y'^2 = 2y^4 - y^2, y''y = y^2 + 2y'^2. - Michael Somos, Feb 03 2004
a(n) = (-1)^floor(n/2) Sum_{k=0..n} 2^k C(n,k) Euler(k). - Peter Luschny, Jul 08 2009
From Peter Bala, Feb 02 2011: (Start)
(1)... a(n) = ((1 + i)/2)^n*B(n,(1 - i)/(1 + i)), where i = sqrt(-1) and {B(n,x)}n>=0 = [1, 1 + x, 1 + 6*x + x^2, 1 + 23*x + 23*x^2 + x^3, ...] is the sequence of type B Eulerian polynomials - see A060187.
This yields the explicit formula
(2)... a(n) = ((1 + i)/2)^n*Sum_{k = 0..n} ((1 - i)/(1 + i))^k * Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(2*j + 1)^n.
The result (2) can be used to find congruences satisfied by the Springer numbers. For example, for odd prime p
(3)
... a(p) = 1 (mod p) when p = 4*n + 1
... a(p) = -1 (mod p) when p = 4*n + 3.
(End)
E.g.f.: 1/Q(0) where Q(k) = 1 - x/((2k+1)-x*(2k+1)/(x+(2k+2)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2011
E.g.f.: 2/U(0) where U(k) = 1 + 1/(1 + x/(2*k + 1 -x - (2*k+1)/(2 - x/(x+ (2*k+2)/U(k+1))))); (continued fraction, 5-step). - Sergei N. Gladkovskii, Sep 24 2012
E.g.f.: 1/G(0) where G(k) = 1 - x/(4*k+1 - x*(4*k+1)/(4*k+2 + x + x*(4*k+2)/(4*k+3 - x - x*(4*k+3)/(x + (4*k+4)/G(k+1) )))); (continued fraction, 3rd kind, 5-step). - Sergei N. Gladkovskii, Oct 02 2012
G.f.: 1/G(0) where G(k) = 1 - x*(2*k+1) - 2*x^2*(k+1)*(k+1)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 11 2013
a(n) = | 2*4^n*lerchphi(-1, -n, 1/4) |. - Peter Luschny, Apr 27 2013
a(n) ~ 4 * n^(n+1/2) * (4/Pi)^n / (sqrt(Pi)*exp(n)). - Vaclav Kotesovec, Oct 07 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)^2/( 2*x^2*(k+1)^2 - (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 15 2013
a(n) = (-1)^C(n+1,2)*2^(3*n+1)*(Zeta(-n,1/8)-Zeta(-n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
E.g.f. A(x) satisfies: A(x) = exp( Integral A(x)/A(-x) dx ). - Paul D. Hanna, Feb 04 2017
E.g.f. A(x) satisfies: A'(x) = A(x)^2/A(-x). - Paul D. Hanna, Feb 04 2017

Extensions

More terms from Vladeta Jovovic, Jan 25 2005

A134264 Coefficients T(j, k) of a partition transform for Lagrange compositional inversion of a function or generating series in terms of the coefficients of the power series for its reciprocal. Enumeration of noncrossing partitions and primitive parking functions. T(n,k) for n >= 1 and 1 <= k <= A000041(n-1), an irregular triangle read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 4, 2, 6, 1, 1, 5, 5, 10, 10, 10, 1, 1, 6, 6, 3, 15, 30, 5, 20, 30, 15, 1, 1, 7, 7, 7, 21, 42, 21, 21, 35, 105, 35, 35, 70, 21, 1, 1, 8, 8, 8, 4, 28, 56, 56, 28, 28, 56, 168, 84, 168, 14, 70, 280, 140, 56, 140, 28, 1, 1, 9, 9, 9, 9, 36, 72
Offset: 1

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Tom Copeland, Jan 14 2008

Keywords

Comments

Coefficients are listed in Abramowitz and Stegun order (A036036).
Given an invertible function f(t) analytic about t=0 (or a formal power series) with f(0)=0 and Df(0) not equal to 0, form h(t) = t / f(t) and let h_n denote the coefficient of t^n in h(t).
Lagrange inversion gives the compositional inverse about t=0 as g(t) = Sum_{j>=1} ( t^j * (1/j) * Sum_{permutations s with s(1) + s(2) + ... + s(j) = j - 1} h_s(1) * h_s(2) * ... * h_s(j) ) = t * T(1,1) * h_0 + Sum_{j>=2} ( t^j * Sum_{k=1..(# of partitions for j-1)} T(j,k) * H(j-1,k ; h_0,h_1,...) ), where H(j-1,k ; h_0,h_1,...) is the k-th partition for h_1 through h_(j-1) corresponding to n=j-1 on page 831 of Abramowitz and Stegun (ordered as in A&S) with (h_0)^(j-m)=(h_0)^(n+1-m) appended to each partition subsumed under n and m of A&S.
Denoting h_n by (n') for brevity, to 8th order in t,
g(t) = t * (0')
+ t^2 * [ (0') (1') ]
+ t^3 * [ (0')^2 (2') + (0') (1')^2 ]
+ t^4 * [ (0')^3 (3') + 3 (0')^2 (1') (2') + (0') (1')^3 ]
+ t^5 * [ (0')^4 (4') + 4 (0')^3 (1') (3') + 2 (0')^3 (2')^2 + 6 (0')^2 (1')^2 (2') + (0') (1')^4 ]
+ t^6 * [ (0')^5 (5') + 5 (0')^4 (1') (4') + 5 (0')^4 (2') (3') + 10 (0')^3 (1')^2 (3') + 10 (0')^3 (1') (2')^2 + 10 (0')^2 (1')^3 (2') + (0') (1')^5 ]
+ t^7 * [ (0')^6 (6') + 6 (0')^5 (1') (5') + 6 (0')^5 (2') (4') + 3 (0')^5 (3')^2 + 15 (0')^4 (1')^2 (4') + 30 (0')^4 (1') (2') (3') + 5 (0')^4 (2')^3 + 20 (0')^3 (1')^3 (3') + 30 (0')^3 (1')^2 (2')^2 + 15 (0')^2 (1')^4 (2') + (0') (1')^6]
+ t^8 * [ (0')^7 (7') + 7 (0')^6 (1') (6') + 7 (0')^6 (2') (5') + 7 (0')^6 (3') (4') + 21 (0')^5 (1')^2* (5') + 42 (0')^5 (1') (2') (4') + 21 (0')^5 (1') (3')^2 + 21 (0')^5 (2')^2 (3') + 35 (0')^4 (1')^3 (4') + 105 (0)^4 (1')^2 (2') (3') + 35 (0')^4 (1') (2')^3 + 35 (0')^3 (1')^4 (3') + 70 (0')^3 (1')^3 (2')^2 + 21 (0')^2 (1')^5 (2') + (0') (1')^7 ]
+ ..., where from the formula section, for example, T(8,1',2',...,7') = 7! / ((8 - (1'+ 2' + ... + 7'))! * 1'! * 2'! * ... * 7'!) are the coefficients of the integer partitions (1')^1' (2')^2' ... (7')^7' in the t^8 term.
A125181 is an extended, reordered version of the above sequence, omitting the leading 1, with alternate interpretations.
If the coefficients of partitions with the same exponent for h_0 are summed within rows, A001263 is obtained, omitting the leading 1.
From identification of the elements of the inversion with those on page 25 of the Ardila et al. link, the coefficients of the irregular table enumerate non-crossing partitions on [n]. - Tom Copeland, Oct 13 2014
From Tom Copeland, Oct 28-29 2014: (Start)
Operating with d/d(1') = d/d(h_1) on the n-th partition polynomial Prt(n;h_0,h_1,..,h_n) in square brackets above associated with t^(n+1) generates n * Prt(n-1;h_0,h_1,..,h_(n-1)); therefore, the polynomials are an Appell sequence of polynomials in the indeterminate h_1 when h_0=1 (a special type of Sheffer sequence).
Consequently, umbrally, [Prt(.;1,x,h_2,..) + y]^n = Prt(n;1,x+y,h_2,..); that is, Sum_{k=0..n} binomial(n,k) * Prt(k;1,x,h_2,..) * y^(n-k) = Prt(n;1,x+y,h_2,..).
Or, e^(x*z) * exp[Prt(.;1,0,h_2,..) * z] = exp[Prt(.;1,x,h_2,..) * z]. Then with x = h_1 = -(1/2) * d^2[f(t)]/dt^2 evaluated at t=0, the formal Laplace transform from z to 1/t of this expression generates g(t), the comp. inverse of f(t), when h_0 = 1 = df(t)/dt eval. at t=0.
I.e., t / (1 - t*(x + Prt(.;1,0,h_2,..))) = t / (1 - t*Prt(.;1,x,h_2,..)) = g(t), interpreted umbrally, when h_0 = 1.
(End)
Connections to and between arrays associated to the Catalan (A000108 and A007317), Riordan (A005043), Fibonacci (A000045), and Fine (A000957) numbers and to lattice paths, e.g., the Motzkin, Dyck, and Łukasiewicz, can be made explicit by considering the inverse in x of the o.g.f. of A104597(x,-t), i.e., f(x) = P(Cinv(x),t-1) = Cinv(x) / (1 + (t-1)*Cinv(x)) = x*(1-x) / (1 + (t-1)*x*(1-x)) = (x-x^2) / (1 + (t-1)*(x-x^2)), where Cinv(x) = x*(1-x) is the inverse of C(x) = (1 - sqrt(1-4*x)) / 2, a shifted o.g.f. for the Catalan numbers, and P(x,t) = x / (1+t*x) with inverse Pinv(x,t) = -P(-x,t) = x / (1-t*x). Then h(x,t) = x / f(x,t) = x * (1+(t-1)Cinv(x)) / Cinv(x) = 1 + t*x + x^2 + x^3 + ..., i.e., h_1=t and all other coefficients are 1, so the inverse of f(x,t) in x, which is explicitly in closed form finv(x,t) = C(Pinv(x,t-1)), is given by A091867, whose coefficients are sums of the refined Narayana numbers above obtained by setting h_1=(1')=t in the partition polynomials and all other coefficients to one. The group generators C(x) and P(x,t) and their inverses allow associations to be easily made between these classic number arrays. - Tom Copeland, Nov 03 2014
From Tom Copeland, Nov 10 2014: (Start)
Inverting in x with t a parameter, let F(x;t,n) = x - t*x^(n+1). Then h(x) = x / F(x;t,n) = 1 / (1-t*x^n) = 1 + t*x^n + t^2*x^(2n) + t^3*x^(3n) + ..., so h_k vanishes unless k = m*n with m an integer in which case h_k = t^m.
Finv(x;t,n) = Sum_{j>=0} {binomial((n+1)*j,j) / (n*j + 1)} * t^j * x^(n*j + 1), which gives the Catalan numbers for n=1, and the Fuss-Catalan sequences for n>1 (see A001764, n=2). [Added braces to disambiguate the formula. - N. J. A. Sloane, Oct 20 2015]
This relation reveals properties of the partitions and sums of the coefficients of the array. For n=1, h_k = t^k for all k, implying that the row sums are the Catalan numbers. For n = 2, h_k for k odd vanishes, implying that there are no blocks with only even-indexed h_k on the even-numbered rows and that only the blocks containing only even-sized bins contribute to the odd-row sums giving the Fuss-Catalan numbers for n=2. And so on, for n > 2.
These relations are reflected in any combinatorial structures enumerated by this array and the partitions, such as the noncrossing partitions depicted for a five-element set (a pentagon) in Wikipedia.
(End)
From Tom Copeland, Nov 12 2014: (Start)
An Appell sequence possesses an umbral inverse sequence (cf. A249548). The partition polynomials here, Prt(n;1,h_1,...), are an Appell sequence in the indeterminate h_1=u, so have an e.g.f. exp[Prt(.;1,u,h_2...)*t] = e^(u*t) * exp[Prt(.;1,0,h2,...)*t] with umbral inverses with an e.g.f e^(-u*t) / exp[Prt(.;1,0,h2,...)*t]. This makes contact with the formalism of A133314 (cf. also A049019 and A019538) and the signed, refined face partition polynomials of the permutahedra (or their duals), which determine the reciprocal of exp[Prt(.,0,u,h2...)*t] (cf. A249548) or exp[Prt(.;1,u,h2,...)*t], forming connections among the combinatorics of permutahedra and the noncrossing partitions, Dyck paths and trees (cf. A125181), and many other important structures isomorphic to the partitions of this entry, as well as to formal cumulants through A127671 and algebraic structures of Lie algebras. (Cf. relationship of permutahedra with the Eulerians A008292.)
(End)
From Tom Copeland, Nov 24 2014: (Start)
The n-th row multiplied by n gives the number of terms in the homogeneous symmetric monomials generated by [x(1) + x(2) + ... + x(n+1)]^n under the umbral mapping x(m)^j = h_j, for any m. E.g., [a + b + c]^2 = [a^2 + b^2 + c^2] + 2 * [a*b + a*c + b*c] is mapped to [3 * h_2] + 2 * [3 * h_1^2], and 3 * A134264(3) = 3 *(1,1)= (3,3) the number of summands in the two homogeneous polynomials in the square brackets. For n=3, [a + b + c + d]^3 = [a^3 + b^3 + ...] + 3 [a*b^2 + a*c^2 + ...] + 6 [a*b*c + a*c*d + ...] maps to [4 * h_3] + 3 [12 * h_1 * h_2] + 6 [4 * (h_1)^3], and the number of terms in the brackets is given by 4 * A134264(4) = 4 * (1,3,1) = (4,12,4).
The further reduced expression is 4 h_3 + 36 h_1 h_2 + 24 (h_1)^3 = A248120(4) with h_0 = 1. The general relation is n * A134264(n) = A248120(n) / A036038(n-1) where the arithmetic is performed on the coefficients of matching partitions in each row n.
Abramowitz and Stegun give combinatorial interpretations of A036038 and relations to other number arrays.
This can also be related to repeated umbral composition of Appell sequences and topology with the Bernoulli numbers playing a special role. See the Todd class link.
(End)
These partition polynomials are dubbed the Voiculescu polynomials on page 11 of the He and Jejjala link. - Tom Copeland, Jan 16 2015
See page 5 of the Josuat-Verges et al. reference for a refinement of these partition polynomials into a noncommutative version composed of nondecreasing parking functions. - Tom Copeland, Oct 05 2016
(Per Copeland's Oct 13 2014 comment.) The number of non-crossing set partitions whose block sizes are the parts of the n-th integer partition, where the ordering of integer partitions is first by total, then by length, then lexicographically by the reversed sequence of parts. - Gus Wiseman, Feb 15 2019
With h_0 = 1 and the other h_n replaced by suitably signed partition polynomials of A263633, the refined face partition polynomials for the associahedra of normalized A133437 with a shift in indices are obtained (cf. In the Realm of Shadows). - Tom Copeland, Sep 09 2019
Number of primitive parking functions associated to each partition of n. See Lemma 3.8 on p. 28 of Rattan. - Tom Copeland, Sep 10 2019
With h_n = n + 1, the d_k (A006013) of Table 2, p. 18, of Jong et al. are obtained, counting the n-point correlation functions in a quantum field theory. - Tom Copeland, Dec 25 2019
By inspection of the diagrams on Robert Dickau's website, one can see the relationship between the monomials of this entry and the connectivity of the line segments of the noncrossing partitions. - Tom Copeland, Dec 25 2019
Speicher has examples of the first four inversion partition polynomials on pp. 22 and 23 with his k_n equivalent to h_n = (n') here with h_0 = 1. Identifying z = t, C(z) = t/f(t) = h(t), and M(z) = f^(-1)(t)/t, then statement (3), on p. 43, of Theorem 3.26, C(z M(z)) = M(z), is equivalent to substituting f^(-1)(t) for t in t/f(t), and statement (4), M(z/C(z)) = C(z), to substituting f(t) for t in f^(-1)(t)/t. - Tom Copeland, Dec 08 2021
Given a Laurent series of the form f(z) = 1/z + h_1 + h_2 z + h_3 z^2 + ..., the compositional inverse is f^(-1)(z) = 1/z + Prt(1;1,h_1)/z^2 + Prt(2;1,h_1,h_2)/z^3 + ... = 1/z + h_1/z^2 + (h_1^2 + h_2)/z^3 + (h_1^3 + 3 h_1 h_2 + h_3)/z^4 + (h_1^4 + 6 h_1^2 h_2 + 4 h_1 h_3 + 2 h_2^2 + h_4)/z^5 + ... for which the polynomials in the numerators are the partition polynomials of this entry. For example, this formula applied to the q-expansion of Klein's j-invariant / function with coefficients A000521, related to monstrous moonshine, gives the compositional inverse with the coefficients A091406 (see He and Jejjala). - Tom Copeland, Dec 18 2021
The partition polynomials of A350499 'invert' the polynomials of this entry giving the indeterminates h_n. A multinomial formula for the coefficients of the partition polynomials of this entry, equivalent to the multinomial formula presented in the first four sentences of the formula section below, is presented in the MathOverflow question referenced in A350499. - Tom Copeland, Feb 19 2022

Examples

			1) With f(t) = t / (t-1), then h(t) = -(1-t), giving h_0 = -1, h_1 = 1 and h_n = 0 for n>1. Then g(t) = -t - t^2 - t^3 - ... = t / (t-1).
2) With f(t) = t*(1-t), then h(t) = 1 / (1-t), giving h_n = 1 for all n. The compositional inverse of this f(t) is g(t) = t*A(t) where A(t) is the o.g.f. for the Catalan numbers; therefore the sum over k of T(j,k), i.e., the row sum, is the Catalan number A000108(j-1).
3) With f(t) = (e^(-a*t)-1) / (-a), h(t) = Sum_{n>=0} Bernoulli(n) * (-a*t)^n / n! and g(t) = log(1-a*t) / (-a) = Sum_{n>=1} a^(n-1) * t^n / n. Therefore with h_n = Bernoulli(n) * (-a)^n / n!, Sum_{permutations s with s(1)+s(2)+...+s(j)=j-1} h_s(1) * h_s(2) * ... * h_s(j) = j * Sum_{k=1..(# of partitions for j-1)} T(j,k) * H(j-1,k ; h_0,h_1,...) = a^(j-1). Note, in turn, Sum_{a=1..m} a^(j-1) = (Bernoulli(j,m+1) - Bernoulli(j)) / j for the Bernoulli polynomials and numbers, for j>1.
4) With f(t,x) = t / (x-1+1/(1-t)), then h(t,x) = x-1+1/(1-t), giving (h_0)=x and (h_n)=1 for n>1. Then g(t,x) = (1-(1-x)*t-sqrt(1-2*(1+x)*t+((x-1)*t)^2)) / 2, a shifted o.g.f. in t for the Narayana polynomials in x of A001263.
5) With h(t)= o.g.f. of A075834, but with A075834(1)=2 rather than 1, which is the o.g.f. for the number of connected positroids on [n] (cf. Ardila et al., p. 25), g(t) is the o.g.f. for A000522, which is the o.g.f. for the number of positroids on [n]. (Added Oct 13 2014 by author.)
6) With f(t,x) = x / ((1-t*x)*(1-(1+t)*x)), an o.g.f. for A074909, the reverse face polynomials of the simplices, h(t,x) = (1-t*x) * (1-(1+t)*x) with h_0=1, h_1=-(1+2*t), and h_2=t*(1+t), giving as the inverse in x about 0 the o.g.f. (1+(1+2*t)*x-sqrt(1+(1+2*t)*2*x+x^2)) / (2*t*(1+t)*x) for signed A033282, the reverse face polynomials of the Stasheff polytopes, or associahedra. Cf. A248727. (Added Jan 21 2015 by author.)
7) With f(x,t) = x / ((1+x)*(1+t*x)), an o.g.f. for the polynomials (-1)^n * (1 + t + ... + t^n), h(t,x) = (1+x) * (1+t*x) with h_0=1, h_1=(1+t), and h_2=t, giving as the inverse in x about 0 the o.g.f. (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2)) / (2*x*t) for the Narayana polynomials A001263. Cf. A046802. (Added Jan 24 2015 by author.)
From _Gus Wiseman_, Feb 15 2019: (Start)
Triangle begins:
   1
   1
   1   1
   1   3   1
   1   4   2   6   1
   1   5   5  10  10  10   1
   1   6   6   3  15  30   5  20  30  15   1
   1   7   7   7  21  42  21  21  35 105  35  35  70  21   1
Row 5 counts the following non-crossing set partitions:
  {{1234}}  {{1}{234}}  {{12}{34}}  {{1}{2}{34}}  {{1}{2}{3}{4}}
            {{123}{4}}  {{14}{23}}  {{1}{23}{4}}
            {{124}{3}}              {{12}{3}{4}}
            {{134}{2}}              {{1}{24}{3}}
                                    {{13}{2}{4}}
                                    {{14}{2}{3}}
(End)
		

References

  • A. Nica and R. Speicher (editors), Lectures on the Combinatorics of Free Probability, London Mathematical Society Lecture Note Series: 335, Cambridge University Press, 2006 (see in particular, Eqn. 9.14 on p. 141, enumerating noncrossing partitions).

Crossrefs

(A001263,A119900) = (reduced array, associated g(x)). See A145271 for meaning and other examples of reduced and associated.
Other orderings are A125181 and A306438.
Cf. A119900 (e.g.f. for reduced W(x) with (h_0)=t and (h_n)=1 for n>0).
Cf. A248927 and A248120, "scaled" versions of this Lagrange inversion.
Cf. A091867 and A125181, for relations to lattice paths and trees.
Cf. A249548 for use of Appell properties to generate the polynomials.
Cf. A133314, A049019, A019538, A127671, and A008292 for relations to permutahedra, Eulerians.
Cf. A006013.

Programs

  • Mathematica
    Table[Binomial[Total[y],Length[y]-1]*(Length[y]-1)!/Product[Count[y,i]!,{i,Max@@y}],{n,7},{y,Sort[Sort/@IntegerPartitions[n]]}] (* Gus Wiseman, Feb 15 2019 *)
  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); n!/((n-#v+1)!*prod(i=1, #S, my(x=S[i]); (#select(y->y==x, v))!))}
    row(n)=[C(Vec(p)) | p<-partitions(n-1)]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 01 2022

Formula

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.
If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.
T(j,k) is [ P(j,m;a...) / j ] for the k-th partition of j-1 as described in the comments.
For example from g(t) above, T(5,4) = (5! / ((5-3)! * 2!)) / 5 = 6 for the 4th partition under n=5-1=4 with m=3 parts in A&S.
From Tom Copeland, Sep 30 2011: (Start)
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}
= [(h_0)-1+:1/(1-h.*x):]^2 / {(h_0)-:[h.x/(1-h.x)]^2:}
= [(h_0)+(h_1)x+(h_2)x^2+...]^2 / [(h_0)-(h_2)x^2-2(h_3)x^3-3(h_4)x^4-...], where :" ": denotes umbral evaluation of the expression within the colons and h. is an umbral coefficient.
Then for the partition polynomials of A134264,
Poly[n;h_0,...,h_(n-1)]=(1/n!)(W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t) = exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)), and g(t) gives A001263 with (h_0)=u and (h_n)=1 for n>0 and A000108 with u=1.
(End)
From Tom Copeland, Oct 20 2011: (Start)
With exp(x* PS(.,t)) = exp(t*g(x)) = exp(x*W(y)d/dy) exp(t*y) eval. at y=0, the raising (creation) and lowering (annihilation) operators defined by R PS(n,t) = PS(n+1,t) and L PS(n,t) = n*PS(n-1,t) are
R = t*W(d/dt) = t*((h_0) + (h_1)d/dt + (h_2)(d/dt)^2 + ...)^2 / ((h_0) - (h_2)(d/dt)^2 - 2(h_3)(d/dt)^3 - 3(h_4)(d/dt)^4 + ...), and
L = (d/dt)/h(d/dt) = (d/dt) 1/((h_0) + (h_1)*d/dt + (h_2)*(d/dt)^2 + ...)
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0 are the row polynomials of A134264. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
(End)
Using the formalism of A263634, the raising operator for the partition polynomials of this array with h_0 = 1 begins as R = h_1 + h_2 D + h_3 D^2/2! + (h_4 - h_2^2) D^3/3! + (h_5 - 5 h_2 h_3) D^4/4! + (h_6 + 5 h_2^3 - 7 h_3^2 - 9 h_2 h_4) D^5/5! + (h_7 - 14 h_2 h_5 + 56 h_2^2 h_3) D^6/6! + ... with D = d/d(h_1). - Tom Copeland, Sep 09 2016
Let h(x) = x/f^{-1}(x) = 1/[1-(c_2*x+c_3*x^2+...)], with c_n all greater than zero. Then h_n are all greater than zero and h_0 = 1. Determine P_n(t) from exp[t*f^{-1}(x)] = exp[x*P.(t)] with f^{-1}(x) = x/h(x) expressed in terms of the h_n (cf. A133314 and A263633). Then P_n(b.) = 0 gives a recursion relation for the inversion polynomials of this entry a_n = b_n/n! in terms of the lower order inversion polynomials and P_j(b.)P_k(b.) = P_j(t)P_k(t)|{t^n = b_n} = d{j,k} >= 0 is the coefficient of x^j/j!*y^k/k! in the Taylor series expansion of the formal group law FGL(x,y) = f[f^{-1}(x)+f^{-1}(y)]. - Tom Copeland, Feb 09 2018
A raising operator for the partition polynomials with h_0 = 1 regarded as a Sheffer Appell sequence in h_1 is described in A249548. - Tom Copeland, Jul 03 2018

Extensions

Added explicit t^6, t^7, and t^8 polynomials and extended initial table to include the coefficients of t^8. - Tom Copeland, Sep 14 2016
Title modified by Tom Copeland, May 28 2018
More terms from Gus Wiseman, Feb 15 2019
Title modified by Tom Copeland, Sep 10 2019

A090582 T(n, k) = Sum_{j=0..n-k} (-1)^j*binomial(n - k + 1, j)*(n - k + 1 - j)^n. Triangle read by rows, T(n, k) for 1 <= k <= n.

Original entry on oeis.org

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

Views

Author

Hugo Pfoertner, Jan 11 2004

Keywords

Comments

Let Q(m, n) = Sum_(k=0..n-1) (-1)^k * binomial(n, k) * (n-k)^m. Then Q(m,n) is the numerator of the probability P(m,n) = Q(m,n)/n^m of seeing each card at least once if m >= n cards are drawn with replacement from a deck of n cards, written in a two-dimensional array read by antidiagonals with Q(m,m) as first row and Q(m,1) as first column.
The sequence is given as a matrix with the first row containing the cases #draws = size_of_deck. The second row contains #draws = 1 + size_of_deck. If "mn" indicates m cards drawn from a deck with n cards then the locations in the matrix are:
11 22 33 44 55 66 77 ...
21 32 43 54 65 76 87 ...
31 42 53 64 75 86 97 ...
41 52 63 74 85 .. .. ...
read by antidiagonals ->:
11, 22, 21, 33, 32, 31, 44, 43, 42, 41, 55, 54, 53, 52, ....
The probabilities are given by Q(m,n)/n^m:
.(m,n):.....11..22..21..33..32..31..44..43..42..41...55...54..53..52..51
.....Q:......1...2...1...6...6...1..24..36..14...1..120..240.150..30...1
...n^m:......1...4...1..27...8...1.256..81..16...1.3125.1024.243..32...1
%.Success:.100..50.100..22..75.100...9..44..88.100....4...23..62..94.100
P(n,n) = n!/n^n which can be approximated by sqrt(Pi*(2n+1/3))/e^n (Gosper's approximation to n!).
Let P[n] be the set of all n-permutations. Build a superset Q[n] of P[n] composed of n-permutations in which some (possibly all or none) ascents have been designated. An ascent in a permutation s[1]s[2]...s[n] is a pair of consecutive elements s[i],s[i+1] such that s[i] < s[i+1]. As a triangular array read by rows T(n,k) is the number of elements in Q[n] that have exactly k distinguished ascents, n >= 1, 0 <= k <= n-1. Row sums are A000670. E.g.f. is y/(1+y-exp(y*x)). For example, T(3,1)=6 because there are four 3-permutations with one ascent, with these we would also count 1->2 3, and 1 2->3 where exactly one ascent is designated by "->". (After Flajolet and Sedgewick.) - Geoffrey Critzer, Nov 13 2012
Sum_(k=1..n) Q(n, k)*binomial(x, k) = x^n such that Sum_{k=1..n} Q(n, i)*binomial(x+1,i+1) = Sum_{k=1..x} k^n. - David A. Corneth, Feb 17 2014
A141618(n,n-k+1) = a(n,k) * C(n,k-1) / k. - Tom Copeland, Oct 25 2014
See A074909 and above g.f.s below for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
For connections to toric varieties and Eulerian polynomials (in addition to those noted below), see the Dolgachev and Lunts and the Stembridge links in A019538. - 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
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

Examples

			For m = 4, n = 2, we draw 4 times from a deck of two cards. Call the cards "a" and "b" - of the 16 possible combinations of draws (each of which is equally likely to occur), only two do not contain both a and b: a, a, a, a and b, b, b, b. So the probability of seeing both a and b is 14/16. Therefore Q(m, n) = 14.
Table starts:
  [1] 1;
  [2] 2,      1;
  [3] 6,      6,       1;
  [4] 24,     36,      14,      1;
  [5] 120,    240,     150,     30,      1;
  [6] 720,    1800,    1560,    540,     62,     1;
  [7] 5040,   15120,   16800,   8400,    1806,   126,    1;
  [8] 40320,  141120,  191520,  126000,  40824,  5796,   254,   1;
  [9] 362880, 1451520, 2328480, 1905120, 834120, 186480, 18150, 510, 1.
		

Crossrefs

Cf. A073593 first m >= n giving at least 50% probability, A085813 ditto for 95%, A055775 n^n/n!, A090583 Gosper's approximation to n!.
Reflected version of A019538.
Cf. A233734 (central terms).

Programs

  • Haskell
    a090582 n k = a090582_tabl !! (n-1) !! (k-1)
    a090582_row n = a090582_tabl !! (n-1)
    a090582_tabl = map reverse a019538_tabl
    -- Reinhard Zumkeller, Dec 15 2013
    
  • Maple
    T := (n, k) -> add((-1)^j*binomial(n - k + 1, j)*(n - k + 1 - j)^n, j = 0..n-k):
    # Or:
    T := (n, k) -> (n - k + 1)!*Stirling2(n, n - k + 1):
    for n from 1 to 9 do seq( T(n, k), k = 1..n) od; # Peter Luschny, May 21 2021
  • Mathematica
    In[1]:= Table[Table[k! StirlingS2[n, k], {k, n, 1, -1}], {n, 1, 6}] (* Victor Adamchik, Oct 05 2005 *)
    nn=6; a=y/(1+y-Exp[y x]); Range[0,nn]! CoefficientList[Series[a, {x,0,nn}], {x,y}]//Grid (* Geoffrey Critzer, Nov 10 2012 *)
  • PARI
    a(n)={m=ceil((-1+sqrt(1+8*n))/2);k=m+1+binomial(m,2)-n;k*sum(i=1,k,(-1)^(i+k)*i^(m-1)*binomial(k-1,i-1))} \\ David A. Corneth, Feb 17 2014

Formula

T(n, k) = (n - k + 1)!*Stirling2(n, n - k + 1), generated by Stirling numbers of the second kind multiplied by a factorial. - Victor Adamchik, Oct 05 2005
Triangle T(n,k), 1 <= k <= n, read by rows given by [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] DELTA [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 10 2006
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 of A090582, the 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 of permutohedra.
G[(t+1)x,-1/(t+1)] = 1 + (1 + t)*x + (1 + 3*t + 2*t^2)*x^2/2! + ... gives row polynomials of A028246. (End)
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f. A(x,t) = G(x,t) - 1, the compositional inverse in x is
B(x,t) = log((t+1)-t/(1+x))/t. Let h(x,t) = 1/(dB/dx) = (1+x)*(1+(1+t)x), then the row polynomial P(n,t) is given by (1/n!)*(h(x,t)*d/dx)^n x, evaluated at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t). (End)
k <= 0 or k > n yields Q(n, k) = 0; Q(1,1) = 1; Q(n, k) = k * (Q(n-1, k) + Q(n-1, k-1)). - David A. Corneth, Feb 17 2014
T = A008292*A007318. - Tom Copeland, Nov 13 2016
With all offsets 0 for this entry, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125 with offsets -1 so that the array becomes A008292; i.e., we ignore the first row and first column of A123125. Then the row polynomials of this entry, A090582, are given by A_n(1 + x;0). Other specializations of A_n(x;y) give A028246, A046802, A119879, A130850, and A248727. - Tom Copeland, Jan 24 2020

A119879 Exponential Riordan array (sech(x),x).

Original entry on oeis.org

1, 0, 1, -1, 0, 1, 0, -3, 0, 1, 5, 0, -6, 0, 1, 0, 25, 0, -10, 0, 1, -61, 0, 75, 0, -15, 0, 1, 0, -427, 0, 175, 0, -21, 0, 1, 1385, 0, -1708, 0, 350, 0, -28, 0, 1, 0, 12465, 0, -5124, 0, 630, 0, -36, 0, 1, -50521, 0, 62325, 0, -12810, 0, 1050, 0, -45, 0, 1
Offset: 0

Views

Author

Paul Barry, May 26 2006

Keywords

Comments

Row sums have e.g.f. exp(x)*sech(x) (signed version of A009006). Inverse of masked Pascal triangle A119467. Transforms the sequence with e.g.f. g(x) to the sequence with e.g.f. g(x)*sech(x).
Coefficients of the Swiss-Knife polynomials for the computation of Euler, tangent and Bernoulli number (triangle read by rows). Another version in A153641. - Philippe Deléham, Oct 26 2013
Relations to Green functions and raising/creation and lowering/annihilation/destruction operators are presented in Hodges and Sukumar and in Copeland's discussion of this sequence and 2020 pdf. - Tom Copeland, Jul 24 2020

Examples

			Triangle begins:
     1;
     0,    1;
    -1,    0,     1;
     0,   -3,     0,   1;
     5,    0,    -6,   0,   1;
     0,   25,     0, -10,   0,   1;
   -61,    0,    75,   0, -15,   0,   1;
     0, -427,     0, 175,   0, -21,   0,  1;
  1385,    0, -1708,   0, 350,   0, -28,  0,  1;
		

Crossrefs

Row sums are A155585. - Johannes W. Meijer, Apr 20 2011
Rows reversed: A081658.

Programs

  • Maple
    T := (n,k) -> binomial(n,k)*2^(n-k)*euler(n-k,1/2): # Peter Luschny, Jan 25 2009
  • Mathematica
    T[n_, k_] := Binomial[n, k] 2^(n-k) EulerE[n-k, 1/2];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 20 2018, after Peter Luschny *)
  • PARI
    {T(n,k) = binomial(n,k)*2^(n-k)*(2/(n-k+1))*(subst(bernpol(n-k+1, x), x, 1/2) - 2^(n-k+1)*subst(bernpol(n-k+1, x), x, 1/4))};
    for(n=0,5, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 25 2019
  • Sage
    @CachedFunction
    def A119879_poly(n, x) :
        return 1 if n == 0  else add(A119879_poly(k, 0)*binomial(n, k)*(x^(n-k)-1+n%2) for k in range(n)[::2])
    def A119879_row(n) :
        R = PolynomialRing(ZZ, 'x')
        return R(A119879_poly(n,x)).coeffs()  # Peter Luschny, Jul 16 2012
    # Alternatively:
    
  • Sage
    # uses[riordan_array from A256893]
    riordan_array(sech(x), x, 9, exp=true) # Peter Luschny, Apr 19 2015
    

Formula

Number triangle whose k-th column has e.g.f. sech(x)*x^k/k!.
T(n,k) = C(n,k)*2^(n-k)*E_{n-k}(1/2) where C(n,k) is the binomial coefficient and E_{m}(x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
The coefficients in ascending order of x^i of the polynomials p{0}(x) = 1 and p{n}(x) = Sum_{k=0..n-1; k even} binomial(n,k)*p{k}(0)*((n mod 2) - 1 + x^(n-k)). - Peter Luschny, Jul 16 2012
E.g.f.: exp(x*z)/cosh(x). - Peter Luschny, Aug 01 2012
Sum_{k=0..n} T(n,k)*x^k = A122045(n), A155585(n), A119880(n), A119881(n) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Oct 27 2013
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of this entry, A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
Triangle equals P*((I + P^2)/2)^(-1), where P denotes Pascal's triangle A007318. - Peter Bala, Mar 07 2024

A073107 Triangle T(n,k) read by rows, where e.g.f. for T(n,k) is exp((1+y)*x)/(1-x).

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 16, 15, 6, 1, 65, 64, 30, 8, 1, 326, 325, 160, 50, 10, 1, 1957, 1956, 975, 320, 75, 12, 1, 13700, 13699, 6846, 2275, 560, 105, 14, 1, 109601, 109600, 54796, 18256, 4550, 896, 140, 16, 1, 986410, 986409, 493200, 164388, 41076, 8190, 1344, 180, 18, 1
Offset: 0

Views

Author

Vladeta Jovovic, Aug 19 2002

Keywords

Comments

Triangle is second binomial transform of A008290. - Paul Barry, May 25 2006
Ignoring signs, n-th row is the coefficient list of the permanental polynomial of the n X n matrix with 2's along the main diagonal and 1's everywhere else (see Mathematica code below). - John M. Campbell, Jul 02 2012

Examples

			exp((1 + y)*x)/(1 - x) =
  1 +
  1/1! * (2 + y) * x +
  1/2! * (5 + 4*y + y^2) * x^2 +
  1/3! * (16 + 15*y + 6*y^2 + y^3) * x^3 +
  1/4! * (65 + 64*y + 30*y^2 + 8*y^3 + y^4) * x^4 +
  1/5! * (326 + 325*y + 160*y^2 + 50*y^3 + 10*y^4 + y^5) * x^5 + ...
Triangle starts:
  [0]     1;
  [1]     2,     1;
  [2]     5,     4,    1;
  [3]    16,    15,    6,    1;
  [4]    65,    64,   30,    8,   1;
  [5]   326,   325,  160,   50,  10,   1;
  [6]  1957,  1956,  975,  320,  75,  12,  1;
  [7] 13700, 13699, 6846, 2275, 560, 105, 14, 1;
		

Crossrefs

Cf. A008290, A008291, A046802, A093375 (unsigned inverse), A094587, A010842 (row sums), A000142 (alternating row sums), A367963 (central terms).
Column k=0..4 give A000522, A007526, A038155, A357479, A357480.

Programs

  • Maple
    T := (n, k) -> binomial(n,k)*KummerU(k-n, k-n, 1);
    seq(seq(simplify(T(n, k)), k = 0..n), n=0..8);  # Peter Luschny, Oct 16 2024
  • Mathematica
    perm[m_List] := With[{v=Array[x,Length[m]]},Coefficient[Times@@(m.v),Times@@v]] ;
    A[q_] := Array[KroneckerDelta[#1,#2] + 1&,{q,q}] ;
    n = 1 ; Print[{1}]; While[n < 10, Print[Abs[CoefficientList[perm[A[n] - IdentityMatrix[n] * k], k]]]; n++] (* John M. Campbell, Jul 02 2012 *)
    A073107[n_, k_] := If[n == k, 1, Floor[E*(n - k)!]*Binomial[n, k]];
    Table[A073107[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Oct 16 2024 *)
  • SageMath
    def T(n, k):
        return sum(binomial(j,k) * factorial(n) // factorial(j) for j in range(n+1))
    for n in range(8): print([T(n, k) for k in range(n+1)])
    # Peter Luschny, Oct 16 2024

Formula

O.g.f. for k-th column is (1/k!)*Sum_{i >= k} i!*x^i/(1-x)^(i+1).
For n > 0, T(n, 0) = floor(n!*exp(1)) = A000522(n), T(n, 1) = floor(n!*exp(1) - 1) = A007526(n), T(n, 2) = 1/2!*floor(n!*exp(1) - 1 - n) = A038155(n), T(n, 3) = 1/3!*floor(n!*exp(1) - 1 - n - n*(n - 1)), T(n, 4) = 1/4!*floor(n!*exp(1) - 1 - n - n*(n - 1) - n*(n - 1)*(n - 2)), ... .
Row sums give A010842.
E.g.f. for k-th column is (x^k/k!)*exp(x)/(1 - x).
O.g.f. for k-th row is n!*Sum_{k = 0..n} (1 + x)^k/k!.
T(n,k) = Sum_{j = 0..n} binomial(j,k)*n!/j!. - Paul Barry, May 25 2006
-exp(-x) * Sum_{k=0..n} T(n,k)*x^k = Integral (x+1)^n*exp(-x) dx = -exp(1)*Gamma(n+1,x+1). - Gerald McGarvey, Mar 15 2009
From Peter Bala, Sep 20 2012: (Start)
Exponential Riordan array [exp(x)/(1-x),x] belonging to the Appell subgroup, which factorizes in the Appell group as [1/1-x,x]*[exp(x),x] = A094587*A007318.
The n-th row polynomial R(n,x) of the triangle satisfies d/dx(R(n,x)) = n*R(n-1,x), as well as R(n,x + y) = Sum {k = 0..n} binomial(n,k)*R(k,x)*y^(n-k). The row polynomials are a Sheffer sequence of Appell type.
Matrix inverse of triangle is a signed version of A093375. (End)
From Tom Copeland, Oct 20 2015: (Start)
The raising operator, with D = d/dx, for the row polynomials is RP = x + d{log[e^D/(1-D)]}/dD = x + 1 + 1/(1-D) = x + 2 + D + D^2 + ..., i.e., RP R(n,x) = R(n+1,x).
This operator is the limit as t tends to 1 of the raising operator of the polynomials p(n,x;t) described in A046802, implying R(n,x) = p(n,x;1). Compare with the raising operator of A094587, x + 1/(1-D), and that of signed A093375, x - 1 - 1/(1-D).
From the Appell formalism, the row polynomials RI(n,x) of signed A093375 are the umbral inverse of this entry's row polynomials; that is, R(n,RI(.,x)) = x^n = RI(n,R(.,x)) under umbral composition. (End)
From Werner Schulte, Sep 07 2020: (Start)
T(n,k) = (n! / k!) * (Sum_{i=k..n} 1 / (n-i)!) for 0 <= k <= n.
T(n,k) = n * T(n-1,k) + binomial(n,k) for 0 <= k <= n with initial values T(0,0) = 1 and T(i,j) = 0 if j < 0 or j > i.
T(n,k) = A000522(n-k) * binomial(n,k) for 0 <= k <= n. (End)

Extensions

More terms from Emeric Deutsch, Feb 23 2004

A142175 Triangle read by rows: T(n,k) = (1/4)*(A007318(n,k) - 6*A008292(n+1,k+1) + 9*A060187(n+1,k+1)).

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 36, 36, 1, 1, 133, 420, 133, 1, 1, 449, 3334, 3334, 449, 1, 1, 1446, 21939, 49364, 21939, 1446, 1, 1, 4534, 130044, 560957, 560957, 130044, 4534, 1, 1, 13991, 724222, 5459561, 10284514, 5459561, 724222, 13991, 1, 1, 42747, 3880014, 48160170, 154214412, 154214412, 48160170, 3880014, 42747, 1
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Sep 16 2008

Keywords

Comments

Row n gives the coefficients in the expansion of (1/4)*(1 + x)^n + (9/4)*2^n*(1 - x)^(1 + n)*Phi(x, -n, 1/2) - (3/2)*(1 - x)^(n + 2)*Phi(x, -1 - n, 1), where Phi is the Lerch transcendant.

Examples

			Triangle begins:
     1;
     1,    1;
     1,    8,      1;
     1,   36,     36,      1;
     1,  133,    420,    133,      1;
     1,  449,   3334,   3334,    449,      1;
     1, 1446,  21939,  49364,  21939,   1446,    1;
     1, 4534, 130044, 560957, 560957, 130044, 4534, 1;
      ... reformatted. - _Franck Maminirina Ramaharo_, Oct 21 2018
		

Crossrefs

Triangles related to Eulerian numbers: A008292, A046802, A060187, A123125.

Programs

  • Magma
    A060187:= func< n,k | (&+[(-1)^(k-j)*Binomial(n, k-j)*(2*j-1)^(n-1): j in [1..k]]) >;
    A142175:= func< n,k | (Binomial(n,k) - 6*EulerianNumber(n+1,k) + 9*A060187(n+1,k+1))/4 >;
    [A142175(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 30 2024
    
  • Mathematica
    p[x_, n_] = 1/4*(1 + x)^n + 9/4*2^n*(1 - x)^(1 + n)*LerchPhi[x, -n, 1/2] - 3/2*(1 - x)^(2 + n)*PolyLog[-1 - n, x]/x;
    Table[CoefficientList[FullSimplify[p[x, n]], x], {n, 0, 10}]// Flatten
  • Maxima
    A008292(n, k) := sum((-1)^j*(k - j)^n*binomial(n + 1, j), j, 0, k)$
    A060187(n, k) := sum((-1)^(k - j)*binomial(n, k - j)*(2*j - 1)^(n - 1), j, 1, k)$
    T(n, k) := (binomial(n, k) - 6*A008292(n + 1, k + 1) + 9*A060187(n + 1, k + 1))/4$
    create_list(T(n, k), n, 0, 10, k, 0, n);
    /* Franck Maminirina Ramaharo, Oct 20 2018 */
    
  • SageMath
    # from sage.all import * # (use for Python)
    from sage.combinat.combinat import eulerian_number
    def A060187(n,k): return sum(pow(-1, k-j)*binomial(n, k-j)*pow(2*j-1, n-1) for j in range(1,k+1))
    def A142175(n,k): return (binomial(n,k) - 6*eulerian_number(n+1,k) +9*A060187(n+1,k+1))//4
    print(flatten([[A142175(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Dec 30 2024

Formula

E.g.f.: (exp((1 + x)*y) - 6*(1 - x)^2*exp(y*(1 - x))/(1 - x*exp(y*(1 - x)))^2 + 9*(1 - x)*exp((1 - x)*y)/(1 - x*exp(2*(1 - x)*y)))/4. - Franck Maminirina Ramaharo, Oct 20 2018

Extensions

Edited, new name, and offset corrected by Franck Maminirina Ramaharo, Oct 19 2018

A130850 Triangle read by rows, 0 <= k <= n, T(n,k) = Sum_{j=0..n} A(n,j)*binomial(n-j,k) where A(n,j) are the Eulerian numbers A173018.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 12, 7, 1, 24, 60, 50, 15, 1, 120, 360, 390, 180, 31, 1, 720, 2520, 3360, 2100, 602, 63, 1, 5040, 20160, 31920, 25200, 10206, 1932, 127, 1, 40320, 181440, 332640, 317520, 166824, 46620, 6050, 255, 1, 362880, 1814400, 3780000, 4233600, 2739240, 1020600, 204630, 18660, 511, 1
Offset: 0

Views

Author

Philippe Deléham, Aug 20 2007

Keywords

Comments

Old name was: Triangle T(n,k), 0<=k<=n, read by rows given by [1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,2,0,3,0,4,0,5,0,6,0,...] where DELTA is the operator defined in A084938.
Vandervelde (2018) refers to this as the Worpitzky number triangle - N. J. A. Sloane, Mar 27 2018 [Named after the German mathematician Julius Daniel Theodor Worpitzky (1835-1895). - Amiram Eldar, Jun 24 2021]
Triangle given by A123125*A007318 (as infinite lower triangular matrices), A123125 = Euler's triangle, A007318 = Pascal's triangle; A007318*A123125 gives A046802.
Taylor coefficients of Eulerian polynomials centered at 1. - Louis Zulli, Nov 28 2015
A signed refinement is A263634. - Tom Copeland, Nov 14 2016
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of this entry (the Worpitsky triangle, A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020

Examples

			Triangle begins:
1
1      1
2      3       1
6      12      7       1
24     60      50      15      1
120    360     390     180     31      1
720    2520    3360    2100    602     63      1
5040   20160   31920   25200   10206   1932    127    1
40320  181440  332640  317520  166824  46620   6050   255   1
362880 1814400 3780000 4233600 2739240 1020600 204630 18660 511 1
...
		

Crossrefs

Programs

  • Mathematica
    Table[(n-k)!*StirlingS2[n+1, n-k+1], {n, 0, 10}, {k, 0, n}] (* G. C. Greubel, Nov 15 2015 *)
  • PARI
    t(n, k) = (n-k)!*stirling(n+1, n-k+1, 2);
    tabl(nn) = for (n=0, 10, for (k=0, n, print1(t(n,k),", ")); print()); \\ Michel Marcus, Nov 16 2015
  • Sage
    from sage.combinat.combinat import eulerian_number
    def A130850(n, k):
        return add(eulerian_number(n, j)*binomial(n-j, k) for j in (0..n))
    for n in (0..7): [A130850(n, k) for k in (0..n)] # Peter Luschny, May 21 2013
    

Formula

T(n,k) = (-1)^k*A075263(n,k).
T(n,k) = (n-k)!*A008278(n+1,k+1).
T(n,n-1) = 2^n - 1 for n > 0. - Derek Orr, Dec 31 2015
E.g.f.: x/(e^(-x*t)*(1+x)-1). - Tom Copeland, Nov 14 2016
Sum_{k=1..floor(n/2)} T(n,2k) = Sum_{k=0..floor(n/2)} T(n,2k+1) = A000670(n). - Jacob Sprittulla, Oct 03 2021

Extensions

New name from Peter Luschny, May 21 2013

A142147 Irregular triangle read by rows: first row is 1, and the n-th row gives the coefficients in the expansion of (1/2*x)*(1 - 2*x*(1 - x))^(n + 1)*Li(-n, 2*x*(1 - x)), where Li(n, z) is the polylogarithm.

Original entry on oeis.org

1, 1, -1, 1, 1, -4, 2, 1, 7, -12, -4, 12, -4, 1, 21, 0, -102, 100, 4, -32, 8, 1, 51, 160, -532, -24, 904, -672, 48, 80, -16, 1, 113, 980, -1094, -5128, 8760, -736, -6224, 3920, -432, -192, 32, 1, 239, 4284, 5276, -43964, 19764, 90272, -114080, 19824, 36304
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Sep 15 2008

Keywords

Examples

			Triangle begins:
     1;
     1, -1;
     1,  1,  -4,    2;
     1,  7, -12,   -4,  12,  -4;
     1, 21,   0, -102, 100,   4,  -32,  8;
     1, 51, 160, -532, -24, 904, -672, 48, 80, -16;
      ... reformatted. - _Franck Maminirina Ramaharo_, Oct 21 2018
		

Crossrefs

Triangles related to Eulerian numbers: A008292, A046802, A060187, A123125.

Programs

  • Mathematica
    p[x_, n_] = If[n == 0, 1, (1 + 2*(-1 + x)*x)^(n + 1)*PolyLog[-n, 2*x*(1 - x)]/(2*x)];
    Table[CoefficientList[FullSimplify[Expand[p[x, n]]], x], {n, 0, 10}]//Flatten

Formula

E.g.f.: ((1 - x)*(1 - 2*x)*exp(t*(1 + 2*x^2)) + x*exp(2*t*x))/(exp(2*t*x) - 2*x*(1 - x)*exp(t*(1 + 2*x^2))). - Franck Maminirina Ramaharo, Oct 22 2018

Extensions

Edited, new name, and offset corrected by Franck Maminirina Ramaharo, Oct 21 2018

A046803 Triangle of numbers related to Eulerian numbers.

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 14, 22, 4, 1, 30, 105, 65, 5, 1, 62, 416, 581, 171, 6, 1, 126, 1491, 3920, 2695, 420, 7, 1, 254, 5034, 22506, 29310, 11180, 988, 8, 1, 510, 16365, 116667, 256317, 188361, 43041, 2259, 9, 1, 1022, 51892, 564667, 1945297, 2419897, 1090135
Offset: 1

Views

Author

Keywords

Examples

			Triangle begins
  1;
  1,   2;
  1,   6,    3;
  1,  14,   22,    4;
  1,  30,  105,   65,    5;
  1,  62,  416,  581,  171,   6;
  1, 126, 1491, 3920, 2695, 420, 7;
  ...
		

References

  • D. Singh, The numbers L(m,n) and their relations with prepared Bernoulli and Eulerian numbers, Math. Student, 20 (1952), 66-70.

Crossrefs

Row sums give A002627.
Cf. A008292 (Eulerian numbers), A046802.

Programs

  • Mathematica
    egf = Exp[x*y]*(Exp[x]-1)*((y-1)/(y*Exp[x] - Exp[x*y])); row[n_] := Last[ CoefficientList[ Series[egf, {x, 0, n}, {y, 0, n}], {x, y}]]*n!; Flatten[ Table[ row[n], {n, 1, 10}]] (* Jean-François Alcover, Dec 20 2012, after Vladeta Jovovic *)
  • PARI
    T(n)={my(A=O(x*x^n)); [Vecrev(p) | p<-Vec(serlaplace(exp(x*y + A)*(exp(x + A)-1)*(y-1)/(y*exp(x + A)-exp(x*y + A))))]}
    { my(A=T(10)); for(n=1, #A, print(A[n])) } \\ Andrew Howroyd, Mar 07 2020
    
  • PARI
    \\ here U(n,k) is A008292.
    U(n, k)={sum(j=0, k, (-1)^j * (k-j)^n * binomial( n+1, j))};
    T(n, k)={sum(i=1, n, binomial(n,i)*U(n-i, k-1))} \\ Andrew Howroyd, Mar 07 2020

Formula

T(n, k) = Sum_{i=1..n} binomial(n,i) * A008292(n-i, k-1).
E.g.f.: exp(x*y)*(exp(x)-1)*(y-1)/(y*exp(x)-exp(x*y)). - Vladeta Jovovic, Sep 20 2003

Extensions

More terms from Vladeta Jovovic, Sep 20 2003

A089249 Triangular array read by rows illustrating the connection between A000522 and A008292.

Original entry on oeis.org

1, 3, 4, 6, 16, 11, 10, 40, 55, 26, 15, 80, 165, 156, 57, 21, 140, 385, 546, 399, 120
Offset: 1

Views

Author

Alford Arnold, Dec 11 2003

Keywords

Comments

The general case corresponding to other diagonals of A046802 can be derived from A046802(m,n) = Sum C(m-1,n-1) * A008292(m-r,n-1).

Examples

			The fifth row of the array is 15 80 165 156 57 resulting from A089249 (1 4 11 26 57 ) times ( 15 20 15 6 1)
		

Crossrefs

Row sums = the third diagonal of A046802.
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