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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A248120 Triangle read by rows: Lagrange (compositional) inversion of a function in terms of the coefficients of the Taylor series expansion of its reciprocal, scaled version of A248927, n >= 1, k = 1..A000041(n-1).

Original entry on oeis.org

1, 2, 6, 3, 24, 36, 4, 120, 360, 60, 80, 5, 720, 3600, 1800, 1200, 300, 150, 6, 5040, 37800, 37800, 16800, 3150, 12600, 3150, 420, 630, 252, 7, 40320, 423360, 705600, 235200, 176400, 352800, 58800, 35280, 23520, 35280, 7056, 1960, 1176, 392, 8
Offset: 1

Views

Author

Tom Copeland, Oct 28 2014

Keywords

Comments

Coefficients are listed in reverse graded colexicographic order (A228100). This is the reverse of Abramowitz and Stegun order (A036036).
Coefficients for Lagrange (compositional) inversion of a function in terms of the Taylor series expansion of its shifted reciprocal. Complementary to A134264 for formal power series and a scaled version of A248927. A refinement of A055302, which enumerates the number of labeled rooted trees with n nodes and k leaves, with row sums A000169.
Given an invertible function f(t) analytic about t=0 with f(0)=0 and df(0)/dt not 0, form h(t) = t / f(t) and denote h_n = (n') as the coefficient of t^n/n! in h(t). Then the compositional inverse of f(t), g(t), as a formal Taylor series, or e.g.f., is given up to the first few orders by
g(t) = [ 1 (0') ] * t
+ [ 2 (0') (1') ] * t^2/2!
+ [ 6 (0') (1')^2 + 3 (0')^2 (2') ] * t^3/3!
+ [24 (0') (1')^3 + 36 (0')^2 (1') (2') + 4 (0')^3 (3')] * t^4/4!
+ [120 (0') (1')^4 + 360 (0')^2 (1')^2 (2') + (0')^3 [60 (2')^2
+ 80 (1') (3')] + 5 (0')^4 (4')] * t^5/5!
+ [720 (0')(1')^5 + 3600 (0')^2 (1')^3(2') + (0')^3 [1800 (1')(2')^2 + 1200 ( 1')^2(3')] + (0')^4 [300 (2')(3') + 150 (1')(4')] + 6 (0')^5 (5')] * t^6/6! + ... .
Operating with [1/(n*(n-1))] d/d(1') = [1/(n*(n-1))] d/d(h_1) on the n-th partition polynomial in square brackets above associated with t^n/n! generates the (n-1)-th partition polynomial.
Each n-th partition polynomial here is n times the (n-1)-th partition polynomial of A248927.
From Tom Copeland, Nov 24 2014: (Start)
The n-th row is a mapping of the homogeneous symmetric monomials generated by [x(1) + x(2) + ... + x(n)]^(n-1) 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 * h_1] = 3 * h_2 + 6 * h_1^2 = A248120(3) with h_0 = 1. (Example corrected Jul 14 2015.)
For another example and relations to A134264 and A036038, see A134264. 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.
The Abramowitz and Stegun reference in A036038 gives 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)
As presented above and in the Copeland link, this entry is related to exponentiation of e.g.f.s and, therefore, to discussions in the Scott and Sokal preprint (see eqn. 3.1 on p. 10 and eqn. 3.62 p. 24). - Tom Copeland, Jan 17 2017

Examples

			Triangle begins
     1;
     2;
     6,     3;
    24,    36,     4;
   120,   360,    60,    80,    5;
   720,  3600,  1800,  1200,  300,   150,    6;
  5040, 37800, 37800, 16800, 3150, 12600, 3150, 420, 630, 252, 7;
  ...
For f(t)= e^t-1, h(t)= t/f(t)= t/(e^t-1), the e.g.f. for the Bernoulli numbers, and plugging the Bernoulli numbers into the Lagrange inversion formula gives g(t)= t - t^2/2 + t^3/3 + ... = log(1+t).
		

Crossrefs

Cf. A134264 and A248927, "scaled" versions of this Lagrange inversion.
Cf. A036038.

Programs

  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); (n+1)*n!^2/((n-#v+1)!*prod(i=1, #S, my(x=S[i], c=#select(y->y==x, v)); x!^c*c!))}
    row(n)=[C(Vec(p)) | p<-Vecrev(partitions(n-1))]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 02 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 (j-1)! P(j,m;a...) / [(2!)^a_2 (3!)^a_3 ... ((j-1)!)^a_(j-1) ] for the k-th partition of j-1. The partitions are in reverse order--from bottom to top--from the order in Abramowitz and Stegun (page 831).
For example, from g(t) above, T(6,3) = 5! * [6!/(3!*2!)]/(2!)^2 = 1800 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = 5! * [6!/4!]/(2!*3!) = 300.
If the initial factorial and final denominator of T(n,k) are removed and the expression divided by j and the partitions reversed in order, then A134264 is obtained, a refinement of the Narayana numbers.
For f(t) = t*e^(-t), g(t) = T(t), the Tree function, which is the e.g.f. of A000169, and h(t) = t/f(t) = e^t, so h_n = 1 for all n in this case; therefore, the row sums are A000169(n) = n^(n-1) = n* A000272(n).
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}=1/[d{x/[h_0+h_1*x+ ...]/dx]. Then the partition polynomials above are given by (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)). See A145271.
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) and L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2/2!+...], which will give a lowering operator associated to the refined f-vectors of permutohedra (cf. A133314 and A049019).
Then [dPS(n,z)/dz]/n eval. at z=0 are the row partition polynomials of this entry. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
Following the notes connected to the Lagrange reversion theorem in A248927, a generator for the n-th partition polynomial P_n of this entry is (d/dx)^(n-1) (h (x))^n, and -log(1-t*P.) = (t*Q.) / (1 - t*Q.), umbrally, where (Q.)^n = Q_n is the n-th partition polynomial of A248927. - Tom Copeland, Nov 25 2016
With h_0 = 1, the n-th partition polynomial is obtained as the n-th element (with initial index 0) of the first column of M^{n+1}, where M is the matrix with M_{i,j}= binomial(i,j) h_{i-j}, i.e., the lower triangular Pascal matrix with its n-th diagonal multiplied by h_n. This follows from the Lagrange inversion theorem and the relation between powers of matrices such as M and powers of formal Taylor series discussed in A133314. This is equivalent to repeated binomial convolution of the coefficients of the Taylor series with itself. - Tom Copeland, Nov 13 2019
T(n,k) = n*A248927(n,k). - Andrew Howroyd, Feb 02 2022

Extensions

Terms a(31) and beyond from Andrew Howroyd, Feb 02 2022

A055140 Triangle read by rows: T(n, k) = number of matchings of 2n people with partners (of either sex) such that exactly k couples are left together.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 8, 6, 0, 1, 60, 32, 12, 0, 1, 544, 300, 80, 20, 0, 1, 6040, 3264, 900, 160, 30, 0, 1, 79008, 42280, 11424, 2100, 280, 42, 0, 1, 1190672, 632064, 169120, 30464, 4200, 448, 56, 0, 1, 20314880, 10716048, 2844288, 507360, 68544, 7560, 672, 72, 0, 1
Offset: 0

Views

Author

Christian G. Bower, May 09 2000

Keywords

Comments

T is an example of the group of matrices outlined in the table in A132382--the associated matrix for aC(1,1). The e.g.f. for the row polynomials is exp(x*t) * exp(-x) * (1-2*x)^(-1/2). T(n,k) = Binomial(n,k)* s(n-k) where s = A053871 with an e.g.f. of exp(-x) * (1-2*x)^(-1/2) which is the reciprocal of the e.g.f. of A055142. The row polynomials form an Appell sequence. Tom Copeland, Sep 10 2008
A231846 provides a refinement of this array. - Tom Copeland, Oct 12 2016

Examples

			Triangle T(n,k) starts:
     1;
     0,    1;
     2,    0,   1;
     8,    6,   0,   1;
    60,   32,  12,   0,  1;
   544,  300,  80,  20,  0, 1;
  6040, 3264, 900, 160, 30, 0, 1;
  ...
		

Crossrefs

First column is A053871.
Row sums are A001147.

Programs

  • Maple
    g[0] := 1: g[1] := 0: for n from 2 to 20 do g[n] := (2*(n-1))*(g[n-1]+g[n-2]) end do: T := proc (n, k) options operator, arrow; g[n-k]*binomial(n, k) end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form; Emeric Deutsch, Jan 24 2009
  • Mathematica
    Table[(-1)^# HypergeometricPFQ[{1/2, -#}, {}, 2] Binomial[n, k] &[n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 10 2019, after Eric W. Weisstein at A053871 *)

Formula

T(n, k) = A053871(n-k)*binomial(n, k).
From Tom Copeland, Oct 12 2016: (Start)
E.g.f.: e^(xt) e^(-t) (1-2t)^(-1/2) = e^(p.(x)*t)(from my 2008 comment).
Row sums are A001147.
L = D = d/dx and R = x + d[log[e^(L)(1-2L)^(-1/2)]]/dL = x - 1 + 1/(1-2D) = x + 2D + (2D)^2 + (2D)^3 + ... are the lowering and raising operators, i.e., L p_n(x) = n * p_(n-1)(x) and R p_n(x) = p_(n+1)(x); e.g., L p_2(x) = D (2 + x^2) = 2 x = 2 p_1(x) and R P_2(x) = (x + 2D + 4D^2 + ...) (2 + x^2) = 2x + x^3 + 4x + 8 = 8 + 6x + x^3 = p_3(x).
Another generator is (1-2D)^(-1/2) e^(-D) x^n = (1-2D)^(-1/2) (x-1)^n = p_n(x). For example, (1-2D)^(-1/2)(x-1)^2 = (1 + D + 3 D^2/2 + ...) (x-1)^2 = (x-1)^2 + 2(x-1) + 3 = 2 + x^2 = p_2(x).
Umbral binomial convolution gives p_n(x) = (a. + x)^n = sum_{k = 0,..,n} C(n,k) a_(n-k) * x^k with (a.)^k = a_k = A053871(k).
The Appell sequence of umbral compositional inverses has the e.g.f. e^(xt) e^t (1-2t)^(1/2) associated with A055142. Cf. A231846 for a definition of umbral compositional inversion.
See A132382 and A133314 for more relations.
(End)

A238390 E.g.f.: x / BesselJ(1, 2*x) (even powers only).

Original entry on oeis.org

1, 1, 4, 35, 546, 13482, 485892, 24108513, 1576676530, 131451399794, 13609184032808, 1712978776719938, 257612765775847132, 45620136452519144700, 9396239458048330569840, 2227147531572856811691105, 601916577165056911293330930, 183994483721828524163677628370
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 01 2014

Keywords

Comments

Aerated, the e.g.f. is e^(a.t) = 1/AC(i*t) = 1/[I_1(2i*t)/(it)] = 1/Sum_{n>=0} (-1)^n t^(2n) / [n!(n+1)!] = a_0 + a_2 t^2/2! + a_4 t^4/4! + ... = 1 + t^2/2! + 4 t^4/4! + 35 t^6/6! + ..., where AC(t) is the e.g.f. for the aerated Catalan numbers c_n of A126120 and I_n(t) are the modified Bessel functions of the first kind (i = sqrt(-1)). The signed, aerated sequence b_n = (i)^n a_n has the e.g.f. e^(b.t) = 1/AC(t) and, therefore, (i*a. + c.)^n = Sum_{k=0..n} binomial(n,k) i^k a_k c_(n-k) vanishes except for n=0 for which it's unity. - Tom Copeland, Jan 23 2016
With q(n) = A126120(n+1) and q(0) = 0, d(2n) = (-1)^n A238390(n) and zero for odd arguments, and r(2n+1) = (-1)^n A180874(n+1) and zero for even arguments, then r(n) = (q. + d.)^n = Sum_{k=0..n} binomial(n,k) q(k) d(n-k), relating these sequences (and A000108) through binomial convolutions. Then also, (r. + c. + d.)^n = r(n). See A180874 for proofs and for relations to A097610. For quick reference, q = (0, 1, 0, 2, 0, 5, 0, 14, ..), d = (1, 0, -1, 0, 4, 0, -35, 0, ..), and r = (0, 1, 0, -1, 0, 5, 0, -56, ..). - Tom Copeland, Jan 28 2016
Aerated and signed, this sequence contains the moments m(n) of the Appell polynomial sequence UMT(n,h1,h2) that is the umbral compositional inverse of the Appell sequence of Motzkin polynomials MT(n,h1,h2) of A097610 with exp[x UMT(.,h1,h2)] = e^(x*h1) / AC(x*y) where y = sqrt(h2) and AC is defined above. UMT(n,h1,h2) = (m.y + h1)^n with (m.)^(2n) = m(2n) = (-1)^n A238390(n) and zero otherwise. Consequently, the associated lower triangular matrices A007318(n,k)*m(n-k) and A007318(n,k)*A126120(n-k) form an inverse pair (cf. also A133314), and MT(n,UMT(.,h1,h2),h2) = h1^n = UMT(n,MT(.,h1,h2),h2). - Tom Copeland, Jan 30 2016

Crossrefs

Programs

  • Maple
    S:= series(x/BesselJ(1,2*x),x,102):
    seq((2*j)!*coeff(S,x,2*j),j=0..50); # Robert Israel, Jan 31 2016
  • Mathematica
    Table[(CoefficientList[Series[x/BesselJ[1, 2*x], {x, 0, 40}], x] * Range[0, 40]!)[[n]], {n, 1, 41, 2}]

Formula

a(n) ~ c * (n!)^2 / (sqrt(n) * r^n), where r = BesselJZero[1, 1]^2/16 = 0.91762316513274332857623611, and c = 1/(Sqrt[Pi]*BesselJ[2, BesselJZero[1, 1]]) = 1.4008104828035425937394082168... - Vaclav Kotesovec, Mar 01 2014, updated Apr 01 2018

A111062 Triangle T(n, k) = binomial(n, k) * A000085(n-k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 6, 3, 1, 10, 16, 12, 4, 1, 26, 50, 40, 20, 5, 1, 76, 156, 150, 80, 30, 6, 1, 232, 532, 546, 350, 140, 42, 7, 1, 764, 1856, 2128, 1456, 700, 224, 56, 8, 1, 2620, 6876, 8352, 6384, 3276, 1260, 336, 72, 9, 1, 9496, 26200, 34380, 27840, 15960, 6552, 2100, 480, 90, 10, 1
Offset: 0

Views

Author

Philippe Deléham, Oct 07 2005

Keywords

Comments

Triangle related to A000085.
Riordan array [exp(x(2+x)/2),x]. - Paul Barry, Nov 05 2008
Array is exp(S+S^2/2) where S is A132440 the infinitesimal generator for Pascal's triangle. T(n,k) gives the number of ways to choose a subset of {1,2,...,n} of size k and then partitioning the remaining n-k elements into sets each of size 1 or 2. Cf. A122832. - Peter Bala, May 14 2012
T(n,k) is equal to the number of R-classes (equivalently, L-classes) in the D-class consisting of all rank k elements of the partial Brauer monoid of degree n. - James East, Aug 17 2015

Examples

			Rows begin:
     1;
     1,    1;
     2,    2,    1;
     4,    6,    3,    1;
    10,   16,   12,    4,    1;
    26,   50,   40,   20,    5,    1;
    76,  156,  150,   80,   30,    6,   1;
   232,  532,  546,  350,  140,   42,   7,  1;
   764, 1856, 2128, 1456,  700,  224,  56,  8, 1;
  2620, 6876, 8352, 6384, 3276, 1260, 336, 72, 9, 1;
From _Paul Barry_, Apr 23 2009: (Start)
Production matrix is:
  1, 1,
  1, 1, 1,
  0, 2, 1, 1,
  0, 0, 3, 1, 1,
  0, 0, 0, 4, 1, 1,
  0, 0, 0, 0, 5, 1, 1,
  0, 0, 0, 0, 0, 6, 1, 1,
  0, 0, 0, 0, 0, 0, 7, 1, 1,
  0, 0, 0, 0, 0, 0, 0, 8, 1, 1 (End)
From _Peter Bala_, Feb 12 2017: (Start)
The infinitesimal generator has integer entries and begins
  0
  1  0
  1  2  0
  0  3  3  0
  0  0  6  4  0
  0  0  0 10  5  0
  0  0  0  0 15  6  0
  ...
and is the generalized exponential Riordan array [x + x^2/2!,x].(End)
		

Crossrefs

Cf. A099174, A133314, A159834 (inverse matrix).

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->(Factorial(n)/Factorial(k))*Sum([0..n-k],j->Binomial(j,n-k-j)/(Factorial(j)*2^(n-k-j)))))); # Muniru A Asiru, Jun 29 2018
  • Mathematica
    a[n_] := Sum[(2 k - 1)!! Binomial[n, 2 k], {k, 0, n/2}]; Table[Binomial[n, k] a[n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 20 2015, after Michael Somos at A000085 *)
  • Sage
    def A111062_triangle(dim):
        M = matrix(ZZ,dim,dim)
        for n in (0..dim-1): M[n,n] = 1
        for n in (1..dim-1):
            for k in (0..n-1):
                M[n,k] = M[n-1,k-1]+M[n-1,k]+(k+1)*M[n-1,k+1]
        return M
    A111062_triangle(9) # Peter Luschny, Sep 19 2012
    

Formula

Sum_{k>=0} T(m, k)*T(n, k)*k! = T(m+n, 0) = A000085(m+n).
Sum_{k=0..n} T(n, k) = A005425(n).
Apparently satisfies T(n,m) = T(n-1,m-1) + T(n-1,m) + (m+1) * T(n-1,m+1). - Franklin T. Adams-Watters, Dec 22 2005 [corrected by Werner Schulte, Feb 12 2025]
T(n,k) = (n!/k!)*Sum_{j=0..n-k} C(j,n-k-j)/(j!*2^(n-k-j)). - Paul Barry, Nov 05 2008
G.f.: 1/(1-xy-x-x^2/(1-xy-x-2x^2/(1-xy-x-3x^2/(1-xy-x-4x^2/(1-... (continued fraction). - Paul Barry, Apr 23 2009
T(n,k) = C(n,k)*Sum_{j=0..n-k} C(n-k,j)*(n-k-j-1)!! where m!!=0 if m is even. - James East, Aug 17 2015
From Tom Copeland, Jun 26 2018: (Start)
E.g.f.: exp[t*p.(x)] = exp[t + t^2/2] e^(x*t).
These polynomials (p.(x))^n = p_n(x) are an Appell sequence with the lowering and raising operators L = D and R = x + 1 + D, with D = d/dx, such that L p_n(x) = n * p_(n-1)(x) and R p_n(x) = p_(n+1)(x), so the formalism of A133314 applies here, giving recursion relations.
The transpose of the production matrix gives a matrix representation of the raising operator R.
exp(D + D^2/2) x^n= e^(D^2/2) (1+x)^n = h_n(1+x) = p_n(x) = (a. + x)^n, with (a.)^n = a_n = A000085(n) and h_n(x) the modified Hermite polynomials of A099174.
A159834 with the e.g.f. exp[-(t + t^2/2)] e^(x*t) gives the matrix inverse for this entry with the umbral inverse polynomials q_n(x), an Appell sequence with the raising operator x - 1 - D, such that umbrally composed q_n(p.(x)) = x^n = p_n(q.(x)). (End)

Extensions

Corrected by Franklin T. Adams-Watters, Dec 22 2005
10th row added by James East, Aug 17 2015

A119502 Triangle read by rows, T(n,k) = (n-k)!, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 6, 2, 1, 1, 24, 6, 2, 1, 1, 120, 24, 6, 2, 1, 1, 720, 120, 24, 6, 2, 1, 1, 5040, 720, 120, 24, 6, 2, 1, 1, 40320, 5040, 720, 120, 24, 6, 2, 1, 1, 362880, 40320, 5040, 720, 120, 24, 6, 2, 1, 1, 3628800, 362880, 40320, 5040, 720, 120, 24, 6, 2, 1, 1, 39916800
Offset: 0

Views

Author

Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006

Keywords

Comments

The reciprocal of each entry in a lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals one (and all other entries are zero). Note all said entries are unit fractions (all numerators are one).
Denominators of unfinished fractional coefficients for polynomials A152650/A152656 = A009998/A119052. - Paul Curtz, Dec 13 2008
Multiplying the n-th diagonal by b_n with b_0 = 1 and then beheading the triangle provides a Gram matrix whose determinant is related to the reciprocal of e.g.f.s as presented in A133314. - Tom Copeland, Dec 04 2016

Examples

			Triangle starts:
   1;
   1, 1;
   2, 1, 1;
   6, 2, 1, 1;
  24, 6, 2, 1, 1;
		

Crossrefs

Cf. A025581.
Cf. A133314.

Programs

  • Magma
    [[Factorial(n-k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Jun 18 2015
  • Mathematica
    Table[Gamma[Binomial[1 + Floor[(1/2) + Sqrt[2*(1 + n)]], 2] - n], {n, 0, 77}]

Formula

T(n,k) = A025581(n,k)!.
a(n) = Gamma(binomial(1 + floor((1/2) + sqrt(2*(1 + n))), 2) - n).

Extensions

Name edited by Peter Luschny, Jun 17 2015

A248927 Triangle read by rows: T(n,k) are the coefficients of the Lagrange (compositional) inversion of a function in terms of the Taylor series expansion of its reciprocal, n >= 1, k = 1..A000041(n-1).

Original entry on oeis.org

1, 1, 2, 1, 6, 9, 1, 24, 72, 12, 16, 1, 120, 600, 300, 200, 50, 25, 1, 720, 5400, 5400, 2400, 450, 1800, 450, 60, 90, 36, 1, 5040, 52920, 88200, 29400, 22050, 44100, 7350, 4410, 2940, 4410, 882, 245, 147, 49, 1, 40320, 564480, 1411200, 376320, 705600, 940800
Offset: 1

Views

Author

Tom Copeland, Oct 16 2014

Keywords

Comments

Coefficients are listed in reverse graded colexicographic order (A228100). This is the reverse of Abramowitz and Stegun order (A036036).
Coefficients for Lagrange (compositional) inversion of a function in terms of the Taylor series expansion of its shifted reciprocal. Complementary to A134264 for formal power series. A refinement of A141618 with row sums A000272.
Given an invertible function f(t) analytic about t=0 with f(0)=0 and df(0)/dt not 0, form h(t) = t / f(t) and denote h_n = (n') as the coefficient of t^n/n! in h(t). Then the compositional inverse of f(t), g(t), as a formal Taylor series, or e.g.f., is given up to the first few orders by
g(t)/t = [ 1 (0') ]
+ [ 1 (0') (1') ] * t
+ [ 2 (0') (1')^2 + 1 (0')^2 (2') ] * t^2/2!
+ [ 6 (0') (1')^3 + 9 (0')^2 (1') (2') + 1 (0')^3 (3') ] * t^3/3!
+ [24 (0') (1')^4 + 72 (0')^2 (1')^2 (2') + (0')^3 [12 (2')^2
+ 16 (1') (3')] + (0')^4 (4')] * t^4/4!
+ [120 (0')(1')^5 + 600 (0')^2 (1')^3(2') + (0')^3 [300 (1')(2')^2 + 200 ( 1')^2(3')] + (0')^4 [50 (2')(3') + 25 (1')(4')] + (0')^5 (5')] * t^5/5! + [720 (0')(1')^6 + (0')^2 (1')^4(2')+(0')^3 [5400 (1')^2(2')^2 + 2400 (1')^3(3')] + (0')^4 [450 (2')^3+ 1800 (1')(2')(3') + 450( 1')^2(4')]+ (0')^5 [60 (3')^2 + 90 (2')(4') + 36 (1')(5')] + (0')^6 (6')] * t^6/6! + ...
..........
From Tom Copeland, Oct 28 2014: (Start)
Expressing g(t) as a Taylor series or formal e.g.f. in the indeterminates h_n generates a refinement of A055302, which enumerates the number of labeled root trees with n nodes and k leaves, with row sum A000169.
Operating with (1/n^2) d/d(1') = (1/n^2) d/d(h_1) on the n-th partition polynomial in square brackets above associated with t^n/n! generates the (n-1)-th partition polynomial.
Multiplying the n-th partition polynomial here by (n + 1) gives the (n + 1)-th partition polynomial of A248120. (End)
These are also the coefficients in the expansion of a series related to the Lagrange reversion theorem presented in Wikipedia of which the Lagrange inversion formula about the origin is a special case. Cf. Copeland link. - Tom Copeland, Nov 01 2016

Examples

			Triangle T(n,k) begins:
    1;
    1;
    2,    1;
    6,    9,    1;
   24,   72,   12,   16,   1;
  120,  600,  300,  200,  50,   25,   1;
  720, 5400, 5400, 2400, 450, 1800, 450, 60, 90, 36, 1;
  ...
For f(t) = e^t-1, h(t) = t/f(t) = t/(e^t-1), the e.g.f. for the Bernoulli numbers, and plugging the Bernoulli numbers into the Lagrange inversion formula gives g(t) = t - t^2/2 + t^3/3 + ... = log(1+t).
		

Crossrefs

Cf. A134264 and A248120, "scaled" versions of this Lagrange inversion.
Cf. A036038.

Programs

  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); n!^2/((n-#v+1)!*prod(i=1, #S, my(x=S[i], c=#select(y->y==x, v)); x!^c*c!))}
    row(n)=[C(Vec(p)) | p<-Vecrev(partitions(n-1))]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 02 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 [(j-1)!/j]* P(j,m;a...) / [(2!)^a_2 (3!)^a_3 ... ((j-1)!)^a_(j-1) ] for the k-th partition of j-1. The partitions are in reverse order--from bottom to top--from the order in Abramowitz and Stegun (page 831).
For example, from g(t) above, T(6,3) = [5!/6][6!/(3!*2!)]/(2!)^2 = 300 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = [5!/6][6!/4!]/(2!*3!) = 50.
If the initial factorial and final denominator are removed and the partitions reversed in order, A134264 is obtained, a refinement of the Narayana numbers.
For f(t) = t*e^(-t), g(t) = T(t), the Tree function, which is the e.g.f. of A000169, and h(t) = t/f(t) = e^t, so h_n = 1 for all n in this case; therefore, the row sums of A248927 are A000169(n)/n = n^(n-2) = A000272(n).
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}=1/{d[x/[h_0+h_1*x+ ...]]/dx}. Then the partition polynomials above are given by (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)). See A145271.
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) and L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2/2!+...], which will give a lowering operator associated to the refined f-vectors of permutohedra (cf. A133314 and A049019).
Then [dPS(n,z)/dz]/n eval. at z=0 are the row partition polynomials of this entry. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
As noted in A248120 and A134264, this entry is given by the Hadamard product by partition of A134264 and A036038. For example, (1,4,2,6,1)*(1,4,6,12,24) = (1,16,12,72,24). - Tom Copeland, Nov 25 2016
T(n,k) = ((n-1)!)^2/((n-j)!*Product_{i>=1} s_i!*(i!)^s_i), where (1*s_1 + 2*s_2 + ... = n-1) is the k-th partition of n-1 and j = s_1 + s_2 ... is the number of parts. - Andrew Howroyd, Feb 02 2022

Extensions

Name edited and terms a(31) and beyond from Andrew Howroyd, Feb 02 2022

A249548 Coefficients of reduced partition polynomials of A134264 for computing Lagrange compositional inversion.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 1, 5, 1, 6, 3, 5, 1, 7, 7, 21, 1, 8, 8, 4, 28, 28, 14, 1, 9, 9, 9, 36, 72, 12, 84, 1, 10, 10, 10, 5, 45, 90, 45, 45, 120, 180, 42, 1, 11, 11, 11, 11, 55, 110, 110, 55, 55, 165, 495, 165, 330, 1, 12, 12, 12, 12, 6, 66, 132, 132, 66, 66, 132, 22, 220, 660, 330, 660, 55, 495, 990, 132
Offset: 0

Views

Author

Tom Copeland, Oct 31 2014

Keywords

Comments

Coefficients of reduced partition polynomials of A134264 for computing the complete partition polynomials for the Lagrange compositional inversion of A134264 (see Oct 2014 comment by Copeland there). Umbrally,
e^(x*t) * exp[Prt(.;1,0,h_2,..) * t] = exp[Prt(.;1,x,h_2,..) * t], where Prt(n;1,0,h_2,..,h_n) are the reduced (h_0 = 1 and h_1 = 0) partition polynomials of the complete polynomials Prt(n;h_0,h_1,h_2,..,h_n) of A134264.
Partitions are given in the order of those on page 831 of Abramowitz and Stegun. Formulas for the coefficients of the partitions are given in A134264.
Row sums are the Motzkin sums or Riordan numbers A005043. - Tom Copeland, Nov 09 2014
From Tom Copeland, Jul 03 2018: (Start)
The matrix and operator formalism for Sheffer Appell sequences leads to the following relations with D = d/dh_1.
Exp[Prt(.;1,0,h_2,..) * D] (h_1)^n = [h_1 + Prt(.;1,0,h_2,...)]^n = Prt(n;1,h_1,h_2,...), the partition polynomials of A134264 for g(t)/t with h_0 = 1.
For the umbral compositional inverses described below,
Exp[UPrt(.;1,0,h_2,..) * D] (h_1)^n = [h_1 + UPrt(.;1,0,h_2,...)]^n = UPrt(n;1,h_1,h_2,...).
The respective e.g.f.s are multiplicative inverses; that is, exp[Prt(.;1,0,h_2,..) * t] = 1/exp[UPrt(.;1,0,h_2,..) * t], so the formalism of A133314 applies.
The raising operator R such that R Prt(n;1,h_1,h_2,...) = Prt(n+1;1,h_1,h_2,...) is R = exp[Prt(.;1,0,h_2,...)*D] h_1 exp[UPrt(.;1,0,h_2,..)*D] since R Prt(n+1;1,h_1,h_2,...) = exp[Prt(.;1,0,h_2,...)*D] h_1 (h_1)^n = Prt(n+1;h_1,h_2,...) from the definition of the umbral compositional inverse. This may also be expressed as R = h_1 + d/dD log[exp[Prt(.;1,0,h_2,...) * D]], so, using A127671, 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 - 9 h_2 h_4 + 5 h_2^3 - 7 h_3^2) D^5/5! + (h_7 - 28 h_3 h_4 - 14 h_2 h_5 + 56 h_2^2 h_3) D^6/6! + ... . (End)

Examples

			Prt(0) = 1
Prt(1;1,0) = 0
Prt(2;1,0,h_2) = 1 h_2
Prt(3;1,0,h_2,h_3) = 1 h_3
Prt(4;1,0,h_2,..,h_4) = 1 h_4 + 2 (h_2)^2
Prt(5;1,0,h_2,..,h_5) = 1 h_5 + 5 h_2 h_3
Prt(6;1,0,h_2,..,h_6) = 1 h_6 + 6 h_2 h_4 + 3 (h_3)^2 + 5 (h_2)^3
Prt(7;1,0,h_2,..,h_7) = 1 h_7 + 7 h_3 h_4 + 7 h_2 h_5 + 21 h_3 (h_2)^2
...
------------
With h_n denoted by (n'), the first seven partition polynomials of A134264 with h_0=1 are given by the first seven coefficients of the truncated Taylor series expansion of the Euler binomial transform
e^[(1') * t] * {1 + 1 (2') *  t^2/2! + 1 (3') *  t^3/3! + [1 (4') + 2 (2')^2] *  t^4/4! + [1 (5') + 5 (2')(3')] *  t^5/5! + [1 (6') + 6 (2')(4') + 3 (3')^2 + 5 (2')^3] *  t^6/6!}, giving the truncated expansion
1 + 1 (1') * t + [1 (2') + 1 (1')^2] * t^2/2! + ... + [1 (6') + 6 (1')(5') + 6 (2')(4') + 3 (3')^2 + 15 (1')^2(4') + 30 (1')(2')(3') + 5 (2')^3 + 20 (1')^3(3') + 30 (1')^2(2')^2 + 15 (1')^4(2') + 1 (1')^6] * t^6/6!.
Extending the number of reduced partition polynomials of the transform allows for further complete polynomials of A134264 to be computed.
		

Crossrefs

Cf. A127671.
Rows lengths are given by A002865 (except for row 1).

Programs

  • Mathematica
    rows[n_] := {{1}, {0}}~Join~Module[
        {g = 1 / D[t / (1 + Sum[h[k] t^k, {k, 2, n}] + O[t]^(n+1)), t], p = t, r},
        r = Reap[Do[p = g D[p, t]/k; Sow[Expand[Normal@p /. {t -> 0}]], {k, n+1}]][[2, 1, 2 ;;]];
        Table[Coefficient[r[[k]], Product[h[t], {t, p}]], {k, 2, n}, {p, Sort[Sort /@ IntegerPartitions[k, k, Range[2, k]]]}]];
    rows[12] // Flatten (* Andrey Zabolotskiy, Feb 18 2024 *)

Formula

From Tom Copeland, Nov 10 2014: (Start)
Terms may be computed symbolically up to order n by using an iterated derivative evaluated at t=0:
with g(t) = 1/{d/dt [t/(1 + 0 t + h_2 t^2 + h_3 t^3 + ... + h_n t^n)]},
evaluate 1/n! * [g(t) d/dt]^n t at t=0, i.e., ask a symbolic math app for the first term in a series expansion of this iterated derivative, to obtain Prt(n-1).
Alternatively, the explicit formula in A134264 for the numerical coefficients of each partition can be used. (End)
From Tom Copeland, Nov 12 2014: (Start)
The first few partitions polynomials formed by taking the reciprocal of the e.g.f. of this entry's e.g.f. (cf. A133314) are
UPrt(0) = 1
UPrt(1;1,0) = 0
UPrt(2;1,0,h_2) = -1 h_2
UPrt(3;1,0,h_2,h_3) = -1 h_3
UPrt(4;1,0,h_2,..,h_4) = -1 h_4 + 4 (h_2)^2
UPrt(5;1,0,h_2,..,h_5) = -1 h_5 + 15 h_2 h_3
UPrt(6;1,0,h_2,..,h_6) = -1 h_6 + 24 h_2 h_4 + 17 (h_3)^2 + -35 (h_2)^3
...
Therefore, umbrally, [Prt(.;1,0,h_2,...) + UPrt(.;1,0,h_2,...)]^n = 0 for n>0 and unity for n=0.
Example of the umbral operation:
(a. + b.)^2 = a.^2 + 2 a.* b. + b.^2 = a_2 + 2 a_1 * b_1 + b_2.
This implies that the umbral compositional inverses (see below) of the partition polynomials of the Lagrange inversion formula (LIF) of A134264 with h_0=1 are given by UPrt(n;1,h_1,h_2,...,h_n) = [UPrt(.;1,0,h_2,...,h_n) + h_1]^n, so that the sequence of polynomials UPrt(n;1,h_1,h_2,...,h_n) is an Appell sequence in the indeterminate h_1. So, if one calculates UPrt(n;1,h_1,...,h_n), the lower order UPrt(n-1;1,h_1,...,h_(n-1)) can be found by taking the derivative w.r.t. h_1 and dividing by n. Same applies for Prt(n;1,h_1,h_2,...,h_n).
This connects the combinatorics of the permutohedra through A133314 and A049019, or their duals, to the noncrossing partitions, Dyck lattice paths, etc. that are isomorphic with the LIF of A134264.
An Appell sequence P(.,x) with the e.g.f. e^(x*t)/w(t) possesses an umbral inverse sequence UP(.,x) with the e.g.f. w(t)e^(x*t), i.e., polynomials such that P(n,UP(.,x))= x^n = UP(n,P(.,x)) through umbral substitution, as in the binomial example. The Bernoulli polynomials with w(t) = t/(e^t - 1) are a good example with the umbral compositional inverse sequence UP(n,x) = [(x+1)^(n+1)-x^(n+1)] / (n+1) (cf. A074909 and A135278). (End)

Extensions

Formula for Prt(7,..) and a(12)-a(15) added by Tom Copeland, Jul 22 2016
Rows 8-12 added by Andrey Zabolotskiy, Feb 18 2024

A356145 Coefficients of the inverse refined Eulerian partition polynomials [E]^{-1}, partitional inverse to A145271. Irregular triangle read by row with lengths A000041.

Original entry on oeis.org

1, 1, -1, 1, 3, -4, 1, -15, 25, -4, -7, 1, 105, -210, 70, 60, -15, -11, 1, -945, 2205, -1120, -630, 70, 350, 126, -15, -26, -16, 1, 10395, -27720, 18900, 7875, -2800, -6930, -1638, 560, 455, 784, 238, -56, -42, -22, 1, -135135, 405405, -346500, -114345, 84700
Offset: 0

Views

Author

Tom Copeland, Jul 27 2022

Keywords

Comments

These are the coefficients of the inverse refined Eulerian partitions polynomials, the substitutional inverse to the refined Eulerian partition polynomials [E] of A145271. [E] and [E]^{-1} are a conjugate dual with respect to the permutahedra polynomials [P] of A133314 (see formula section).

Examples

			The first few rows of coefficients with monomials in reverse order to the partitions of Abramowitz and Stegun (link in A000041, pp. 831-2) are
0)       1;
1)       1;
2)      -1,      1;
3)       3,     -4,       1;
4)     -15,     25,      -4,      -7,     1;
5)     105,   -210,      70,      60,   -15,    -11,     1;
6)    -945,   2205,   -1120,    -630,    70,    350,   126,   -15,    -26,    -16,      1;
7)   10395, -27720,   18900,    7875, -2800,  -6930, -1638,   560,    455,    784,    238,   -56,  -42,  -22,    1;
8) -135135, 405405, -346500, -114345, 84700, 138600, 24255, -2800, -27300, -11025, -18900, -3780, 1575, 1344, 2142, 1596, 414, -56, -98, -64, -29, 1;
    ...
The first few partition polynomials are
E_0^{(-1)} = 1,
E_1^{(-1)} = a1,
E_2^{(-1)} = -a1^2 + a2,
E_3^{(-1)} = 3 a1^3 - 4 a1 a2 + a3,
E_4^{(-1)} = -15 a1^4 + 25 a1^2 a2 - 4 a2^2 - 7 a1 a3 + a4,
E_5^{(-1)} = 105 a1^5 - 210 a1^3 a2 + 70 a1 a2^2 + 60 a1^2 a3 - 15 a2 a3 - 11 a1 a4 + a5,
E_6^{(-1)} = -945 a1^6 + 2205 a1^4 a2 - 1120 a1^2 a2^2 - 630 a1^3 a3 + 70 a2^3 + 350 a1 a2 a3 + 126 a1^2 a4 - 15 a3^2 - 26 a2 a4 - 16 a1 a5 + a6,
E_7^{(-1)} = 10395 a1^7 - 27720 a1^5 a2 + 18900 a1^3 a2^2 + 7875 a1^4 a3 - 2800 a1 a2^3 - 6930 a1^2 a2 a3 - 1638 a1^3 a4 + 560 a2^2 a3 + 455 a1 a3^2 + 784 a1 a2 a4 + 238 a1^2 a5 - 56 a3 a4 - 42 a2 a5 - 22 a1 a6 + a7,
E_8^{(-1)} = -135135 a1^8 + 405405 a1^6 a2 - 346500 a1^4 a2^2 - 114345 a1^5 a3 + 84700 a1^2 a2^3 + 138600 a1^3 a2 a3 + 24255 a1^4 a4 - 2800 a2^4 - 27300 a1 a2^2 a3 - 11025 a1^2 a3^2 - 18900 a1^2 a2 a4 - 3780 a1^3 a5 + 1575 a2 a3^2 + 1344 a2^2 a4 + 2142 a1 a3 a4 + 1596 a1 a2 a5 + 414 a1^2 a6 - 56 a4^2 - 98 a3 a5 - 64 a2 a6 - 29 a1ma7 + a8,
... .
Example substitution identities:
With the permutahedra polynomials
P_1 = -a_1,
P_2 = 2*a_1^2 - a_2,
P_3 = -6*a_1^3 + 6*a_2*a_1 - a_3,
the refined Eulerian polynomials
E_1 = a_1,
E_2 = a_1^2 + a_2,
E_3 = a_1^3 + 4*a_1*a_2 + a_3,
the reciprocal tangent polynomials
RT_1 = -a_1,
RT_2 = -a_2 + a_1^2,
RT_3 = -a_3 + 2*a_1*a_2 - a_1^3,
the Lagrange inversion polynomials
L_1 = -a_1,
L_2 = 3*a_1^2 - a_2,
L_3 = -15*a_1^3 + 10*a_1a_2 - a_3,
then
E^{-1}_3 = P_3(L_1,L_2,L_3) = -6*(-a_1)^3 + 6*(3*a_1^2 - a_2)*(-a_1) - (-15*a_1^3 + 10*a_1*a_2 - a_3) = 3*a_1^3 - 4*a_2*a_1 + a_3,
E^{-1}_3 = RT_3(P_1,P_2,P_3) = -(-6*a_1^3 + 6*a_2*a_1 - a_3) + 2*(-a_1)*(2*a_1^2 - a_2) - (-a_1)^3 = 3*a_1^3 - 4*a_2*a_1 + a_3,
E{-1}_3(E_1,E_2,E_3) = 3*a_1^3 - 4*a_1*(a_1^2 + a_2) + (a_1^3 + 4*a_1*a_2 + a_3) = a_3.
		

Crossrefs

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = 1/D[InverseSeries[x + Sum[c[k - 1] x^k/k!, {k, 2, nn}] + O[x]^(nn + 1)], x]}, Table[Coefficient[n! s, x^n Product[c[t], {t, p}]], {n, nn-1}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
    rows[8] // Flatten (* Andrey Zabolotskiy, Feb 17 2024 *)
  • SageMath
    B. = PolynomialRing(ZZ)
    A. = PowerSeriesRing(B)
    f =  x + a1*x^2/factorial(2) + a2*x^3/factorial(3) + a3*x^4/factorial(4) + a4*x^5/factorial(5) + a5*x^6/factorial(6) + a6*x^7/factorial(7) + a7*x^8/factorial(8) + a8*x^9/factorial(9) + a9*x^10/factorial(10)
    g = f.reverse()
    w = derivative(g,x)
    I = 1 / w
    # Added by Peter Luschny, Feb 17 2024:
    for n, c in enumerate(I.list()[:9]):
        print(f"E[{n}]", (factorial(n)*c).coefficients())

Formula

Given the formal Taylor series or e.g.f. f(x) = x + a_1 x^2/2! + a_2 x^3/3! + ...,
E_n^{-1}(a_1,a_2,...,a_n) = D_{x=0}^n 1 / (D_x f^{(-1)}(x)), where D_x is the derivative w.r.t. x and f^{(-1)}(x) is the (possibly formal) compositional inverse of f(x) about the origin.
E_n^{-1}(a_1,a_2,...,a_n) = D_{x=0}^n 1 f'(f^{(-1)}(x)) by the inverse function theorem, where the prime indicates differentiation w.r.t. the argument of the function f. Note the correspondence to the analytic definitions of the reciprocal tangents [RT] of A356144, consistent with the following algebraic identities.
[E]^{-1} = [P][L] = [P][E][P] = [RT][P], representing, e.g., the substitution of the permutahedra polynomials [P] of A133314 for the indeterminates of the reciprocal tangent polynomials [RT] of A356144. [E] are the refined Eulerian polynomials of A145271, and [L], the classic Lagrange inversion polynomials of A134685.
Since [P]^2 = [L]^2 = [RT]^2 = [I], the substitutional identity, i.e., [P], [L], and [RT] are involutive transformations, many identities follow from the basic ones above, e.g., [L] = [P][E]^{-1} gives an inversion formula for a formal e.g.f. f(x) = x + a_1 x^2/2! + a_2 x^3/3! + ..., and we can identify [E] and [E]^{-1} as a conjugate dual.
With a_n = -x, [E]^{-1} reduces to a signed version of A112493 with an additional initial row, with the row sums of the unsigned coefficients being (1, A006351). A112493 is also given by the diagonals of A124324. See my link above on the reduced polynomials and associated arrays for more detail.
The sequence of row sums of the signed coefficients, i.e., E^{-1}(1,1,...,1), is the sequence (1, 1, 0, 0, 0, 0, ...).
Conjecture: row polynomials are R(n,1) for n > 0 where R(n,k) = R(n-1,k+1) - Sum_{j=1..n-1} binomial(n-1,j-1)*R(j,k)*R(n-j,1) for n > 1, k > 0 with R(1,k) = a_k for k > 0. - Mikhail Kurkov, Mar 22 2025

A132681 Infinitesimal generator matrix for a diagonally-shifted Pascal matrix, binomial(n+m,k+m), for m=1, related to Laguerre(n,x,m).

Original entry on oeis.org

0, 2, 0, 0, 3, 0, 0, 0, 4, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12, 0
Offset: 0

Views

Author

Tom Copeland, Nov 15 2007, Nov 16 2007, Nov 27 2007

Keywords

Comments

Analogous to the infinitesimal Pascal matrix (m=0), A132440.
In general the matrix T begins (here m=1)
0;
m+1,0;
0, m+2, 0;
0, 0, m+3, 0;
0, 0, 0, m+4, 0;
Let LM(t) = exp(t*T) = limit [1 + t*T/n]^n as n tends to infinity.
Laguerre matrix(m) = [bin(n+m,k+m)] = LM(1) = exp(T) = [ revert of A074909 for m=1 ]. Truncating the series gives the n X n submatrices. In fact, the submatrices of T are nilpotent with [Tsub_n]^(n+1) = 0 for n=0,1,2,....
Inverse Lag matrix(m) = LM(-1) = exp(-T)
Umbrally LM[b(.)] = exp(b(.)*T) = [ bin(n+m,k+m) * b(n-k) ]
A(j) = T^j / j! equals the matrix [bin(n+m,k+m) * delta(n-k-j)] where delta(n) = 1 if n=0 and vanishes otherwise (Kronecker delta); i.e. A(j) is a matrix with all the terms 0 except for the j-th lower (or main for j=0) diagonal which equals that of the Laguerre(m) matrix. Hence the A(j)'s form a linearly independent basis for all matrices of the form [bin(n+m,k+m) d(n-k)].
For sequences with b(0) = 1, umbrally,
LM[b(.)] = exp(b(.)*T) = [ bin(n+m,k+m)] * b(n-k) ] .
[LM[b(.)]]^(-1) = exp(c(.)*T) = [ bin(n+m,k+m)] * c(n-k) ] where c = LPT(b) with LPT the list partition transform of A133314. Or,
[LM[b(.)]]^(-1) = exp[LPT(b(.))*T] = LPT[LM(b(.))] = LM[LPT(b(.))] = LM[c(.)] .
The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or e.g.f.'s EA(x) and EB(x).
1) b(0) = 0, b(n) = (n+m) * a(n-1),
2) B(x) = x^(-m) (xDx) x^m A(x)
3) B(x) = x * Lag(1,-:xD:,m) A(x) = x * [(m+1) + xD] A(x)
4) EB(x) = D^(m) * (x) * D^(-m) EA(x) where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j, Lag(n,x,m) is the associated Laguerre polynomial and D^(-m) x^n / n! = x^(m+n) / (m+n)! are Riemann-Liouville integrals.
So the exponentiated operator can be characterized (with loose notation) as
5) exp(t*T) A(x) = x^(-m) exp(t*xDx) x^m A(x) = [sum(n=0,1,...) (t*x)^n * Lag(n,-:xD:m)] A(x) = [exp{[t*u/(1-t*u)]*:xD:} / (1-t*u)^(m+1) ] A(x) (eval. at u=x) = A[x/(1-t*x)]/(1-t*x)^(m+1), a generalized Euler transformation for an o.g.f.,
6) exp(t*T) EA(x) = D^(m) exp(t*x) D^(-m) EA(x) = [D/(D-1)]^m exp[(t+a(.))*x] = exp(t*x) [(t+D)/D]^m EA(x)
7) exp(t*T) * a = LM(t) * a = [sum(k=0,...,n) bin(n+m,k+m)* t^(n-k) * a(k)], a vector array.
With t=1 and a(k) = (-x)^k / k!, then LM(1) * a = [Laguerre(n,x,m)], a vector array with index n and the o.g.f. A(x) becomes transformed into the e.g.f. for the associated Laguerre polynomials of order m.
The exponential formulas can be umbrally extended as in A132440. And, the formulas can be extended to non-integer m.

Programs

Formula

Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising operators L and R defined by L P_n(x) = n * P_(n-1)(x) and
R P_n(x) = P_(n+1)(x), the matrix T represents the action of
R[(m+1)+ RL] in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x) = x^n/n!, L = DxD and R = D^(-1). - Tom Copeland, Oct 25 2012

Extensions

Missing 0 added to array by Tom Copeland, Aug 02 2013

A132710 Infinitesimal generator for a diagonally-shifted Lah matrix, unsigned A105278, related to n! Laguerre(n,-x,1).

Original entry on oeis.org

0, 2, 0, 0, 6, 0, 0, 0, 12, 0, 0, 0, 0, 20, 0, 0, 0, 0, 0, 30, 0, 0, 0, 0, 0, 0, 42, 0, 0, 0, 0, 0, 0, 0, 56, 0, 0, 0, 0, 0, 0, 0, 0, 72, 0, 0, 0, 0, 0, 0, 0, 0, 0, 90, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 110, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 132, 0
Offset: 0

Views

Author

Tom Copeland, Nov 15 2007, Nov 16 2007, Nov 27 2007

Keywords

Comments

Analogous to the infinitesimal generators of A132681 and A132792.
The matrix T begins
0;
2, 0;
0, 6, 0;
0, 0, 12 0;
0, 0, 0, 20, 0;
Along the nonvanishing diagonal the n-th term is (n+2)*(n+1).
Let LM(t) = exp(t*T) = lim_{n->infinity} (1 + t*T/n)^n.
Shifted Lah matrix = [bin(n+1,k+1)*(n)!/(k)! ] = LM(1) = exp(T). Truncating the series gives the n X n submatrices. In fact, the submatrices of T are nilpotent with [Tsub_n]^(n+1) = 0 for n=0,1,2,....
Inverse shifted Lah matrix = LM(-1) = exp(-T)
Umbrally shifted Lah[b(.)] = exp(b(.)*T) = [ binomial(n+1,k+1)*(n)!/(k)! * b(n-k) ]
A(j) = T^j / j! equals the matrix [binomial(n+1,k+1)*(n)!/(k)! * delta(n-k-j)] where delta(n) = 1 if n=0 and vanishes otherwise (Kronecker delta); i.e. A(j) is a matrix with all the terms 0 except for the j-th lower (or main for j=0) diagonal which equals that of the Lah matrix. Hence the A(j)'s form a linearly independent basis for all matrices of the form [binomial(n+1,k+1) * (n)! / (k)! * d(n-k)].
For sequences with b(0) = 1, umbrally,
LM[b(.)] = exp(b(.)*T) = [ bin(n+1,k+1)*(n)!/(k)! * b(n-k) ] .
[LM[b(.)]]^(-1) = exp(c(.)*T) = [ bin(n+1,k+1)*(n)!/(k)! * c(n-k) ] where c = LPT(b) with LPT the list partition transform of A133314. Or,
[LM[b(.)]]^(-1) = exp[LPT(b(.))*T] = LPT[LM(b(.))] = LM[LPT(b(.))] = LM[c(.)] .
The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or e.g.f.'s EA(x) and EB(x).
1) b(0) = 0, b(n) = (n+1)*(n) * a(n-1),
2) B(x) = x * D^2 * x^2 A(x)
3) B(x) = x * 2 *Lag(2,-:xD:,0) A(x)
4) EB(x) = D * x^2 EA(x)
where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j and Lag(n,x,m) is the associated Laguerre polynomial of order m.
The exponentiated operator can be characterized (with loose notation) as
5) exp(t*T) * a = LM(t) * a = [sum(k=0,...,n) bin(n+1,k+1) * n!/k! t^(n-k) * a(k)] = [ t^n * n! * Lag(n,-a(.)/t,1) ], a vector array.
With t=1 and a(k) = (-x)^k, then LM(1) * a = [ n! * Laguerre(n,x,1) ], a vector array with index n .
6) exp(t*T) EA(x) = EB(x) = EA[ x / (1-x*t) ] / (1-x*t)^2

Programs

  • Mathematica
    Table[PadLeft[{n*(n-1), 0}, n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Apr 30 2014 *)

Formula

Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising operators L and R defined by L P_n(x) = n * P_(n-1)(x) and R P_n(x) = P_(n+1)(x), the matrix T represents the action of R*L^2*R^2 in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x) = x^n/n!, L = DxD and R = D^(-1). - Tom Copeland, Oct 25 2012
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