cp's OEIS Frontend

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

Showing 1-10 of 10 results.

A000522 Total number of ordered k-tuples (k=0..n) of distinct elements from an n-element set: a(n) = Sum_{k=0..n} n!/k!.

Original entry on oeis.org

1, 2, 5, 16, 65, 326, 1957, 13700, 109601, 986410, 9864101, 108505112, 1302061345, 16926797486, 236975164805, 3554627472076, 56874039553217, 966858672404690, 17403456103284421, 330665665962404000, 6613313319248080001, 138879579704209680022, 3055350753492612960485, 70273067330330098091156
Offset: 0

Views

Author

Keywords

Comments

Total number of permutations of all subsets of an n-set.
Also the number of one-to-one sequences that can be formed from n distinct objects.
Old name "Total number of permutations of a set with n elements", or the same with the word "arrangements", both sound too much like A000142.
Related to number of operations of addition and multiplication to evaluate a determinant of order n by cofactor expansion - see A026243.
a(n) is also the number of paths (without loops) in the complete graph on n+2 vertices starting at one vertex v1 and ending at another v2. Example: when n=2 there are 5 paths in the complete graph with 4 vertices starting at the vertex 1 and ending at the vertex 2: (12),(132),(142),(1342),(1432) so a(2) = 5. - Avi Peretz (njk(AT)netvision.net.il), Feb 23 2001; comment corrected by Jonathan Coxhead, Mar 21 2003
Also row sums of Table A008279, which can be generated by taking the derivatives of x^k. For example, for y = 1*x^3, y' = 3x^2, y" = 6x, y'''= 6 so a(4) = 1 + 3 + 6 + 6 = 16. - Alford Arnold, Dec 15 1999
a(n) is the permanent of the n X n matrix with 2s on the diagonal and 1s elsewhere. - Yuval Dekel, Nov 01 2003
(A000166 + this_sequence)/2 = A009179, (A000166 - this_sequence)/2 = A009628.
Stirling transform of A006252(n-1) = [1,1,1,2,4,14,38,...] is a(n-1) = [1,2,5,16,65,...]. - Michael Somos, Mar 04 2004
Number of {12,12*,21*}- and {12,12*,2*1}-avoiding signed permutations in the hyperoctahedral group.
a(n) = b such that Integral_{x=0..1} x^n*exp(-x) dx = a-b*exp(-1). - Sébastien Dumortier, Mar 05 2005
a(n) is the number of permutations on [n+1] whose left-to-right record lows all occur at the start. Example: a(2) counts all permutations on [3] except 231 (the last entry is a record low but its predecessor is not). - David Callan, Jul 20 2005
a(n) is the number of permutations on [n+1] that avoid the (scattered) pattern 1-2-3|. The vertical bar means the "3" must occur at the end of the permutation. For example, 21354 is not counted by a(4): 234 is an offending subpermutation. - David Callan, Nov 02 2005
Number of deco polyominoes of height n+1 having no reentrant corners along the lower contour (i.e., no vertical step that is followed by a horizontal step). In other words, a(n)=A121579(n+1,0). A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: a(1)=2 because the only deco polyominoes of height 2 are the vertical and horizontal dominoes, having no reentrant corners along their lower contours. - Emeric Deutsch, Aug 16 2006
Unreduced numerators of partial sums of the Taylor series for e. - Jonathan Sondow, Aug 18 2006
a(n) is the number of permutations on [n+1] (written in one-line notation) for which the subsequence beginning at 1 is increasing. Example: a(2)=5 counts 123, 213, 231, 312, 321. - David Callan, Oct 06 2006
a(n) is the number of permutations (written in one-line notation) on the set [n + k], k >= 1, for which the subsequence beginning at 1,2,...,k is increasing. Example: n = 2, k = 2. a(2) = 5 counts 1234, 3124, 3412, 4123, 4312. - Peter Bala, Jul 29 2014
a(n) and (1,-2,3,-4,5,-6,7,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Nov 01 2007
Consider the subsets of the set {1,2,3,...,n} formed by the first n integers. E.g., for n = 3 we have {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}. Let the variable sbst denote a subset. For each subset sbst we determine its number of parts, that is nprts(sbst). The sum over all possible subsets is written as Sum_{sbst=subsets}. Then a(n) = Sum_{sbst=subsets} nprts(sbst)!. E.g., for n = 3 we have 1!+1!+1!+1!+2!+2!+2!+3!=16. - Thomas Wieder, Jun 17 2006
Equals row sums of triangle A158359(unsigned). - Gary W. Adamson, Mar 17 2009
Equals eigensequence of triangle A158821. - Gary W. Adamson, Mar 30 2009
For positive n, equals 1/BarnesG(n+1) times the determinant of the n X n matrix whose (i,j)-coefficient is the (i+j)th Bell number. - John M. Campbell, Oct 03 2011
a(n) is the number of n X n binary matrices with i) at most one 1 in each row and column and ii) the subset of rows that contain a 1 must also be the columns that contain a 1. Cf. A002720 where restriction ii is removed. - Geoffrey Critzer, Dec 20 2011
Number of restricted growth strings (RGS) [d(1),d(2),...,d(n)] such that d(k) <= k and d(k) <= 1 + (number of nonzero digits in prefix). The positions of nonzero digits determine the subset, and their values (decreased by 1) are the (left) inversion table (a rising factorial number) for the permutation, see example. - Joerg Arndt, Dec 09 2012
Number of a restricted growth strings (RGS) [d(0), d(1), d(2), ..., d(n)] where d(k) >= 0 and d(k) <= 1 + chg([d(0), d(1), d(2), ..., d(k-1)]) and chg(.) gives the number of changes of its argument. Replacing the function chg(.) by a function asc(.) that counts the ascents in the prefix gives A022493 (ascent sequences). - Joerg Arndt, May 10 2013
The sequence t(n) = number of i <= n such that floor(e*i!) is a square is mentioned in the abstract of Luca & Shparlinski. The values are t(n) = 0 for 0 <= n <= 2 and t(n) = 1 for at least 3 <= n <= 300. - R. J. Mathar, Jan 16 2014
a(n) = p(n,1) = q(n,1), where p and q are polynomials defined at A248664 and A248669. - Clark Kimberling, Oct 11 2014
a(n) is the number of ways at most n people can queue up at a (slow) ticket counter when one or more of the people may choose not to queue up. Note that there are C(n,k) sets of k people who quene up and k! ways to queue up. Since k can run from 0 to n, a(n) = Sum_{k=0..n} n!/(n-k)! = Sum_{k=0..n} n!/k!. For example, if n=3 and the people are A(dam), B(eth), and C(arl), a(3)=16 since there are 16 possible lineups: ABC, ACB, BAC, BCA, CAB, CBA, AB, BA, AC, CA, BC, CB, A, B, C, and empty queue. - Dennis P. Walsh, Oct 02 2015
As the row sums of A008279, Motzkin uses the abbreviated notation $n_<^\Sigma$ for a(n).
The piecewise polynomial function f defined by f(x) = a(n)*x^n/n! on each interval [ 1-1/a(n), 1-1/a(n+1) ) is continuous on [0,1) and lim_{x->1} f(x) = e. - Luc Rousseau, Oct 15 2019
a(n) is composite for 3 <= n <= 2015, but a(2016) is prime (or at least a strong pseudoprime): see Johansson link. - Robert Israel, Jul 27 2020 [a(2016) is prime, ECPP certificate generated with CM 0.4.3 and checked by factordb. - Jason H Parker, Jun 15 2025]
In general, sequences of the form a(0)=a, a(n) = n*a(n-1) + k, n>0, will have a closed form of n!*a + floor(n!*(e-1))*k. - Gary Detlefs, Oct 26 2020
From Peter Bala, Apr 03 2022: (Start)
a(2*n) is odd and a(2*n+1) is even. More generally, a(n+k) == a(n) (mod k) for all n and k. It follows that for each positive integer k, the sequence obtained by reducing a(n) modulo k is periodic, with the exact period dividing k. Various divisibility properties of the sequence follow from this; for example, a(5*n+2) == a(5*n+4) == 0 (mod 5), a(25*n+7) == a(25*n+19) == 0 (mod 25) and a(13*n+4) == a(13*n+10)== a(13*n+12) == 0 (mod 13). (End)
Number of possible ranking options on a typical ranked choice voting ballot with n candidates (allowing undervotes). - P. Christopher Staecker, May 05 2024
From Thomas Scheuerle, Dec 28 2024: (Start)
Number of decorated permutations of size n.
Number of Le-diagrams with bounding box semiperimeter n, for n > 0.
By counting over all k = 1..n and n > 0, the number of positroid cells for the totally nonnegative real Grassmannian Gr(k, n), equivalently the number of Grassmann necklaces of type (k, n). (End)

Examples

			G.f. = 1 + 2*x + 5*x^2 + 16*x^3 + 65*x^4 + 326*x^5 + 1957*x^6 + 13700*x^7 + ...
With two objects we can form 5 sequences: (), (a), (b), (a,b), (b,a), so a(2) = 5.
From _Joerg Arndt_, Dec 09 2012: (Start)
The 16 arrangements of the 3-set and their RGS (dots denote zeros) are
  [ #]       RGS        perm. of subset
  [ 1]    [ . . . ]      [  ]
  [ 2]    [ . . 1 ]      [ 3 ]
  [ 3]    [ . 1 . ]      [ 2 ]
  [ 4]    [ . 1 1 ]      [ 2 3 ]
  [ 5]    [ . 1 2 ]      [ 3 2 ]
  [ 6]    [ 1 . . ]      [ 1 ]
  [ 7]    [ 1 . 1 ]      [ 1 3 ]
  [ 8]    [ 1 . 2 ]      [ 3 1 ]
  [ 9]    [ 1 1 . ]      [ 1 2 ]
  [10]    [ 1 1 1 ]      [ 1 2 3 ]
  [11]    [ 1 1 2 ]      [ 1 3 2 ]
  [12]    [ 1 1 3 ]      [ 2 3 1 ]
  [13]    [ 1 2 . ]      [ 2 1 ]
  [14]    [ 1 2 1 ]      [ 2 1 3 ]
  [15]    [ 1 2 2 ]      [ 3 1 2 ]
  [16]    [ 1 2 3 ]      [ 3 2 1 ]
(End)
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 75, Problem 9.
  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 65, p. 23, Ellipses, Paris 2008.
  • J. M. Gandhi, On logarithmic numbers, Math. Student, 31 (1963), 73-83.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E11.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 16.
  • D. Singh, The numbers L(m,n) and their relations with prepared Bernoulli and Eulerian numbers, Math. Student, 20 (1952), 66-70.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Average of n-th row of triangle in A068424 [Corrected by N. J. A. Sloane, Feb 29 2024].
Row sums of A008279 and A094816.
First differences give A001339. Second differences give A001340.
Partial sums are in A001338, A002104.
A row of the array in A144502.
See also A370973, Nearest integer to e*n!.

Programs

  • Haskell
    import Data.List (subsequences, permutations)
    a000522 = length . choices . enumFromTo 1 where
    choices = concat . map permutations . subsequences
    -- Reinhard Zumkeller, Feb 21 2012, Oct 25 2010
    
  • Magma
    [1] cat [n eq 1 select (n+1) else n*Self(n-1)+1: n in [1..25]]; // Vincenzo Librandi, Feb 15 2015
    
  • Maple
    a(n):= exp(1)*int(x^n*exp(-x)*Heaviside(x-1), x=0..infinity); # Karol A. Penson, Oct 01 2001
    A000522 := n->add(n!/k!,k=0..n);
    G(x):=exp(x)/(1-x): f[0]:=G(x): for n from 1 to 26 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..20);
    # Zerinvary Lajos, Apr 03 2009
    G:=exp(z)/(1-z): Gser:=series(G,z=0,21):
    for n from 0 to 20 do a(n):=n!*coeff(Gser,z,n): end do
    # Paul Weisenhorn, May 30 2010
    k := 1; series(hypergeom([1,k],[],x/(1-x))/(1-x), x=0, 20); # Mark van Hoeij, Nov 07 2011
    # one more Maple program:
    a:= proc(n) option remember;
          `if`(n<0, 0, 1+n*a(n-1))
        end:
    seq(a(n), n=0..23);  # Alois P. Heinz, Sep 13 2019
    seq(simplify(KummerU(-n, -n, 1)), n = 0..23); # Peter Luschny, May 10 2022
  • Mathematica
    Table[FunctionExpand[Gamma[n + 1, 1]*E], {n, 0, 24}]
    nn = 20; Accumulate[Table[1/k!, {k, 0, nn}]] Range[0, nn]! (* Jan Mangaldan, Apr 21 2013 *)
    FoldList[#1*#2 + #2 &, 0, Range@ 23] + 1 (* or *)
    f[n_] := Floor[E*n!]; f[0] = 1; Array[f, 20, 0] (* Robert G. Wilson v, Feb 13 2015 *)
    RecurrenceTable[{a[n + 1] == (n + 1) a[n] + 1, a[0] == 1}, a, {n, 0, 12}] (* Emanuele Munarini, Apr 27 2017 *)
    nxt[{n_,a_}]:={n+1,a(n+1)+1}; NestList[nxt,{0,1},30][[All,2]] (* Harvey P. Dale, Jan 29 2023 *)
  • Maxima
    a(n) := if n=0 then 1 else n*a(n-1)+1; makelist(a(n),n,0,12); /* Emanuele Munarini, Apr 27 2017 */
  • PARI
    {a(n) = local(A); if( n<0, 0, A = vector(n+1); A[1]=1; for(k=1, n, A[k+1] = k*A[k] + 1); A[n+1])}; /* Michael Somos, Jul 01 2004 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( x +x * O(x^n)) / (1 - x), n))}; /* Michael Somos, Mar 06 2004 */
    
  • PARI
    a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=1/(1-x)^2+x^2*deriv(A)/(1-x)); polcoeff(A, n) \\ Paul D. Hanna, Sep 03 2008
    
  • PARI
    {a(n)=local(X=x+x*O(x^n));polcoeff(sum(m=0,n,(m+2)^m*x^m/(1+(m+1)*X)^(m+1)),n)} /* Paul D. Hanna */
    
  • PARI
    a(n)=sum(k=0,n,binomial(n,k)*k!); \\ Joerg Arndt, Dec 14 2014
    
  • Sage
    # program adapted from Alois P. Heinz's Maple code in A022493
    @CachedFunction
    def b(n, i, t):
        if n <= 1:
            return 1
        return sum(b(n - 1, j, t + (j == i)) for j in range(t + 2))
    def a(n):
        return b(n, 0, 0)
    v000522 = [a(n) for n in range(33)]
    print(v000522)
    # Joerg Arndt, May 11 2013
    

Formula

a(n) = n*a(n-1) + 1, a(0) = 1.
a(n) = A007526(n-1) + 1.
a(n) = A061354(n)*A093101(n).
a(n) = n! * Sum_{k=0..n} 1/k! = n! * (e - Sum_{k>=n+1} 1/k!). - Michael Somos, Mar 26 1999
a(0) = 1; for n > 0, a(n) = floor(e*n!).
E.g.f.: exp(x)/(1-x).
a(n) = 1 + Sum_{n>=k>=j>=0} (k-j+1)*k!/j! = a(n-1) + A001339(n-1) = A007526(n) + 1. Binomial transformation of n!, i.e., A000142. - Henry Bottomley, Jun 04 2001
a(n) = floor(2/(n+1))*A009578(n+1)-1. - Emeric Deutsch, Oct 24 2001
Integral representation as n-th moment of a nonnegative function on a positive half-axis: a(n) = e*Integral_{x>=0} x^n*e^(-x)*Heaviside(x-1) dx. - Karol A. Penson, Oct 01 2001
Formula, in Mathematica notation: Special values of Laguerre polynomials, a(n)=(-1)^n*n!*LaguerreL[n, -1-n, 1], n=1, 2, ... . This relation cannot be checked by Maple, as it appears that Maple does not incorporate Laguerre polynomials with second index equal to negative integers. It does check with Mathematica. - Karol A. Penson and Pawel Blasiak ( blasiak(AT)lptl.jussieu.fr), Feb 13 2004
G.f.: Sum_{k>=0} k!*x^k/(1-x)^(k+1). a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*k^(n-k)*(k+1)^k. - Vladeta Jovovic, Aug 18 2002
a(n) = Sum_{k=0..n} A008290(n, k)*2^k. - Philippe Deléham, Dec 12 2003
a(n) = Sum_{k=0..n} A046716(n, k). - Philippe Deléham, Jun 12 2004
a(n) = e*Gamma(n+1,1) where Gamma(z,t) = Integral_{x>=t} e^(-x)*x^(z-1) dx is incomplete gamma function. - Michael Somos, Jul 01 2004
a(n) = Sum_{k=0..n} P(n, k). - Ross La Haye, Aug 28 2005
Given g.f. A(x), then g.f. A059115 = A(x/(1-x)). - Michael Somos, Aug 03 2006
a(n) = 1 + n + n*(n-1) + n*(n-1)*(n-2) + ... + n!. - Jonathan Sondow, Aug 18 2006
a(n) = Sum_{k=0..n} binomial(n,k) * k!; interpretation: for all k-subsets (sum), choose a subset (binomial(n,k)), and permutation of subset (k!). - Joerg Arndt, Dec 09 2012
a(n) = Integral_{x>=0} (x+1)^n*e^(-x) dx. - Gerald McGarvey, Oct 19 2006
a(n) = Sum_{k=0..n} A094816(n, k), n>=0 (row sums of Poisson-Charlier coefficient matrix). - N. J. A. Sloane, Nov 10 2007
From Tom Copeland, Nov 01 2007, Dec 10 2007: (Start)
Act on 1/(1-x) with 1/(1-xDx) = Sum_{j>=0} (xDx)^j = Sum_{j>=0} x^j*D^j*x^j = Sum_{j>=0} j!*x^j*L(j,-:xD:,0) where Lag(n,x,0) are the Laguerre polynomials of order 0, D the derivative w.r.t. x and (:xD:)^j = x^j*D^j. Truncating the operator series at the j = n term gives an o.g.f. for a(0) through a(n) consistent with Jovovic's.
These results and those of Penson and Blasiak, Arnold, Bottomley and Deleham are related by the operator A094587 (the reverse of A008279), which is the umbral equivalent of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j)*j!*x^(n-j) = Sum_{j=0..n} (n!/j!)*x^j. Umbral substitution of b(.) for x and then letting b(n)=1 for all n connects the results. See A132013 (the inverse of A094587) for a connection between these operations and 1/(1-xDx).
(End)
From Peter Bala, Jul 15 2008: (Start)
a(n) = n!*e - 1/(n + 1/(n+1 + 2/(n+2 + 3/(n+3 + ...)))).
Asymptotic result (Ramanujan): n!*e - a(n) ~ 1/n - 1/n^3 + 1/n^4 + 2/n^5 - 9/n^6 + ..., where the sequence [1,0,-1,1,2,-9,...] = [(-1)^k*A000587(k)], for k>=1.
a(n) is a difference divisibility sequence, that is, the difference a(n) - a(m) is divisible by n - m for all n and m (provided n is not equal to m). For fixed k, define the derived sequence a_k(n) = (a(n+k)-a(k))/n, n = 1,2,3,... . Then a_k(n) is also a difference divisibility sequence.
For example, the derived sequence a_0(n) is just a(n-1). The set of integer sequences satisfying the difference divisibility property forms a ring with term-wise operations of addition and multiplication.
Recurrence relations: a(0) = 1, a(n) = (n-1)*(a(n-1) + a(n-2)) + 2, for n >= 1. a(0) = 1, a(1) = 2, D-finite with recurrence: a(n) = (n+1)*a(n-1) - (n-1)*a(n-2) for n >= 2. The sequence b(n) := n! satisfies the latter recurrence with the initial conditions b(0) = 1, b(1) = 1. This leads to the finite continued fraction expansion a(n)/n! = 1/(1-1/(2-1/(3-2/(4-...-(n-1)/(n+1))))), n >= 2.
Limit_{n->oo} a(n)/n! = e = 1/(1-1/(2-1/(3-2/(4-...-n/((n+2)-...))))). This is the particular case m = 0 of the general result m!/e - d_m = (-1)^(m+1) *(1/(m+2 -1/(m+3 -2/(m+4 -3/(m+5 -...))))), where d_m denotes the m-th derangement number A000166(m).
For sequences satisfying the more general recurrence a(n) = (n+1+r)*a(n-1) - (n-1)*a(n-2), which yield series acceleration formulas for e/r! that involve the Poisson-Charlier polynomials c_r(-n;-1), refer to A001339 (r=1), A082030 (r=2), A095000 (r=3) and A095177 (r=4).
For the corresponding results for the constants log(2), zeta(2) and zeta(3) refer to A142992, A108625 and A143007 respectively.
(End)
G.f. satisfies: A(x) = 1/(1-x)^2 + x^2*A'(x)/(1-x). - Paul D. Hanna, Sep 03 2008
From Paul Barry, Nov 27 2009: (Start)
G.f.: 1/(1-2*x-x^2/(1-4*x-4*x^2/(1-6*x-9*x^2/(1-8*x-16*x^2/(1-10*x-25*x^2/(1-... (continued fraction);
G.f.: 1/(1-x-x/(1-x/(1-x-2*x/(1-2*x/(1-x-3*x/(1-3*x/(1-x-4*x/(1-4*x/(1-x-5*x/(1-5*x/(1-... (continued fraction).
(End)
O.g.f.: Sum_{n>=0} (n+2)^n*x^n/(1 + (n+1)*x)^(n+1). - Paul D. Hanna, Sep 19 2011
G.f. hypergeom([1,k],[],x/(1-x))/(1-x), for k=1,2,...,9 is the generating function for A000522, A001339, A082030, A095000, A095177, A096307, A096341, A095722, and A095740. - Mark van Hoeij, Nov 07 2011
G.f.: 1/U(0) where U(k) = 1 - x - x*(k+1)/(1 - x*(k+1)/U(k+1)); (continued fraction). - Sergei N. Gladkovskii, Oct 14 2012
E.g.f.: 1/U(0) where U(k) = 1 - x/(1 - 1/(1 + (k+1)/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 16 2012
G.f.: 1/(1-x)/Q(0), where Q(k) = 1 - x/(1-x)*(k+1)/(1 - x/(1-x)*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 19 2013
G.f.: 2/(1-x)/G(0), where G(k) = 1 + 1/(1 - x*(2*k+2)/(x*(2*k+3) - 1 + x*(2*k+2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
G.f.: (B(x)+ 1)/(2-2*x) = Q(0)/(2-2*x), where B(x) be g.f. A006183, Q(k) = 1 + 1/(1 - x*(k+1)/(x*(k+1) + (1-x)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 08 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(k+1) - x^2*(k+1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 30 2013
E.g.f.: e^x/(1-x) = (1 - 12*x/(Q(0) + 6*x - 3*x^2))/(1-x), where Q(k) = 2*(4*k+1)*(32*k^2 + 16*k + x^2 - 6) - x^4*(4*k-1)*(4*k+7)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2013
G.f.: conjecture: T(0)/(1-2*x), where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 - (1 - 2*x*(k+1))*(1 - 2*x*(k+2))/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2013
0 = a(n)*(+a(n+1) - 3*a(n+2) + a(n+3)) + a(n+1)*(+a(n+1) - a(n+3)) + a(n+2)*(+a(n+2)) for all n>=0. - Michael Somos, Jul 04 2014
From Peter Bala, Jul 29 2014: (Start)
a(n) = F(n), where the function F(x) := Integral_{0..infinity} e^(-u)*(1 + u)^x du smoothly interpolates this sequence to all real values of x. Note that F(-1) = G and for n = 2,3,... we have F(-n) = (-1)^n/(n-1)! *( A058006(n-2) - G ), where G = 0.5963473623... denotes Gompertz's constant - see A073003.
a(n) = n!*e - e*( Sum_{k >= 0} (-1)^k/((n + k + 1)*k!) ).
(End)
a(n) = hypergeometric_U(1, n+2, 1). - Peter Luschny, Nov 26 2014
a(n) ~ exp(1-n)*n^(n-1/2)*sqrt(2*Pi). - Vladimir Reshetnikov, Oct 27 2015
a(n) = A038155(n+2)/A000217(n+1). - Anton Zakharov, Sep 08 2016
a(n) = round(exp(1)*n!), n > 1 - Simon Plouffe, Jul 28 2020
a(n) = KummerU(-n, -n, 1). - Peter Luschny, May 10 2022
a(n) = (e/(2*Pi))*Integral_{x=-oo..oo} (n+1+i*x)!/(1+i*x) dx. - David Ulgenes, Apr 18 2023
Sum_{i=0..n} (-1)^(n-i) * binomial(n, i) * a(i) = n!. - Werner Schulte, Apr 03 2024

Extensions

Additional comments from Michael Somos

A094587 Triangle of permutation coefficients arranged with 1's on the diagonal. Also, triangle of permutations on n letters with exactly k+1 cycles and with the first k+1 letters in separate cycles.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 6, 6, 3, 1, 24, 24, 12, 4, 1, 120, 120, 60, 20, 5, 1, 720, 720, 360, 120, 30, 6, 1, 5040, 5040, 2520, 840, 210, 42, 7, 1, 40320, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 362880, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 3628800, 3628800
Offset: 0

Views

Author

Paul Barry, May 13 2004

Keywords

Comments

Also, table of Pochhammer sequences read by antidiagonals (see Rudolph-Lilith, 2015). - N. J. A. Sloane, Mar 31 2016
Reverse of A008279. Row sums are A000522. Diagonal sums are A003470. Rows of inverse matrix begin {1}, {-1,1}, {0,-2,1}, {0,0,-3,1}, {0,0,0,-4,1} ... The signed lower triangular matrix (-1)^(n+k)n!/k! has as row sums the signed rencontres numbers Sum_{k=0..n} (-1)^(n+k)n!/k!. (See A000166). It has matrix inverse 1 1,1 0,2,1 0,0,3,1 0,0,0,4,1,...
Exponential Riordan array [1/(1-x),x]; column k has e.g.f. x^k/(1-x). - Paul Barry, Mar 27 2007
From Tom Copeland, Nov 01 2007: (Start)
T is the umbral extension of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! * Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j) * j! * x^(n-j) = Sum_{j=0..n} (n!/j!) x^j. The inverse operator is A132013 with generalizations discussed in A132014.
b = T*a can be characterized several ways in terms of a(n) and b(n) or their o.g.f.'s A(x) and B(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0], umbrally,
2) b(n) = (-1)^n n! Lag(n,a(.),-1-n)
3) b(n) = Sum_{j=0..n} (n!/j!) a(j)
4) B(x) = (1-xDx)^(-1) A(x), formally
5) B(x) = Sum_{j=0,1,...} (xDx)^j A(x)
6) B(x) = Sum_{j=0,1,...} x^j * D^j * x^j A(x)
7) B(x) = Sum_{j=0,1,...} j! * x^j * L(j,-:xD:,0) A(x) where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (0!,1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314 so T(n,k) = binomial(n,k)*c(n-k). The reciprocal sequence is d = (1,-1,0,0,0,...). (End)
From Peter Bala, Jul 10 2008: (Start)
This array is the particular case P(1,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
n\k|0.....................1...............2.......3......4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
The entries A(n,k) of this array satisfy the recursion A(n,k) = (a+b*(n-k-1))*A(n-1,k) + A(n-1,k-1), which reduces to the Pascal formula when a = 1, b = 0.
Various cases are recorded in the database, including: P(1,0) = Pascal's triangle A007318, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159, P(1,3) = A136215 and P(2,3) = A136216.
When b <> 0 the array P(a,b) has e.g.f. exp(x*y)/(1-b*y)^(a/b) = 1 + (a+x)*y + (a*(a+b)+2a*x+x^2)*y^2/2! + (a*(a+b)*(a+2b) + 3a*(a+b)*x + 3a*x^2+x^3)*y^3/3! + ...; the array P(a,0) has e.g.f. exp((x+a)*y).
We have the matrix identities P(a,b)*P(a',b) = P(a+a',b); P(a,b)^-1 = P(-a,b).
An analog of the binomial expansion for the row entries of P(a,b) has been proved by [Echi]. Introduce a (generally noncommutative and nonassociative) product ** on the ring of polynomials in two variables by defining F(x,y)**G(x,y) = F(x,y)G(x,y) + by^2*d/dy(G(x,y)).
Define the iterated product F^(n)(x,y) of a polynomial F(x,y) by setting F^(1) = F(x,y) and F^(n)(x,y) = F(x,y)**F^(n-1)(x,y) for n >= 2. Then (x+a*y)^(n) = x^n + C(n,1)*a*x^(n-1)*y + C(n,2)*a*(a+b)*x^(n-2)*y^2 + ... + C(n,n)*a*(a+b)*(a+2b)*...*(a+(n-1)b)*y^n. (End)
(n+1) * n-th row = reversal of triangle A068424: (1; 2,2; 6,6,3; ...) - Gary W. Adamson, May 03 2009
Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n,k,p) = G(n-1,n-k,p) then T(n, k, 1) is this sequence, T(n, k, 2) = A112292(n, k) and T(n, k, 3) = A136214. - Peter Luschny, Jun 01 2009, revised Jun 18 2019
The higher order exponential integrals E(x,m,n) are defined in A163931. For a discussion of the asymptotic expansions of the E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) see A130534. The asymptotic expansion of E(x,m=1,n) leads for n >= 1 to the left hand columns of the triangle given above. Triangle A165674 is generated by the asymptotic expansions of E(x,m=2,n). - Johannes W. Meijer, Oct 07 2009
T(n,k) = n!/k! = number of permutations of [n+1] with exactly k+1 cycles and with elements 1,2,...,k+1 in separate cycles. See link and example below. - Dennis P. Walsh, Jan 24 2011
T(n,k) is the number of n permutations that leave some size k subset of {1,2,...,n} fixed. Sum_{k=0..n}(-1)^k*T(n,k) = A000166(n) (the derangements). - Geoffrey Critzer, Dec 11 2011
T(n,k) = A162995(n-1,k-1), 2 <= k <= n; T(n,k) = A173333(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012
The row polynomials form an Appell sequence. The matrix is a special case of a group of general matrices sketched in A132382. - Tom Copeland, Dec 03 2013
For interpretations in terms of colored necklaces, see A213936 and A173333. - Tom Copeland, Aug 18 2016
See A008279 for a relation of this entry to the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016
Also, T(n,k) is the number of ways to arrange n-k nonattacking rooks on the n X (n-k) chessboard. - Andrey Zabolotskiy, Dec 16 2016
The infinitesimal generator of this triangle is the generalized exponential Riordan array [-log(1-x), x] and equals the unsigned version of A238363. - Peter Bala, Feb 13 2017
Formulas for exponential and power series infinitesimal generators for this triangle T are given in Copeland's 2012 and 2014 formulas as T = unsigned exp[(I-A238385)] = 1/(I - A132440), where I is the identity matrix. - Tom Copeland, Jul 03 2017
If A(0) = 1/(1-x), and A(n) = d/dx(A(n-1)), then A(n) = n!/(1-x)^(n+1) = Sum_{k>=0} (n+k)!/k!*x^k = Sum_{k>=0} T(n+k, k)*x^k. - Michael Somos, Sep 19 2021

Examples

			Rows begin {1}, {1,1}, {2,2,1}, {6,6,3,1}, ...
For n=3 and k=1, T(3,1)=6 since there are exactly 6 permutations of {1,2,3,4} with exactly 2 cycles and with 1 and 2 in separate cycles. The permutations are (1)(2 3 4), (1)(2 4 3), (1 3)(2 4), (1 4)(2 3), (1 3 4)(2), and (1 4 3)(2). - _Dennis P. Walsh_, Jan 24 2011
Triangle begins:
     1,
     1,    1,
     2,    2,    1,
     6,    6,    3,    1,
    24,   24,   12,    4,    1,
   120,  120,   60,   20,    5,    1,
   720,  720,  360,  120,   30,    6,    1,
  5040, 5040, 2520,  840,  210,   42,    7,    1
The production matrix is:
      1,     1,
      1,     1,     1,
      2,     2,     1,    1,
      6,     6,     3,    1,    1,
     24,    24,    12,    4,    1,   1,
    120,   120,    60,   20,    5,   1,   1,
    720,   720,   360,  120,   30,   6,   1,   1,
   5040,  5040,  2520,  840,  210,  42,   7,   1,   1,
  40320, 40320, 20160, 6720, 1680, 336,  56,   8,   1,   1
which is the exponential Riordan array A094587, or [1/(1-x),x], with an extra superdiagonal of 1's.
Inverse begins:
   1,
  -1,  1,
   0, -2,  1,
   0,  0, -3,  1,
   0,  0,  0, -4,  1,
   0,  0,  0,  0, -5,  1,
   0,  0,  0,  0,  0, -6,  1,
   0,  0,  0,  0,  0,  0, -7,  1
		

Crossrefs

Programs

  • Haskell
    a094587 n k = a094587_tabl !! n !! k
    a094587_row n = a094587_tabl !! n
    a094587_tabl = map fst $ iterate f ([1], 1)
       where f (row, i) = (map (* i) row ++ [1], i + 1)
    -- Reinhard Zumkeller, Jul 04 2012
    
  • Maple
    T := proc(n, m): n!/m! end: seq(seq(T(n, m), m=0..n), n=0..9);  # Johannes W. Meijer, Oct 07 2009, revised Nov 25 2012
    # Alternative: Note that if you leave out 'abs' you get A021009.
    T := proc(n, k) option remember; if n = 0 and k = 0 then 1 elif k < 0 or k > n then 0 else abs((n + k)*T(n-1, k) - T(n-1, k-1)) fi end: #  Peter Luschny, Dec 30 2021
  • Mathematica
    Flatten[Table[Table[n!/k!, {k,0,n}], {n,0,10}]] (* Geoffrey Critzer, Dec 11 2011 *)
  • Sage
    def A094587_row(n): return (factorial(n)*exp(x).taylor(x,0,n)).list()
    for n in (0..7): print(A094587_row(n)) # Peter Luschny, Sep 28 2017

Formula

T(n, k) = n!/k! if n >= k >= 0, otherwise 0.
T(n, k) = Sum_{i=k..n} |S1(n+1, i+1)*S2(i, k)| * (-1)^i, with S1, S2 the Stirling numbers.
T(n,k) = (n-k)*T(n-1,k) + T(n-1,k-1). E.g.f.: exp(x*y)/(1-y) = 1 + (1+x)*y + (2+2*x+x^2)*y^2/2! + (6+6*x+3*x^2+x^3)*y^3/3!+ ... . - Peter Bala, Jul 10 2008
A094587 = 1 / ((-1)*A129184 * A127648 + I), I = Identity matrix. - Gary W. Adamson, May 03 2009
From Johannes W. Meijer, Oct 07 2009: (Start)
The o.g.f. of right hand column k is Gf(z;k) = (k-1)!/(1-z)^k, k => 1.
The recurrence relations of the right hand columns lead to Pascal's triangle A007318. (End)
Let f(x) = (1/x)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+1)*R(n,x)-x*R'(n,x). Cf. A132159. - Peter Bala, Oct 28 2011
A padded shifted version of this lower triangular matrix with zeros in the first column and row except for a one in the diagonal position is given by integral(t=0 to t=infinity) exp[-t(I-P)] = 1/(I-P) = I + P^2 + P^3 + ... where P is the infinitesimal generator matrix A218234 and I the identity matrix. The non-padded version is given by P replaced by A132440. - Tom Copeland, Oct 25 2012
From Peter Bala, Aug 28 2013: (Start)
The row polynomials R(n,x) form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x).
Let P(n,x) = Product_{k=0..n-1} (x + k) denote the rising factorial polynomial sequence with the convention that P(0,x) = 1. Then this is triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (6, 6, 3, 1) so P(3,x + 1) = (x + 1)*(x + 2)*(x + 3) = 6 + 6*x + 3*x*(x + 1) + x*(x + 1)*(x + 2). (End)
From Tom Copeland, Apr 21 & 26, and Aug 13 2014: (Start)
T-I = M = -A021009*A132440*A021009 with e.g.f. y*exp(x*y)/(1-y). Cf. A132440. Dividing the n-th row of M by n generates the (n-1)th row of T.
T = 1/(I - A132440) = {2*I - exp[(A238385-I)]}^(-1) = unsigned exp[(I-A238385)] = exp[A000670(.)*(A238385-I)] = , umbrally, where I = identity matrix.
The e.g.f. is exp(x*y)/(1-y), so the row polynomials form an Appell sequence with lowering operator d/dx and raising operator x + 1/(1-D).
With L(n,m,x)= Laguerre polynomials of order m, the row polynomials are (-1)^n*n!*L(n,-1-n,x) = (-1)^n*(-1!/(-1-n)!)*K(-n,-1-n+1,x) = n!* K(-n,-n,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
Operationally, (-1)^n*n!*L(n,-1-n,-:xD:) = (-1)^n*x^(n+1)*:Dx:^n*x^(-1-n) = (-1)^n*x*:xD:^n*x^(-1) = (-1)^n*n!*binomial(xD-1,n) = n!*K(-n,-n,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706 and A132159.
The n-th row of signed M has the coefficients of d[(-:xD:)^n]/d(:Dx:)= f[d/d(-:xD:)](-:xD:)^n with f(y)=y/(y-1), :Dx:^n= n!L(n,0,-:xD:), and (-:xD:)^n = n!L(n,0,:Dx:). M has the coefficients of [D/(1-D)]x^n. (End)
From Tom Copeland, Nov 18 2015: (Start)
Coefficients of the row polynomials of the e.g.f. Sum_{n>=0} P_n(b1,b2,..,bn;t) x^n/n! = e^(P.(..;t) x) = e^(xt) / (1-b.x) = (1 + b1 x + b2 x^2 + b3 x^3 + ...) e^(xt) = 1 + (b1 + t) x + (2 b2 + 2 b1 t + t^2) x^2/2! + (6 b3 + 6 b2 t + 3 b1 t^2 + t^3) x^3/3! + ... , with lowering operator L = d/dt, i.e., L P_n(..;t) = n * P_(n-1)(..;t), and raising operator R = t + d[log(1 + b1 D + b2 D^2 + ...)]/dD = t - Sum_{n>=1} F(n,b1,..,bn) D^(n-1), i.e., R P_n(..,;t) = P_(n+1)(..;t), where D = d/dt and F(n,b1,..,bn) are the Faber polynomials of A263916.
Also P_n(b1,..,bn;t) = CIP_n(t-F(1,b1),-F(2,b1,b2),..,-F(n,b1,..,bn)), the cycle index polynomials A036039.
(End)
The raising operator R = x + 1/(1-D) = x + 1 + D + D^2 + ... in matrix form acting on an o.g.f. (formal power series) is the transpose of the production matrix M below. The linear term x is the diagonal of ones after transposition. The other transposed diagonals come from D^m x^n = n! / (n-m)! x^(n-m). Then P(n,x) = (1,x,x^2,..) M^n (1,0,0,..)^T is a matrix representation of R P(n-1,x) = P(n,x). - Tom Copeland, Aug 17 2016
The row polynomials have e.g.f. e^(xt)/(1-t) = exp(t*q.(x)), umbrally. With p_n(x) the row polynomials of A132013, q_n(x) = v_n(p.(u.(x))), umbrally, where u_n(x) = (-1)^n v_n(-x) = (-1)^n Lah_n(x), the Lah polynomials with e.g.f. exp[x*t/(t-1)]. This has the matrix form [T] = [q] = [v]*[p]*[u]. Conversely, p_n(x) = u_n (q.(v.(x))). - Tom Copeland, Nov 10 2016
From the Appell sequence formalism, 1/(1-b.D) t^n = P_n(b1,b2,..,bn;t), the generalized row polynomials noted in the Nov 18 2015 formulas, consistent with the 2007 comments. - Tom Copeland, Nov 22 2016
From Peter Bala, Feb 18 2017: (Start)
G.f.: Sum_{n >= 1} (n*x)^(n-1)/(1 + (n - t)*x)^n = 1 + (1 + t)*x + (2 + 2*t + t^2)*x^2 + ....
n-th row polynomial R(n,t) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^k*(x + k - t)^(n-k) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^(n-k)*(x + k + t)^k, for arbitrary x. The particular case of the latter sum when x = 0 and t = 1 is identity 10.35 in Gould, Vol.4. (End)
Rodrigues-type formula for the row polynomials: R(n, x) = -exp(x)*Int(exp(-x)* x^n, x), for n >= 0. Recurrence: R(n, x) = x^n + n*R(n-1, x), for n >= 1, and R(0, x) = 1. d/dx(R(n, x)) = R(n, x) - x^n, for n >= 0 (compare with the formula from Peter Bala, Aug 28 2013). - Wolfdieter Lang, Dec 23 2019
T(n, k) = Sum_{i=0..n-k} A048994(n-k, i) * n^i for 0 <= k <= n. - Werner Schulte, Jul 26 2022

Extensions

Edited by Johannes W. Meijer, Oct 07 2009
New description from Dennis P. Walsh, Jan 24 2011

A132440 Infinitesimal Pascal matrix: generator (lower triangular matrix representation) of the Pascal matrix, the classical operator xDx, iterated Laguerre transforms, associated matrices of the list partition transform and general Euler transformation for sequences.

Original entry on oeis.org

0, 1, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0
Offset: 0

Views

Author

Tom Copeland, Nov 13 2007, Nov 15 2007, Nov 22 2007, Dec 02 2007

Keywords

Comments

Let M(t) = exp(t*T) = lim_{n->oo} (1 + t*T/n)^n.
Pascal matrix = [ binomial(n,k) ] = M(1) = exp(T), truncating the series gives the n X n submatrices.
Inverse Pascal matrix = M(-1) = exp(-T) = matrix for inverse binomial transform.
A(j) = T^j / j! equals the matrix [binomial(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 Pascal triangle. Hence the A(j)'s form a linearly independent basis for all matrices of the form [binomial(n,k) * d(n-k)] which include as a subset the invertible associated matrices of the list partition transform (LPT) of A133314.
For sequences with b(0) = 1, umbrally,
M[b(.)] = exp(b(.)*T) = [ binomial(n,k) * b(n-k) ] = matrices associated to b by LPT.
[M[b(.)]]^(-1) = exp(c(.)*T) = [ binomial(n,k) * c(n-k) ] = matrices associated to c, where c = LPT(b) . Or,
[M[b(.)]]^(-1) = exp[LPT(b(.))*T] = LPT[M(b(.))] = M[LPT(b(.))]= M[c(.)].
This is related to xDx, the iterated Laguerre transform and the general Euler transformation of a sequence through the comments in A132013 and A132014 and the relation [Sum_{k=0..n} binomial(n,k) * b(n-k) * d(k)] = M(b)*d, (n-th term). See also A132382.
If b(n,x) is a binomial type Sheffer sequence, then M[b(.,x)]*s(y) = s(x+y) when s(y) = (s(0,y),s(1,y),s(2,y),...) is an array for a Sheffer sequence with the same delta operator as b(n,x) and [M[b(.,x)]]^(-1) is given by the formulas above with b(n) replaced by b(n,x) as b(0,x)=1 for a binomial-type Sheffer sequence.
T = I - A132013 and conversely A132013 = I - T, which is the matrix representation for the iterated mixed order Laguerre transform characterized in A132013 (and A132014).
(I-T)^m generates the group [A132013]^m for m = 0,1,2,... discussed in A132014.
The inverse is 1/(I-T) = I + T + T^2 + T^3 + ... = [A132013]^(-1) = A094587 with the associated sequence (0!,1!,2!,3!,...) under the LPT.
And 1/(I-T)^2 = I + 2*T + 3*T^2 + 4*T^3 + ... = [A132013]^(-2) = A132159 with the associated sequence (1!,2!,3!,4!,...) under the LPT.
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 * a(n-1),
2) B(x) = xDx A(x)
3) B(x) = x * Lag(1,-:xD:) A(x)
4) EB(x) = x * EA(x) where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j and Lag(n,x) is the Laguerre polynomial.
So the exponentiated operator can be characterized as
5) exp(t*T) A(x) = exp(t*xDx) A(x) = [Sum_{n=0,1,...} (t*x)^n * Lag(n,-:xD:)] A(x) = [exp{[t*u/(1-t*u)]*:xD:} / (1-t*u) ] A(x) (eval. at u=x) = A[x/(1-t*x)]/(1-t*x), a generalized Euler transformation for an o.g.f.,
6) exp(t*T) EA(x) = exp(t*x)*EA(x) = exp[(t+a(.))*x], gen. Euler trf. for an e.g.f.
7) exp(t*T) * a = M(t) * a = [Sum_{k=0..n} binomial(n,k) * t^(n-k) * a(k)].
The umbral extension of formulas 5, 6 and 7 gives formally
8) exp[c(.)*T] A(x) = exp(c(.)*xDx) A(x) = [Sum_{n>=0} (c(.)*x)^n * Lag(n,-:xD:)] A(x) = [exp{[c(.)*u/(1-c(.)*u)]*:xD:} / (1-c(.)*u) ] A(x) (eval. at u=x) = A[x/(1-c(.)*x)]/(1-c(.)*x), where the umbral evaluation should be applied only after a power series in c is obtained,
9) exp[c(.)*T] EA(x) = exp(c(.)*x)*EA(x) = exp[(c(.)+a(.))*x]
10) exp[c(.)*T] * a = M[c(.)] * a = [Sum_{k=0..n} binomial(n,k) * c(n-k) * a(k)] .
The n X n principal submatrix of T is nilpotent, in particular, [Tsub_n]^(n+1) = 0, n=0,1,2,3,....
Note (xDx)^n = x^n D^n x^n = x^n n! (:Dx:)^n/n! = x^n n! Lag(n,-:xD:).
The operator xDx is an important, classical operator explored by among others Dattoli, Al-Salam, Carlitz and Stokes and even earlier investigators.
For a recent treatment of xDx, DxD and more general operators see the paper "Laguerre-type derivatives: Dobinski relations and combinatorial identities". - Karol A. Penson, Sep 15 2009
See Copeland's link for generalized Laguerre functions and connection to fractional differ-integrals in exercises through (:Dx:)^a/a!=(D^a x^a)/a!. - Tom Copeland, Nov 17 2011
From Tom Copeland, Apr 25 2014: (Start)
Conjugation or "similarity" transformations of [dP]=A132440 have an operator interpretation (cf. A074909 and A238363):
In general, select two operators A and B such that A^n = F1(n,B) and B^n = F2(n,A); then A^n = F1(n,F2(.,A)) and B^n = F2(n,F1(.,B)), evaluated umbrally, i.e., F1(n,F2(.,x))=F2(n,F1(.,x))=x^n, implying the polynomials F1 and F2 are an umbral compositional inverse pair.
One such pair are the Bell polynomials Bell(n,x) and falling factorials (x)_n with Bell(n,:xD:)=(xD)^n and (xD)_n=:xD:^n (cf. A074909). Another are the Laguerre polynomials LN(n,x)= n!*Lag(n,x) (A021009), which are umbrally self-inverse, with LN(n,-:xD:)=:Dx:^n and LN(n,:Dx:)= (-:xD:)^n with :Dx:^n=D^n*x^n.
Evaluating, for n>=0, the operator derivative d(B^n)/dA = d(F2(n,A))/dA in the basis B^n, i.e., with A^n finally replaced by F1(n,B), or A^n=F1(.,B)^n=F1(n,B), is equivalent to the matrix conjugation
A) [F2]*[dP]*[F1]
B) = [F2]*[dP]*[F2]^(-1)
C) = [F1]^(-1)*[dP]*[F1],
where [F1] is the lower triangular matrix with the n-th row the coefficients of F1(n,x) and analogously for [F2].
So, given the row vector Rv=(c0 c1 c2 c3 ...) and the column vector Cv(x)=(1 x x^2 x^3 ...)^Transpose, form the power series V(x)=Rv*Cv(x).
D) dV(B)/dA = Rv * [F2]*[dP]*[F1] * Cv(B).
E) With A=D and B=D, F1(n,x)=F2(n,x)=x^n and [F1]=[F2]=I. Then d(B^n)/dA = d(D^n)/dD = n * D^(n-1); therefore, consistently [F2]*[dP]*[F1] = [dP] and dV(D)/dD = Rv * [dP] * Cv(D). (End)

Examples

			Matrix T begins
  0;
  1,0;
  0,2,0;
  0,0,3,0;
  0,0,0,4,0;
  ...
		

References

  • T. Mansour and M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, Chapman and Hall/CRC, 2015, (x^n D^n x^n on p. 187).

Programs

Formula

T = log(P) with the Pascal matrix P:=A007318. This should be read as T_N = log(P_N) with P_N the N X N matrix P, N>=2. Because P_N is lower triangular with all diagonal elements 1, the series log(1_N-(1_N-P_N)) stops after N-1 terms because (1_N-P_N)^N is the 0_N-matrix. - Wolfdieter Lang, Oct 14 2010
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*R 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
From Tom Copeland, Apr 26 2014: (Start)
A) T = exp(A238385-I) - I
B) = [St1]*P*[St2] - I
C) = [St1]*P*[St1]^(-1) - I
D) = [St2]^(-1)*P*[St2] - I
E) = [St2]^(-1)*P*[St1]^(-1) - I
where P=A007318, [St1]=padded A008275 just as [St2]=A048993=padded A008277, and I=identity matrix. (End)
From Robert Israel, Oct 02 2015: (Start)
G.f. Sum_{k >= 1} k x^((k+3/2)^2/2 - 17/8) is related to Jacobi theta functions.
If 8*n+17 = y^2 is a square, then a(n) = (y-3)/2, otherwise a(n) = 0. (End)

Extensions

Missing zero added in table by Tom Copeland, Feb 25 2014

A238363 Coefficients for the commutator for the logarithm of the derivative operator [log(D),x^n D^n]=d[(xD)!/(xD-n)!]/d(xD) expanded in the operators :xD:^k.

Original entry on oeis.org

1, -1, 2, 2, -3, 3, -6, 8, -6, 4, 24, -30, 20, -10, 5, -120, 144, -90, 40, -15, 6, 720, -840, 504, -210, 70, -21, 7, -5040, 5760, -3360, 1344, -420, 112, -28, 8, 40320, -45360, 25920, -10080, 3024, -756, 168, -36, 9, -362880, 403200, -226800, 86400, -25200, 6048, -1260, 240, -45, 10
Offset: 1

Views

Author

Tom Copeland, Feb 25 2014

Keywords

Comments

Let D=d/dx and [A,B]=A·B-B·A. Then each row corresponds to the coefficients of the operators :xD:^k = x^k D^k in the expansion of the commutator [log(D),:xD:^n]=[-log(x),:xD:^n]=sum(k=0 to n-1, a(n,k) :xD:^k). The e.g.f. is derived from [log(D), exp(t:xD:)]=[-log(x), exp(t:xD:)]= log(1+t)exp(t:xD:), using the shift property exp(t:xD:)f(x)=f((1+t)x).
The reversed unsigned array is A111492.
See the mathoverflow link and link therein to an associated mathstackexchange question for other formulas for log(D). In addition, R_x = log(D) = -log(x) + c - sum[n=1 to infnty, (-1)^n 1/n :xD:^n/n!]=
-log(x) + Psi(1+xD) = -log(x) + c + Ein(:xD:), where c is the Euler-Mascheroni constant, Psi(x), the digamma function, and Ein(x), a breed of the exponential integrals (cf. Wikipedia). The :xD:^k ops. commute; therefore, the commutator reduces to the -log(x) term.
Also the n-th row corresponds to the expansion of d[(xD)!/(xD-n)!]/d(xD) = d[:xD:^n]/d(xD) in the operators :xD:^k, or, equivalently, the coefficients of x in d[z!/(z-n)!]/dz=d[St1(n,z)]]/dz evaluated umbrally with z=St2(.,x), i.e., z^n replaced by St2(n,x), where St1(n,x) and St2(n,x) are the signed and unsigned Stirling polynomials of the first (A008275) and second (A008277) kinds. The derivatives of the unsigned St1 are A028421. See examples. This formalism follows from the relations between the raising and lowering operators presented in the MathOverflow link and the Pincherle derivative. The results can be generalized through the operator relations in A094638, which are related to the celebrated Witt Lie algebra and pseudodifferential operators / symbols, to encompass other integral arrays.
A002741(n)*(-1)^(n+1) (row sums), A002104(n)*(-1)^(n+1) (alternating row sums). Column sequences: A133942(n-1), A001048(n-1), A238474, ... - Wolfdieter Lang, Mar 01 2014
Add an additional head row of zeros to the lower triangular array and denote it as T (with initial indexing in columns and rows being 0). Let dP = A132440, the infinitesimal generator for the Pascal matrix, and I, the identity matrix, then exp(T)=I+dP, i.e., T=log(I+dP). Also, (T_n)^n=0, where T_n denotes the n X n submatrix, i.e., T_n is nilpotent of order n. - Tom Copeland, Mar 01 2014
Any pair of lowering and raising ops. L p(n,x) = n·p(n-1,x) and R p(n,x) = p(n+1,x) satisfy [L,R]=1 which implies (RL)^n = St2(n,:RL:), and since (St2(·,u))!/(St2(·,u)-n)!= u^n, when evaluated umbrally, d[(RL)!/(RL-n)!]/d(RL) = d[:RL:^n]/d(RL) is well-defined and gives A238363 when the LHS is reduced to a sum of :RL:^k terms, exactly as for L=d/dx and R=x above. (Note that R_x above is a raising op. different from x, with associated L_x=-xD.) - Tom Copeland, Mar 02 2014
For relations to colored forests, disposition of flags on flagpoles, and the colorings of the vertices of the complete graphs K_n, encoded in their chromatic polynomials, see A130534. - Tom Copeland, Apr 05 2014
The unsigned triangle, omitting the main diagonal, gives A211603. See also A092271. Related to the infinitesimal generator of A008290. - Peter Bala, Feb 13 2017

Examples

			The first few row polynomials are
p(1,x)=  1
p(2,x)= -1 + 2x
p(3,x)=  2 - 3x + 3x^2
p(4,x)= -6 + 8x - 6x^2 + 4x^3
p(5,x)= 24 -30x +20x^2 -10x^3 + 5x^4
...........
For n=3: z!/(z-3)!=z^3-3z^2+2z=St1(3,z) with derivative 3z^2-6z+2, and
3·St2(2,x)-6·St2(1,x)+2=3(x^2+x)-6x+2=3x^2-3x+2=p(3,x). To see the relation to the operator formalism, note that (xD)^k=St2(k,:xD:) and (xD)!/(xD-k)!=[St2(·,:xD:)]!/[St2(·,:xD:)-k]!= :xD:^k.
The triangle a(n,k) begins:
n\k       0       1       2      3      4     5      6    7   8   9 ...
1:        1
2:       -1       2
3:        2      -3       3
4:       -6       8      -6      4
5:       24     -30      20    -10      5
6:     -120     144     -90     40    -15     6
7:      720    -840     504   -210     70   -21      7
8:    -5040    5760   -3360   1344   -420   112    -28    8
9:    40320  -45360   25920 -10080   3024  -756    168  -36   9
10: -362880  403200 -226800  86400 -25200  6048  -1260  240 -45  10
... formatted by _Wolfdieter Lang_, Mar 01 2014
-----------------------------------------------------------------------
		

Crossrefs

Programs

  • Mathematica
    a[n_, k_] := (-1)^(n-k-1)*n!/((n-k)*k!); Table[a[n, k], {n, 1, 10}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Jul 09 2015 *)

Formula

a(n,k) = (-1)^(n-k-1)*n!/((n-k)*k!) for k=0 to (n-1).
E.g.f.: log(1+t)*exp(x*t).
E.g.f.for unsigned array: -log(1-t)*exp(x*t).
The lowering op. for the row polynomials is L=d/dx, i.e., L p(n,x) = n*p(n-1,x).
An e.g.f. for an unsigned related version is -log(1+t)*exp(x*t)/t= exp(t*s(·,x)) with s(n,x)=(-1)^n * p(n+1,-x)/(n+1). Let L=d/dx and R= x-(1/((1-D)log(1-D))+1/D),then R s(n,x)= s(n+1,x) and L s(n,x)= n*s(n-1,x), defining a special Sheffer sequence of polynomials, an Appell sequence. So, R (-1)^(n-1) p(n,-x)/n = (-1)^n p(n+1,-x)/(n+1).
From Tom Copeland, Apr 17 2014: (Start)
Dividing each diagonal by its first element (-1)^(n-1)*(n-1)! yields the reverse of A104712.
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x). Then with dP = A132440, M = padded A238363 = A238385-I, I = identity matrix, and (B(.,x))^n = B(n,x) = the n-th Bell polynomial Bell(n,x) of A008277,
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x), and
B) P(:xD:)=exp(dP:xD:)=exp[(e^M-I):xD:]=exp[M*B(.,:xD:)]=exp[M*xD]=
(1+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x].
C) P(x)^m = P(m*x). P(2x) = A038207(x) = exp[M*B(.,2x)], face vectors of n-D hypercubes. (End)
From Tom Copeland, Apr 26 2014: (Start)
M = padded A238363 = A238385-I
A) = [St1]*[dP]*[St2] = [padded A008275]*A132440*A048993
B) = [St1]*[dP]*[St1]^(-1)
C) = [St2]^(-1)*[dP]*[St2]
D) = [St2]^(-1)*[dP]*[St1]^(-1),
where [St1]=padded A008275 just as [St2]=A048993=padded A008277.
E) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I + dP)^x*[St1].
F) exp(x*M) = [St1]*P(x)*[St2] = (I + dP)^x,
where (I + dP)^x = sum(k>=0, C(x,k)*dP^k).
Let the row vector Rv=(c0 c1 c2 c3 ...) and the column vector Cv(x)=(1 x x^2 x^3 ...)^Transpose. Form the power series V(x)= Rv * Cv(x) and W(y) := V(x.) evaluated umbrally with (x.)^n = x_n = (y)_n = y!/(y-n)!. Then
G) U(:xD:) = dV(:xD:)/d(xD) = dW(xD)/d(xD) evaluated with (xD)^n = Bell(n,:xD:),
H) U(x) = dV(x.)/dy := dW(y)/dy evaluated with y^n=y_n=Bell(n,x), and
I) U(x) = Rv * M * Cv(x). (Cf. A132440, A074909.) (End)
The Bernoulli polynomials Ber_n(x) are related to the polynomials q_n(x) = p(n+1,x) / (n+1) with the e.g.f. [log(1+t)/t] e^(xt) (cf. s_n (x) above) as Ber_n(x) = St2_n[q.(St1.(x))], umbrally, or [St2]*[q]*[St1], in matrix form. Since q_n(x) is an Appell sequence of polynomials, q_n(x) = [log(1+D_x)/D_x]x^n. - Tom Copeland, Nov 06 2016

Extensions

Pincherle formalism added by Tom Copeland, Feb 27 2014

A132382 Lower triangular array T(n,k) generator for group of arrays related to A001147 and A102625.

Original entry on oeis.org

1, -1, 1, -1, -2, 1, -3, -3, -3, 1, -15, -12, -6, -4, 1, -105, -75, -30, -10, -5, 1, -945, -630, -225, -60, -15, -6, 1, -10395, -6615, -2205, -525, -105, -21, -7, 1, -135135, -83160, -26460, -5880, -1050, -168, -28, -8, 1, -2027025, -1216215, -374220, -79380, -13230, -1890, -252, -36, -9, 1
Offset: 0

Views

Author

Tom Copeland, Nov 11 2007, Nov 12 2007, Nov 19 2007, Dec 04 2007, Dec 06 2007

Keywords

Comments

Let b(n) = LPT[ A001147 ] = -A001147(n-1) for n > 0 and 1 for n=0, where LPT represents the action of the list partition transform described in A133314.
Then T(n,k) = binomial(n,k) * b(n-k) .
Form the matrix of polynomials TB(n,k,t) = T(n,k) * t^(n-k) = binomial(n,k) * b(n-k) * t^(n-k) = binomial(n,k) * Pb(n-k,t),
beginning as
1;
-1, 1;
-1*t, -2, 1;
-3*t^2, -3*t, -3, 1;
-15*t^3, -12*t^2, -6*t, -4, 1;
-105*t^4, -75*t^3, -30*t^2, -10*t, -5, 1;
Let Pc(n,t) = LPT(Pb(.,t)).
Then [TB(t)]^(-1) = TC(t) = [ binomial(n,k) * Pc(n-k,t) ] = LPT(TB),
whose first column is
Pc(0,t) = 1
Pc(1,t) = 1
Pc(2,t) = 2 + t
Pc(3,t) = 6 + 6*t + 3*t^2
Pc(4,t) = 24 + 36*t + 30*t^2 + 15*t^3
Pc(5,t) = 120 + 240*t + 270*t^2 + 210*t^3 + 105*t^4.
The coefficients of these polynomials are given by the reverse of A102625 with the highest order coefficients given by A001147 with an additional leading 1.
Note this is not the complete matrix TC. The complete matrix is formed by multiplying along the diagonal of the lower triangular Pascal matrix by these polynomials, embedding trees of coefficients in the matrix.
exp[Pb(.,t)*x] = 1 + [(1-2t*x)^(1/2) - 1] / (t-0) = [1 + a finite diff. of [(1-2t*x)^(1/2)] with step t] = e.g.f. of the first column of TB.
exp[Pc(.,t)*x] = 1 / { 1 + [(1-2t*x)^(1/2) - 1] / t } = 1 / exp[Pb(.,t)*x) = e.g.f. of the first column of TC.
TB(t) and TC(t), being inverse to each other, are the generators of an Abelian group.
TB(0) and TC(0) are generators for a subgroup representing the iterated Laguerre operator described in A132013 and A132014.
Let sb(t,m) and sc(t,m) be the associated sequences under the LPT to TB(t)^m = B(t,m) and TC(t)^m = C(t,m).
Let Esb(t,m) and Esc(t,m) be e.g.f.'s for sb(t,m) and sc(t,m), rB(t,m) and rC(t,m) be the row sums of B(t,m) and C(t,m) and aB(t,m) and aC(t,m) be the alternating row sums.
Then B(t,m) is the inverse of C(t,m), Esb(t,m) is the reciprocal of Esc(t,m) and sb(t,m) and sc(t,m) form a reciprocal pair under the LPT. Similar relations hold among the row sums and the alternating sign row sums and associated quantities.
All the group members have the form B(t,m) * C(u,p) = TB(t)^m * TC(u)^p = [ binomial(n,k) * s(n-k) ]
with associated e.g.f. Es(x) = exp[m * Pb(.,t) * x] * exp[p * Pc(.,u) * x] for the first column of the matrix, with terms s(n), so group multiplication is isomorphic to matrix multiplication and to multiplication of the e.g.f.'s for the associated sequences (see examples).
These results can be extended to other groups of integer-valued arrays by replacing the 2 by any natural number in the expression for exp[Pb(.,t)*x].
More generally,
[ G.f. for M = Product_{i=0..j} B[s(i),m(i)] * C[t(i),n(i)] ]
= exp(u*x) * Product_{i=0..j} { exp[m(i) * Pb(.,s(i)) * x] * exp[n(i) * Pc(.,t(i)) * x] }
= exp(u*x) * Product_{i=0..j} { 1 + [ (1 - 2*s(i)*x)^(1/2) - 1 ] / s(i) }^m(i) / { 1 + [ (1 - 2*t(i)*x)^(1/2) - 1 ] / t(i) }^n(i)
= exp(u*x) * H(x)
[ E.g.f. for M ] = I_o[2*(u*x)^(1/2)] * H(x).
M is an integer-valued matrix for m(i) and n(i) positive integers and s(i) and t(i) integers. To invert M, change B to C in Product for M.
H(x) is the e.g.f. for the first column of M and diagonally multiplying the Pascal matrix by the terms of this column generates M. See examples.
The G.f. for M, i.e., the e.g.f. for the row polynomials of M, implies that the row polynomials form an Appell sequence (see Wikipedia and Mathworld). - Tom Copeland, Dec 03 2013

Examples

			Some group members and associated arrays are
(t,m) :: Array :: Asc. Matrix :: Asc. Sequence :: E.g.f. for sequence
..............................................................................
(0,1).::.B..::..A132013.::.(1,-1,0,0,0,0,...).....::.s(x).=.1-x
(0,1).::.C..::..A094587.::.(0!,1!,2!,3!,...)......::.1./.s(x)
(0,1).::.rB.::.~A055137.::.(1,0,-1,-2,-3,-4,...)..::.exp(x).*.s(x)
(0,1).::.rC.::....-.....::..A000522...............::.exp(x)./.s(x)
(0,1).::.aB.::....-.....::.(1,-2,3,-4,5,-6,...)...::.exp(-x).*.s(x)
(0,1).::.aC.::..A008290.::..A000166...............::.exp(-x)./.s(x)
..............................................................................
(0,2).::.B..::..A132014.::.(1,-2,2,0,0,0,0...)....::.s(x).=.(1-x)^2
(0,2).::.C..::..A132159.::.(1!,2!,3!,4!,...)......::..1./.s(x).
(0,2).::.rB.::...-......::.(1,-1,-1,1,5,11,19,29,)::.exp(x).*.s(x).
(0,2).::.rC.::...-......::..A001339...............::.exp(x)./.s(x).
(0,2).::.aB.::...-......::.(-1)^n.A002061(n+1)....::.exp(-x).*.s(x).
(0,2).::.aC.::...-......::..A000255...............::.exp(-x)./.s(x).
..............................................................................
(1,1).::.B..::..T.......::.(1,-A001147(n-1))......::.s(x).=.(1-2x)^(1/2)
(1,1).::.C..::.~A113278.::..A001147...............::.1./.s(x)...
(1,1).::.rB.::...-......::..A055142...............::.exp(x).*.s(x).
(1,1).::.rC.::...-......::..A084262...............::.exp(x)./.s(x).
(1,1).::.aB.::...-......::.(1,-2,2,-4,-4,-56,...).::.exp(-x).*.s(x).
(1,1).::.aC.::...-......::..A053871...............::.exp(-x)./.s(x).
..............................................................................
(2,1).::.B..::...-......::.(1,-A001813)...........::.s=[1+(1-4x)^(1/2)]/2....
(2,1).::.C..::...-......::..A001761...............::.1./.s(x)..
(2,1).::.rB.::...-......::.(1,0,-3,-20,-183,...)..::.exp(x).*.s(x)..
(2,1).::.rC.::...-......::.(1,2,7,46,485,...).....::.exp(x)./.s(x).
(2,1).::.aB.::...-......::.(1,-2,1,-10,-79,...)...::.exp(-x).*.s(x).
(2,1).::.aC.::...-......::.(1,0,3,20,237,...).....::.exp(-x)./.s(x)
..............................................................................
(1,2).::.B..::.~A134082.::.(1,-2,0,0,0,0,...).....::.s(x).=.1.-.2x
(1,2).::.C..::....-.....::..A000165...............::.1./.s(x)..
(1,2).::.rB.::....-.....::.(1,-1,-3,-5,-7,-9,...).::.exp(x).*.s(x).
(1,2).::.rC.::....-.....::..A010844...............::.exp(x)./.s(x)..
(1,2).::.aB.::....-.....::.(1,-3,5,-7,9,-11,...)..::.exp(-x).*.s(x).
(1,2).::.aC.::....-.....::..A000354...............::.exp(-x)./.s(x).
..............................................................................
(The tilde indicates the match is not exact--specifically, there are differences in signs from the true matrices.)
Note the row sums correspond to binomial transforms of s(x) and the alternating row sums, to inverse binomial transforms, or, finite differences.
Some additional examples:
C(1,2)*B(0,1) = B(1,-2)*C(0,-1) = [ binomial(n,k)*A002866(n-k) ] with asc. e.g.f. (1-x) / (1-2x).
B(1,2)*C(0,1) = C(1,-2)*B(0,-1) = 2I - A094587 with asc. e.g.f. (1-2x) / (1-x).
		

Formula

[G.f. for TB(n,k,t)] = GTB(u,x,t) = exp(u*x) * { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } = exp[(u+Pb(.,t))*x] where TB(n,k,t) = (D_x)^n (D_u)^k /k! GTB(u,x,t) eval. at u=x=0.
[G.f. for TC(n,k,t)] = GTC(u,x,t) = exp(u*x) / { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } = exp[(u+Pc(.,t))*x] where TC(n,k,t) = (D_x)^n (D_u)^k /k! GTC(u,x,t) eval. at u=x=0.
[E.g.f. for TB(n,k,t)] = I_o[2*(u*x)^(1/2)] * { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } and
[E.g.f. for TC(n,k,t)] = I_o[2*(u*x)^(1/2)] / { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t }
where I_o is the zeroth modified Bessel function of the first kind, i.e.,
I_o[2*(u*x)^(1/2)] = Sum_{j>=0} (u^j/j!) * (x^j/j!).
So [e.g.f. for TB(n,k)] = I_o[2*(u*x)^(1/2)] * (1 - 2x)^(1/2).

Extensions

More terms from Tom Copeland, Dec 05 2007

A132159 Lower triangular matrix T(n,j) for double application of an iterated mixed order Laguerre transform inverse to A132014. Coefficients of Laguerre polynomials (-1)^n * n! * L(n,-2-n,x).

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 24, 18, 6, 1, 120, 96, 36, 8, 1, 720, 600, 240, 60, 10, 1, 5040, 4320, 1800, 480, 90, 12, 1, 40320, 35280, 15120, 4200, 840, 126, 14, 1, 362880, 322560, 141120, 40320, 8400, 1344, 168, 16, 1, 3628800, 3265920, 1451520, 423360, 90720, 15120
Offset: 0

Views

Author

Tom Copeland, Nov 01 2007

Keywords

Comments

The matrix operation b = T*a can be characterized several ways in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or their e.g.f.'s EA(x) and EB(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a1(.),-1],0], umbrally,
where a1(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0]
2) b(n) = (-1)^n * n! * Lag(n,a(.),-2-n)
3) b(n) = Sum_{j=0..n} (-1)^j * binomial(n,j) * binomial(-2,j) * j! * a(n-j)
4) b(n) = Sum_{j=0..n} binomial(n,j) * (j+1)! * a(n-j)
5) B(x) = (1-xDx))^(-2) A(x), formally
6) B(x) = Sum_{j>=0} (-1)^j * binomial(-2,j) * (xDx)^j A(x)
= Sum_{j>=0} (j+1) * (xDx)^j A(x)
7) B(x) = Sum_{j>=0} (j+1) * x^j * D^j * x^j A(x)
8) B(x) = Sum_{j>=0} (j+1)! * x^j * Lag(j,-:xD:,0) A(x)
9) EB(x) = Sum_{j>=0} x^j * Lag[j,(.)! * Lag[.,a1(.),-1],0]
10) EB(x) = Sum_{j>=0} Lag[j,a1(.),-1] * (-x)^j / (1-x)^(j+1)
11) EB(x) = Sum_{j>=0} x^n * Sum_{j=0..n} (j+1)!/j! * a(n-j) / (n-j)!
12) EB(x) = Sum_{j>=0} (-x)^j * Lag[j,a(.),-2-j]
13) EB(x) = exp(a(.)*x) / (1-x)^2 = (1-x)^(-2) * EA(x)
14) T = A094587^2 = A132013^(-2) = A132014^(-1)
where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the D operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and associated operations described in A133314. Thus T(n,k) = binomial(n,k)*c(n-k) . c are also the coefficients in formulas 4 and 8.
The reciprocal sequence to c is d = (1,-2,2,0,0,0,...), so the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A132014. (A121757 is the reverse of T.)
These formulas are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),-1],0] by replacing 2 by m in formulas 2, 3, 5, 6, 12, 13 and 14, or (j+1)! by (m-1+j)!/(m-1)! in 4, 8 and 11. For further discussion of repeated applications of T, see A132014.
The row sums of T = [formula 4 with a(n) all 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = A001339. Therefore the e.g.f. of A001339 = [formula 13 with a(n) all 1] = exp(x)*(1-x)^(-2) = exp(x)*exp[c(.)*x)] = exp[(1+c(.))*x].
Note the reciprocal is 1/{exp[(1+c(.))*x]} = exp(-x)*(1-x)^2 = e.g.f. of signed A002061 with leading 1 removed], which makes A001339 and the signed, shifted A002061 reciprocal arrays under the list partition transform of A133314.
The e.g.f. for the row polynomials (see A132382) implies they form an Appell sequence (see Wikipedia). - Tom Copeland, Dec 03 2013
As noted in item 12 above and reiterated in the Bala formula below, the e.g.f. is e^(x*t)/(1-x)^2, and the Poisson-Charlier polynomials P_n(t,y) have the e.g.f. (1+x)^y e^(-xt) (Feinsilver, p. 5), so the row polynomials R_n(t) of this entry are (-1)^n P_n(t,-2). The associated Appell sequence IR_n(t) that is the umbral compositional inverse of this entry's polynomials has the e.g.f. (1-x)^2 e^(xt), i.e., the e.g.f. of A132014 (noted above), and, therefore, the row polynomials (-1)^n PC(t,2). As umbral compositional inverses, R_n(IR.(t)) = t^n = IR_n(R.(t)), where, by definition, P.(t)^n = P_n(t), is the umbral evaluation. - Tom Copeland, Jan 15 2016
T(n,k) is the number of ways to place (n-k) rooks in a 2 x (n-1) Ferrers board (or diagram) under the Goldman-Haglund i-row creation rook mode for i=2. Triangular recurrence relation is given by T(n,k) = T(n-1,k-1) + (n+1-k)*T(n-1,k). - Ken Joffaniel M. Gonzales, Jan 21 2016

Examples

			First few rows of the triangle are
    1;
    2,  1;
    6,  4,  1;
   24, 18,  6, 1;
  120, 96, 36, 8, 1;
		

Crossrefs

Columns: A000142 (k=0), A001563 (k=1), A001286 (k=2), A005990 (k=3), A061206 (k=4), A062199 (k=5), A062148 (k=6).

Programs

  • Haskell
    a132159 n k = a132159_tabl !! n !! k
    a132159_row n = a132159_tabl !! n
    a132159_tabl = map reverse a121757_tabl
    -- Reinhard Zumkeller, Mar 06 2014
    
  • Magma
    /* As triangle */ [[Binomial(n,k)*Factorial(n-k+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 10 2016
    
  • Maple
    T := proc(n,k) return binomial(n,k)*factorial(n-k+1): end: seq(seq(T(n,k),k=0..n),n=0..10); # Nathaniel Johnston, Sep 28 2011
  • Mathematica
    nn=10;f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[Series[Exp[y x]/(1-x)^2,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Feb 15 2013 *)
  • Sage
    flatten([[binomial(n,k)*factorial(n-k+1) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, May 19 2021

Formula

T(n,k) = binomial(n,k)*c(n-k).
From Peter Bala, Jul 10 2008: (Start)
T(n,k) = binomial(n,k)*(n-k+1)!.
T(n,k) = (n-k+1)*T(n-1,k) + T(n-1,k-1).
E.g.f.: exp(x*y)/(1-y)^2 = 1 + (2+x)*y + (6+4*x+x^2)*y^2/2! + ... .
This array is the particular case P(2,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
n\k|0....................1...............2.........3.....4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
See A094587 for some general properties of these arrays.
Other cases recorded in the database include: P(1,0) = Pascal's triangle A007318, P(1,1) = A094587, P(2,0) = A038207, P(3,0) = A027465, P(1,3) = A136215 and P(2,3) = A136216. (End)
Let f(x) = (1/x^2)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+2)*R(n,x)-x*R'(n,x). Cf. A094587. - Peter Bala, Oct 28 2011
Exponential Riordan array [1/(1 - y)^2, y]. The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x). Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = Product_{k = 0..n-1} (2*x + k) with the convention that P(0,x) = 1. Then the present triangle is the triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (24, 18, 6, 1) so P(3,x + 1) = (2*x + 2)*(2*x + 3)*(2*x + 4) = 24 + 18*(2*x) + 6*(2*x)*(2*x + 1) + (2*x)*(2*x + 1)*(2*x + 2). Matrix square of triangle A094587. - Peter Bala, Aug 29 2013
From Tom Copeland, Apr 21 2014: (Start)
T = (I-A132440)^(-2) = {2*I - exp[(A238385-I)]}^(-2) = unsigned exp[2*(I-A238385)] = exp[A005649(.)*(A238385-I)], umbrally, where I = identity matrix.
The e.g.f. is exp(x*y)*(1-y)^(-2), so the row polynomials form an Appell sequence with lowering operator D=d/dx and raising operator x+2/(1-D).
With L(n,m,x) = Laguerre polynomials of order m, the row polynomials are (-1)^n * n! * L(n,-2-n,x) = (-1)^n*(-2!/(-2-n)!)*K(-n,-2-n+1,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
Operationally, (-1)^n*n!*L(n,-2-n,-:xD:) = (-1)^n*x^(n+2)*:Dx:^n*x^(-2-n) = (-1)^n*x^2*:xD:^n*x^(-2) = (-1)^n*n!*binomial(xD-2,n) = (-1)^n*n!*binomial(-2,n)*K(-n,-2-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706.
The generalized Pascal triangle Bala mentions is a special case of the fundamental generalized factorial matrices in A133314. (End)
From Peter Bala, Jul 26 2021: (Start)
O.g.f: 1/y * Sum_{k >= 0} k!*( y/(1 - x*y) )^k = 1 + (2 + x)*y + (6 + 4*x + x^2)*y^2 + ....
First-order recurrence for the row polynomials: (n - x)*R(n,x) = n*(n - x + 1)*R(n-1,x) - x^(n+1) with R(0,x) = 1.
R(n,x) = (x + n + 1)*R(n-1,x) - (n - 1)*x*R(n-2,x) with R(0,x) = 1 and R(1,x) = 2 + x.
R(n,x) = A087981 (x = -2), A000255 (x = -1), A000142 (x = 0), A001339 (x = 1), A081923 (x = 2) and A081924 (x = 3). (End)

Extensions

Formula 3) in comments corrected by Tom Copeland, Apr 20 2014
Title modified by Tom Copeland, Apr 23 2014

A132014 T(n,j) for double application of an iterated mixed order Laguerre transform: Coefficients of Laguerre polynomial (-1)^n*n!*L(n,2-n,x).

Original entry on oeis.org

1, -2, 1, 2, -4, 1, 0, 6, -6, 1, 0, 0, 12, -8, 1, 0, 0, 0, 20, -10, 1, 0, 0, 0, 0, 30, -12, 1, 0, 0, 0, 0, 0, 42, -14, 1, 0, 0, 0, 0, 0, 0, 56, -16, 1, 0, 0, 0, 0, 0, 0, 0, 72, -18, 1, 0, 0, 0, 0, 0, 0, 0, 0, 90, -20, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 110, -22, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 132, -24, 1
Offset: 0

Views

Author

Tom Copeland, Oct 30 2007, Nov 05 2007, Nov 11 2007

Keywords

Comments

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) = a(0), b(1) = a(n) - 2 a(0), b(n) = a(n) - 2n a(n-1) + n(n-1) a(n-2) for n > 0.
2) b(n) = n! Lag{n,(.)!*Lag[.,a1(.),0],-1}, umbrally, where a1(n) = n! Lag{n,(.)!*Lag[.,a(.),0],-1}.
3) b(n) = n! Sum_{j=0..min(2,n)} (-1)^j * binomial(n,j)*a(n-j)/(n-j)!
4) b(n) = (-1)^n n! Lag(n,a(.),2-n)
5) B(x) = (1-xDx)^2 A(x)
6) B(x) = Sum_{j=0..2} {(-1)^j * binomial(2,j)*j!*x^j*Lag(j,-:xD:,0)} A(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.
7) EB(x) = (1-x)^2 EA(x)
8) T = S^2 = A132013^2 = A094587^(-2) = A132159^(-1).
c = (1,-2,2,0,0,...) is the sequence associated to T under the list partition transform and associated operations described in A133314. c are also the coefficients in formula 6. Thus T(n,k) = binomial(n,k)*c(n-k).
The reciprocal sequence to c is d = (1!,2!,3!,4!,...), so the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A132159.
These formulas are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),0],-1] by replacing 2 with m in formulas 3, 4, 5, 6 and 7.
The generalized c are given by the generalized coefficients of 6, i.e.,
c(n) = (-1)^n * binomial(m,n)*n! = (-1)^n * m!/(m-n)!.
The generalized d are given by the array at and below the term SI(m-1,m-1) in SI(n,k) = binomial(n,k) * (n-k)!, the inverse of S; i.e.,
d(n) = SI(m-1+n,m-1) = binomial(m-1+n,m-1) * n! = (m-1+n)!/(m-1)!.
As an aside, this shows that the signed falling factorials and the rising factorials form reciprocal pairs under the list partition transform of A133314.
Row sums of T = [formula 3 with all a(n) = 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = (1,-1,-1,1,5,11,19,...),
with e.g.f. = [EB(x) with EA(x) = exp(x)] = (1-x)^2 * exp(x) = exp(x)*exp(c(.)*x) = exp[(1+c(.))*x].
Alternating row sums of T = [formula 3 with all a(n) = (-1)^n] = [finite differences of c] = [coefficients of B(x) with A(x) = 1/(1+x)] = (1,-3,7,-13,21,-31,...) = (-1)^n A002061(n+1),
with e.g.f. = [EB(x) with EA(x) = exp(-x)] = (1-x)^2 * exp(-x) = exp(- x)*exp(c(.)*x) = exp[-(1-c(.))*x].
See A132159 for a relation to the Poisson-Charlier polynomials. - Tom Copeland, Jan 15 2016

Examples

			First few rows of the triangle are
   1;
  -2,   1;
   2,  -4,   1;
   0,   6,  -6,   1;
   0,   0,  12,  -8,   1;
   0,   0,   0,  20, -10,   1;
		

Crossrefs

Programs

  • Mathematica
    m = 12; s = Exp[x*y]*(1 - x)^2 + O[x]^(m + 2) + O[y]^(m + 2); T[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {y, 0, k}]*n!; T[0, 0] = 1; Table[T[n, k], {n, 0, m}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 09 2015 *)
  • PARI
    row(n) = Vecrev((-1)^n*n!*pollaguerre(n, 2-n)); \\ Michel Marcus, Feb 06 2021

Formula

T(n,k) = binomial(n,k)*c(n-k).
E.g.f. for row polynomials: exp(x*y)(1-x)^2. Implies the row polynomials form an Appell sequence (see Wikipedia). - Tom Copeland, Dec 03 2013
From Tom Copeland, Apr 21 2014: (Start)
Change notation letting L(n,m,x) = Lag(n,x,m).
Row polynomials: (-1)^n*n!*L(n,2-n,x) = (-1)^n*(-x)^(n-2)*2!*L(2,n-2,x) =
(-1)^n*(2!/(2-n)!)*K(-n,2-n+1,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
For the row polynomials, the lowering operator = d/dx and the raising operator = x - 2/(1-D).
T = (I - A132440)^2 = [2*I - exp(A238385-I)]^2 = signed exp[2*(A238385-I)], where I = identity matrix.
Operationally, (-1)^n*n!*L(n,2-n,-:xD:) = (-1)^n*x^(n-2)*:Dx:^n*x^(2-n) = (-1)^n*x^(-2)*:xD:^n*x^2 = (-1)^n*n!*binomial(xD+2,n) = (-1)^n*n!*binomial(2,n)*K(-n,2-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706. (End)
n-th row polynomial: n!*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*Lag(k,2,x). - Peter Bala, Jul 25 2021

Extensions

Title modified by Tom Copeland, Apr 21 2014
Missing term -18 inserted in 10th row by Jean-François Alcover, Jul 09 2015

A235706 (I + A132440)^3: Coefficients for normalized generalized Laguerre polynomials n!*Lag(n, 3-n, -x).

Original entry on oeis.org

1, 3, 1, 6, 6, 1, 6, 18, 9, 1, 0, 24, 36, 12, 1, 0, 0, 60, 60, 15, 1, 0, 0, 0, 120, 90, 18, 1, 0, 0, 0, 0, 210, 126, 21, 1, 0, 0, 0, 0, 0, 336, 168, 24, 1, 0, 0, 0, 0, 0, 0, 504, 216, 27, 1, 0, 0, 0, 0, 0, 0, 0, 720, 270, 30, 1
Offset: 0

Views

Author

Tom Copeland, Apr 20 2014

Keywords

Comments

The associated Laguerre polynomials n!*Lag(n,3-n,-x) are related to the rook polynomials of a rectangular 3 X n chessboard by R(3,n,x) = n!*x^n*Lag(n,3-n,-1/x), which are also the matching polynomials, or generating function of the number of k-edge matchings, of the complete bipartite graph K(m,n), or biclique (cf. Wikipedia for details).
The formulas here and below can be naturally extended with 3 replaced by any positive integer m. For m = 1 and 2, see unsigned A132013 and A132014. The formulas there can be extrapolated to apply to this matrix.

Examples

			Triangle begins:
  1;
  3,  1;
  6,  6,  1;
  6, 18,  9,  1;
  0, 24, 36, 12,  1;
  0,  0, 60, 60, 15, 1;
  ...
		

Crossrefs

Cf. A007318, A008306 for a generalization, A132013, A132014, A132440, A238363, A238385.
....................................
With 0th row: 1
n-th row: n!*Lag(n,3-n,-x)
....................................
1st: 1!*Lag(1,2,-x) = A062139(1,k,-x)
2nd: 2!*Lag(2,1,-x) = A105278(2,k,x)
3rd: 3!*Lag(3,0,-x) = A021009(3,k,-x)
4th: 4!*Lag(4,-1,-x) = A111596(4,k,-x)
5th: 5!*Lag(5,-2,-x) = cf. x^2*A062139(3,k,x)
6th: 6!*Lag(6,-3,-x) = cf. x^3*A062137(3,k,-x)
....................................
n-th row: x^(n-3)*3!*Lag(3,n-3,-x)
....................................
1st: x^(-2)*3!Lag(3,-2,-x) = cf. x^(-2)*[x^2*A062139(1,k,x)]
2nd: x^(-1)*3!Lag(3,-1,-x) = x^(-1)*A111596(3,k,-x)
3rd: x^0*3!Lag(3,0,-x) = x^0*A021009(3,k,-x)
4th: x^1*3!Lag(3,1,-x) = x^1*A105278(3,k,x)
5th: x^2*3!Lag(3,2,-x) = x^2*A062139(3,k,-x)
6th: x^3*3!Lag(3,3,-x) = x^3*A062137(3,k,-x)

Programs

  • Magma
    /* As triangle */ [[Binomial(3, n-k)*Factorial(n)/Factorial(k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 28 2017
  • Mathematica
    Table[Binomial[3, n - k] n! / k!, {n, 0, 9}, {k, 0, n}]//Flatten (* Vincenzo Librandi, Jul 28 2017 *)
  • PARI
    T(n,k) = binomial(3,n-k)*n!/k!
    tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 28 2017
    
  • PARI
    row(n) = Vecrev(n!*pollaguerre(n, 3-n, -x)); \\ Michel Marcus, Feb 06 2021
    

Formula

T(n,k) = binomial(3,n-k)*n!/k! = binomial(n,k)*3!/(3-n+k)!.
E.g.f.: exp(y*x)(1+y)^3, so this is an Appell sequence of polynomials with lowering operator L= D= d/dx and raising operator R = x + 3/(1+D).
E.g.f. of inverse matrix is exp(x*y)/(1+y)^3.
Multiply the n-th diagonal of the Pascal matrix A007318 by d(0)=1, d(1)=3, d(2)=6, d(3)=6, and d(n)=0 for n>3 to obtain T.
Row polynomials: n!*Lag(n,3-n,-x) = x^(n-3)*3!*Lag(3,n-3,-x) =
(3!/(3-n)!)*K(-n,3-n+1,-x) where K is Kummer's confluent hypergeometric function (as a limit of n+c as c tends to zero).
T = (I + A132440)^3 = exp[3*(A238385-I)]. I = identity matrix.
Operationally, n!Lag(n,3-n,-:xD:) = x^(n-3)*:Dx:^n*x^(3-n) = x^(-3)*:xD:^n*x^3 = n!*binomial(xD+3,n) = n!*binomial(3,n)*K(-n,3-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators.
n-th row polynomial: n!*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*Lag(k,3,-x). - Peter Bala, Jul 25 2021

A131202 A coefficient tree from the list partition transform relating A111884, A084358, A000262, A094587, A128229 and A131758.

Original entry on oeis.org

1, -1, 3, 1, -8, 13, 1, 11, -61, 73, -19, 66, 66, -494, 501, 151, -993, 2102, -298, -4293, 4051, -1091, 9528, -33249, 52816, -21069, -39528, 37633, 7841, -82857, 378261, -929101, 1207299, -560187, -375289, 394353, -56519, 692422, -3832928, 12255802, -23834210, 26643994, -12620672, -3481562, 4596553
Offset: 1

Views

Author

Tom Copeland, Oct 22 2007, Nov 30 2007

Keywords

Comments

Construct the infinite array of polynomials
a(0,t) = 1
a(1,t) = 1
a(2,t) = -1 + 3*t
a(3,t) = 1 - 8*t + 13*t^2
a(4,t) = 1 + 11*t - 61*t^2 + 73*t^3
a(5,t) = -19 + 66*t + 66*t^2 - 494*t^3 + 501*t^4
a(6,t) = 151 - 993*t + 2102*t^2 - 298*t^3 - 4293*t^4 + 4051*t^5
This array is the reciprocal array of the following array b(n,t) under the list partition transform and its associated operations described in A133314.
b(0,t) = 1 and b(n,t) = -A000262(n)*(t-1)^(n-1) for n > 0.
Then A111884(n) = a(n,0).
Lower triangular matrix A094587 = binomial(n,k)*a(n-k,1).
A084358(n) = a(n,2).
Signed A128229 = matrix inverse of binomial(n,k)*a(n-k,1) = binomial(n,k)*b(n-k,1) = A132013.
As t tends to infinity, a(n,t)/t^(n-1) tends to A000262(n) for n > 0.
The P(n,t) of A131758 can be constructed from T(n,k,t) = binomial(n,k)*a(n-k,t) by letting T(n,k,t) multiply the column vector c(n,t) given by c(0,t) = 0! and c(n,t) = n!*(t-1)^(n-1) for n > 0. The P(n,t) have rich associations to other sequences.

Programs

  • Mathematica
    CoefficientList[#, t] & /@ (# Range@Length@#!) &@ Rest@CoefficientList[(t-1) / (t - Exp[x(t-1)/(1-x(t-1))]) + O[x]^10 // Simplify, x] // Flatten (* Andrey Zabolotskiy, Feb 19 2024 *)
  • PARI
    T(n) = [Vecrev(p) | p<-Vec(-1 + serlaplace((y-1) / (y - exp(x*(y-1)/(1-x*(y-1)) + O(x*x^n) ))))]
    { my(A=T(7)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Feb 19 2024

Formula

E.g.f. for the row polynomials, which are a(n, t) for n > 0, is:
(t-1) / (t - exp(x*(t-1)/(1-x*(t-1)))).
E.g.f. for the polynomials b(n, t), introduced above, is the reciprocal of that.

Extensions

Rows 7-9 added and offset changed by Andrey Zabolotskiy, Feb 19 2024

A360753 Matrix inverse of A360657.

Original entry on oeis.org

1, 0, 1, 0, -2, 1, 0, 1, -5, 1, 0, 1, 8, -9, 1, 0, 2, 4, 29, -14, 1, 0, 6, 4, -10, 75, -20, 1, 0, 24, 4, -41, -115, 160, -27, 1, 0, 120, -8, -147, -196, -490, 301, -35, 1, 0, 720, -136, -624, -392, -231, -1484, 518, -44, 1
Offset: 0

Views

Author

Werner Schulte, Feb 21 2023

Keywords

Examples

			Triangle T(n, k) for 0 <= k <= n starts:
n\k :  0    1     2     3     4     5      6    7    8  9
=========================================================
  0 :  1
  1 :  0    1
  2 :  0   -2     1
  3 :  0    1    -5     1
  4 :  0    1     8    -9     1
  5 :  0    2     4    29   -14     1
  6 :  0    6     4   -10    75   -20      1
  7 :  0   24     4   -41  -115   160    -27    1
  8 :  0  120    -8  -147  -196  -490    301  -35    1
  9 :  0  720  -136  -624  -392  -231  -1484  518  -44  1
  etc.
		

Crossrefs

Cf. A132013, A215534, A354794, A354795, A360657 (matrix inverse).

Programs

  • PARI
    tabl(m) = {my(n=2*m, A = matid(n), B, C, T); for( i = 2, n, for( j = 2, i, A[i, j] = A[i-1, j-1] + j * A[i-1, j] ) ); B = A^(-1); C = matrix( m, m, i, j, if( j == 1, 0^(i-1), sum( r = 0, i-j, B[i-j+1, r+1] * A[i-1+r, i-1] ) ) ); T = 1/C; }

Formula

Conjectured formulas:
1. Matrix product of A354794 and T without column 0 equals A215534.
2. Matrix product of T and A354794 without column 0 equals A132013.
3. E.g.f. of column k > 0: Sum_{n >= k} T(n, k) * t^(n-1) / (n-1)! = (1 - t) * (Sum_{n >= k} A354795(n, k) * t^(n-1) / (n-1)!).
Showing 1-10 of 10 results.