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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

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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;

Links

Crossrefs

Cf. A008277, A094587, A132013, A132382.
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

A165674 Triangle generated by the asymptotic expansions of the E(x,m=2,n).

Original entry on oeis.org

1, 3, 1, 11, 5, 1, 50, 26, 7, 1, 274, 154, 47, 9, 1, 1764, 1044, 342, 74, 11, 1, 13068, 8028, 2754, 638, 107, 13, 1, 109584, 69264, 24552, 5944, 1066, 146, 15, 1, 1026576, 663696, 241128, 60216, 11274, 1650, 191, 17, 1
Offset: 1

Views

Author

Johannes W. Meijer, Oct 05 2009

Keywords

Comments

The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the E(x,m=2,n) ~ (exp(-x)/x^2)*(1 - (1+2*n)/x + (2+6*n+3*n^2)/x^2 - (6+22*n+18*n^2+ 4*n^3)/x^3 + ... ) is discussed in A028421. The formula for the asymptotic expansion leads for n = 1, 2, 3, .., to the left hand columns of the triangle given above.
The recurrence relations of the right hand columns of this triangle lead to Pascal's triangle A007318, their a(n) formulas lead to Wiggen's triangle A028421 and their o.g.f.s lead to Wood's polynomials A126671; cf. A080663, A165676, A165677, A165678 and A165679.
The row sums of this triangle lead to A093344. Surprisingly the e.g.f. of the row sums Egf(x) = (exp(1)*Ei(1,1-x) - exp(1)*Ei(1,1))/(1-x) leads to the exponential integrals in view of the fact that E(x,m=1,n=1) = Ei(n=1,x). We point out that exp(1)*Ei(1,1) = A073003.
The Maple programs generate the coefficients of the triangle given above. The first one makes use of a relation between the triangle coefficients, see the formulas, and the second one makes use of the asymptotic expansions of the E(x,m=2,n).
Amarnath Murthy discovered triangle A093905 which is the reversal of our triangle.
A165675 is an extended version of this triangle. Its reversal is A105954.
Triangle A094587 is generated by the asymptotic expansions of E(x,m=1,n).

Crossrefs

A093905 is the reversal of this triangle.
A000254, A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562, A051564 are the first ten left hand columns.
A080663, n>=2, is the third right hand column.
A165676, A165677, A165678 and A165679 are the next right hand columns, A093344 gives the row sums.
A073003 is Gompertz's constant.
A094587 is generated by the asymptotic expansions of E(x, m=1, n).
Cf. A165675, A105954 (Quet) and A067176 (Bottomley).
Cf. A007318 (Pascal), A028421 (Wiggen), A126671 (Wood).

Programs

  • Maple
    nmax:=9; for n from 1 to nmax do a(n, n) := 1 od: for n from 2 to nmax do a(n, 1) := n*a(n-1, 1) + (n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do a(n, m) := (n-m+1)*a(n-1, m) + a(n-1, m-1) od: od: seq(seq(a(n, m), m = 1..n), n = 1..nmax);
# End program 1
nmax := nmax+1: m:=2; with(combinat): EA := proc(x, m, n) local E, i; E:=0: for i from m-1 to nmax+2 do E := E + sum((-1)^(m+k1+1) * binomial(k1, m-1) * n^(k1-m+1) * stirling1(i, k1), k1=m-1..i) / x^(i-m+1) od: E:= exp(-x)/x^(m) * E: return(E); end: for n1 from 1 to nmax do f(n1-1) := simplify(exp(x) * x^(nmax+3) * EA(x, m, n1)); for m1 from 0 to nmax+2 do b(n1-1, m1) := coeff(f(n1-1), x, nmax+2-m1) od: od: for n1 from 0 to nmax-1 do for m1 from 0 to n1-m+1 do a(n1-m+2, m1+1) := abs(b(m1, n1-m1)) od: od: seq(seq(a(n, m), m = 1..n),n = 1..nmax-1);
# End program 2
# Maple programs revised by Johannes W. Meijer, Sep 22 2012

Formula

a(n,m) = (n-m+1)*a(n-1,m) + a(n-1,m-1), for 2 <= m <= n-1, with a(n,n) = 1 and a(n,1) = n*a(n-1,1) + (n-1)!.
a(n,m) = product(i, i= m..n)*sum(1/i, i = m..n).

A132013 T(n,j) for an iterated mixed order Laguerre transform. Coefficients of the normalized generalized Laguerre polynomials (-1)^n*n!*L(n,1-n,x).

Original entry on oeis.org

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, 0, 0, 0, 0, 0, 0, 0, -8, 1, 0, 0, 0, 0, 0, 0, 0, 0, -9, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -10, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -11, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -12, 1
Offset: 0

Views

Author

Tom Copeland, Oct 30 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(n) = a(n) - n*a(n-1) for n > 0
2) b(n) = n! Lag{n,(.)!*Lag[.,a(.),0],-1}, umbrally
3) b(n) = n! Sum_{j=0..min(1,n)} (-1)^j * binomial(n,j)*a(n-j)/(n-j)!
4) b(n) = (-1)^n n! Lag(n,a(.),1-n)
5) B(x) = (1-xDx) A(x) = [1-x*Lag(1,-xD:,0)] A(x)
6) EB(x) = (1-x) EA(x),
where D is the derivative w.r.t. x and Lag(n,x,m) is the associated Laguerre polynomial of order m. These formulas are easily generalized for repeated applications of the operator.
c = (1,-1,0,0,0,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314. The reciprocal sequence is d = (0!,1!,2!,3!,4!,...).
Consequently, the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A094587, which has the property that the terms at and below TI(m,m) are the associated sequence under the list partition transform for the inverse for T^(m+1) for m=0,1,2,3,... .
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)] = A024000 = (1,0,-1,-2,-3,...), with e.g.f. = [EB(x) with EA(x) = exp(x)] = (1-x) * 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,-2,3,-4,...), with e.g.f. = [EB(x) with EA(x) = exp(-x)] = (1-x) * exp(-x) = exp(-x)*exp(c(.)*x) = exp[-(1-c(.))*x].
An e.g.f. for the o.g.f.s for repeated applications of T on A(x) is given by
exp[t*(1-xDx)] A(x) = e^t * Sum_{n=0,1,...} (-t*x)^n * Lag(n,-:xD:,0) A(x)
= e^t * exp{[-t*u/(1+t*u)]*:xD:} / (1+t*u) A(x) (eval. at u=x)
= e^t * A[x/(1+t*x)]/(1+t*x) .
See A132014 for more notes on repeated applications.

Examples

			First few rows of the triangle are
   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;

Links

Crossrefs

Cf. A235706, A132382, A132159, A094587, A132014, A154955, A235706.

Programs

  • Maple
    c := n -> `if`(n=0,1,`if`(n=1,-1,0)):
T := (n,k) -> binomial(n,k)*c(n-k); # Peter Luschny, Nov 14 2016
  • Mathematica
    Table[PadLeft[{-n, 1}, n+1], {n, 0, 13}] // Flatten (* Jean-François Alcover, Apr 29 2014 *)
  • PARI
    row(n) = Vecrev((-1)^n*n!*pollaguerre(n, 1-n)); \\ Michel Marcus, Jul 26 2021

Formula

T(n,k) = binomial(n,k)*c(n-k), with the sequence c defined in the comments.
E.g.f.: exp(x*y)(1-x), which implies the row polynomials form an Appell sequence. More relations can be found in A132382. - 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,1-n,x) = -x^(n-1)*L(1,n-1,x) =
(-1)^n*(1/(1-n)!)*K(-n,1-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 - 1/(1-D).
T = I - A132440 = 2*I - exp[A238385-I] = signed exp[A238385-I], where I = identity matrix.
Operationally, (-1)^n*n!*L(n,1-n,-:xD:) = -x^(n-1)*:Dx:^n*x^(1-n) = (-1)^n*x^(-1)*:xD:^n*x = (-1)^n*n!*binomial(xD+1,n) = (-1)^n*n!*binomial(1,n)*K(-n,1-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706. (End)
The unsigned row polynomials have e.g.f. (1+t)e^(xt) = exp(t*p.(x)), umbrally, and p_n(x) = (1+D) x^n. With q_n(x) the row polynomials of A094587, p_n(x) = u_n(q.(v.(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 unsigned [T] = [p] = [u]*[q]*[v]. Conversely, q_n(x) = v_n (p.(u.(x))). - Tom Copeland, Nov 10 2016
n-th row polynomial: n!*Sum_{k = 0..n} (-1)^k*binomial(n,k)*Lag(k,1,x). - Peter Bala, Jul 25 2021

Extensions

Title modified by Tom Copeland, Apr 21 2014

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

Original entry on oeis.org

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

Views

Author

Vladeta Jovovic, Aug 19 2002

Keywords

Comments

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

Examples

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

Links

Crossrefs

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

Programs

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

Formula

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

Extensions

More terms from Emeric Deutsch, Feb 23 2004

A133932 Coefficients of a partition transform for Lagrange inversion of -log(1 - u(.)*t), complementary to A134685 for an e.g.f.

Original entry on oeis.org

1, -1, 3, -2, -15, 20, -6, 105, -210, 40, 90, -24, -945, 2520, -1120, -1260, 420, 504, -120, 10395, -34650, 25200, 18900, -2240, -15120, -9072, 1260, 2688, 3360, -720, -135135, 540540, -554400, -311850, 123200, 415800, 166320, -50400, -56700, -120960, -75600, 18144, 20160, 25920, -5040
Offset: 1

Views

Author

Tom Copeland, Jan 27 2008

Keywords

Comments

Let f(t) = -log(1 - u(.)*t) = Sum_{n>=1} (u_n / n) * t^n.
If u_1 is not equal to 0, then the compositional inverse for f(t) is given by g(t) = Sum_{j>=1} P(n,t) where, with u_n denoted by (n'),
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -1 (2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 3 (2')^2 - 2 (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -15 (2')^3 + 20 (1')(2')(3') - 6 (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 105 (2')^4 - 210 (1') (2')^2 (3') + 40 (1')^2 (3')^2 + 90 (1')^2 (2') (4') - 24 (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -945 (2')^5 + 2520 (1') (2')^3 (3') - 1120 (1')^2 (2') (3')^2 - 1260 (1')^2 (2')^2 (4') + 420 (1')^3 (3')(4') + 504 (1')^3 (2')(5') - 120 (1')^4 (6') ] * t^6 / 6!
See A134685 for more information.
From Tom Copeland, Sep 28 2016: (Start)
P(7,t) = (1')^(-13) * [ 10395 (2')^6 - 34650 (1')(2')^4(3') + (1')^2 [25200 (2')^2(3')^2 + 18900 (2')^3(4')] - (1')^3 [2240 (3')^3 + 15120 (2')(3')(4') + 9072 (2')^2(5')] + (1')^4 [1260 (4')^2 + 2688 (3')(5') + 3360 (2')(6')] - 720 (1')^5(7')] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -135135 (2')^7 + 540540 (1')(2')^5(3') - (1')^2 [554400 (2')^3(3')^2 + 311850 (2')^4(4')] + (1')^3 [123200 (2')(3')^3 + 415800 (2')^2(3')(4') + 166320 (2')^3(5')] - (1')^4 [50400 (3')^2(4') + 56700 (2')(4')^2 + 120960 (2')(3')(5') + 75600 (2')^2(6')] + (1')^5 [18144 (4')(5') + 20160 (3')(6') + 25920 (2')(7')] - 5040 (1')^6(8')] * t^8 / 8! (End)

Examples

			From _Tom Copeland_, Sep 18 2014: (Start)
Let f(x) = log((1-ax)/(1-bx))/(b-a) = -log(1-u.*x) = x + (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 + (a^3+a^2b+ab^2+a^3)x^4/4 + ... . with (u.)^n = u_n = h_(n-1)(a,b) the complete homogeneous polynomials in two indeterminates.
Then the inverse g(x) = (e^(ax)-e^(bx))/(a*e^(ax)-b*e^(bx)) = x - (a+b)x^2/2! + (a^2+4ab+b^2)x^3/3! - (a^3+11a^2b+11ab^2+b^3)x^4/4! + ... , where the bivariate polynomials are the Eulerian polynomials of A008292.
The inversion formula gives, e.g., P(3,x) = 3(u_2)^2 - 2u_3 = 3(h_1)^2 - 2h_2 = 3(a+b)^2 - 2(a^2 + ab + b^2) = a^2 + 4ab + b^2. (End)

Links

Crossrefs

Cf. A145271 (A111999, A007318) = (reduced array, associated g(x)).
Cf. A008292, A086810, A094587, A133314, A134685, A139605, A263633.
Cf. A036039, A133437.

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = InverseSeries[t (1 + Sum[u[k] t^k/(k+1), {k, nn}] + O[t]^(nn+1))]}, Table[(n+1)! Coefficient[s, t^(n+1) Product[u[w], {w, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
rows[7] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Formula

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [ 2^e(2) (e(2))! * 3^e(3) (e(3))! * ... n^e(n) * (e(n))! ].
From Tom Copeland, Sep 06 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u./(1-u.t)]
= 1/((u_1) + (u_2)*t + (u_3)*t^2 + (u_4)*t^3 + ...),
where Ev denotes umbral evaluation.
Then for the partition polynomials of A133932,
n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
and the compositional inverse of f(t) is
g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
Also, dg(t)/dt = h(g(t)). (End)
From Tom Copeland, Oct 20 2011: (Start)
With exp[x* PS(.,t)] = exp[t*g(x)] = exp[x*h(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*h(d/dt) = t* 1/[(u_1) + (u_2)*d/dt + (u_3)*(d/dt)^2 + ...], and
L = f(d/dt) = (u_1)*d/dt + (u_2)*(d/dt)^2/2 + (u_3)*(d/dt)^3/3 + ....
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)
The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x/2 + u_3 * x^2/3 + ... + u_n * x^(n-1)/n]^n evaluated at x=0. - Tom Copeland, Jul 07 2015
From Tom Copeland, Sep 19 2016: (Start)
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix A007318 by f_m = (m-1)! u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) f_{n+1-k}, or equivalently, multiply the diagonals of A094587 by u_m. Then P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314.
With u_1 = 1, the first column of UP^(-1) with u_1 = 1 (with initial indices [0,0]) is composed of the row polynomials n! * OP_n(-u_2,...,-u_(n+1)), where OP_n(x[1],...,x[n]) are the row polynomials of A263633 for n > 0 and OP_0 = 1, which are related to those of A133314 as reciprocal o.g.f.s are related to reciprocal e.g.f.s; e.g., UP^(-1)[0,0] = 1, Up^(-1)[1,0] = -u_2, UP^(-1)[2,0] = 2! * (-u_3 + u_2^2) = 2! * OP_2(-u_2,-u_3).
Also, P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. (End)
From Tom Copeland, Sep 20 2016: (Start)
Let PS(n,u1,u2,...,un) = P(n,t) / (t^n/n!), i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.
Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -24 u5, b2 = 90 u2 u4 + 40 u3^2, b3 = -210 u2^2 u3, and b4 = 105 u2^4.
The relation between solutions of the inviscid Burgers's equation and compositional inverse pairs (cf. link and A086810) implies, for n > 2, PB(n, 0 * b1, 1 * b2, ..., (K-1) * bK, ...) = (1/2) * Sum_{k = 2..n-1} binomial(n+1,k) * PS(n-k+1, u_1=1, u_2, ..., u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).
For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 105 u2^4 - 2 * 210 u2^2 u3 + 1 * 90 u2 u4 + 1 * 40 u3^2 - 0 * -24 u5 = 315 u2^4 - 420 u2^2 u3 + 90 u2 u4 + 40 u3^2 = (1/2) [2 * 6!/(4!*2!) * PS(2,1,u2) * PS(4,1,u2,...,u4) + 6!/(3!*3!) * PS(3,1,u2,u3)^2] = (1/2) * [ 2 * 6!/(4!*2!) * (-u2) (-15 u2^3 + 20 u2 u3 - 6 u4) + 6!/(3!*3!) * (3 u2^2 - 2 u3)^2].
Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.
(End)
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the refined Stirling polynomials of the first kind A036039 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

Extensions

Terms ordered according to the reversed Abramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by Andrey Zabolotskiy, Mar 08 2024

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;

Links

Crossrefs

Cf. A132382, A110330, A132159, A094587, A132013, A235706.

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

A136214 Triangle U, read by rows, where U(n,k) = Product_{j=k..n-1} (3*j+1) for n > k with U(n,n) = 1.

Original entry on oeis.org

1, 1, 1, 4, 4, 1, 28, 28, 7, 1, 280, 280, 70, 10, 1, 3640, 3640, 910, 130, 13, 1, 58240, 58240, 14560, 2080, 208, 16, 1, 1106560, 1106560, 276640, 39520, 3952, 304, 19, 1, 24344320, 24344320, 6086080, 869440, 86944, 6688, 418, 22, 1
Offset: 0

Views

Author

Paul D. Hanna, Feb 07 2008

Keywords

Comments

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) = A094587(n, k), T(n, k, 2) = A112292(n, k) and T(n, k, 3) is this sequence. - Peter Luschny, Jun 01 2009, revised Jun 18 2019

Examples

			Triangle begins:
        1;
        1,       1;
        4,       4,      1;
       28,      28,      7,     1;
      280,     280,     70,    10,    1;
     3640,    3640,    910,   130,   13,   1;
    58240,   58240,  14560,  2080,  208,  16,  1;
  1106560, 1106560, 276640, 39520, 3952, 304, 19, 1; ...
Matrix inverse begins:
   1;
  -1,   1;
   0,  -4,   1;
   0,   0,  -7,   1;
   0,   0,   0, -10,   1;
   0,   0,   0,   0, -13,   1; ...

Links

Crossrefs

Cf. A094587, A112333, A136216, A136239; A007559, A136212, A136213.

Programs

  • Magma
    [[n eq 0 select 1 else k eq n select 1 else (&*[3*j+1: j in [k..n-1]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Jun 14 2019
  • Maple
    nmax:=8; for n from 0 to nmax do U(n, n):=1 od: for n from 0 to nmax do for k from 0 to n do if n > k then U(n, k) := mul((3*j+1), j = k..n-1) fi: od: od: for n from 0 to nmax do seq(U(n, k), k=0..n) od: seq(seq(U(n, k), k=0..n), n=0..nmax); # Johannes W. Meijer, Jul 04 2011, revised Nov 23 2012
  • Mathematica
    Table[Product[3*j+1, {j,k,n-1}], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 14 2019 *)
  • PARI
    T(n,k)=if(n==k,1,prod(j=k,n-1,3*j+1))
  • Sage
    def T(n, k):
    if (k==n): return 1
    else: return product(3*j+1 for j in (k..n-1))
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jun 14 2019

Formula

Matrix powers: column 0 of U^(k+1) = column k of A136216 for k >= 0; simultaneously, column k = column 0 of A136216^(3k+1) for k >= 0. Element in column 0, row n, of matrix power U^(k+1) = A007559(n)*C(n+k,k), where A007559 are triple factorials found in column 0 of this triangle.

A094588 a(n) = n*F(n-1) + F(n), where F = A000045.

Original entry on oeis.org

0, 1, 3, 5, 11, 20, 38, 69, 125, 223, 395, 694, 1212, 2105, 3639, 6265, 10747, 18376, 31330, 53277, 90385, 153011, 258523, 436010, 734136, 1234225, 2072043, 3474029, 5817515, 9730748, 16258910, 27139509, 45258917, 75408775, 125538539
Offset: 0

Views

Author

Paul Barry, May 13 2004

Keywords

Comments

This is the transform of the Fibonacci numbers under the inverse of the signed permutations matrix (see A094587).

Links

Crossrefs

Cf. A000045, A007502, A045925, A088209, A354044.

Programs

  • Haskell
    a094588 n = a094588_list !! n
a094588_list = 0 : zipWith (+) (tail a000045_list)
                               (zipWith (*) [1..] a000045_list)
-- Reinhard Zumkeller, Mar 04 2012
  • Julia
    # The function 'fibrec' is defined in A354044.
function A094588(n)
    n == 0 && return BigInt(0)
    a, b = fibrec(n - 1)
    a*n + b
end
println([A094588(n) for n in 0:34]) # Peter Luschny, May 16 2022
  • Magma
    [n*Fibonacci(n-1)+Fibonacci(n): n in [0..60]]; // Vincenzo Librandi, Apr 23 2011
  • Mathematica
    CoefficientList[Series[x (1+x-2x^2)/(1-x-x^2)^2,{x,0,40}],x]  (* Harvey P. Dale, Apr 16 2011 *)
  • PARI
    Vec((1+x-2*x^2)/(1-x-x^2)^2+O(x^99)) \\ Charles R Greathouse IV, Mar 04 2012

Formula

G.f. : x*(1 + x - 2*x^2)/(1 - x - x^2)^2.
a(n) = A101220(n, 0, n) - Ross La Haye, Jan 28 2005
a(n) = A109754(n, n). - Ross La Haye, Aug 20 2005
a(n) = (sin(c*n)*i - n*sin(c*(n - 1)))/(i^n*sqrt(5/4)) where c = arccos(i/2). - Peter Luschny, May 16 2022

A112292 An invertible triangle of ratios of double factorials.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 15, 15, 5, 1, 105, 105, 35, 7, 1, 945, 945, 315, 63, 9, 1, 10395, 10395, 3465, 693, 99, 11, 1, 135135, 135135, 45045, 9009, 1287, 143, 13, 1, 2027025, 2027025, 675675, 135135, 19305, 2145, 195, 15, 1, 34459425, 34459425, 11486475, 2297295, 328185, 36465, 3315, 255, 17, 1
Offset: 0

Views

Author

Paul Barry, Sep 01 2005

Keywords

Comments

As a square array read by antidiagonals, column k has e.g.f. (1/(1-2x)^(1/2))*(1/(1-2x))^k. - Paul Barry, Sep 04 2005
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) = A094587(n, k), T(n, k, 2) is this sequence and T(n, k, 3) = A136214. - Peter Luschny, Jun 01 2009, revised Jun 18 2019

Examples

			Triangle begins
      1;
      1,     1;
      3,     3,    1;
     15,    15,    5,  1;
    105,   105,   35,  7,  1;
    945,   945,  315, 63,  9,  1;
  10395, 10395, 3465,693, 99, 11, 1;
Inverse is A112295, which begins
   1;
  -1,  1;
   0, -3,  1;
   0,  0, -5,  1;
   0,  0,  0, -7,  1;
   0,  0,  0,  0, -9,  1;
Similar results arise for higher factorials.

Crossrefs

Columns include A001147, A051577, A051579.
Row sums are A112293.
Diagonal sums are A112294.
Cf. A094587 (p=1), this sequence (p=2), A136214 (p=3).

Programs

  • Mathematica
    T[n_, k_] := If[k <= n, (2n-1)!!/(2k-1)!!, 0];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 13 2019 *)

Formula

T(n, k)=if(k<=n, (2n-1)!!/(2k-1)!!, 0);
T(n, k)=if(k<=n, n!*C(2n, n)2^(k-n)/(k!*C(2k, k)), 0);
T(n, k)=if(k<=n, 2^(n-k)(n-1/2)!/(k-1/2)!, 0);
T(n, k)=if(k<=n, (n+1)!*C(n)2^(k-n)/((k+1)!*C(k)), 0).
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