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

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

0,5

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

Links

Table of n, a(n) for n=0..54.

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).

Example

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).

Crossrefs

Sequence in context: A136018 A138022 A113278 * A366627 A325495 A048865

Adjacent sequences: A132379 A132380 A132381 * A132383 A132384 A132385

Keyword

sign,tabl

Author

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

Extensions

More terms from Tom Copeland, Dec 05 2007

Status

approved