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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.
22
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
1,3
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
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
EXAMPLE
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
-----------------------------------------------------------------------
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 *)
KEYWORD
sign,tabl
AUTHOR
Tom Copeland, Feb 25 2014
EXTENSIONS
Pincherle formalism added by Tom Copeland, Feb 27 2014
STATUS
approved