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A134685 Irregular triangle read by rows: coefficients C(j,k) of a partition transform for direct Lagrange inversion . 17
1, -1, 3, -1, -15, 10, -1, 105, -105, 15, 10, -1, -945, 1260, -280, -210, 21, 35, -1, 10395, -17325, 6300, 3150, -1260, -378, -280, 28, 56, 35, -1, -135135, 270270, -138600, -51975, 15400, 34650, 6930, -1575, -2520, -630, -2100, 36, 84, 126, -1 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

Let f(t) = u(t) - u(0) = Ev[exp(u.* t) - u(0)] = log{Ev[(exp(z.* t))/z_0]} = Ev[-log(1- a.* t)], where the operator Ev denotes umbral evaluation of the umbral variables u., z. or a., e.g., Ev[a.^n + a.^m] = a_n + a_m . The relation between z_n and u_n is given in reference in A127671 and u_n = (n-1)! * a_n .

If u_1 is not equal to 0, then the compositional inverse for these expressions is given by g(t) = sum(j=1,...) P(j,t) where, with u_n denoted by (n') for brevity,

P(1,t) = (1')^(-1) * [ 1 ] * t

P(2,t) = (1')^(-3) * [ -(2') ] * t^2 / 2!

P(3,t) = (1')^(-5) * [ 3 (2')^2 - (1')(3') ] * t^3 / 3!

P(4,t) = (1')^(-7) * [ -15 (2')^3 + 10 (1')(2')(3') - (1')^2 (4') ] * t^4 / 4!

P(5,t) = (1')^(-9) * [ 105 (2')^4 - 105 (1') (2')^2 (3') + 15 (1')^2 (2') (4') + 10 (1')^2 (3')^2 - (1')^3 (5') ] * t^5 / 5!

P(6,t) = (1')^(-11) * [ -945 (2')^5 + 1260 (1') (2')^3 (3') - 280 (1')^2 (2') (3')^2 - 210 (1')^2 (2')^2 (4') + 21 (1')^3 (2')(5') + 35 (1')^3 (3')(4') - (1')^4 (6') ] * t^6 / 6!

P(7,t) = (1')^(-13) * [ 10395 (2')^6 - 17325 (1') (2')^4 (3') +  (1')^2 [ 6300 (2')^2 (3')^2 + 3150  (2')^3 (4')] - (1')^3 [1260  (2')(3')(4') + 378  (2')^2(5') + 280 (3')^3] + (1')^4 [28 (2')(6') + 56 (3')(5') + 35 (4')^2]  - (1')^5 (7') ] * t^7 / 7!

P(8,t) = (1')^(-15) * [ -135135 (2')^7 + 270270 (1') (2')^5 (3') -  (1')^2 [ 138600 (2')^3 (3')^2 +  51975 (2')^4 (4')] + (1')^3 [15400 (2')(3')^2 + 34650  (2')^2(3')(4') + 6930  (2')^3(5')] - (1')^4 [1575 (2')(4')^2 + 2520 (2')(3')(5') + 630 (2')^2(6') + 2100 (3')^2(4') ]  + (1')^5 [36 (2')(7') + 84 (3')6') + 126 (4')(5')] - (1')^6 (8') ] * t^8 / 8!

...

Substituting ((m-1)') for (m') in each partition and ignoring the (0') factors, the partitions in the brackets of P(n,t) become those of n-1 listed in Abramowitz and Stegun on page 831 and the number of partitions in P(n,t) is given by A000041(n-1).

Combinatorial interpretations are given in the link.

REFERENCES

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 831.

LINKS

Table of n, a(n) for n=1..45.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

T. Copeland, Generators, Inversion, and Matrix, Binomial, and Integral Transforms

T. Copeland, Compositional inverse pairs, the Burgers-Hopf equation, and the Stasheff associahedra,

T. Copeland, Lagrange a la Lah

T. Copeland, Short Note on Lagrange Inversion, (posted Sept 2008)

E. Getzler, The semi-classical approximation for modular operads, arXiv:alg-geom/9612005, 1996 (see pg. 2).

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 05 2011: (Start)

Let h(t) = 1/(df(t)/dt)

  = 1/Ev[u.*exp(u.*t)]

  = 1/(u_1+(u_2)*t+(u_3)*t^2/2!+(u_4)*t^3/3!+...),

  an e.g.f. for the partition polynomials of A133314

  (signed A049019) with an index shift.

  Then for the partition polynomials of A134685,

  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)). (Cf. A000311 and A134991)(End)

From Tom Copeland, Oct 30 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)^2/2!+...],  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

Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix by u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) u_{n+1-k}. 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. - Tom Copeland, Jul 15 2016

Also, P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. - Tom Copeland, Aug 27 2016

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 = -u5, b2 = 15 u2 u4 + 10 u3^2, b3 = -105 u2^2 u3,  and b4 = 105 u2^4.

The relation between solutions of the inviscid Burgers' 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 * 105 u2^2 u3 + 1 * 15 u2 u4 + 1 * 10 u3^2 - 0 * u5 = 315 u2^4 - 210 u2^2 u3 + 15 u2 u4 + 10 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 + 10 u2 u3 - u4) + 6!/(3!*3!) * (3 u2^2 - 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)

EXAMPLE

Examples and checks:

1) Let u_1 = -1 and u_n = 1 for n>1,

then f(t) = exp(u.*t) - u(0) = exp(t)-2t-1

and g(t) = [e.g.f. of signed A000311];

therefore, the row sums of unsigned [C(j,k)] are A000311 =

(0,1,1,4,26,236,2752,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),4!*P(4,1),...).

2) Let u_1 = -1 and u_n = (n-1)! for n>1,

then f(t) = -log(1-t)-2t

and g(t) = [e.g.f. of signed (0,A032188)]

with (0,A032188) = (0,1,1,5,41,469,6889,...) = (0,-P(1,1),2!*P(2,1),-3!P(3,1),...).

3) Let u_1 = -1 and u_n = (-1)^n (n-2)! for n>1, then

f(t) = (1+t) log(1+t) - 2t

and g(t) = [e.g.f. of signed (0,A074059)]

with (0,A074059) = (0,1,1,2,7,34,213,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),...).

4) Let u_1 = 1, u_2 = -1 and u_n = 0 for n>2,

then f(t) = t(1-t/2)

and g(t) = [e.g.f. of (0,A001147)] = 1 - (1-2t)^(1/2)

with (0,A001147) = (0,1,1,3,15,105,945...) =(0,P(1,1),2!*P(2,1),3!*P(3,1),...).

5) Let u_1 = 1, u_2 = -2 and u_n = 0 for n>2,

then f(t)= t(1-t)

and g(t) = t * [o.g.f. of A000108] = [1 - (1-4t)^(1/2)] / 2

with (0,A000108) = (0,1,1,2,5,14,42,...) = (0,P(1,1),P(2,1),P(3,1),...).

CROSSREFS

Cf. A145271, (A134991, A019538) = (reduced array, associated g(x)).

Cf. A000108, A000311, A001147, A032188, A049019, A074059, A129185, A133314, A139605.

Sequence in context: A039815 A147453 A147020 * A130757 A181996 A014621

Adjacent sequences:  A134682 A134683 A134684 * A134686 A134687 A134688

KEYWORD

sign,tabf,more

AUTHOR

Tom Copeland, Jan 26 2008, Sep 13 2008

EXTENSIONS

P(7,t) and P(8,t) data added by Tom Copeland, Jan 14 2016

STATUS

approved

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Last modified December 6 11:07 EST 2016. Contains 278776 sequences.