

A134685


Irregular triangle read by rows: coefficients C(j,k) of a partition transform for direct Lagrange inversion.


19



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
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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 = (n1)! * 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 ((m1)') for (m') in each partition and ignoring the (0') factors, the partitions in the brackets of P(n,t) become those of n1 listed in Abramowitz and Stegun on page 831 and the number of partitions in P(n,t) is given by A000041(n1).
Combinatorial interpretations are given in the link.
From Tom Copeland, Jul 10 2018: (Start)
Coefficients occurring in prolongation for the special Euclidean group SE(2) and special affine group SA(2) in the Olver presentation on moving frames (MFP) in slides 33 and 42. These are a result of applying an iterated derivative of the form h(x)d/dx = d/dy as in this entry (more generally as g(x) d/dx as discussed in A145271). See also p. 6 of Olver's paper on contact forms, but note that the 12 should be a 15 in the formula for the compositional inverse of S(t).
Change variables in the MFP to obtain connections to the partition polynomials Prt_n = n! * P(n,1) above. Let delta and beta in the formulas for the equiaffine curves in MFP be L and B, respectively, and D_y = (1/(LB*u_x)) d/dx = (1/w_x) d/dx. Then v_(yy) = (1/B) [w_(xx)/(w_x)^3] in MFP (there is an overall sign error in MFP for v_(yy) and higher derivatives w.r.t. y), and (d/dy)^n v = v_n = (1/B)* [(1/w_1)*d/dx)^(n2) [w_2/(w_1)^3] for n > 1, with w_n = (d/dx)^n w. Consequently, in the partition polynomials Prt_n for n > 1 here substitute (n') = B*u_n = w_n for n > 1 and (1') = LB*u_1 = w_1, where u_n = (d/dx)^n u, and then divide by B. For example, v_4 = (1/B)*Prt_4 = (1/B)*4!*P(4,1) = (1/B) (LB*u_n)^(7) [15*(B*u_2)^3 + 10 (LB*u_1)(B*u_2)(B*u_3)  (LB*u_1)^2 (B*u_4)], agreeing with v_4 in MFP except for the overall sign.
For the SE(2) transformation formulas in MFP, let w_x = cos(phi) + sin(phi)*u_x, and then the same transformations apply as above with cos(phi) and sin(phi) substituted for L and B, respectively. (End)


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].
O. Arnaldsson, Élie Cartan’s Theory of Moving Frames, slide presentation, 2014.
T. Copeland, Generators, Inversion, and Matrix, Binomial, and Integral Transforms, 2015.
T. Copeland, Important formulas in combinatorics, MathOverflow answer, 2015.
T. Copeland, Compositional inversion and generating functions in algebraic geometry, MathOverflow question, 2014.
T. Copeland, Compositional inverse pairs, the BurgersHopf equation, and the Stasheff associahedra,
T. Copeland, Lagrange a la Lah, 2011.
T. Copeland, Short Note on Lagrange Inversion, 2008.
T. Copeland, Formal group laws and binomial Sheffer sequences, 2018.
E. Getzler, The semiclassical approximation for modular operads, arXiv:alggeom/9612005, 1996 (see p. 2).
P. Olver, The canonical contact form.
P. Olver, Moving Frames, slide presentation, 2009.
J. Taylor, Formal group laws and hypergraph colorings, doctoral thesis, Univ. of Wash., 2016, p. 95.


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)^(n1+e(1)) * [2*(n1)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(n1,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)^(n1) 1/[u_1 + u_2 * x/2! + u_3 * x^2/3! + ... + u_n * x^(n1)/n!]^n evaluated at x=0.  Tom Copeland, Jul 07 2015
Equivalent matrix computation: Multiply the mth 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+1k}. Then P(n,t) = (1, 0, 0, 0, ...) [UP^(1) * S]^(n1) 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 squarebracketed 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,..., (K1) * bK, ...) = (1/2) * Sum_{k = 2..n1} binomial(n+1,k) * PS(nk+1,u_1=1,u_2,...,u_(nk+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,...,(K1)*bK,...) = d/dt t^(n2)*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 Bell polynomials of A036040 is presented in the blog entry "Formal group laws and binomial Sheffer sequences."  Tom Copeland, Feb 06 2018


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)2t1
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 = (n1)! for n>1,
then f(t) = log(1t)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 (n2)! 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(1t/2)
and g(t) = [e.g.f. of (0,A001147)] = 1  (12t)^(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(1t)
and g(t) = t * [o.g.f. of A000108] = [1  (14t)^(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.
Cf. A036040.
Sequence in context: A318392 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



