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%I #67 Nov 23 2020 06:04:26
%S 1,1,2,1,9,6,1,28,72,24,1,75,500,600,120,1,186,2700,7800,5400,720,1,
%T 441,12642,73500,117600,52920,5040,1,1016,54096,571536,1764000,
%U 1787520,564480,40320,1,2295,217800,3916080,21019824,40007520,27941760,6531840,362880,1,5110,839700,24555600,214326000
%N Triangle read by rows: number of nilpotent partial transformations (of an n-element set) of height r (height(alpha) = |Im(alpha)|), 0 <= r < n.
%C The sum of each row of the sequence (as a triangular array) is A000272. Second left-downward diagonal is A058877.
%C From _Tom Copeland_, Oct 26 2014: (Start)
%C With T(x,t) the e.g.f. for A055302 for the number of labeled rooted trees with n nodes and k leaves, the mirror of the row polynomials of this array are given by e^T(x,t) = exp[ t * x + (2t) * x^2/2! + (6t + 3t^2) * x^3/3! + ...] = 1 + t * x + (2t + t^2) * x^2/2! + (6t + 9t^2 + t^3) * x^3/3! + ... = 1 + Nr(x,t).
%C Equivalently, e^x-1 = Nr[Tinv(x,t),t] = t * N[t*Tinv(x,t),1/t], where N(x,t) is the e.g.f. of this array and Tinv(x,t) is the comp. inverse in x of T(x,t). Note that Nr(x,t) = t * N(x*t,1/t), and N(x,t) = t * Nr(x*t,1/t). Also, log[1 + Nr(x,t)]= x * [t + Nr(x,t)] = T(x,t).
%C E.g.f. is N(x,t)= t * {exp[T(x*t,1/t)] - 1}, and log[1 + N(x,t)/t] = T(x*t,1/t) = x + (2t) * x^2/2! + (3t + 6t^2) * x^3/3! + (4t + 36t^2 + 24t^3) * x^4/4! + ... = x + (t) * x^2 + (t + 2t^2) * x^3/2! + (t + 9t^2 + 6t^3) * x^4/3! + ... is the comp. inverse in x of x / [1 + t * (e^x - 1)].
%C The exp/log transforms (A036040/A127671) generally give associations between enumerations of sets of connected graphs/objects (in this case, trees) and sets of disconnected (or not necessarily connected) graphs/objects (in this case, bipartite graphs of the nilpotent transformations). The transforms also relate formal cumulants and moments so that Nr(x,t) is then the e.g.f. for the formal moments associated to the formal cumulants whose e.g.f. is T(x,t). (End)
%H A. Laradji and A. Umar, <a href="http://dx.doi.org/10.1081/AGB-120038637">On the number of nilpotents in the partial symmetric semigroup</a>, Comm. Algebra 32 (2004), 3017-3023.
%H A. Laradji and A. Umar, <a href="https://mathfiles.kfupm.edu.sa/data/files/mathonly/TechnicalReportsData/305.pdf">On the number of nilpotents in the partial symmetric semigroup</a>, Tech. Report TR305, King Fahd Univ. of Petroleum and Minerals, (2003).
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Cumulant#Cumulants_of_some_discrete_probability_distributions">Cumulant</a>
%F N(J(n,r)) = C(n,r)*S(n,r+1)*r! where S(n, r + 1) is a Stirling number of the second kind (given by A048993 with zeros removed); generating function = (x+1)^(n-1).
%F From _Peter Bala_, Oct 22 2008: (Start)
%F Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim_{n -> infinity} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).
%F Let f(x) = 1 + a*x + a*x^2/2! + a*x^3/3! + ... . Then the e.g.f. for this table is I(f(x)) = 1 + a*x +(a + 2*a^2)*x^2/2! + (a + 9*a^2 + 6*a^3)*x^3/3! + (a + 28*a^2 + 72*a^3 + 24*a^4)*x^4/4! + ... . Note, if we take f(x) = 1 + a*x + a*x^2 + a*x^3 + ... then I(f(x)) is the o.g.f. of the Narayana triangle A001263. (End)
%F A generator for this array is given by the inverse, g(x,t), of f(x,t)= x/(1 + t * (e^x-1)). Then A248927 gives h(x,t)= x / f(x,t) = 1 + t*(e^x-1)= 1 + t * (x + x^2/2! + x^3/3! + ...) and g(x,t)= x * (1 + t * x + (t + 2 t^2) * x^2/2! + (t + 9 t^2 + 6 t^3) * x^3/3! + ...), so by Bala's arguments A248927 is a refinement of A141618 with row sums A000272. The connection to Narayana numbers is reflected in the relation between A248927 and A134264. See A145271 for more relations that g(x,t) and f(x,t) must satisfy. - _Tom Copeland_, Oct 17 2014
%F T(n,k) = C(n,k-1) * A028246(n,k) = C(n,k-1) * A019538(n,k)/k = A055302(n+1,n+1-k) / (n+1). - _Tom Copeland_, Oct 25 2014
%F E.g.f. is the series reversion of log(1 + x)/(1 + t*x) with respect to x. Cf. A198204. - _Peter Bala_, Oct 21 2015
%e N(J(4,2)) = 6*6*2 = 72.
%e From _Peter Bala_, Oct 22 2008: (Start)
%e Triangle begins
%e n\k|..0.....1.....2.....3.....4....5
%e =====================================
%e .1.|..1
%e .2.|..1.....2
%e .3.|..1.....9.....6
%e .4.|..1....28....72....24
%e .5.|..1....75...500...600...120
%e .6.|..1...186..2700..7800..5400...720
%e ...
%e (End)
%p A048993 := proc(n,k)
%p combinat[stirling2](n,k) ;
%p end proc:
%p A141618 := proc(n,k)
%p binomial(n,k)*k!*A048993(n,k+1) ;
%p end proc:
%t Flatten[CoefficientList[CoefficientList[InverseSeries[Series[Log[1 + x]/(1 + t*x),{x,0,9}]],x]*Table[n!, {n,0,9}],t]] (* _Peter Luschny_, Oct 24 2015, after _Peter Bala_ *)
%o (PARI)
%o A055302(n,k)=n!/k!*stirling(n-1, n-k,2);
%o T(n,k)=A055302(n+1,n+1-k) / (n+1);
%o for(n=1,10,for(k=1,n,print1(T(n,k),", "));print());
%o \\ _Joerg Arndt_, Oct 27 2014
%Y Cf. A000272, A007318, A036040, A048993, A055302, A058877, A127671, A198204.
%K nonn,tabl
%O 1,3
%A _Abdullahi Umar_, Aug 23 2008
%E More terms from _Joerg Arndt_, Oct 27 2014
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