login
Expansion of e.g.f.: (1+x)/(1-x*exp(x)).
2

%I #14 Feb 27 2024 03:04:53

%S 1,2,6,33,232,2045,21636,267043,3766848,59776137,1053986500,

%T 20442543671,432537117552,9914571191005,244742376434388,

%U 6473030199939675,182614314495736576,5473825396372806545,173728330742517310980

%N Expansion of e.g.f.: (1+x)/(1-x*exp(x)).

%C The n-th term of the n-th inverse binomial transform of this sequence equals n! for n>=0.

%F E.g.f. (1+x)/(1-x*exp(x)) - _Olivier Gérard_, Sep 15 2016

%F O.g.f. satisfies: [x^n] A( x/(1+n*x) )/(1+n*x) = n! for n>=0.

%F E.g.f. satisfies: [x^n] A(x)*exp(-n*x) = 1 for n>=0.

%F a(n) = n! + Sum_{k=0..n-1} k!*C(n,k)*n*k^(n-k-1) for n>1 with a(0)=1.

%F Equivalent to the sum above by properties of the binomial triangle:

%F a(n) = A006153(n)+n*A006153(n-1).

%F a(n) = n! ( Sum_{k=0..n-1} ((n-1-k)^k + (n-k)^k)/k!) for n>1 with a(0)=1.

%F a(n) ~ n! / LambertW(1)^n. - _Vaclav Kotesovec_, Oct 30 2017

%e If the successive inverse binomial transforms are placed in a table,

%e then we see that the diagonal consists of the factorials:

%e n=0:[(1),2,6,33,232,2045,21636,267043,3766848,59776137,1053986500,...];

%e n=1:[1,(1),3,20,129,1164,12265,151458,2136337,33901640,597761361,...];

%e n=2:[1,0,(2),13,64,693,6856,86175,1210896,19228825,339012304,...];

%e n=3:[1,-1,3,(6),25,482,3429,50908,678465,10937430,192150469,...];

%e n=4:[1,-2,6,-7,(24),381,844,36291,341728,6433865,107801436,...];

%e n=5:[1,-3,11,-32,97,(120),-839,37158,55953,4638052,54573025,...];

%e n=6:[1,-4,18,-75,304,-811,(720),40783,-262608,5542425,6069736,...];

%e n=7:[1,-5,27,-142,729,-3282,11941,(5040),-497279,9166130,...];

%e n=8:[1,-6,38,-239,1480,-8643,45844,-178557,(40320),12301705,...];

%e n=9:[1,-7,51,-372,2689,-18844,125289,-741974,3354513,(362880),...].

%t With[{nn=20},CoefficientList[Series[(1+x)/(1-x*Exp[x]),{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, Jun 09 2018 *)

%o (PARI) {a(n)=local(A=[1]);for(k=1,n,A=concat(A,0); A[k+1]=k!-polcoeff(subst(Ser(A),x,x/(1+k*x+x*O(x^k)))/(1+k*x),k));A[n+1]}

%o (PARI) {a(n)=n!+sum(k=0,n-1,k!*binomial(n,k)*n*k^(n-k-1))}

%Y Cf. A006153.

%Y Cf. A138910 (inverse binomial transform); variants: A138911, A138737.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Apr 05 2008, Apr 06 2008

%E Name change and e.g.f. by _Olivier Gérard_, Sep 15 2016