0,2

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

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

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

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

then we see that the diagonal consists of the factorials:

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

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

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

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

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

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

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

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

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

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

(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]}

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

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

Sequence in context: A018952 A019028 A127114 this_sequence A138983 A121774 A053042

Adjacent sequences: A138906 A138907 A138908 this_sequence A138910 A138911 A138912

nonn

Paul D. Hanna (pauldhanna(AT)juno.com), Apr 05 2008, Apr 06 2008