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A138737 The n-th term of the n-th inverse binomial transform of this sequence equals (n+1)^(n-1) for n>=0.

%I A138737 #6 Oct 30 2017 08:17:14
%S A138737 1,2,7,52,541,7446,127939,2641192,63746169,1762380010,54938528191,
%T A138737 1906911695580,72949449568021,3049813346508670,138352912908850683,
%U A138737 6769028553912294736,355311287187804226033,19918243846821103623378
%N A138737 The n-th term of the n-th inverse binomial transform of this sequence equals (n+1)^(n-1) for n>=0.
%C A138737 Related to LambertW(-x)/(-x) = Sum_{n>=0} (n+1)^(n-1)*x^n/n!.
%H A138737 Vaclav Kotesovec, <a href="/A138737/b138737.txt">Table of n, a(n) for n = 0..300</a>
%F A138737 O.g.f. satisfies: [x^n] A( x/(1+n*x) )/(1+n*x) = (n+1)^(n-1) for n>=0.
%F A138737 E.g.f. satisfies: [x^n] A(x)*exp(-n*x) = (n+1)^(n-1)/n! for n>=0.
%F A138737 a(n) ~ (1 + LambertW(exp(-1)))^(3/2) * n^(n-1) / (exp(n-1) * LambertW(exp(-1))^n). - _Vaclav Kotesovec_, Oct 30 2017
%e A138737 If the successive inverse binomial transforms are placed in a table,
%e A138737 then we see that the diagonal consists of terms (n+1)^(n-1):
%e A138737 n=0:[(1),2,7,52,541,7446,127939,2641192,63746169,1762380010,...];
%e A138737 n=1:[1,(1),4,36,368,5200,90432,1884736,45817088,1273874688,...];
%e A138737 n=2:[1, 0,(3),26,245,3684,64087,1349214,33003945,922386824,...];
%e A138737 n=3:[1,-1, 4,(16),160,2688,45184,970240,23814144,668975104,...];
%e A138737 n=4:[1,-2,7, 0,(125),2002,31203,705268,17177273,486100710,...];
%e A138737 n=5:[1,-3,12,-28, 176,(1296),21184,524352,12305664,354510080,...];
%e A138737 n=6:[1,-4,19,-74,373, 0,(16807),395866,8645673,260994628,...];
%e A138737 n=7:[1,-5,28,-144,800,-2816, 24192,(262144),5980160,195969024,...];
%e A138737 n=8:[1,-6,39,-244,1565,-8562,56419, 0,(4782969),149083874,...];
%e A138737 n=9:[1,-7,52,-380,2800,-19248,136768,-638912, 6966528,(100000000),..];
%e A138737 n=10:[1,-8,67,-558,4661,-37604,302679,-2112938,17204009, 0,...].
%e A138737 Notice the occurrence of zeros in the secondary diagonal = A138734.
%o A138737 (PARI) {a(n)=local(A=[1]);for(k=1,n,A=concat(A,0); A[k+1]=(k+1)^(k-1)-Vec(subst(Ser(A),x,x/(1+k*x+x*O(x^k)))/(1+k*x))[k+1]);A[n+1]}
%Y A138737 Cf. A138736 (inverse binomial transform), A138734; variants: A138909, A138911.
%K A138737 nonn
%O A138737 0,2
%A A138737 _Paul D. Hanna_, Apr 05 2008