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A138911 The n-th term of n-th inverse binomial transform of this sequence is 1 for n>=0.

%I A138911 #6 Jul 27 2025 18:53:54
%S A138911 1,2,5,19,81,401,2233,13721,91969,666145,5174001,42827225,375850225,
%T A138911 3481908353,33923685097,346468507201,3698848172289,41173895716289,
%U A138911 476826776487649,5733422156706473,71449460236945201,921311262658989217,12274120192571550105,168717589565046739409
%N A138911 The n-th term of n-th inverse binomial transform of this sequence is 1 for n>=0.
%H A138911 Alois P. Heinz, <a href="/A138911/b138911.txt">Table of n, a(n) for n = 0..541</a>
%F A138911 O.g.f. satisfies: [x^n] A( x/(1+n*x) )/(1+n*x) = 1 for n>=0.
%F A138911 E.g.f. satisfies: [x^n] A(x)*exp(-n*x) = 1/n! for n>=0.
%F A138911 a(n) = 1 + Sum_{k=0..n-1} C(n,k)*n*k^(n-k-1) for n>0 with a(0)=1.
%e A138911 If the successive inverse binomial transforms are placed in a table,
%e A138911 then we see that the diagonal consists of all 1's:
%e A138911 n=0:[(1), 2, 5, 19, 81, 401, 2233, 13721, 91969, 666145,...];
%e A138911 n=1:[1, (1), 2, 9, 28, 145, 726, 4249, 27000, 186561,...];
%e A138911 n=2:[1, 0, (1), 5, 1, 79, 121, 1511, 6721, 50975,...];
%e A138911 n=3:[1, -1, 2, (1), -12, 113, -422, 2441, -6584, 44929,...];
%e A138911 n=4:[1, -2, 5, -9, (1), 157, -1263, 8173, -45087, 243261,...];
%e A138911 n=5:[1, -3, 10, -31, 76, (1), -1922, 19841, -153896, 1068065,...];
%e A138911 n=6:[1, -4, 17, -71, 273, -805, (1), 29339, -359135, 3316411,...];
%e A138911 n=7:[1, -5, 26, -135, 676, -3071, 10626, (1), -525144, 7462305,...];
%e A138911 n=8:[1, -6, 37, -229, 1393, -8087, 42313, -167839, (1), 10811609,...];
%e A138911 n=9:[1, -7, 50, -359, 2556, -17743, 116902, -691447, 3091768, (1),...].
%o A138911 (PARI) {a(n)=local(A=[1]);for(k=1,n,A=concat(A,0); A[k+1]=1-polcoeff(subst(Ser(A),x,x/(1+k*x+x*O(x^k)))/(1+k*x),k));A[n+1]}
%o A138911 (PARI) {a(n)=1 + sum(k=0,n-1,binomial(n,k)*n*k^(n-k-1) )}
%Y A138911 Cf. A138912 (inverse binomial transform).
%K A138911 nonn
%O A138911 0,2
%A A138911 _Paul D. Hanna_, Apr 05 2008, Apr 06 2008