A305133 - OEIS

%I #21 Aug 15 2022 13:59:46

%S 1,1,3,16,113,996,10537,130054,1834513,29111896,513307601,9955832514,

%T 210652214665,4828548335092,119193293536969,3152465052989326,

%U 88935973854834593,2665836978234855984,84608363388300429601,2834484567764492239354,99956558270008377397081,3701159405682998540166796,143571313108884280622221913,5822409005523822986360056326

%N E.g.f.: (1-x) / (exp(-x) - x).

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

%F a(n) ~ n! * (1 - LambertW(1)) / ((1 + LambertW(1)) * LambertW(1)^(n+1)). - _Vaclav Kotesovec_, Jun 16 2018

%F a(n) = 1 + n * Sum_{k=1..n-1} binomial(n-1,k) * a(k). - _Ilya Gutkovskiy_, Aug 08 2020

%e E.g.f.: A(x) = 1 + x + 3*x^2/2! + 16*x^3/3! + 113*x^4/4! + 996*x^5/5! + 10537*x^6/6! + 130054*x^7/7! + 1834513*x^8/8! + 29111896*x^9/9! + ...

%e RELATED TABLE.

%e The table of coefficients of x^k in exp(n*x) * A(x) begins:

%e n=0: [1, (1), 3/2, 8/3, 113/24, 83/10, 10537/720, 65027/2520, ...];

%e n=1: [(1), 2, (3), 29/6, 25/3, 1757/120, 929/36, 45863/1008, ...];

%e n=2: [1, (3), 11/2, (9), 361/24, 1559/60, 729/16, 101107/1260, ...];

%e n=3: [1, 4, (9), 97/6, (82/3), 1863/40, 3637/45, 714319/5040, ...];

%e n=4: [1, 5, 27/2, (82/3), 1169/24, (251/3), 103801/720, 632897/2520, ...];

%e n=5: [1, 6, 19, 87/2, (251/3), 17821/120, (5147/20), 2250499/5040, ...];

%e n=6: [1, 7, 51/2, 197/3, 3305/24, (5147/20), 65633/144, (14293/18), ...];

%e n=7: [1, 8, 33, 569/6, 652/3, 51893/120, (14293/18), 7078303/5040, ...]; ...

%e in which terms along the diagonals (enclosed in parenthesis) are equal:

%e [x^n] exp((n+1)*x) * A(x) = [x^(n+1)] exp(n*x) * A(x) for n >= 0.

%t With[{nn=30},CoefficientList[Series[(1-x)/(Exp[-x]-x),{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, Aug 15 2022 *)

%o (PARI) {a(n) = n!*polcoeff( (1-x) / (exp(-x +x*O(x^n)) - x), n)}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A072597, A305990.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jun 15 2018