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A155456 Write (1+1/x)*log(1+x) = Sum c(n)*x^n; then a(n) = (n+1)!*c(n).

%I A155456 #19 Aug 09 2022 14:17:25
%S A155456 -1,-1,1,-2,6,-24,120,-720,5040,-40320,362880,-3628800,39916800,
%T A155456 -479001600,6227020800,-87178291200,1307674368000,-20922789888000,
%U A155456 355687428096000,-6402373705728000,121645100408832000
%N A155456 Write (1+1/x)*log(1+x) = Sum c(n)*x^n; then a(n) = (n+1)!*c(n).
%C A155456 Apart from initial terms and signs, identical to A000142.
%C A155456 a(n-1), n >= 0, is the negative of the alternating row sum of A048994 (Stirling1) with e.g.f. -1/(1+x). - _Wolfdieter Lang_, May 09 2017
%H A155456 P. W. Anderson, D. J. Thouless, E. Abrahams and D. S. Fisher, <a href="http://scholar.google.com/scholar?hl=en&amp;lr=&amp;safe=off&amp;cluster=17075446093910447147">New method for a scaling theory of localization</a>, Physical Review B, 1980.
%F A155456 G.f.: -1-x+x^2/(G(0)+x) where G(k)= 1 + (k+1)*x/(1 + x*(k+2)/G(k+1)); (continued fraction, 2-step). - _Sergei N. Gladkovskii_, Aug 14 2012
%F A155456 G.f.: conjecture: T(0)*x^2/(1+2*x) - 1 - x, where T(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - (1+2*x*(k+1))*(1+2*x*(k+2))/T(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Nov 19 2013
%t A155456 p[x] = -(1 + 1/x)*Log[1 + x];
%t A155456 Table[ (n + 1)!*SeriesCoefficient[ Series[p[x], {x, 0, 30}], n], {n, 0, 30}]
%Y A155456 Cf. A048994.
%K A155456 sign,easy
%O A155456 0,4
%A A155456 _Roger L. Bagula_, Jan 22 2009
%E A155456 Edited by _N. J. A. Sloane_, Jun 02 2009