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A121922 The result of the integration Integral_{t=0..oo} -rho*exp(-rho*s*t)*t^j*s*log(1+t) dt can be written as (F(u,j)*exp(u)*Ei(1,u) + G(u,j))/u^j, where rho>0, s>0, and u=rho*s. Sequence is the regular triangle corresponding to G(u,j).

%I A121922 #23 Jan 26 2025 17:25:48
%S A121922 -1,1,-3,-1,4,-11,1,-5,18,-50,-1,6,-27,96,-274,1,-7,38,-168,600,-1764,
%T A121922 -1,8,-51,272,-1200,4320,-13068,1,-9,66,-414,2200,-9720,35280,-109584,
%U A121922 -1,10,-83,600,-3750,19920,-88200,322560,-1026576,1,-11,102,-836,6024,-37620,199920,-887040,3265920,-10628640
%N A121922 The result of the integration Integral_{t=0..oo} -rho*exp(-rho*s*t)*t^j*s*log(1+t) dt can be written as (F(u,j)*exp(u)*Ei(1,u) + G(u,j))/u^j, where rho>0, s>0, and u=rho*s. Sequence is the regular triangle corresponding to G(u,j).
%e A121922 At j=7, the result of the integration Integral_{t=0..oo} -rho*exp(-rho*s*t)*t^j*s*log(1+t) dt
%e A121922 can be written as (F(u,7)*exp(u)*Ei(1,u) + G(u,7))/u^7, where
%e A121922 F(u,7) = u^7 - 7*u^6 + 42*u^5 - 210*u^4 + 840*u^3 -2520*u^2 + 5040*u - 5040,
%e A121922 G(u,7) = - u^6 + 8*u^5 - 51*u^4 + 272*u^3 - 1200*u^2 + 4320*u - 13068,
%e A121922 and u=rho*s.
%e A121922 The coefficients of F(u,7), i.e., (1, -7, 42, -210, 840, 2520, 5040, -5040), comprise the 7th row of A008279 (see also A068424). The coefficients of G(u,7), i.e., (-1, 8, -51, 272, -1200, 4320, -13068) give the 7th row of the triangle below.
%e A121922 Triangle begins:
%e A121922   -1
%e A121922   1, -3
%e A121922   -1, 4, -11
%e A121922   1, -5, 18, -50
%e A121922   -1, 6, -27, 96, -274
%e A121922   1, -7, 38, -168, 600, -1764
%e A121922   -1, 8, -51, 272, -1200, 4320, -13068
%Y A121922 The right-hand diagonal is A000254, the one before that is A001563.
%Y A121922 Cf. A008279, A068424.
%K A121922 sign,tabl
%O A121922 0,3
%A A121922 _Arie Harel_, Sep 09 2006
%E A121922 Edited by _Jon E. Schoenfield_, Oct 20 2013