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%I #26 Dec 19 2023 17:45:11
%S 1,1,-1,2,1,-5,7,-1,9,-30,37,1,-14,81,-229,266,-1,20,-175,835,-2165,
%T 2431,1,-27,330,-2330,9990,-24576,27007,-1,35,-567,5495,-34300,137466,
%U -326515,353522,1,-44,910,-11522,97405,-561386,2148139,-4976315,5329837
%N Triangle read by rows: coefficients of polynomials arising from the recurrence A[n](x) = A[n-1]'(x)/(1-x) with A[0] = exp(x).
%H Seiichi Manyama, <a href="/A144505/b144505.txt">Rows n = 0..140, flattened</a>
%H Ling Gao, <a href="http://hdl.handle.net/20.500.12680/h989rb533">Graph assembly for spider and tadpole graphs</a>, Master's Thesis, Cal. State Poly. Univ. (2023). See pp. 42, 63.
%H N. J. A. Sloane, <a href="/A144505/a144505.txt">Rows 0 through 25 of the triangle, together with the corresponding polynomials P[n](x).</a>
%F Let A[0](x) = exp(x), A[n](x) = A[n-1]'(x)/(1-x) for n>0 and let P[n](x) = A[n](x)*(1-x)^(2n-1)/exp(x). Row n of triangle gives coefficients of P[n](x) with exponents of x in decreasing order.
%F From _Vladeta Jovovic_, Dec 15 2008: (Start)
%F P[n] = Sum_{k=0..n} ((n+k)!/((n-k)!*k!*2^k)) * (1-x)^(n-k).
%F E.g.f.: exp((1-x)*(1-sqrt(1-2*y)))/sqrt(1-2*y). (End)
%e The first few polynomials P[n] (n >= 0) are:
%e P[0] = 1;
%e P[1] = 1;
%e P[2] = -x +2;
%e P[3] = x^2 -5*x +7;
%e P[4] = -x^3 + 9*x^2 - 30*x +37;
%e P[5] = x^4 -14*x^3 + 81*x^2 - 229*x +266;
%e P[6] = -x^5 +20*x^4 -175*x^3 + 835*x^2 -2165*x +2431;
%e P[7] = x^6 -27*x^5 +330*x^4 -2330*x^3 +9990*x^2 -24576*x +27007;
%e ...
%e Triangle of coefficients begins:
%e 1;
%e 1;
%e -1, 2;
%e 1, -5, 7;
%e -1, 9, -30, 37;
%e 1, -14, 81, -229, 266;
%e -1, 20, -175, 835, -2165, 2431;
%e 1, -27, 330, -2330, 9990, -24576, 27007;
%e -1, 35, -567, 5495, -34300, 137466, -326515, 353522;
%e 1, -44, 910, -11522, 97405, -561386, 2148139, -4976315, 5329837;
%e ...
%p A[0]:=exp(x);
%p P[0]:=1;
%p for n from 1 to 12 do
%p A[n]:=sort(simplify( diff(A[n-1],x)/(1-x)));
%p P[n]:=sort(simplify(A[n]*(1-x)^(2*n-1)/exp(x)));
%p t1:=simplify(x^(degree(P[n],x))*subs(x=1/x,P[n]));
%p t2:=series(t1,x,2*n+3);
%p lprint(P[n]);
%p lprint(seriestolist(t2));
%p od:
%t f[n_, x_]:= x^n*Sum[((n+j)!/((n-j)!*j!*2^j))*(1-1/x)^(n-j), {j,0,n}];
%t t[n_, k_]:= Coefficient[Series[f[n,x], {x,0,30}], x, k];
%t Join[{1}, Table[t[n,k], {n,0,12}, {k,0,n}]//Flatten] (* _G. C. Greubel_, Oct 02 2023 *)
%o (Magma)
%o R<x>:=PowerSeriesRing(Integers(), 50);
%o f:= func< n,x | x^n*(&+[Binomial(n,j)*Factorial(n+j)*(1-1/x)^(n-j)/(2^j*Factorial(n)) : j in [0..n]]) >;
%o T:= func< n,k | Coefficient(R!( f(n,x) ), k) >;
%o [1] cat [T(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Oct 02 2023
%o (SageMath)
%o P.<x> = PowerSeriesRing(QQ, 50)
%o def f(n,x): return x^n*sum(binomial(n,j)*rising_factorial(n+1,j)*(1-1/x)^(n-j)/2^j for j in range(n+1))
%o def T(n,k): return P( f(n,x) ).list()[k]
%o [1] + flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # _G. C. Greubel_, Oct 02 2023
%Y Columns give A001515 (really A144301), A144498, A001514, A144506, A144507.
%Y Row sums give A001147.
%Y Alternating row sums give A043301.
%K sign,tabf
%O 0,4
%A _N. J. A. Sloane_, Dec 14 2008
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