function [ a, val ] = i4poly ( n, a, x0, iopt )
%*****************************************************************************80
%
%% i4poly() performs operations on integer polynomials in power or factorial form.
%
% Discussion:
%
% The power sum form of a polynomial is
%
% P(X) = A1 + A2*X + A3*X^2 + ... + (AN+1)*X^N
%
% The Taylor expansion at C has the form
%
% P(X) = A1 + A2*(X-C) + A3*(X-C)^2 + ... + (AN+1)*(X-C)^N
%
% The factorial form of a polynomial is
%
% P(X) = A1 + A2*X + A3*(X)*(X-1) + A4*(X)*(X-1)*(X-2)+...
% + (AN+1)*(X)*(X-1)*...*(X-N+1)
%
% Licensing:
%
% This code is distributed under the MIT license.
%
% Modified:
%
% 05 July 2004
%
% Author:
%
% Original FORTRAN77 version by Albert Nijenhuis, Herbert Wilf.
% This version by John Burkardt.
%
% Reference:
%
% Albert Nijenhuism, Herbert Wilf,
% Combinatorial Algorithms,
% Academic Press, 1978, second edition,
% ISBN 0-12-519260-6.
%
% Input:
%
% integer N, the number of coefficients in the polynomial
% (in other words, the polynomial degree + 1)
%
% integer A(N), the coefficients of the polynomial.
%
% integer X0, for IOPT = -1, 0, or positive, the value of the
% argument at which the polynomial is to be evaluated, or the
% Taylor expansion is to be carried out.
%
% integer IOPT, a flag describing which algorithm is to
% be carried out:
% -3: Reverse Stirling. Input the coefficients of the polynomial in
% factorial form, output them in power sum form.
% -2: Stirling. Input the coefficients in power sum form, output them
% in factorial form.
% -1: Evaluate a polynomial which has been input in factorial form.
% 0: Evaluate a polynomial input in power sum form.
% 1 or more: Given the coefficients of a polynomial in
% power sum form, compute the first IOPT coefficients of
% the polynomial in Taylor expansion form.
%
% Output:
%
% integer A(N), the coefficients of the output polynomial.
% Depending on the option chosen, these coefficients are the input values,
% or those of a different form of the polynomial.
%
% integer VAL, for IOPT = -1 or 0, the value of the
% polynomial at the point X0.
%
val = 0;
n1 = min ( n, iopt );
n1 = max ( 1, n1 );
if ( iopt < -1 )
n1 = n;
end
delta = mod ( max ( -iopt, 0 ), 2 );
w = - n * delta;
if ( -2 < iopt )
w = w + x0;
end
for m = 1 : n1
val = 0;
z = w;
for i = m : n
z = z + delta;
val = a(n+m-i) + z * val;
if ( iopt ~= 0 && iopt ~= -1 )
a(n+m-i) = val;
end
end
if ( iopt < 0 )
w = w + 1;
end
end
return
end