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%I #5 May 10 2021 22:15:25
%S 1,1,1,1,1,1,1,1,1,2,3,2,1,1,1,1,3,6,8,8,8,6,3,1,1,1,1,4,10,18,26,33,
%T 38,38,33,26,18,10,4,1,1,1,1,5,15,33,59,92,129,166,195,211,211,195,
%U 166,129,92,59,33,15,5,1,1,1,1,6,21,54,113,205,334,499,693,899,1095,1257,1364,1401,1364,1257,1095,899,693,499,334,205,113,54,21,6,1,1
%N Irregular triangle T(n, k) = coefficients of p(n, x), where p(n, x) = p(n-1, x)*Sum_{j=0..n-1} x^i - x*Sum_{j=0..binomial(n,2)-2} x^i, read by rows.
%D J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 146.
%H G. C. Greubel, <a href="/A171457/b171457.txt">Rows n = 1..20 of the irregular triangle, flattened</a>
%F T(n, k) = coefficients of p(n, x), where p(n, x) = p(n-1, x)*Sum_{j=0..n-1} x^i - x*Sum_{j=0..binomial(n,2)-2} x^i, p(1,x) = 1, p(2,x) = 1+x, p(3,x) = 1+x+x^2+x^3.
%e The irregular triangle begins as:
%e 1;
%e 1, 1;
%e 1, 1, 1, 1;
%e 1, 1, 2, 3, 2, 1, 1;
%e 1, 1, 3, 6, 8, 8, 8, 6, 3, 1, 1;
%e 1, 1, 4, 10, 18, 26, 33, 38, 38, 33, 26, 18, 10, 4, 1, 1;
%t p[n_, x_]:= p[n, x]= If[n<4, (1-x^2^(n-1))/(1-x), ((1-x^n)*p[n-1, x] - x*(1 - x^(Binomial[n, 2] -1)))/(1-x)];
%t T[n_]:= CoefficientList[p[n, x], x];
%t Table[T[n], {n, 1, 10}] (* modified by _G. C. Greubel_, May 10 2021 *)
%o (Sage)
%o def p(n,x): return (1-x^2^(n-1))/(1-x) if (n<4) else ((1-x^n)*p(n-1,x) -x*(1-x^(binomial(n,2)-1)))/(1-x)
%o def T(n): return ( p(n,x) ).full_simplify().coefficients(sparse=False)
%o [T(n) for n in (1..10)] # _G. C. Greubel_, May 10 2021
%Y Cf. A171456.
%K nonn,tabf
%O 1,10
%A _Roger L. Bagula_, Dec 09 2009
%E Edited by _G. C. Greubel_, May 10 2021
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