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%I #12 Mar 07 2024 07:54:06
%S 1,1,1,1,1,1,1,1,2,3,4,3,2,1,1,3,6,10,13,14,13,10,6,3,1,1,4,10,20,33,
%T 47,59,66,66,59,47,33,20,10,4,1,1,5,15,35,68,115,174,239,301,350,377,
%U 377,350,301,239,174,115,68,35,15,5,1,1,6,21,56,124,239,413,652,952,1297,1659,2001,2283,2469,2534,2469,2283,2001,1659,1297,952,652,413,239,124,56,21,6,1
%N Triangle T(n, k) = coefficients of p(n, x) where p(n, x) = ((1-x^n)/(1-x))*p(n-1, x), read by rows.
%D J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 146.
%H G. C. Greubel, <a href="/A171456/b171456.txt">Rows n = 1..20 of the irregular triangle, flattened</a>
%F T(n, k) = coefficients of p(n, x) where p(n, x) = ((1-x^n)/(1-x))*p(n-1, x), 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, 2, 3, 4, 3, 2, 1;
%e 1, 3, 6, 10, 13, 14, 13, 10, 6, 3, 1;
%e 1, 4, 10, 20, 33, 47, 59, 66, 66, 59, 47, 33, 20, 10, 4, 1;
%t p[x_, n_]:= p[x, n]= If[n<4, (1-x^2^(n-1))/(1-x), ((1-x^4)/(1-x))*Product[(1 - x^j)/(1-x), {j, 4, n}]];
%t T[n_]:= CoefficientList[p[x, n], x];
%t Table[T[n], {n, 1, 10}]//Flatten (* 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^4)/(1-x))*product( (1-x^j)/(1-x) for j in (4..n) )
%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
%K nonn,tabf
%O 1,9
%A _Roger L. Bagula_, Dec 09 2009
%E Edited by _G. C. Greubel_, May 10 2021
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