%I #9 Apr 08 2022 08:15:00
%S 1,1,1,1,0,0,1,1,1,1,0,0,0,1,0,0,1,1,1,0,0,1,1,0,0,1,1,0,0,0,1,0,0,1,
%T 1,1,1,0,0,0,1,0,0,1,2,0,0,1,1,0,0,1,1,0,0,0,1,0,0,1,1,1,0,0,1,1,0,0,
%U 1,1,0,0,1,2,0,0,1,2,0,0,1,2,0,0,1,1,0,0,1,1,0,0,0,1,0,0,1,1
%N Irregular triangle T(n, k) = [x^k]( p(n, x) ), where p(n, x) = x^(2*n-1)*p(n-1, x) + p(n-2, x) with p(0, x) = 1 and p(1, x) = 1 + x, read by rows.
%C There are n^2 + 1 terms in row n, for n >= 0. - _G. C. Greubel_, Apr 07 2022
%H G. C. Greubel, <a href="/A172099/b172099.txt">Rows n = 0..30 of the irregular triangle, flattened</a>
%F T(n, k) = [x^k]( p(n, x) ), where p(n, x) = x^(2*n-1)*p(n-1, x) + p(n-2, x) with p(0, x) = 1 and p(1, x) = 1 + x.
%F Sum_{k=0..n^2} T(n, k) = Fibonacci(n+2) (row sums). - _G. C. Greubel_, Apr 07 2022
%e Irregular triangle begins as:
%e 1;
%e 1, 1;
%e 1, 0, 0, 1, 1;
%e 1, 1, 0, 0, 0, 1, 0, 0, 1, 1;
%e 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1;
%e 1, 1, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1;
%t p[n_, x_]:= p[n, x]= If[n<2, n*x+1, x^(2*n-1)*p[n-1, x] + p[n-2, x]];
%t Table[CoefficientList[p[n, x], x], {n,0,10}]//Flatten
%o (SageMath)
%o @CachedFunction
%o def p(n,x):
%o if (n<2): return n*x+1
%o else: return x^(2*n-1)*p(n-1, x) + p(n-2, x)
%o def T(n,k): return ( p(n,x) ).series(x, n^2+2).list()[k]
%o flatten([[T(n,k) for k in (0..n^2)] for n in (0..12)]) # _G. C. Greubel_, Apr 07 2022
%Y Cf. A000045, A002522, A172098.
%K nonn,tabf
%O 0,45
%A _Roger L. Bagula_, Jan 25 2010
%E Edited by _G. C. Greubel_, Apr 07 2022