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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.
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%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