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%I #12 Jul 11 2022 20:48:41
%S 1,1,1,1,1,1,1,3,3,1,1,1,1,1,1,1,5,1,1,5,1,1,1,15,1,15,1,1,1,7,21,35,
%T 35,21,7,1,1,1,1,15,1,15,1,1,1,1,9,1,5,1,1,5,1,9,1,1,1,45,1,1,1,1,1,
%U 45,1,1,1,11,55,165,1,3,3,1,165,55,11,1,1,1,1,1,495,1,1,1,495,1,1,1,1
%N Table T(n, k), n >= 0, k = 0..n, read by rows; row n is obtained by replacing in row n of Pascal's triangle (A007318) runs of k consecutive even numbers by the terms of row k+1 of the present triangle.
%C This triangle has fractal features: even terms of Pascal's triangle are clustered as wXwXw subtriangles; these subtriangles are replaced by the first w rows (flipped upside-down) of the present triangle.
%H Rémy Sigrist, <a href="/A355604/b355604.txt">Table of n, a(n) for n = 0..8384</a> (rows for n = 0..128 flattened)
%H Rémy Sigrist, <a href="/A355604/a355604.png">Colored representation of the first 2^10 rows</a> (where the hue is function of T(n, k), black pixels correspond to 1's)
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%e Triangle T(n, k) begins (stars indicate replacements):
%e n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12
%e ---+-----------------------------------------------------------------
%e 0| 1
%e 1| 1 1
%e 2| 1 1* 1
%e 3| 1 3 3 1
%e 4| 1 1* 1* 1* 1
%e 5| 1 5 1* 1* 5 1
%e 6| 1 1* 15 1* 15 1* 1
%e 7| 1 7 21 35 35 21 7 1
%e 8| 1 1* 1* 15* 1* 15* 1* 1* 1
%e 9| 1 9 1* 5* 1* 1* 5* 1* 9 1
%e 10| 1 1* 45 1* 1* 1* 1* 1* 45 1* 1
%e 11| 1 11 55 165 1* 3* 3* 1* 165 55 11 1
%e 12| 1 1* 1* 1* 495 1* 1* 1* 495 1* 1* 1* 1
%o (PARI) row(n) = { my (r=binomial(n)); for (i=1, #r, if (r[i]%2==0, for (w=1, oo, if (r[i+w]%2==1, my (t=row(w-1)); for (j=1, #t, r[i-1+j]=t[j]); i+=w; break)))); return (r) }
%Y Cf. A007318, A065040, A014421, A143333, A348648.
%K nonn,look,tabl
%O 0,8
%A _Rémy Sigrist_, Jul 09 2022