login
A355604
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.
1
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, 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, 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
OFFSET
0,8
COMMENTS
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.
LINKS
Rémy Sigrist, Table of n, a(n) for n = 0..8384 (rows for n = 0..128 flattened)
Rémy Sigrist, Colored representation of the first 2^10 rows (where the hue is function of T(n, k), black pixels correspond to 1's)
EXAMPLE
Triangle T(n, k) begins (stars indicate replacements):
n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12
---+-----------------------------------------------------------------
0| 1
1| 1 1
2| 1 1* 1
3| 1 3 3 1
4| 1 1* 1* 1* 1
5| 1 5 1* 1* 5 1
6| 1 1* 15 1* 15 1* 1
7| 1 7 21 35 35 21 7 1
8| 1 1* 1* 15* 1* 15* 1* 1* 1
9| 1 9 1* 5* 1* 1* 5* 1* 9 1
10| 1 1* 45 1* 1* 1* 1* 1* 45 1* 1
11| 1 11 55 165 1* 3* 3* 1* 165 55 11 1
12| 1 1* 1* 1* 495 1* 1* 1* 495 1* 1* 1* 1
PROG
(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) }
CROSSREFS
KEYWORD
nonn,look,tabl
AUTHOR
Rémy Sigrist, Jul 09 2022
STATUS
approved