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%I #32 Dec 17 2021 02:55:45
%S 1,1,1,1,2,1,1,3,3,1,1,4,6,4,1,1,5,10,10,5,1,1,6,15,4,15,6,1,1,7,5,3,
%T 3,5,7,1,1,8,12,8,6,8,12,8,1,1,9,4,4,14,14,4,4,9,1,1,10,13,8,2,12,2,8,
%U 13,10,1,1,11,7,5,10,14,14,10
%N Triangle, read by rows, formed by reading Pascal's triangle (A007318) mod 16.
%C T(n+1,k) = (T(n,k) + T(n,k-1)) mod 16. - _Reinhard Zumkeller_, Mar 14 2015
%H Reinhard Zumkeller, <a href="/A034932/b034932.txt">Rows n = 0..120 of triangle, flattened</a>
%H Ilya Gutkovskiy, <a href="/A275198/a275198.pdf">Illustrations (triangle formed by reading Pascal's triangle mod m)</a>
%H James G. Huard, Blair K. Spearman, and Kenneth S. Williams, <a href="https://doi.org/10.1006/eujc.1997.0146">Pascal's triangle (mod 8)</a>, European Journal of Combinatorics 19:1 (1998), pp. 45-62.
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F T(i, j) = binomial(i, j) mod 16.
%e Triangle begins:
%e 1
%e 1 1
%e 1 2 1
%e 1 3 3 1
%e 1 4 6 4 1
%e 1 5 10 10 5 1
%e 1 6 15 4 15 6 1
%e 1 7 5 3 3 5 7 1
%e 1 8 12 8 6 8 12 8 1
%e 1 9 4 4 14 14 4 4 9 1
%e 1 10 13 8 2 12 2 8 13 10 1
%e 1 11 7 5 10 14 14 10 5 7 11 1
%e .
%e Written in hexadecimal (with a=10, b=11, ..., f=15), rows 0..32 are
%e .
%e 1
%e 1 1
%e 1 2 1
%e 1 3 3 1
%e 1 4 6 4 1
%e 1 5 a a 5 1
%e 1 6 f 4 f 6 1
%e 1 7 5 3 3 5 7 1
%e 1 8 c 8 6 8 c 8 1
%e 1 9 4 4 e e 4 4 9 1
%e 1 a d 8 2 c 2 8 d a 1
%e 1 b 7 5 a e e a 5 7 b 1
%e 1 c 2 c f 8 c 8 f c 2 c 1
%e 1 d e e b 7 4 4 7 b e e d 1
%e 1 e b c 9 2 b 8 b 2 9 c b e 1
%e 1 f 9 7 5 b d 3 3 d b 5 7 9 f 1
%e 1 0 8 0 c 0 8 0 6 0 8 0 c 0 8 0 1
%e 1 1 8 8 c c 8 8 6 6 8 8 c c 8 8 1 1
%e 1 2 9 0 4 8 4 0 e c e 0 4 8 4 0 9 2 1
%e 1 3 b 9 4 c c 4 e a a e 4 c c 4 9 b 3 1
%e 1 4 e 4 d 0 8 0 2 8 4 8 2 0 8 0 d 4 e 4 1
%e 1 5 2 2 1 d 8 8 2 a c c a 2 8 8 d 1 2 2 5 1
%e 1 6 7 4 3 e 5 0 a c 6 8 6 c a 0 5 e 3 4 7 6 1
%e 1 7 d b 7 1 3 5 a 6 2 e e 2 6 a 5 3 1 7 b d 7 1
%e 1 8 4 8 2 8 4 8 f 0 8 0 c 0 8 0 f 8 4 8 2 8 4 8 1
%e 1 9 c c a a c c 7 f 8 8 c c 8 8 f 7 c c a a c c 9 1
%e 1 a 5 8 6 4 6 8 3 6 7 0 4 8 4 0 7 6 3 8 6 4 6 8 5 a 1
%e 1 b f d e a a e b 9 d 7 4 c c 4 7 d 9 b e a a e d f b 1
%e 1 c a c b 8 4 8 9 4 6 4 b 0 8 0 b 4 6 4 9 8 4 8 b c a c 1
%e 1 d 6 6 7 3 c c 1 d a a f b 8 8 b f a a d 1 c c 3 7 6 6 d 1
%e 1 e 3 c d a f 8 d e 7 4 9 a 3 0 3 a 9 4 7 e d 8 f a d c 3 e 1
%e 1 f 1 f 9 7 9 7 5 b 5 b d 3 d 3 3 d 3 d b 5 b 5 7 9 7 9 f 1 f 1
%e 1 0 0 0 8 0 0 0 c 0 0 0 8 0 0 0 6 0 0 0 8 0 0 0 c 0 0 0 8 0 0 0 1
%t Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 16] (* _Robert G. Wilson v_, May 26 2004 *)
%o (Haskell)
%o a034932 n k = a034932_tabl !! n !! k
%o a034932_row n = a034932_tabl !! n
%o a034932_tabl = iterate
%o (\ws -> zipWith ((flip mod 16 .) . (+)) ([0] ++ ws) (ws ++ [0])) [1]
%o -- _Reinhard Zumkeller_, Mar 14 2015
%Y Cf. A007318, A047999, A083093, A034931, A034930, A008975.
%Y Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), A034931 (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930 (m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), (this sequence) (m = 16).
%K nonn,tabl
%O 0,5
%A _N. J. A. Sloane_
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