%I #10 Mar 29 2023 08:58:29

%S 1,1,1,1,1,2,2,1,3,5,1,4,9,9,1,5,14,23,23,1,6,20,43,66,1,7,27,70,136,

%T 136,1,8,35,105,241,377,377,1,9,44,149,390,767,1144,1,10,54,203,593,

%U 1360,2504,2504,1,11,65,268,861,2221,4725,7229,7229,1,12,77,345,1206

%N Irregular triangle read by rows: T(0,0) = 1, T(n,k) = T(n,k-1) + T(n-1,k) for n > 0, 0 < k <= f(n), where f(n) = floor((2*n+3)/3), and entries outside triangle are 0.

%C There are f(n) = floor((2*n+3)/3) = A004396(n+1) terms in row n.

%e Triangle begins:

%e k=0 1 2 3 4 5 6 7 8

%e n=0: 1

%e n=1: 1

%e n=2: 1, 1

%e n=3: 1, 2, 2

%e n=4: 1, 3, 5

%e n=5: 1, 4, 9, 9

%e n=6: 1, 5, 14, 23, 23

%e n=7: 1, 6, 20, 43, 66

%e n=8: 1, 7, 27, 70, 136, 136

%e n=9: 1, 8, 35, 105, 241, 377, 377

%e n=10: 1, 9, 44, 149, 390, 767, 1144

%e n=11: 1, 10, 54, 203, 593, 1360, 2504, 2504

%e n=12: 1, 11, 65, 268, 861, 2221, 4725, 7229, 7229

%e n=13: 1, 12, 77, 345, 1206, 3427, 8152, 15381, 22610

%e ...

%o (PARI) f(n) = floor((2*(n-1)+3)/3); s=14; M=matrix(s,s); for(n=1,s,M[n,1]=1); for(n=2,s,for(k=2,f(n),M[n,k]=M[n,k-1]+M[n-1,k])); for(n=1,s,for(k=1,f(n),print1(M[n,k],", ")))

%Y Cf. A004396 (row lengths).

%Y Cf. A060941 (diagonal T(3*n, 2*n)).

%K nonn,tabf

%O 0,6

%A _Gerald McGarvey_, Oct 03 2009