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%I #23 Feb 19 2023 16:08:26
%S 1,2,1,3,3,1,4,0,2,1,5,4,2,3,1,6,1,6,1,2,1,7,7,7,7,3,3,1,8,0,0,0,4,0,
%T 2,1,9,8,0,0,4,4,2,3,1,10,1,8,0,4,0,6,1,2,1,11,11,9,8,4,4,6,7,3,3,1,
%U 12,0,2,1,12,0,2,1,4,0,2,1,13,12,2,3,13,12,2,3,5,4,2,3,1
%N Table with T(n,1) = n, otherwise T(n,k) = xor(T(n-1,k-1), T(n-1,k)).
%C We take T(n,n+1) = 0 in the calculation (so T(n,n) = 1).
%C It appears that, if 2^j > n, T(2^j+n, 2^j+k) = T(n,k). This is equivalent to the periodicity conjecture in A195916.
%H John Tyler Rascoe, <a href="/A195915/b195915.txt">Rows n = 1..140 of triangle, flattened</a>
%e The table starts:
%e 1;
%e 2, 1;
%e 3, 3, 1;
%e 4, 0, 2, 1;
%e 5, 4, 2, 3, 1;
%e 6, 1, 6, 1, 2, 1;
%e ...
%o (PARI) anrow(n)=local(r,v);r=v=[1];for(k=2,n,v=vector(#v+1,j,if(j==1,k,bitxor(v[j-1],if(j==k,0,v[j]))));r=concat(r,v));r
%o (Python)
%o def A195915_list(row_n):
%o T = []
%o for n in range(1,row_n+1):
%o T.append([])
%o for k in range(1,n+1):
%o if k == 1: x = n
%o elif k == n: x = 1
%o else: x = T[n-2][k-2]^T[n-2][k-1]
%o T[n-1].append(x)
%o return T # _John Tyler Rascoe_, Feb 11 2023
%Y Row reversals of A195916. For xor, see A003987.
%K nonn,easy,look,tabl
%O 1,2
%A _Franklin T. Adams-Watters_, Sep 25 2011
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