OFFSET
1,2
COMMENTS
We take T(n,n+1) = 0 in the calculation (so T(n,n) = 1).
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.
LINKS
John Tyler Rascoe, Rows n = 1..140 of triangle, flattened
EXAMPLE
The table starts:
1;
2, 1;
3, 3, 1;
4, 0, 2, 1;
5, 4, 2, 3, 1;
6, 1, 6, 1, 2, 1;
...
PROG
(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
(Python)
def A195915_list(row_n):
T = []
for n in range(1, row_n+1):
T.append([])
for k in range(1, n+1):
if k == 1: x = n
elif k == n: x = 1
else: x = T[n-2][k-2]^T[n-2][k-1]
T[n-1].append(x)
return T # John Tyler Rascoe, Feb 11 2023
CROSSREFS
AUTHOR
Franklin T. Adams-Watters, Sep 25 2011
STATUS
approved