A195915 Table with T(n,1) = n, otherwise T(n,k) = xor(T(n-1,k-1), T(n-1,k)).

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, 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, 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

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

Row reversals of A195916. For xor, see A003987.

Sequence in context: A209334 A180975 A210216 * A219158 A049834 A134625

Adjacent sequences: A195912 A195913 A195914 * A195916 A195917 A195918

Keyword

nonn,easy,look,tabl

Author

Franklin T. Adams-Watters, Sep 25 2011

Status

approved