0,3

The sequence may be the binomial transform of A024493 interleaved with zeros. A024493 = (1, 1, 1, 2, 5, 11, 22,...); so the conjecture succeeds through a(12) = A007318 * [1, 0, 1, 0, 1, 0, 2, 0, 5, 0, 11,...].

Infinite product of aerated Pascal's triangles by rows. Let a = A007318, b = an aerated version with alternate rows (1,3,5,...) = (0, 0, 0,...); c = two adjacent rows with (0, 0, 0,...) and so on. The result of a*b*c...*...: = a one column matrix shifted to the left, = A161869.

First few rows = left borders of a, a*b, a*c,...:

1,...1,...1,...1,...1,...1,...1,...1,... = a, left border of A007318

1,...1,...2,...4,...8,..16,..32,..64,... = a*b, left border.

1,...1,...2,...4,...8,..16,..33,..71,... = a*b*c, left border.

1,...1,...2,...4,...8,..16,..33,..71,... = a*b*c*d, left border

...converging to A161869, a one column matrix with the rest zeros.

A024493

Sequence in context: A129986 A110334 A084636 this_sequence A088325 A006210 A096812

Adjacent sequences: A161866 A161867 A161868 this_sequence A161870 A161871 A161872

nonn

Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 20 2009