From Daniel Forgues, Jun 2021 2011: (Start)
In the following formulas, [...]_2 means converted to base 2.
a(n) = [Sum_{i=0..n} (binomial(n,i) mod 2) 2^i]_2, n >= 0.
From row n, 0 <= n <= 2^k  1, k >= 0, being
a(n) = [Product_{i=0..k1} (F_i)^(alpha_i)]_2, alpha_i in {0, 1},
where for k = 0, we get the empty product, i.e., 1, giving a(0) = 1,
we induce from the triangle that row 2^k + n, 0 <= n <= 2^k  1, is
a(2^k + n) = a(n)*[F_k]_2, k >= 0.
Denton Hewgill's identity: (Cf. links)
a(n) = [Product_{i>=0} (F_i)^(floor(n/2^i) mod 2)]_2, F_i = 2^(2^i)+1.
a(0) = 1; a(n) = [Product_{i=0..floor(log_2(n))} (F_i)^(floor(n/2^i) mod 2)]_2, F_i = 2^(2^i)+1, n >= 1. (End)
From Vladimir Shevelev, Dec 2627 2013: (Start)
sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(10^(2^k)+1)^r),
sum_{n>=0} (1)^A000120(n)/a(n)^r = Product_{k>=0} (1  1/(10^(2^k)+1)^r), where r>0 is a real number.
In particular,
sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(10^(2^k)+1)) = 1.10182034...;
sum_{n>=0} (1)^A000120(n)/a(n) = 0.9
a(2^n) = 10^(2^n)+1, n>=0.
Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations a(2^t*n+2^(t1)) = 99*(10^(2^(t1)+1))/(10^(2^(t1))1) * a(2^t*n+2^(t1)2), t>=2. In particular, for t=2,3,4, we have the following formulas:
a(4*n+2) = 101*a(4*n);
a(8*n+4) = 10001/101*a(8*n+2);
a(16*n+8)= 100000001/1010101*(16*n+6), etc. (End)
From Tom Edgar, Oct 11 2015: (Start)
a(2*n+1) = 11*a(2*n).
a(n) = Product_{b_j != 0}a(2^j) where n = Sum_{j>=0}b_j*2^j is the binary representation of n.
(End)
