
FORMULA

a(0) = 1 and for n >= 1, a(n) = 1/n*Sum_{i = 0..n1} R(i+1,3)*a(n1i), where R(n,x) denotes the nth row polynomial of A145901.
O.g.f.: A(z) = 1 + 7*z + 73*z^2 + 1071*z^3 + 21249*z^4 + ... satisfies A^4(z) = 1/(1  z)*1/(1  2*z)^3*A^3(z/(1  2*z)).
O.g.f.: A(z) = exp( Sum_{k >= 1} R(k,3)*z^k/k ).
From Peter Bala, Dec 06 2017: (Start)
a(n) appears to be always odd. Calculation suggests that for k = 1,2,3,..., the sequence a(n) (mod 2^k) is purely periodic with period length a divisor of 2^(k1). For example, a(n) (mod 4) = (1,3,1,3,...) seems to be purely periodic with period 2, a(n) (mod 8) = (1,7,1,7,...) seems to be purely periodic also with period 2 while a(n) (mod 16) = (1,7,9,15,1,7,9,15,...) seems to be purely periodic with period 4 (all three checked up to n = 1000).
The sequences a(n) (mod k), for other values of k, appear to have interesting but more complicated patterns. An example is given below.
(End)


EXAMPLE

a(n) (mod 3) = (1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0, 0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0, 0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0, 0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,...).  Peter Bala, Dec 06 2017
