%I
%S 1,2,2,3,2,4,2,4,3,4,2,6,2,4,4,5,2,6,2,6,4,4,2,8,3,4,4,6,2,7,2,6,4,4,
%T 4,9,2,4,4,8,2,8,2,6,6,4,2,10,3,6,4,6,2,8,4,8,4,4,2,10,2,4,6,7,4,8,2,
%U 6,4,8,2,12,2,4,6,6,4,8,2,10,5,4,2,12,4,4,4,8,2,10,4,6,4,4,4,12,2,6,6,9,2,8
%N Number of divisors of n which are not multiples of 3 consecutive primes.
%C If a number is a product of any number of consecutive primes, the number of its divisors which are not multiples of n consecutive primes is always a Fibonacci nstep number. See also A073485, A166469.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonaccinStepNumber.html">Fibonacci nStep Number</a>
%F a) If n has no prime gaps in its factorization (cf. A073491), then, if the canonical factorization of n into prime powers is the product of p_i^(e_i), a(n) is the sum of all products of exponents which do not include 3 consecutive exponents, plus 1. For example, if A001221(n)=3, a(n)=e_1*e_2 +e_1*e_3 +e_2*e_3 +e_1 +e_2 +e_3 +1. If A001221(n)=k, the total number of terms always equals A000073(k+3).
%F The answer can also be computed in k steps, by finding the answers for the products of the first i powers for i=1 to i=k. Let the result of the ith step be called r(i). r(1)=e_1+1; r(2)=e_1*e_2+e_1+e_2+1; r(3)=e_1*e_2+e_1*e_3+e_2*e_3+e_1+e_2+e_3+1; for i>3, r(i)=r(i1)+e_i*r(i2)+e_i*e(i1)*r(i3).
%F b) If n has prime gaps in its factorization, express it as a product of the minimum number of A073491's members possible. Then apply either of the above methods to each of those members, and multiply the results to get a(n). a(n)=A000005(n) iff n has no triplet of consecutive primes as divisors.
%e Since 2 of 60's 12 divisors (30 and 60) are multiples of at least 3 consecutive primes, a(60) = 12  2 = 10.
%Y A(A002110(n))=A000073(n+2).
%Y See also A000040, A046301.
%K nonn
%O 1,2
%A _Matthew Vandermast_, Nov 05 2009
