

A167447


Number of divisors of n which are not multiples of 3 consecutive primes.


3



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, 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, 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
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OFFSET

1,2


COMMENTS

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.


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..65537
Eric Weisstein's World of Mathematics, Fibonacci nStep Number


FORMULA

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).
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).
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 triple of consecutive primes as divisors.
a(A002110(n)) = A000073(n+2).
a(n) = Sum_{dn} [A300820(d) < 3].  Antti Karttunen, Mar 21 2018


EXAMPLE

Since 2 of 60's 12 divisors (30 and 60) are multiples of at least 3 consecutive primes, a(60) = 12  2 = 10.


PROG

(PARI)
A300820(n) = if(omega(n)<=1, omega(n), my(pis=apply(p>primepi(p), factor(n)[, 1]), el=1, m=1); for(i=2, #pis, if(pis[i] == (1+pis[i1]), el++; m = max(m, el), el=1)); (m));
A167447(n) = sumdiv(n, d, (A300820(d)<3)); \\ Antti Karttunen, Mar 21 2018


CROSSREFS

Cf. A000040, A000073, A002110, A046301, A073485, A166469, A300820.
Sequence in context: A325560 A318412 A322986 * A134687 A184395 A329484
Adjacent sequences: A167444 A167445 A167446 * A167448 A167449 A167450


KEYWORD

nonn


AUTHOR

Matthew Vandermast, Nov 05 2009


STATUS

approved



