# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/

Search: id:a166469
Showing 1-1 of 1

%I A166469 #13 Dec 18 2017 11:56:16
%S A166469 1,2,2,3,2,3,2,4,3,4,2,4,2,4,3,5,2,4,2,6,4,4,2,5,3,4,4,6,2,5,2,6,4,4,
%T A166469 3,5,2,4,4,8,2,6,2,6,4,4,2,6,3,6,4,6,2,5,4,8,4,4,2,7,2,4,6,7,4,6,2,6,
%U A166469 4,6,2,6,2,4,4,6,3,6,2,10,5,4,2,8,4,4,4,8,2,6,4,6,4,4,4,7,2,6,6,9,2,6,2,8,5
%N A166469 Number of divisors of n which are not multiples of consecutive primes.
%C A166469 Links various subsequences of A025487 with an unusual number of important sequences, including the Fibonacci, Lucas, and other generalized Fibonacci sequences (see cross-references).
%C A166469 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 n-step number. See also A073485, A167447.
%H A166469 Antti Karttunen, <a href="/A166469/b166469.txt">Table of n, a(n) for n = 1..16384</a>
%H A166469 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>
%F A166469 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 one or more nonadjacent exponents, plus 1. For example, if A001221(n) = 3, a(n) = e_1*e_3 + e_1 + e_2 + e_3 + 1. If A001221(n) = k, the total number of terms always equals A000045(k+2).
%F A166469 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 i-th step be called r(i). r(1) = e_1 + 1; r(2) = e_1 + e_2 +1; for i > 2, r(i) = r(i-1) + e_i * r(i-2).
%F A166469 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 pair of consecutive primes as divisors.
%F A166469 a(n) = Sum_{d|n} (1-A296210(d)). - _Antti Karttunen_, Dec 15 2017
%e A166469 Since 3 of 30's 8 divisors (6, 15, and 30) are multiples of 2 or more consecutive primes, a(30) = 8 - 3 = 5.
%t A166469 Array[DivisorSum[#, 1 &, FreeQ[Differences@ PrimePi@ FactorInteger[#][[All, 1]], 1] &] &, 105] (* _Michael De Vlieger_, Dec 16 2017 *)
%o A166469 (PARI)
%o A166469 A296210(n) = { if(1==n,return(0)); my(ps=factor(n)[,1], pis=vector(length(ps),i,primepi(ps[i])), diffsminusones = vector(length(pis)-1,i,(pis[i+1]-pis[i])-1)); !factorback(diffsminusones); };
%o A166469 A166469(n) = sumdiv(n,d,!A296210(d)); \\ _Antti Karttunen_, Dec 15 2017
%Y A166469 Cf. A000040, A006094, A104210, A296210.
%Y A166469 A(A002110(n)) = A000045(n+2); A(A097250(n)) = A000032(n+1). For more relationships involving Fibonacci and Lucas numbers, see A166470-A166473, comment on A081341.
%Y A166469 A(A061742(n)) = A001045(n+2); A(A006939(n)) = A000085(n+1); A(A212170(n)) = A000142(n+1). A(A066120(n)) = A166474(n+1).
%K A166469 nonn
%O A166469 1,2
%A A166469 _Matthew Vandermast_, Nov 05 2009
%E A166469 Edited by _Matthew Vandermast_, May 24 2012

# Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE