login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A166469 Number of divisors of n which are not multiples of consecutive primes. 12
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, 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, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Links various subsequences of A025487 with an unusual number of important sequences, including the Fibonacci, Lucas, and other generalized Fibonacci sequences (see cross-references).
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.
LINKS
Eric Weisstein's World of Mathematics, Fibonacci n-Step 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 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).
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).
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.
a(n) = Sum_{d|n} (1-A296210(d)). - Antti Karttunen, Dec 15 2017
EXAMPLE
Since 3 of 30's 8 divisors (6, 15, and 30) are multiples of 2 or more consecutive primes, a(30) = 8 - 3 = 5.
MATHEMATICA
Array[DivisorSum[#, 1 &, FreeQ[Differences@ PrimePi@ FactorInteger[#][[All, 1]], 1] &] &, 105] (* Michael De Vlieger, Dec 16 2017 *)
PROG
(PARI)
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); };
A166469(n) = sumdiv(n, d, !A296210(d)); \\ Antti Karttunen, Dec 15 2017
CROSSREFS
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.
A(A061742(n)) = A001045(n+2); A(A006939(n)) = A000085(n+1); A(A212170(n)) = A000142(n+1). A(A066120(n)) = A166474(n+1).
Sequence in context: A135981 A135615 A348856 * A080226 A060741 A125747
KEYWORD
nonn
AUTHOR
Matthew Vandermast, Nov 05 2009
EXTENSIONS
Edited by Matthew Vandermast, May 24 2012
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified April 24 17:29 EDT 2024. Contains 371962 sequences. (Running on oeis4.)