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A166469
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Number of divisors of n which are not multiples of consecutive primes.
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12
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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FORMULA
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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.
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EXAMPLE
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Since 3 of 30's 8 divisors (6, 15, and 30) are multiples of 2 or more consecutive primes, a(30) = 8 - 3 = 5.
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MATHEMATICA
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Array[DivisorSum[#, 1 &, FreeQ[Differences@ PrimePi@ FactorInteger[#][[All, 1]], 1] &] &, 105] (* Michael De Vlieger, Dec 16 2017 *)
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PROG
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(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); };
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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