A286472 Compound filter (for counting prime gaps): a(1) = 1, a(n) = 2*A032742(n) + (1 if n is composite and spf(A032742(n)) > nextprime(spf(n)), and 0 otherwise). Here spf is the smallest prime factor, A020639.

1, 2, 2, 4, 2, 6, 2, 8, 6, 11, 2, 12, 2, 15, 10, 16, 2, 18, 2, 20, 15, 23, 2, 24, 10, 27, 18, 28, 2, 30, 2, 32, 23, 35, 14, 36, 2, 39, 27, 40, 2, 42, 2, 44, 30, 47, 2, 48, 14, 51, 35, 52, 2, 54, 23, 56, 39, 59, 2, 60, 2, 63, 42, 64, 27, 66, 2, 68, 47, 71, 2, 72, 2, 75, 50, 76, 22, 78, 2

Offset: 1

Author

Antti Karttunen, May 11 2017

Keywords

• nonn

Comments

For n > 1, a(n) is odd if and only if n is a composite with its smallest prime factor occurring only once and with a gap of at least one between the smallest and the next smallest prime factor.

For all i, j: a(i) = a(j) => A073490(i) = A073490(j). This follows because A073490(n) can be computed by recursively invoking a(n), without needing any other information.

Examples

			For n = 4 = 2*2, the two smallest prime factors (taken with multiplicity) are 2 and 2, and the difference between their indices is 0, thus a(4) = 2*A032742(4) + 0 = 2*(4/2) + 0 = 2.
For n = 6 = 2*3 = prime(1)*prime(2), the difference between the indices of two smallest prime factors is 1 (which is less than required 2), thus a(6) = 2*A032742(6) + 0 = 2*(6/2) + 0 = 6.
For n = 10 = 2*5 = prime(1)*prime(3), the difference between the indices of two smallest prime factors is 2, thus a(10) = 2*A032742(10) + 1 = 2*(10/2) + 1 = 11.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A032742, A285729, A286471, A286473, A286474, A286475, A286476.

Cf. A000040 (primes give the positions of 2's).

Cf. A073490 (one of the matched sequences).

Programs

Mathematica

Table[Function[{p, d}, 2 d + If[And[CompositeQ@ n, FactorInteger[d][[1, 1]] > NextPrime[p]], 1, 0] - Boole[n == 1]] @@ {#, n/#} &@ FactorInteger[n][[1, 1]], {n, 98}] (* Michael De Vlieger, May 12 2017 *)

Python

from sympy import primefactors, divisors, nextprime
def ok(n): return 1 if isprime(n)==0 and min(primefactors(divisors(n)[-2])) > nextprime(min(primefactors(n))) else 0
def a(n): return 1 if n==1 else 2*divisors(n)[-2] + ok(n) # Indranil Ghosh, May 12 2017

Scheme

(define (A286472 n) (if (= 1 n) n (+ (* 2 (A032742 n)) (if (> (A286471 n) 2) 1 0))))

Formula

a(n) = 2*A032742(n) + [A286471(n) > 2], a(1) = 1.