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A286470
a(n) = maximal gap between indices of successive primes in the prime factorization of n.
48
0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 3, 1, 0, 0, 1, 0, 2, 2, 4, 0, 1, 0, 5, 0, 3, 0, 1, 0, 0, 3, 6, 1, 1, 0, 7, 4, 2, 0, 2, 0, 4, 1, 8, 0, 1, 0, 2, 5, 5, 0, 1, 2, 3, 6, 9, 0, 1, 0, 10, 2, 0, 3, 3, 0, 6, 7, 2, 0, 1, 0, 11, 1, 7, 1, 4, 0, 2, 0, 12, 0, 2, 4, 13, 8, 4, 0, 1, 2, 8, 9, 14, 5, 1, 0, 3, 3, 2, 0, 5, 0, 5, 1, 15, 0, 1, 0, 2, 10, 3, 0, 6, 6, 9, 4, 16, 3, 1
OFFSET
1,10
LINKS
FORMULA
a(1) = 0, for n > 1, if A001221(n) = 1 [when n is a prime power], a(n) = 0, otherwise a(n) = max((A055396(A032742(n))-A055396(n)), a(A032742(n))).
For all n >= 1, a(n) <= A243055(n).
EXAMPLE
For n = 70 = 2*5*7 = prime(1)*prime(3)*prime(4), the largest index difference occurs between prime(1) and prime(3), thus a(70) = 3-1 = 2.
MATHEMATICA
Table[If[Or[n == 1, PrimeNu@ n == 1], 0, Max@ Differences@ PrimePi[FactorInteger[n][[All, 1]]]], {n, 120}] (* Michael De Vlieger, May 16 2017 *)
PROG
(Scheme) (define (A286470 n) (cond ((or (= 1 n) (= 1 (A001221 n))) 0) (else (max (- (A055396 (A032742 n)) (A055396 n)) (A286470 (A032742 n))))))
(Python)
from sympy import primepi, isprime, primefactors, divisors
def a049084(n): return primepi(n)*(1*isprime(n))
def a055396(n): return 0 if n==1 else a049084(min(primefactors(n)))
def x(n): return 1 if n==1 else divisors(n)[-2]
def a(n): return 0 if n==1 or len(primefactors(n))==1 else max(a055396(x(n)) - a055396(n), a(x(n))) # Indranil Ghosh, May 17 2017
CROSSREFS
Cf. A286469 (version which considers the index of the smallest prime as the initial gap).
Cf. A000961 (positions of zeros).
Differs from A242411 for the first time at n=70, where a(70) = 2, while A242411(70) = 1.
Sequence in context: A087073 A297173 A242411 * A243055 A359358 A318371
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 13 2017
EXTENSIONS
Definition corrected by Zak Seidov, May 16 2017
STATUS
approved