A286470 a(n) = maximal gap between indices of successive primes in the prime factorization of n.
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
Keywords
Examples
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.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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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 *)
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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
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Scheme
(define (A286470 n) (cond ((or (= 1 n) (= 1 (A001221 n))) 0) (else (max (- (A055396 (A032742 n)) (A055396 n)) (A286470 (A032742 n))))))
Formula
Extensions
Definition corrected by Zak Seidov, May 16 2017
Comments