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A117370
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Number of primes between smallest prime divisor of n and largest prime divisor of n.
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2
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0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 1, 3, 0, 0, 0, 4, 0, 2, 0, 1, 0, 0, 2, 5, 0, 0, 0, 6, 3, 1, 0, 2, 0, 3, 0, 7, 0, 0, 0, 1, 4, 4, 0, 0, 1, 2, 5, 8, 0, 1, 0, 9, 1, 0, 2, 3, 0, 5, 6, 2, 0, 0, 0, 10, 0, 6, 0, 4, 0, 1, 0, 11, 0, 2, 3, 12, 7, 3, 0, 1, 1, 7, 8, 13, 4, 0, 0, 2, 2, 1, 0, 5, 0
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OFFSET
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1,14
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COMMENTS
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This sequence first differs from sequence A117371 at the 30th term.
Records in a(n) are for n = 2*prime(k), for which a(n) = k-2. Examples: a(14) = a(2*prime(4)) = 4-2 = 2; a(22) = a(2*prime(5)) = 5-2 = 3; a(26) = a(2*prime(6)) = 6-2 = 4; a(74) = a(2*prime(12)) = 12-2= 10. Those records are each repeated for n = 2*(prime(k)^e_1)*(prime(m)^e_2)*(prime(n)^e_3)...*(prime(x)^e_y) where e_i are positive integers and prime(m), ..., prime(x) are between 2 and prime(k). Minima a(n) = 0 iff least spf(n)=gpf(n) iff n is 1 or a prime power (A000961), or a product of powers of consecutive primes (prime(k)^e_1)*(prime(k+1)^e_2). Here gpf(n) = greatest prime factor = A006530(n) and spf(n) = smallest prime factor = A020639(n). - Jonathan Vos Post, Mar 11 2006
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LINKS
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FORMULA
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EXAMPLE
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a(30) is 1 because there is one prime (which is 3) between the smallest prime dividing 30 (which is 2) and the largest prime dividing 30 (which is 5).
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PROG
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(PARI) A117370(n) = if(1>=omega(n), 0, my(f = factor(n), lpf = f[1, 1], gpf = f[#f~, 1]); -1+(primepi(gpf)-primepi(lpf))); \\ Antti Karttunen, Sep 10 2018
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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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