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 A158378 a(1) = 0, a(n) = gcd(A051904(n), A051903(n)) for n >= 2. 3
 0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS a(n) for n >= 2 equals GCD of minimal and maximal exponents in prime factorization of n. For n >= 2 holds: a(n)*A157754(n) = A051904(n)*A051903(n). a(n) for n >= 2 it deviates from A052409(n), first different term is a(10800) = a(2^4*3^3*5^2), a(10800) = gcd(2,4) = 2, A052409(10800) = gcd(2,3,4) = 1. LINKS Antti Karttunen, Table of n, a(n) for n = 1..65536 FORMULA a(1) = 0, a(p) = 1, a(pq) = 1, a(pq...z) = 1, a(p^k) = k, for p = primes (A000040), pq = product of two distinct primes (A006881), pq...z = product of k (k > 2) distinct primes p, q, ..., z (A120944), p^k = prime powers (A000961(n) for n > 1), k = natural numbers (A000027). EXAMPLE For n = 12 = 2^2 * 3^1 we have a(12) = gcd(2,1) = 1. MATHEMATICA Table[GCD @@ {Min@ #, Max@ #} - Boole[n == 1] &@ FactorInteger[n][[All, -1]], {n, 100}] (* Michael De Vlieger, Jul 12 2017 *) PROG (PARI) A051903(n) = if((1==n), 0, vecmax(factor(n)[, 2])); A051904(n) = if((1==n), 0, vecmin(factor(n)[, 2])); A158378(n) = gcd(A051903(n), A051904(n)); \\ Antti Karttunen, Jul 12 2017 CROSSREFS Cf. A157754, A051904, A051903, A052409. Sequence in context: A327503 A158052 A253641 * A052409 A051904 A070012 Adjacent sequences:  A158375 A158376 A158377 * A158379 A158380 A158381 KEYWORD nonn AUTHOR Jaroslav Krizek, Mar 17 2009 STATUS approved

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Last modified June 1 13:11 EDT 2020. Contains 334762 sequences. (Running on oeis4.)