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A327265
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a(n) is the smallest k such that A309981(k) = n.
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3
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1, 2, 5, 11, 19, 31, 51, 89, 123, 151, 179, 181, 180, 365, 634, 657, 656, 655
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OFFSET
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0,2
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COMMENTS
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Consider the problem of deducing the identity of an unknown positive integer k given only its number of divisors tau(k) and those of its first m successors, i.e., tau(k+1), tau(k+2), ... tau(k+m). A309981(k) is the minimum m that allows unique identification of k, and a(n) is the smallest k for which that minimum m is n.
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LINKS
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EXAMPLE
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n = 0: Using 0 successors, the fact that tau(k) = 1 is sufficient to deduce that k = 1; there is no other k whose identity can be deduced given only tau(k), so a(0) = 1.
n = 1: tau(k) = 2 for all primes k, but given tau(k) = 2 and tau(k+1) = 2, the only solution is k = 2 (since k = 2 and k + 1 = 3 are the only two consecutive integers that are both prime). Other than k = 1, whose identity can be deduced given only tau(k), k = 2 is the smallest k whose identity can be deduced given only tau(k) and tau(k+1), so a(1) = 2.
n = 2: There are many integers k such that tau(k) = 2 and tau(k+1) = 4, but the only k such that tau(k), tau(k+1), and tau(k+2) are 2, 4, and 2, respectively, is k = 5. Thus A309981(5) = 2. There is no number k < 5 for which A309981(k) = 2, so a(2) = 5.
There are numbers m != 89 such that tau(m+j) = tau(89+j) for all j in 0..6 (the first such number is 242510633), but there is no number m such that tau(m+j) = tau(89+j) for all j in 0..7, and 89 is the smallest k such that A309981(k)=7, so a(7)=89.
a(8)=123; the smallest m such that tau(m+j) = tau(123+j) for all j in 0..7 is apparently 476129486151666513937.
a(9)=151; the smallest m such that tau(m+j) = tau(151+j) for all j in 0..8 is 3579145012951.
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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