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A345730
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Variation on the Inventory Sequence A342585: record the number of occurrences of previous terms with an incrementing number of given divisors until 0 is recorded, then restart the divisor count from 0. See the Comments.
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6
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0, 1, 1, 0, 2, 2, 2, 0, 3, 2, 5, 0, 4, 2, 7, 1, 0, 5, 3, 10, 1, 1, 0, 6, 5, 11, 1, 2, 0, 7, 6, 14, 1, 4, 0, 8, 7, 15, 2, 6, 0, 9, 7, 17, 3, 7, 0, 10, 7, 21, 3, 9, 0, 11, 7, 24, 4, 9, 0, 12, 7, 25, 7, 9, 0, 13, 7, 28, 8, 10, 0, 14, 7, 29, 8, 13, 0, 15, 7, 32, 8, 15, 0, 16, 7, 33, 8, 18, 1, 4
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OFFSET
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0,5
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COMMENTS
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This sequence is a variation of A342585. Instead of iteratively counting the occurrences of each number starting from 0 and then repeating when 0 is recorded, we record the number of previous terms with a given number of divisors, where the number of divisors starts at 0 and increments by one until 0 is recorded, after which the divisor count restarts at 0. It is assumed 0 has zero divisors while the number of divisors for all other numbers is given by A000005. See the Examples below.
After 10 million terms, numbers with four, eight and two divisors are the most common, and in general terms with an even number of divisors are more common than those with an odd number. However like A342585 the graph of the sequence shows that the number of terms with a given number of divisors increases somewhat haphazardly and thus it is unclear if terms with four divisors stay the most numerous as n -> infinity.
Alternative definition: Record the number of existing terms which are divisible by any m > 0, then record the number of terms having 1 divisor, then 2, 3 and so on until a zero is recorded. Repeat the process indefinitely. - David James Sycamore, Sep 25 2021
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LINKS
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Scott R. Shannon, Colored image of the first 1000 terms. In this and other colored images the colors are graduated across the spectrum to show which line corresponds to the divisor count it indicates. See the colored key at the top-left. Up to 1000 terms the most common divisor count is 2 followed by 4.
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EXAMPLE
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The sequence begins 0, 1, 1, 0, 2, 2, 2, 0. After the initial 0, a(1) counts the terms with 0 divisors (i.e., the 0's), which is 1. a(2) then counts the terms with one divisor (i.e., the 1's), which is 1, and a(3) counts the terms with two divisors (i.e., the primes), which is 0. So the divisor count then resets to 0 and a(4) counts the terms with 0 divisors, which is 2. a(5) counts the terms with one divisor, which is 2, and a(6) counts the terms with two divisors, which is 2. There are no terms with three divisors so a(7) = 0 and the divisor count then resets to 0.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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