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A266226
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a(n) = floor(Sum_{d|n} 1 / tau(d)).
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1
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1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 2, 3, 2, 2, 1, 4, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 4, 2, 2, 2, 3
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OFFSET
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1,6
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COMMENTS
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a(n) = floor(Sum_{d|n} 1 / A000005(d)).
Sequences of numbers n such that floor(Sum_{d|n} 1/tau(d)) = k for k = 1..6:
k=1: 1, 2, 3, 4, 5, 7, 9, 11, 13, 17, 19, 23, 25, 29, 31, 37, 41, ... (A166684);
k=2: 6, 8, 10, 12, 14, 15, 16, 18, 20, 21, 22, 26, 27, 28, 32, 33, 34, 35, ...;
k=3: 24, 30, 36, 40, 42, 48, 54, 56, 66, 70, 72, 78, 80, 88, 96, 100, ...;
k=4: 60, 84, 90, 120, 126, 132, 140, 144, 150, 156, 168, 198, 204, 216, ...;
k=5: 180, 210, 240, 252, 300, 330, 336, 360, 390, 396, 450, 462, 468, ...;
k=6: 420, 630, 660, 720, 780, 900, 924, 990, 1008, 1020, 1050, 1080, ....
See A265393 - the smallest number n such that a(n) = k for k>= 1.
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LINKS
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EXAMPLE
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For n = 6; a(6) = floor(Sum_{d|6} 1/tau(d)) = floor(1/1 + 1/2 + 1/2 + 1/4) = floor(9/4) = 2.
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MATHEMATICA
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Table[Floor[Sum[1/DivisorSigma[0, d], {d, Divisors[ n]}]], {n, 1, 100}] (* G. C. Greubel, Dec 24 2015 *)
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PROG
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(Magma) [Floor(&+[1/NumberOfDivisors(d): d in Divisors(n)]): n in [1..100]]
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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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