A266226 a(n) = floor(Sum_{d|n} 1 / tau(d)).

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

Offset

1,6

Comments

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.

Links

G. C. Greubel, Table of n, a(n) for n = 1..5000

Example

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.

Mathematica

Table[Floor[Sum[1/DivisorSigma[0, d], {d, Divisors[ n]}]], {n, 1, 100}] (* G. C. Greubel, Dec 24 2015 *)

Prog

(Magma) [Floor(&+[1/NumberOfDivisors(d): d in Divisors(n)]): n in [1..100]]

Crossrefs

Cf. A000005, A237350, A253139, A265390, A265391, A265392, A265393

Sequence in context: A244145 A371734 A086435 * A099305 A334461 A338652

Adjacent sequences: A266223 A266224 A266225 * A266227 A266228 A266229

Keyword

nonn

Author

Jaroslav Krizek, Dec 24 2015

Status

approved