The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A061704 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A061704

Number of cubes dividing n.
(history; published version)

#49 by R. J. Mathar at Wed May 10 07:37:05 EDT 2023

STATUS

editing

approved

#48 by R. J. Mathar at Wed May 10 07:36:58 EDT 2023

MAPLE

# alternative

A061704 := proc(n)

local a, pe ;

a := 1 ;

for pe in ifactors(n)[2] do

op(2, pe) ;

a := a*(1+floor(%/3)) ;

end do:

a ;

end proc:

seq(A061704(n), n=1..80) ; # R. J. Mathar, May 10 2023

STATUS

approved

editing

#47 by Alois P. Heinz at Fri Dec 16 17:15:59 EST 2022

STATUS

proposed

approved

#46 by Geoffrey Critzer at Fri Dec 16 16:58:43 EST 2022

STATUS

editing

proposed

Discussion

Fri Dec 16

17:15

Alois P. Heinz: yes ... thanks!

#45 by Geoffrey Critzer at Fri Dec 16 16:58:04 EST 2022

EXAMPLE

a(128) = 4 3 since 128 is divisible by 1^3 = 1, 2^3 = 8 and 4^3 = 64.

STATUS

approved

editing

Discussion

Fri Dec 16

16:58

Geoffrey Critzer: I corrected a typo

#44 by Wesley Ivan Hurt at Mon Feb 01 21:52:38 EST 2021

STATUS

editing

approved

#43 by Wesley Ivan Hurt at Mon Feb 01 21:52:36 EST 2021

FORMULA

a(n) = Sum_{k=1..n} chi(1 - ceiling(n/k^3) + floor(n/k^3), where chi is the integer characteristic). - Wesley Ivan Hurt, Jan 28 2021

STATUS

approved

editing

#42 by Wesley Ivan Hurt at Sat Jan 30 21:39:13 EST 2021

STATUS

proposed

approved

#41 by Wesley Ivan Hurt at Thu Jan 28 23:49:28 EST 2021

STATUS

editing

proposed

#40 by Wesley Ivan Hurt at Thu Jan 28 23:49:20 EST 2021

FORMULA

a(n) = Sum_{k=1..n} chi(n/k^3), where chi is the integer characteristic. - Wesley Ivan Hurt, Jan 28 2021

STATUS

approved

editing