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Revision History for A369169

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Showing entries 1-10 | older changes
Terms k of A025487 such that A000005(k) = A000688(k).
(history; published version)
#14 by OEIS Server at Mon Jan 15 10:03:21 EST 2024
LINKS

Amiram Eldar, <a href="/A369169/b369169_1.txt">Table of n, a(n) for n = 1..377</a>

#13 by Michael De Vlieger at Mon Jan 15 10:03:21 EST 2024
STATUS

reviewed

approved

Discussion
Mon Jan 15
10:03
OEIS Server: Installed first b-file as b369169.txt.
#12 by Joerg Arndt at Mon Jan 15 08:53:53 EST 2024
STATUS

proposed

reviewed

#11 by Amiram Eldar at Mon Jan 15 06:45:51 EST 2024
STATUS

editing

proposed

Discussion
Mon Jan 15
06:48
David A. Corneth: From tuples in prime signatures we could add and remove fours as we like and keep retrieving signatures of terms. So like signature [9,4,2,1] produces a term and so does [9,2,1] and [9,4,4,2,1], [9,4,4,4,2,1] and so on
06:54
Amiram Eldar: Yes. This is because 4 is the solution e to A000041(e) = e+1.
#10 by Amiram Eldar at Mon Jan 15 06:45:48 EST 2024
CROSSREFS
STATUS

proposed

editing

#9 by David A. Corneth at Mon Jan 15 06:44:25 EST 2024
STATUS

editing

proposed

#8 by David A. Corneth at Mon Jan 15 06:43:54 EST 2024
COMMENTS

This sequence contains A002110(n)^4. - _From _David A. Corneth_, Jan 15 2024: (Start)

16 | a(n) for n > 1.

This sequence contains A002110(n)^4. (End)

#7 by David A. Corneth at Mon Jan 15 06:39:58 EST 2024
COMMENTS

This sequence contains A002110(n)^4. - David A. Corneth, Jan 15 2024

EXAMPLE

16 is in the sequence as 16 has 5 divisors (1, 2, 4, 8, 16) and 5 factorizations into prime powers (16 = 2*8 = 4*4 = 2*2*4 = 2*2*2*2).

STATUS

proposed

editing

#6 by Amiram Eldar at Mon Jan 15 03:17:28 EST 2024
STATUS

editing

proposed

#5 by Amiram Eldar at Mon Jan 15 03:15:17 EST 2024
COMMENTS

Because Since both A000005(k) and A000688(k) depend only on the prime signature of k (A124832), if k is a term of this sequence then every number m such that A046523(m) = k is a term of A369168.