login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A370513 Complement of A323599. 0
2, 5, 29, 39, 53, 59, 73, 95, 119, 123, 125, 129, 137, 145, 147, 149, 157, 159, 163, 173, 179, 191, 199, 207, 209, 213, 219, 221, 235, 251, 257, 263, 265, 269, 271, 279, 291, 293, 299, 303, 305, 325, 327, 329, 343, 345, 347, 359, 365, 367, 369, 375, 385, 395, 397 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Terms of this sequence are not solutions of Sum_{d|k} A069359(d), k >= 1.
Proof that 2 is not a solution of Sum_{d|k} A069359(d), k >= 1: (Start)
If 2 is a solution then the only summands of the above are either (0,2) or (0,1,1).
(0,2) cannot be the only summands. If 2 is a summand then it is also a divisor of a(n) and A069359(2) = 1. If 2 is a summand then so must 1 be a summand.
(0,1,1) cannot be the only summands. There must exist an additional summand A069359(p_1*p_2) where p_1 and p_2 (primes) contribute to each 1 in (0,1,1).
(End)
To prove that 5 is not a solution of Sum_{d|k} A069359(d), k >= 1 we need to show that each of the following summands cannot exist: (0,5), (0,1,4), (0,1,2,2), (0,1,1,3), (0,1,1,1,2). (0,1,1,1,1,1). Following from the above proof this is elementary.
LINKS
EXAMPLE
2 is a term because it is not a solution of Sum_{d|k} A069359(d), k >= 1. See proof in Comments.
CROSSREFS
Sequence in context: A165161 A098858 A213995 * A134449 A103579 A161500
KEYWORD
nonn
AUTHOR
Torlach Rush, Feb 20 2024
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified August 20 21:50 EDT 2024. Contains 375339 sequences. (Running on oeis4.)