The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A143206

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A143206

Product of the n-th cousin prime pair.
(history; published version)

#58 by Michael De Vlieger at Sat Jul 13 10:13:13 EDT 2024

STATUS

reviewed

approved

#57 by Stefano Spezia at Sat Jul 13 08:48:51 EDT 2024

STATUS

proposed

reviewed

Discussion

Sat Jul 13

08:50

Stefano Spezia: You are welcome

#56 by Bill McEachen at Sat Jul 13 08:21:41 EDT 2024

STATUS

editing

proposed

#55 by Bill McEachen at Sat Jul 13 08:20:56 EDT 2024

FORMULA

a(n) = (A023200(n) + 2)^2 - 4. - Bill McEachen, Jul 10 2024

STATUS

proposed

editing

Discussion

Sat Jul 13

08:21

Bill McEachen: @Stefano I understand. Thank you, done

#54 by Bill McEachen at Wed Jul 10 12:59:12 EDT 2024

STATUS

editing

proposed

Discussion

Wed Jul 10

13:27

Michel Marcus: (p+2)^2-4 = p^2+4*p+4 - 4 = p^2+4*p = p*(p+4)

15:46

Bill McEachen: @Michel Can you just revert this edit completely? Not so useful

Sat Jul 13

03:09

Stefano Spezia: We could not revert since we will keep Michel edit (formula move). If you desire, you could delete your formula and to wait for review

#53 by Bill McEachen at Wed Jul 10 12:57:07 EDT 2024

COMMENTS

a(n) is a member of the following (n>1): gcd(m^2-4 , 30)==1, with m an odd >3. Precisely twenty percent of such m^2-4 meet this criteria due to a cycle of 15. - Bill McEachen, Jul 06 2024

FORMULA

a(n) = (A023200(n) + 2)^2 - 4. - Bill McEachen, Jul 10 2024

STATUS

proposed

editing

Discussion

Wed Jul 10

12:57

Bill McEachen: @Michel I just switched to simpler formula ...

#52 by Bill McEachen at Tue Jul 09 13:05:39 EDT 2024

STATUS

editing

proposed

Discussion

Tue Jul 09

14:51

Michel Marcus: for m=221, gcd(m^2-4 , 30) gives 3,   so it is not 1; what am I missing ??

16:35

Bill McEachen: @Michel for 221, m=15. 15^2-4=13*17. Gcd(221,30)==1. Sorry if I'm not explaining well

Wed Jul 10

08:57

Michel Marcus: What about  ?   It appears that a(n) is of the form k^2-4; and for n>1 a(n) is coprime to 30.

#51 by Bill McEachen at Tue Jul 09 07:41:44 EDT 2024

COMMENTS

Conjecture: a(n) is a member of the following (n>1): gcd(m^2-4 , 30)==1, with m an odd >3. Precisely ten twenty percent of such m^2-4 meet this criteria, implying infinitude due to a cycle of cousin primes15. - Bill McEachen, Jul 06 2024

STATUS

proposed

editing

Discussion

Tue Jul 09

07:42

Bill McEachen: I realized it was straightfwd to prove ...

#50 by Michel Marcus at Sun Jul 07 03:40:54 EDT 2024

STATUS

editing

proposed

Discussion

Sun Jul 07

06:07

Bill McEachen: @Michel  this seq is subset of the superset

06:08

Bill McEachen: @Michel: please wait, rechecking ...

06:18

Bill McEachen: @Michel: 221=15^2-4  and looks to satisfy the gcd test ? 13*17

#49 by Michel Marcus at Sun Jul 07 03:37:01 EDT 2024

COMMENTS

a(n) = A023200(n)*A046132(n);

intersection Intersection of A143203 and A001358.

Conjecture: a(n) is a member of the following (n>1): gcd(m^2-4 , 30)==1, m>3. Precisely ten percent of m^2-4 meet this criteria, implying infinitude of Cousin cousin primes. - Bill McEachen, Jul 06 2024

FORMULA

a(n) = A023200(n)*A046132(n).

STATUS

proposed

editing

Discussion

Sun Jul 07

03:40

Michel Marcus: 221 does not satisfy gcd(m^2-4 , 30)==1, so your comment does not seem to work ?