A172483 - OEIS

%I #16 Nov 03 2022 10:08:54

%S 2,1,1,2,5,4,4,2,6,4,7,7,5,9,12,13,14,14,9,12,10,11,13,20,16,15,16,15,

%T 23,19,22,26,27,28,26,22,20,27,25,27,28,26,35,29,29,29,30,45,30,36,22,

%U 30,39,39,40,44,44,43,34,38,36,48,54,43,38,43,49,45,47,53,38,51,51,62,56

%N a(n) is the number of cousin primes between p^2 and p*(p+4) where p is the n-th cousin prime A023200(n).

%C If you graph the order of the consecutive cousin primes along the x-axis (i.e., first pair of cousin primes, second, third,...) and the number of cousin primes in the sequence given above along the y-axis, a clear pattern emerges. As you go farther along the x-axis, greater are the number of consecutive cousin primes, on average, within the interval obtained. If one can prove that there's at least one consecutive cousin prime within each interval, this would imply that cousin primes are infinite. I suspect the number of consecutive primes within each interval will never be zero. Can you prove it?

%D C. C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Perseus Books, 1999.

%D M. D. Sautoy, The Music of the Primes: Searching to Solve the Greatest Mystery in Mathematics, HarperCollins Publishers Inc., 2004.

%H J. S. Cheema, <a href="/A172483/b172483.txt">Table of n, a(n) for n = 1..1104</a> (2 prepended by Michael De Vlieger)

%e The 1st pair of cousin primes is (3, 7), between 3^2=9 and 3*7=21 there is 2 cousin primes: 13 and 19. So a(1) = 2.

%e The 2nd pair of cousin primes is (7, 11), between 7^2=49 and 7*11=77 there is 1 cousin prime: 67. So a(2) = 1.

%o (PARI) vcp(nn) = my(list=List(), p=3); listput(list, p); p=7; forprime(q=11, nn, if(q-p==4, listput(list, p)); p=q); Vec(list); \\ A023200

%o nbcp(p) = my(nb=0); forprime(q=p^2, p*(p+4), if (isprime(q+4), nb++)); nb;

%o lista(nn) = my(v=vcp(nn)); vector(#v, n, nbcp(v[n])); \\ _Michel Marcus_, Nov 02 2022

%Y Cf. A023200, A046132, A087679.

%K nonn

%O 1,1

%A _Jaspal Singh Cheema_, Feb 04 2010

%E New name and a(1)=2 prepended by _Michel Marcus_, Nov 02 2022