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A174222
Number of symmetric primes in the interval [prime(n)^2, prime(n)*prime(n+1)].
0
1, 2, 2, 6, 4, 7, 5, 10, 18, 6, 24, 18, 10, 21, 35, 29, 14, 33, 27, 14, 44, 32, 43, 64, 36, 16, 36, 17, 38, 133, 41, 71, 16, 123, 21, 71, 72, 49, 90, 85, 36, 158, 34, 66, 31, 190, 184, 73, 39, 73, 109, 33, 188, 109, 117, 110, 35, 126, 85, 36, 221, 298, 99, 41, 95, 320, 136, 237
OFFSET
1,2
COMMENTS
If you graph a(n) vs. n, an interesting pattern with random-looking fluctuations begins to emerge.
As you go farther along the n-axis, greater are the number of symmetric primes, on average.
The smallest count of a(.)=1 occurs only once at the very beginning.
I suspect all a(n) are > 0. If one could prove this, it would imply that Symmetric primes are infinite.
FORMULA
#{ A090190(j): A001248(n) < A090190(j) < A006094(n)}.
EXAMPLE
The square of the first prime is 2^2=4 and the product of the first and second prime is 2*3=6. Within this interval, there is 1 symmetric prime, which is 5. Hence a(1)=1.
The second term, a(2)=2, refers to the two symmetric primes 11 and 13 within the interval (9,15).
PROG
(PARI) issym(p) = fordiv(p-1, d, if(isprime(p-d) || isprime(p+d), return(1))); 0; \\ A090190
a(n) = my(p=prime(n), nb=0); forprime(q=p^2, p*nextprime(p+1), if (issym(q), nb++)); nb; \\ Michel Marcus, Nov 03 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Jaspal Singh Cheema, Mar 18 2010
EXTENSIONS
Edited by R. J. Mathar, Mar 31 2010
STATUS
approved