A359113 a(n) counts the bases b in the interval 2 to p = prime(n), where p if written in base b gives again a prime number in base b if all digits are written in reverse order.
0, 1, 3, 5, 7, 10, 12, 9, 14, 15, 20, 19, 23, 26, 24, 33, 22, 30, 38, 36, 40, 39, 38, 33, 54, 49, 43, 52, 37, 60, 65, 53, 59, 57, 50, 52, 85, 52, 79, 76, 57, 77, 69, 103, 90, 83, 84, 106, 80, 68, 90, 85, 89, 94, 75, 100, 108, 87, 128, 97, 119, 99, 118, 139, 105, 96
Offset: 1
Examples
a(3) = 3: prime(3) = 5 in bases 2..5: 5 = 101_2; reversing digits gives 101_2 = 5 (prime). 5 = 12_3; reversing digits gives 21_3 = 7 (prime). 5 = 11_4; reversing digits gives 11_4 = 5 (prime). 5 = 10_5; reversing digits gives 01_5 = 1 (nonprime).
Links
- Thomas Scheuerle, Table of n, a(n) for n = 1..4000
- Thomas Scheuerle, Scatterplot a(n)/(n-1) for the first 2000 values. Will this quotient remain inside the interval 1..2.5 for any n? For n = 2..2000 a mean value of approximately 1.65... (orange line in plot) was observed.
- Rémy Sigrist, Scatterplot of (n, b) such that the reversal of prime(n) in base b gives a prime number with n, b <= 2000 and 2 <= b <= prime(n)
Programs
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PARI
revprime(p, b)=my(q, t=p); while(t, q=b*q+t%b; t\=b); isprime(q) a(n) = sum(b = 2, prime(n), revprime(prime(n), b))
Comments