A328040 a(n) is the number of integers b with 1 < b < p such that p = prime(n) is a base-b nonrepunit circular prime with at least two base-b digits.

0, 0, 1, 3, 4, 7, 9, 7, 11, 12, 15, 14, 18, 23, 20, 28, 18, 24, 30, 31, 35, 34, 32, 29, 48, 41, 40, 45, 35, 54, 58, 50, 56, 54, 47, 43, 78, 47, 74, 70, 50, 69, 63, 93, 82, 78, 78, 103, 69, 62, 82, 79, 82, 87, 68, 92, 100, 80, 120, 89, 117, 91, 112, 132, 97, 93

Offset: 1

Author

Felix Fröhlich, Oct 03 2019

Keywords

• nonn
• base

Comments

Conjecture: a(n) > 0 for n > 2.

Examples

			For n = 4: 7 is the 4th prime and in base 3, 7 is 21, with 12 equal to 5 in decimal, which is prime, in base 4, 7 is 13, with 31 equal to 13 in decimal, which is prime and in base 5, 7 is 12, with 21 equal to 11 in decimal, which is prime. Altogether, there are 3 such bases, so a(4) = 3.
		

Crossrefs

Cf. A293142.

Programs

PARI

rot(n) = if(#Str(n)==1, v=vector(1), v=vector(#n-1)); for(i=2, #n, v[i-1]=n[i]); u=vector(#n); for(i=1, #n, u[i]=n[i]); v=concat(v, u[1]); v
decimal(v, base) = my(w=[]); for(k=0, #v-1, w=concat(w, v[#v-k]*base^k)); sum(i=1, #w, w[i])
is_circularprime(p, base) = my(db=digits(p, base), r=rot(db), i=0); if(vecmin(db)==0, return(0), while(1, dec=decimal(r, base); if(!ispseudoprime(dec), return(0)); r=rot(r); if(r==db, return(1))))
count_bases(n) = my(i=0); for(b=3, n-1, if(vecmin(digits(n, b))!=vecmax(digits(n, b)), if(is_circularprime(n, b), i++))); i
forprime(p=1, 400, print1(count_bases(p), ", "))