A293663 Circular primes that are not repunits.

2, 3, 5, 7, 13, 17, 31, 37, 71, 73, 79, 97, 113, 131, 197, 199, 311, 337, 373, 719, 733, 919, 971, 991, 1193, 1931, 3119, 3779, 7793, 7937, 9311, 9377, 11939, 19391, 19937, 37199, 39119, 71993, 91193, 93719, 93911, 99371, 193939, 199933, 319993, 331999, 391939

Offset: 1

Author

Felix Fröhlich, Dec 30 2017

Keywords

• nonn
• base

Comments

Relative complement of A004022 in A068652.

Conjecture: The sequence is finite.

From Michael De Vlieger, Dec 30 2017: (Start)

Primes > 5 in this sequence must only have digits that are in the reduced residue system modulo 10, i.e., {1, 3, 7, 9}.

There are 54 terms that have 6 or fewer decimal digits, the largest of which is 999331.

a(55) must be larger than 10^11. (End) [Corrected by Felix Fröhlich, Mar 15 + 24 2019]

From Felix Fröhlich, Mar 16 2019: (Start)

a(55) > 10^23 if it exists (cf. De Geest link).

Numbers k such that A262988(k) = A055642(k). (End)

Examples

			The numbers resulting from cyclic permutations of the digits of 1193 are 1931, 9311 and 3119, respectively and all those numbers are prime, so 1193, 1931, 3119 and 9311 are terms of the sequence.
		

Links

• Felix Fröhlich, Table of n, a(n) for n = 1..54
• P. De Geest, Circular Primes

Crossrefs

Cf. A004022, A055642, A068652, A262988, A293142.

Cf. base-b nonrepunit circular primes: A293657 (b=4), A293658 (b=5), A293659 (b=6), A293660 (b=7), A293661 (b=8), A293662 (b=9).

Programs

Mathematica

Select[Prime@ Range[10^5], Function[w, And[AllTrue[Array[FromDigits@ RotateRight[w, #] &, Length@ w - 1], PrimeQ], Union@ w != {1} ]]@ IntegerDigits@ # &] (* or *)
Select[Flatten@ Array[FromDigits /@ Most@ Rest@ Tuples[{1, 3, 7, 9}, #] &, 9, 2], Function[w, And[AllTrue[Array[FromDigits@ RotateRight[w, #] &, Length@ w], PrimeQ], Union@ w != {1} ]]@ IntegerDigits@ # &] (* Michael De Vlieger, Dec 30 2017 *)

PARI

eva(n) = subst(Pol(n), x, 10)
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
is_circularprime(p) = my(d=digits(p), r=rot(d)); if(vecmin(d)==0, return(0), while(1, if(!ispseudoprime(eva(r)), return(0)); r=rot(r); if(r==d, return(1))))
forprime(p=1, , if(vecmax(digits(p)) > 1, if(is_circularprime(p), print1(p, ", "))))

PARI

/* The following is a much faster program that only tests numbers whose decimal expansion consists of digits from the set {1, 3, 7, 9}. */
eva(n) = subst(Pol(n), x, 10)
next_v(vec) = my(k=#vec); if(vecmin(vec)==9, vec=concat(vector(#vec, t, 1), [3]); return(vec)); while(k > 0, if(vec[k]==9, vec[k]=1, if(vec[k]==3, vec[k]=7; return(vec), vec[k]=vec[k]+2, return(vec))); k--)
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
search(n) = my(d=digits(n), e=[], ed=0); while(1, e=rot(d); while(1, if(!ispseudoprime(eva(e)), break, e=rot(e); if(e==d && ispseudoprime(eva(e)), print1(eva(d), ", "); break))); d=next_v(d))
searchfrom(n) = if(n < 12, forprime(p=n, 10, print1(p, ", ")); search(13), my(d=digits(n)); for(k=1, #d, if(d[k]%2==0, d[k]++, if(d[k]==5, d[k]=7))); search(eva(d)))
/* Start a search from 1 upwards as follows: */
searchfrom(1) \\ Felix Fröhlich, Mar 23 2019