OFFSET
1,1
COMMENTS
Any term with two or more digits contains exactly two different digits from the set {1, 3, 7, 9} (cf. Erdős et al., 1977, Solution 953).
Conjecture: The sequence is finite, with 991 being the last term.
The known terms are those terms of A293663 where membership in A293663 trivially implies membership in this sequence, i.e., the numbers resulting from all cyclic permutations of the digits of these terms are the same as the numbers resulting from all permutations of the digits of these terms. This is the case only for terms with less than four digits.
LINKS
P. Erdős et al., Problems, Mathematics Magazine, Vol. 50, No. 2 (1977), 99-104.
Arkadii Slinko, Absolute Primes, arXiv:1811.08613 [math.NT], 2018.
Wikipedia, Permutable prime
EXAMPLE
The other numbers resulting from all possible permutations of the digits of 113 are 131 and 311. 113, 131 and 311 are all primes, so all three numbers are terms of this sequence.
PROG
(PARI) eva(n) = subst(Pol(n), x, 10)
find_index_a(vec) = my(r=#vec-1); while(1, if(vec[r] < vec[r+1], return(r)); r--; if(r==0, return(-1)))
find_index_b(r, vec) = my(s=#vec); while(1, if(vec[r] < vec[s], return(s)); s--; if(s==r, return(-1)))
switch_elements(vec, firstpos, secondpos) = my(g); g=vec[secondpos]; vec[secondpos]=vec[firstpos]; vec[firstpos] = g; vec
reverse_order(vec, r) = my(v=[], w=[]); for(x=1, r, v=concat(v, vec[x])); for(y=r+1, #vec, w=concat(w, vec[y])); w=Vecrev(w); concat(v, w)
next_permutation(vec) = my(r=find_index_a(vec)); if(r==-1, return(0), my(s=find_index_b(r, vec)); vec=switch_elements(vec, r, s); vec=reverse_order(vec, r)); vec
is(n) = if(n < 10, return(1)); my(d=vecsort(digits(n))); if(vecmin(d)==0 || vecmax(d)==1, return(0)); while(1, if(!ispseudoprime(eva(d)), return(0)); d=next_permutation(d); if(d==0, return(1)))
forprime(p=1, , if(is(p), print1(p, ", ")))
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Felix Fröhlich, Aug 04 2018
STATUS
approved