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A293659
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Base-6 circular primes that are not base-6 repunits.
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7
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OFFSET
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1,1
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COMMENTS
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Conjecture: The sequence is finite, with 211 being the last term (see A293142).
This sequence may be particularly constrained to few terms since only {1, 5} are coprime to 6, and any senary circular prime involves just these 2 senary digits. This is because all primes aside from {2, 3} are congruent to {1, 5} (mod 6). Since a senary number consisting of all 5's is divisible by 5 and since we have disqualified prime repunits, the sequence is probably finite.
a(6), if it exists, must be larger than 6^21 = 21936950640377856. (End)
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LINKS
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EXAMPLE
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71 written in base 6 is 155. The base-6 numbers 155, 515, 551 written in base 10 are 71, 191, 211, respectively and all those numbers are prime, so 71, 191 and 211 are terms of the sequence.
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MATHEMATICA
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With[{b = 6}, Select[Prime@ Range[PrimePi@ b + 1, 10^6], Function[w, And[AllTrue[Array[FromDigits[RotateRight[w, #], b] &, Length@ w - 1], PrimeQ], Union@ w != {1} ]]@ IntegerDigits[#, b] &]] (* or *)
With[{b = 6}, Select[Flatten@ Array[FromDigits[#, 6] & /@ Most@ Rest@ Tuples[{1, 5}, #] &, 18, 2], Function[w, And[ AllTrue[ Array[ FromDigits[ RotateRight[w, #], b] &, Length@ w], PrimeQ], Union@ w != {1} ]]@ IntegerDigits[#, b] &]] (* Michael De Vlieger, Dec 30 2017 *)
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PROG
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(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))))
forprime(p=1, , if(vecmin(digits(p, 6))!=vecmax(digits(p, 6)), if(is_circularprime(p, 6), print1(p, ", "))))
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CROSSREFS
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KEYWORD
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nonn,base,more
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AUTHOR
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STATUS
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approved
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