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A346026
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Primes that are the first in a run of exactly 6 emirps.
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4
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10039, 14891, 39791, 119773, 149561, 162293, 163781, 176903, 181751, 197383, 336689, 392911, 393361, 714361, 715361, 779003, 971141, 995443, 996539, 1165037, 1284487, 1307729, 1447151, 1611877, 1640539, 1789621, 1891147, 3136909, 3150557, 3284447, 3339943
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OFFSET
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1,1
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COMMENTS
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There are large gaps in this sequence because all terms need to begin with 1, 3, 7, or 9 otherwise the reversal is composite.
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LINKS
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EXAMPLE
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a(1) = 10039 because of the eight consecutive primes 10037, 10039, 10061, 10067, 10069, 10079, 10091, 10093 all except 10037 and 10093 are emirps and this is the first such occurrence.
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MATHEMATICA
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EmQ[n_]:=(s=IntegerReverse@n; PrimeQ@s&&n!=s);
Select[Prime@Range[2, 50000], Boole[EmQ/@NextPrime[#, Range[-1, 6]]]=={0, 1, 1, 1, 1, 1, 1, 0}&] (* Giorgos Kalogeropoulos, Jul 27 2021 *)
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PROG
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(Python)
from sympy import isprime, nextprime, prime, primerange
def isemirp(p): s = str(p); return s != s[::-1] and isprime(int(s[::-1]))
def aupto(limit, runlength=6):
alst = []
pvec = list(primerange(1, prime(runlength+2)+1))
evec = [int(isemirp(p)) for p in pvec]
target = [0] + [1 for i in range(runlength)] + [0]
p = nextprime(pvec[-1])
while pvec[1] <= limit:
if evec == target: alst.append(pvec[1])
pvec = pvec[1:] + [p]; evec = evec[1:] + [isemirp(p)]; p = nextprime(p)
strp = str(p)
if strp[0] in "24568": # skip large gaps (p is a prime, not an emirp)
evec = [0 for i in range(runlength+2)]
pvec = [0 for i in range(runlength+2)]
p = nextprime(int(str(int(strp[0])+1)+'0'*(len(strp)-1)))
return alst
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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