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A321210
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Palindromic primes in base 10 of the form 6*p+1, where p is a palindromic prime in base 10.
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2
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787, 72781818727, 787812181218787, 72781878787818727, 78121818781812187, 78181272727218187, 7278781818181878727, 7872787812187872787, 727872187212781278727, 781272121818121272187, 781272727818727272187, 781872727212727278187, 72121872127872127812127, 72727272187278127272727
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OFFSET
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1,1
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COMMENTS
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Conjecture 1: There are infinitely many primes of this form.
Conjecture 2: 6 is the only integer k for which palindrome primes of the form k*p+1 exist, where p is a palindromic prime.
Conjecture 2 is false. A counterexample: 180 * 101 + 1 = 18181 where 101 and 18181 are palindromic primes and k = 180. - David A. Corneth, Nov 01 2018
p must have an odd number of digits (since 11 is the only palindrome prime with an even number of digits) and the leading digit of p must be 1, otherwise 6p+1 has an even number of digits. So p must have the form 1...1 and 6p+1 the form 7...7.
Some other terms of this sequence: 727872187212781278727, 781272121818121272187, 781272727818727272187, 781872727212727278187, 72121872127872127812127, 72727272187278127272727, 72781872727872727818727, 78127212127872121272187, 78181218181218181218187, 78727272121212127272787, 78787272787278727278787. - Giovanni Resta, Oct 31 2018
Suppose q = 6p+1 is a term written as q = a0a1a2...a2a1a0 and p = b0b1b2...b2b1b0.
The discussion above shows that a0 = 7 and b0 = 1.
b_i is either 2 or 3 for i odd and b_i is either 0 or 1 for i even (see MathExchange link).
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LINKS
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EXAMPLE
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787 is in the sequence because 787 = 6*131 + 1, where 131 is a palindromic prime.
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PROG
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(PARI) print1(787", " ); for(u=0, 20, z=[[1, 1], [2, 3]]; for(j=1, u-1, z=concat(z, [[0, 9]])); forvec(j=z, for(k=0, 9, n=0; for(m=1, length(j), n=n*10+j[m]); n=n*10+k; forstep(m=length(j), 1, -1, n=n*10+j[m]); if(digits(6*n+1)==Vecrev(digits(6*n+1)), if(ispseudoprime(n)==1, if(ispseudoprime(6*n+1)==1, print1(6*n+1", " )))))))
(Python)
from sympy import isprime
for i in range(2**20):
s = bin(i)[2:]
s += s[-2::-1]
p = int(s) + int('02'*(len(s)//2)+'0')
q = 6*p+1
t = str(q)
if t == t[::-1] and isprime(p) and isprime(q):
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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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EXTENSIONS
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STATUS
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approved
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