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A350832
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a(n) is the least integer > 9 not listed earlier whose first two digits together with a(n-1)'s last two digits form two 2-digit primes when "read vertically in the same direction"; a(1) = 11.
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1
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11, 13, 12, 19, 15, 33, 14, 17, 16, 31, 21, 37, 23, 91, 24, 93, 18, 39, 22, 99, 25, 331, 26, 97, 29, 371, 34, 71, 36, 77, 41, 73, 32, 79, 35, 131, 27, 311, 43, 111, 44, 112, 132, 133, 28, 332, 134, 113, 38, 135, 136, 114, 115, 137, 46, 116, 117, 49, 171, 47
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OFFSET
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1,1
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COMMENTS
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To explain "read vertically in the same direction": let b, c be the last two digits of a(n-1), and d, e the first two digits of a(n), then either both of 10b + d and 10c + e, or both of 10d + a and 10e + b must be primes > 9.
We are tempted to conjecture that this sequence contains all integers > 10 which don't have '0' as second or one of the last two digits.
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LINKS
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Éric Angelini, Prime bayonets, personal web site "Cinquante signes" on blogspot.com, May 03 2022.
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EXAMPLE
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a(2) together with a(3) form the two 2-digit primes 11 and 23, when we write
a(2) = 13 and read the two last columns
a(3) = 12 both from bottom to top.
a(11) = 21 is followed by 37 and not 34 since we cannot write
a(11) = 21 and read one column downwards to get the prime 23
next = 34 and the other column upwards to get the prime 41.
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PROG
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(PARI) u=[a=11]; (ok(x, y)=isprime(x[1]*10+y[1])&&isprime(x[2]*10+y[2])); vector(99, n, if(a>u[1]+1, u=setunion(u, [a]), u[1]=a; while(#u>1 && u[2]==u[1]+1+(u[1]%10==9), u=u[^1])); a=digits(a)[-2..-1]; my(k=u[1], d); n>1&& until(k++%10 && k%100>9&& !setsearch(u, k) && (d=divrem(k\10^logint(k\10, 10), 10))[2]&& (ok(a, d)||ok(d, a)), ); a=k)
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CROSSREFS
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Cf. A350831 for a simpler single-digit variant.
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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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