A173291 -id:A173291 - cp's OEIS frontend

A176781 Smallest prime prime(i) such that concatenation 2//0_(n)//prime(i) is prime.

3, 11, 3, 17, 3, 3, 3, 11, 89, 41, 257, 3, 29, 131, 353, 3, 3, 11, 89, 521, 257, 3, 17, 3, 467, 89, 149, 17, 71, 47, 293, 17, 191, 47, 3, 41, 23, 11, 401, 41, 443, 41, 293, 479, 311, 23, 587, 41, 1289, 1013, 29, 41, 59, 293, 1031, 17, 23, 17, 347, 401, 599, 11, 227, 827, 401

Offset: 0

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Apr 26 2010

We search for the prime such that the first prime (=2) concatenated with n zeros and concatenated with that prime is again a prime number.
If p = prime(i) is a d(i)-digit prime: q = 2 * 10^(n+d(i)) + p has to be prime.
Necessarily prime(i) is congruent to 2 (mod 3).
It is conjectured that prime(i) = 3 occurs infinitely often: at n= 0, 2, 4, 5, 6, 11, 15, 16, 21, 23, 34, 114, 119,...

			n = 0: 2//3 = 23 = prime(9), 3 = prime(2) is first term
n = 1: 2//0//11 = 2011 = prime(305), 11 = prime(5) is 2nd term
n = 2: 2//00//3 = 2003 = prime(304), 3 = prime(2) is 3rd term

E. I. Ignatjew, Mathematische Spielereien, Urania Verlag Leipzig/Jena/ Berlin 1982

Cf. A164968, A173291, A176316

Offset corrected and sequence extended by R. J. Mathar, Apr 28 2010

A202997 a(1) = 3 ; a(n+1) is the prime number obtained by the concatenation {p, a(n)} where p is the smallest prime prefix.

Original entry on oeis.org

3, 23, 223, 31223, 231223, 31231223, 2931231223, 372931231223, 17372931231223, 1317372931231223, 1971317372931231223, 1571971317372931231223, 891571971317372931231223, 79891571971317372931231223, 25179891571971317372931231223, 4325179891571971317372931231223

Offset: 1

Michel Lagneau, Dec 27 2011

By Xylouris' version of Linnik's theorem, a(n) << 3^(6.2^n + cn) for some constant c. [Charles R Greathouse IV, Dec 27 2011]

			a(1) = 3;
a(2) = 23 because 2 is the smallest prime prefix and 23 is prime;
a(3) = 223 because 2 is the smallest prime prefix and 223 is prime;
a(4) = 31223 because 31 is the smallest prime prefix and 31223 is prime.

Cf. A173291, A379354, A379355, A379782.

a0:=3: printf(`%d, `,a0):for it from 1 to 20 do: i:=0:for n from 1 to 1000 while(i=0)  do:p0:=ithprime(n):n0:=length(a0):x:=p0*10^n0+a0: if type(x,prime)=true then printf(`%d, `,x):i:=1:a0:=x:else fi:od:od:

A176833 Smallest prime p = prime(i) such that concatenation q(i) = 13//0_(k)//prime(i) (k = 0, 1, 2, ...) is prime.

Original entry on oeis.org

7, 3, 3, 3, 151, 61, 7, 3, 19, 3, 109, 109, 19, 19, 37, 409, 109, 97, 61, 19, 73, 109, 139, 139, 619, 31, 127, 31, 193, 3, 43, 19, 337, 7, 73, 367, 109, 373, 139, 139

Offset: 1

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Apr 27 2010

See comments in A176781
Necessarily p = 3 or p of form 3 * n + 1
In recreational mathematics some authors call a prime that is composed of mostly naughts, i.e. zeros, a naughty prime

			q(0) = 13//7 = 137 = prime(33), 7 = prime(4) is 1st term
q(1) = 13//0//3 = 1303 = prime(213), 3 = prime(2) is 2nd term
q(26) = 13000000000000000000000000031 is a palindromic prime

A164968, A173291, A176316, A176781

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A176781 Smallest prime prime(i) such that concatenation 2//0_(n)//prime(i) is prime.

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A202997 a(1) = 3 ; a(n+1) is the prime number obtained by the concatenation {p, a(n)} where p is the smallest prime prefix.

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A176833 Smallest prime p = prime(i) such that concatenation q(i) = 13//0_(k)//prime(i) (k = 0, 1, 2, ...) is prime.

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