A380943 Primes written in decimal representation by the concatenation of primes p and q such that the concatenation of q and p also forms a prime.

37, 73, 113, 173, 197, 311, 313, 317, 331, 337, 359, 367, 373, 593, 617, 673, 719, 733, 761, 797, 977, 1093, 1097, 1123, 1277, 1319, 1361, 1373, 1783, 1913, 1931, 1979, 1997, 2293, 2297, 2311, 2347, 2389, 2713, 2837, 2971, 3109, 3119, 3137, 3191, 3229, 3271

Offset: 1

Author

James C. McMahon, Apr 03 2025

Keywords

• nonn
• base

Comments

Member primes could be named Saint Patrick primes because the date of Saint Patrick's Day, March 17 (3/17), produces the terms 173 and 317.

The primes 37, 73, and 373 uniquely satisfy the requirements if cleaved at every possible position. - Robert G. Wilson v, May 03 2025

Examples

			Primes 173 and 317 are members because they are formed by the concatenation of 17 & 3  and 3 & 17, respectively.
While the concatenation of 13 and 7 forms the prime 137, it is not a member because the concatenation of 7 and 13 is 713, which is not prime.
		

Links

• James C. McMahon, Table of n, a(n) for n = 1..1000
• Michael De Vlieger, Plot a(n) = p<>q at (x,y) = (pi(p), pi(q)) showing all terms with pi(p) <= 40 and pi(q) <= 40, labeling p in red and q in blue.
• Michael De Vlieger, Plot a(n) = p<>q at (x,y) = (pi(p), pi(q)) showing all terms with pi(p) <= 2000 and pi(q) <= 2000.

Crossrefs

Subsequence of A019549 and A105184.

Cf. A133187.

Programs

Mathematica

lim=3300; plim=Max[FromDigits[Rest[IntegerDigits[lim]]], FromDigits[Drop[IntegerDigits[lim], -1]]]; p=Prime[Range[PrimePi[plim]]];p2=Subsets[p,{2}];fp[{a_,b_}]:=FromDigits[Join[IntegerDigits[a],IntegerDigits[b]]];rfp[{a_,b_}]:=FromDigits[Join[IntegerDigits[b],IntegerDigits[a]]];pabba=Select[p2,PrimeQ[fp[#]]&&PrimeQ[rfp[#]]&];pp1=fp/@pabba;pp2=rfp/@pabba;Select[Union[Join[pp1,pp2]],#<=lim&]

Python

from sympy import isprime
def ok(n):
    if not isprime(n): return False
    s = str(n)
    return any(s[i]!='0' and isprime(int(s[:i])) and isprime(int(s[i:])) and isprime(int(s[i:]+s[:i])) for i in range(1, len(s)))
print([k for k in range(3300) if ok(k)]) # Michael S. Branicky, Apr 05 2025