A385536 Integers k formed by the concatenation of 3 consecutive prime numbers such that k is divisible by at least one of those 3 primes.

235, 357, 71113, 374143, 616771, 167173179, 143914471451, 909071909089909091, 104651162787110465116278991046511627907

Offset: 1

Author

Gonzalo Martínez, Jul 02 2025

Keywords

• nonn
• base
• more

Comments

Subsequence of A132903.

If p is A255669, as it divides the concatenation of the next two primes, then p divides the concatenation of p with the next two primes. Thus, the first 4 terms of A255669 give rise to a(2), a(3), a(5) and a(6). In this sequence the number formed by concatenating 3 consecutive primes is allowed to be divisible by at least one of those 3, which generates more possibilities.

a(9) has p > 10^11 and thus >= 36 digits. - Michael S. Branicky, Jul 02 2025

a(9) has p <= A258182(11) - 36 = 1046511627871.

a(10) has p <= A258182(22) - 226.

Examples

			235 is a term, since it is the concatenation of the consecutive primes 2, 3, 5 and is divisible by 5.
2, 3 and 5: 235 = 5*47
3, 5, and 7: 357 = 3*119 = 7*51
7, 11 and 13: 71113 = 7*10159
37, 41 and 43: 374143 = 43*8701
61, 67 and 71: 616771 = 61*10111
167, 173 and 179: 167173179 = 167*1001037
1439, 1447 and 1451: 143914471451 = 1447*99457133
909071, 909089 and 909091: 909071909089909091 = 909091*999979000001
		

Crossrefs

Cf. A132903, A255669, A258182.

Programs

Maple

tcat:= proc(a,b,c)
   c + (b + 10^(1+ilog10(b))*a)*10^(1+ilog10(c))
end proc:
R:= NULL: count:= 0:
q:= 2: r:= 3:
while count < 8 do
  p:= q; q:= r; r:= nextprime(r);
  x:= tcat(p,q,r);
  if igcd(x,p*q*r)>1 then
    R:= R,x; count:= count+1;
  fi
od:
R; # Robert Israel, Jul 06 2025

Mathematica

cat[s_] := FromDigits[Flatten[IntegerDigits[s]]]; q[s_] := AnyTrue[s, Divisible[cat[s], #] &]; cat /@ Select[Partition[Prime[Range[72000]], 3, 1], q] (* Amiram Eldar, Jul 03 2025 *)

Extensions

a(9) from Michael S. Branicky, Jul 03 2025