1,1

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.

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

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

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 *)

Cf. A132903, A255669, A258182.

Sequence in context: A049857 A218038 A132903 * A247474 A147148 A173629

Adjacent sequences: A385533 A385534 A385535 * A385537 A385538 A385539

nonn,base,more

Gonzalo Martínez, Jul 02 2025

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

approved