A395103 Numbers with exactly k prime factors, counting repeats, with k>1, that are formed by concatenating k primes.

22, 25, 33, 35, 55, 57, 77, 115, 133, 177, 213, 217, 219, 222, 235, 237, 247, 253, 255, 259, 267, 273, 275, 289, 295, 319, 322, 323, 325, 329, 332, 333, 341, 357, 361, 371, 377, 413, 415, 417, 437, 473, 511, 517, 519, 529, 533, 535, 537, 543, 553, 555, 559, 573, 575, 579, 583, 589, 597

Offset

1,1

Comments

Conjecture: The sequence has infinitely many terms.

The conjecture is true if A117360 is infinite. - Yehune Jeong, May 24 2026

Links

Table of n, a(n) for n=1..59.

Example

237 has 2 prime factors (3, 79) and is the concatenation of 2 primes (23, 7).
332 has 3 prime factors (2, 2, 83) and is the concatenation of 3 primes (3, 3, 2).
3255 has 4 prime factors (3, 5, 7, 31) and is the concatenation of 4 primes (3, 2, 5, 5).

Maple

filter:= n -> not isprime(n) and g(n, NumberTheory:-Omega(n)):
g:= proc(n, t)
  local d, i, a;
  if t = 1 then return isprime(n) fi;
  d:= 1+ilog10(n);
  if d < t then return false fi;
  for i from 1 to d-t+1 do
    a:= n mod 10^i;
    if a > 10^(i-1) and isprime(a) and procname((n-a)/10^i, t-1) then return true fi;
  od;
  false
end proc:
select(filter, [$10..1000]); # Robert Israel, May 18 2026

Prog

(PARI) is(n) = {my(b = bigomega(n)); if(b <= 1, return(0)); isConcatenationOfbPrimes(n, b)}
isConcatenationOfbPrimes(n, b) = {if(b == 1, return(isprime(n))); my(qd = #digits(n), cn, res = 0, g = n%10); if(gcd(g, 10) > 1, if(g == 2 || g == 5, return(isConcatenationOfbPrimes(n\10, b-1))); return); for(i = 1, qd - b + 1, cn = n\10^i; if((isprime(n - 10^i*cn) && n - 10^i*cn >= 10^(i-1)), if(isConcatenationOfbPrimes(cn, b-1), res = 1; return(res)))); res} \\ David A. Corneth, May 19 2026

Crossrefs

Cf. A001222, A117360.

Sequence in context: A338603 A322124 A178423 * A108632 A045096 A227408

Adjacent sequences: A395100 A395101 A395102 * A395104 A395105 A395106

Keyword

nonn,base

Author

Philip Jameson, May 10 2026

Status

approved