A227815 Composite numbers n divisible by their concatenated exponents in prime factorization.

4, 16, 22, 27, 33, 55, 63, 77, 143, 187, 209, 222, 248, 253, 256, 319, 341, 407, 451, 473, 484, 517, 555, 583, 649, 656, 671, 737, 777, 781, 803, 837, 869, 913, 979, 1067, 1111, 1133, 1152, 1177, 1199, 1221, 1243, 1397, 1441, 1443, 1507, 1529, 1639, 1661, 1727

Offset: 1

Author

Michel Lagneau, Jul 31 2013

Keywords

• nonn
• base

Comments

The numbers 2^(2^m), m = 1, 2,... are in the sequence. A majority of semiprimes of the form 11*p where is p prime different from 11 are in the sequence. The numbers of the form p*111 = p*3*37 where p is prime different from 3 or 37 are in the sequence. In the general case, the numbers of the form n = p_1*p_2*...*p_m*R where the p_k are prime numbers and R is a repunit number (A002275) with q digits "1" and (q-m) prime divisors are in the sequence, for example the numbers of the form n = p_1*p_2*p_3*p_4*11111111 = p_1*p_2*p_3*p_4*11*73*101*137 are in the sequence if the primes p_k are different from 11, 73, 101 or 137. So, 11111111 divides n.

Examples

			248 = 2^3*31 => 31 is the concatenate exponents 3 and 1, so 31 divides 248.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A002275, A037916, A070529.

Programs

Maple

with(numtheory):for n from 1 to 10000 do:x:=ifactors(n):y:=x[2];n1:=nops(y):s:=0:for i from 1 to n1 do:z:=y[i][2]:s:=s+z*10^(n1-i):od:if type(n,prime)=false and irem(n,s)=0 then printf(`%d, `, n):else fi:od:

Mathematica

With[{predicate = And[CompositeQ[#], Divisible[#, FromDigits[Join @@ IntegerDigits@(Last /@ FactorInteger[#])]]] &},
Select[Range[10000], predicate]] (* Sidney Cadot, Feb 19 2023 *)

Python

from sympy import isprime, factorint
def ok(n): return n > 1 and not isprime(n) and n%int("".join(str(e) for e in factorint(n).values())) == 0
print([k for k in range(1728) if ok(k)]) # Michael S. Branicky, Feb 19 2023