login
A227815
Composite numbers n divisible by their concatenated exponents in prime factorization.
1
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
1,1
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.
LINKS
EXAMPLE
248 = 2^3*31 => 31 is the concatenate exponents 3 and 1, so 31 divides 248.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Michel Lagneau, Jul 31 2013
STATUS
approved