A372046 Composite numbers that divide the concatenation of the reverse of their ascending order prime factors, with repetition.

998, 1636, 9998, 15584, 49447, 99998, 1639964, 2794612, 9999998, 15842836, 1639360636, 1968390098, 27879461212, 65226742928

Offset: 1

Author

Scott R. Shannon, Apr 17 2024

Keywords

• nonn
• base
• more

Comments

A number 999...9998 will be a term if it has two prime factors 2 and 4999...999. Therefore 999999999999998 and 999...9998 (with 54 9's) are both terms. See A056712.

100000000000 < a(15) <= 999999999999998. Robert P. P. McKone, May 07 2024

Examples

			998 is a term as 998 = 2 * 499 = "2" * "994" when each prime factor is reversed. This gives "2994", and 2994 is divisible by 998.
15584 is a term as 15584 = 2 * 2 * 2 * 2 * 2 * 487 = "2" * "2" * "2" * "2" * "2" * "784" when each prime factor is reversed. This gives "22222784", and 22222784 is divisible by 15584.
		

Crossrefs

Cf. A371696, A371695, A371641, A027746, A056712.

Programs

Mathematica

a[n_Integer] := Module[{f}, f = Flatten[ConstantArray @@@ FactorInteger[n]]; If[Length[f] < 2, Return[False]]; Mod[FromDigits[StringJoin[StringReverse[IntegerString[#, 10]] & /@ f], 10], n] == 0];
Select[Range[2, 10^5], a] (* Robert P. P. McKone, May 03 2024 *)

Python

from itertools import count, islice
from sympy import factorint
def A372046_gen(startvalue=4): # generator of terms >= startvalue
    for n in count(max(startvalue,4)):
        f = factorint(n)
        if sum(f.values()) > 1:
            c = 0
            for p in sorted(f):
                a = pow(10,len(s:=str(p)),n)
                q = int(s[::-1])
                for _ in range(f[p]):
                    c = (c*a+q)%n
            if not c:
                yield n
A372046_list = list(islice(A372046_gen(),5)) # Chai Wah Wu, Apr 24 2024

Extensions

a(13)-a(14) from Robert P. P. McKone, May 05 2024