A384539 Zeroless positive integers k for which for every pair of nonempty substrings that concatenate to give k one substring divides the other.

11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 24, 26, 28, 31, 33, 36, 39, 41, 42, 44, 48, 51, 55, 61, 62, 63, 66, 71, 77, 81, 82, 84, 88, 91, 93, 99, 111, 121, 122, 123, 124, 126, 131, 141, 142, 147, 151, 153, 155, 161, 162, 164, 168, 171, 181, 182, 183, 186, 189

Offset: 1

Author

Felix Huber, Jun 09 2025

Keywords

• nonn
• base

Comments

It is conjectured that this sequence is finite and contains 132 terms.

From David A. Corneth, Jun 19 2025: (Start)

A term with at least 3 digits cannot end in 14. Else we can split it in 10*k + 1 and 4 where k >= 1. So we'd need 4 | 10*k + 1. A contradiction.

If a term with at least 5 digits ends in 89 then it is 100*k + 89 where k >= 1000.

We'd need 9 | 1000*k + 8, 89 | 100*k so 89 | k. This largely restrict the possibilities for k. (End)

a(133) > 10^11, if it exists. - Michael S. Branicky, Jun 18 2025

a(133) > 10^25, if it exists. - David A. Corneth, Jun 19 2025

Examples

			168 is a term because 1 divides 68 and 8 divides 16.
4284 is a term because 4 divides 284, 42 divides 84 and 4 divides 428.
222222 is a term because 2 divides 22222, 22 divides 2222, 222 divides 222 and vice versa.
		

Links

• Felix Huber, Table of n, a(n) for n = 1..132
• David A. Corneth, PARI program

Crossrefs

Subsequence of A052382.

Subsequence of A384538.

Cf. A102766, A228103.

Programs

Maple

A384539:=proc(n)
    option remember;
    local i,j,k,p,m,q,L;
    if n=1 then
        11
    else
        for k from procname(n-1)+1 do
            L:=ListTools:-Reverse(convert(k,'base',10));
            if not member(0,L) then
	        m:=length(k)-1;
	        for j to m do
	            p:=parse(cat(seq(L[i],i=1..j)));
	            q:=k-p*10^(m+1-length(p));
	            if max(p,q) mod min(p,q)<>0 then
	                break
	            elif j=m then
	                return k
	            fi
	        od
	    fi
        od
    fi;
end proc;	
seq(A384539(n),n=1..60);

Python

def c(k, m): return m%k == 0 or k%m == 0
def ok(n):
    s = str(n)
    return n > 9 and "0" not in s and all(c(int(s[:i]), int(s[i:])) for i in range(1, len(s)))
print([k for k in range(200) if ok(k)]) # Michael S. Branicky, Jun 18 2025