cp's OEIS Frontend

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

%I A384539 #30 Aug 17 2025 02:08:15
%S A384539 11,12,13,14,15,16,17,18,19,21,22,24,26,28,31,33,36,39,41,42,44,48,51,
%T A384539 55,61,62,63,66,71,77,81,82,84,88,91,93,99,111,121,122,123,124,126,
%U A384539 131,141,142,147,151,153,155,161,162,164,168,171,181,182,183,186,189
%N A384539 Zeroless positive integers k for which for every pair of nonempty substrings that concatenate to give k one substring divides the other.
%C A384539 It is conjectured that this sequence is finite and contains 132 terms.
%C A384539 From _David A. Corneth_, Jun 19 2025: (Start)
%C A384539 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.
%C A384539 If a term with at least 5 digits ends in 89 then it is 100*k + 89 where k >= 1000.
%C A384539 We'd need 9 | 1000*k + 8, 89 | 100*k so 89 | k. This largely restrict the possibilities for k. (End)
%C A384539 a(133) > 10^11, if it exists. - _Michael S. Branicky_, Jun 18 2025
%C A384539 a(133) > 10^25, if it exists. - _David A. Corneth_, Jun 19 2025
%H A384539 Felix Huber, <a href="/A384539/b384539.txt">Table of n, a(n) for n = 1..132</a>
%H A384539 David A. Corneth, <a href="/A384539/a384539.gp.txt">PARI program</a>
%e A384539 168 is a term because 1 divides 68 and 8 divides 16.
%e A384539 4284 is a term because 4 divides 284, 42 divides 84 and 4 divides 428.
%e A384539 222222 is a term because 2 divides 22222, 22 divides 2222, 222 divides 222 and vice versa.
%p A384539 A384539:=proc(n)
%p A384539     option remember;
%p A384539     local i,j,k,p,m,q,L;
%p A384539     if n=1 then
%p A384539         11
%p A384539     else
%p A384539         for k from procname(n-1)+1 do
%p A384539             L:=ListTools:-Reverse(convert(k,'base',10));
%p A384539             if not member(0,L) then
%p A384539 	        m:=length(k)-1;
%p A384539 	        for j to m do
%p A384539 	            p:=parse(cat(seq(L[i],i=1..j)));
%p A384539 	            q:=k-p*10^(m+1-length(p));
%p A384539 	            if max(p,q) mod min(p,q)<>0 then
%p A384539 	                break
%p A384539 	            elif j=m then
%p A384539 	                return k
%p A384539 	            fi
%p A384539 	        od
%p A384539 	    fi
%p A384539         od
%p A384539     fi;
%p A384539 end proc;	
%p A384539 seq(A384539(n),n=1..60);
%o A384539 (Python)
%o A384539 def c(k, m): return m%k == 0 or k%m == 0
%o A384539 def ok(n):
%o A384539     s = str(n)
%o A384539     return n > 9 and "0" not in s and all(c(int(s[:i]), int(s[i:])) for i in range(1, len(s)))
%o A384539 print([k for k in range(200) if ok(k)]) # _Michael S. Branicky_, Jun 18 2025
%Y A384539 Subsequence of A052382.
%Y A384539 Subsequence of A384538.
%Y A384539 Cf. A102766, A228103.
%K A384539 nonn,base
%O A384539 1,1
%A A384539 _Felix Huber_, Jun 09 2025