A308237 - OEIS

%I #34 Jul 01 2019 01:22:17

%S 11,12,13,14,15,16,17,18,19,22,24,26,28,33,36,39,44,48,55,66,77,88,99,

%T 105,108,121,132,135,143,154,165,176,187,192,195,198,225,231,242,253,

%U 264,275,286,297,315,341,352,363,374,385,396,405,451,462,473,484,495,561,572,583,594,671,682,693

%N Numbers m not ending with 0 that contain a digit, other than the leftmost digit, that can be removed such that the resulting number d divides m.

%C When m is a term, then, necessarily, the digit that is removed is the second from the left.

%C This sequence is finite with 95 integers and the greatest term is 180625. The number of terms with respectively 2, 3, 4, 5, 6 digits is 23, 44, 10, 17, 1.

%C The obtained quotients m/d belong to: { 6, 7, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19 } (all proofs in Diophante link).

%H Marius A. Burtea, <a href="/A308237/b308237.txt">Table of n, a(n) for n = 1..95</a>

%H Diophante, <a href="http://www.diophante.fr/problemes-par-themes/arithmetique-et-algebre/a3-nombres-remarquables/2110-a333-un-chiffre-a-la-trappe">A333. Un chiffre à la trappe</a>, Oct. 2011 (in French).

%e 264 is a term because 264/24 = 11.

%e 34875 is a term because 34875/3875 = 9.

%t Select[Range[700], With[{m = #}, And[Mod[#, 10] != 0, AnyTrue[FromDigits@ Delete[IntegerDigits[m], #] & /@ Range[2, IntegerLength@ m], Mod[m, #] == 0 &]]] &] (* _Michael De Vlieger_, Jun 09 2019 *)

%o (MATLAB) m=1;

%o for u=10:700 digit=dec2base(u,10)-'0';

%o    if digit(length(digit))~=0 aa=str2num(strrep(num2str(digit), ' ', ''));

%o       digit(2)=[]; a=str2num(strrep(num2str(digit), ' ', ''));

%o if mod(aa,a)==0 sol(m)=u;  m=m+1;  end; end; end;

%o sol % _Marius A. Burtea_, May 16 2019

%o (PARI) isok(m) = {if (m % 10, my(d=digits(m)); for (k=2, #d, mk = fromdigits(vector(#d-1, i, if (i<k, d[i], d[i+1]))); if (!(m % mk), return(1));););} \\ _Michel Marcus_, Jun 21 2019

%Y Cf. A034837, A178157.

%K nonn,base,fini

%O 1,1

%A _Bernard Schott_, May 16 2019