A308237 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.

11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 24, 26, 28, 33, 36, 39, 44, 48, 55, 66, 77, 88, 99, 105, 108, 121, 132, 135, 143, 154, 165, 176, 187, 192, 195, 198, 225, 231, 242, 253, 264, 275, 286, 297, 315, 341, 352, 363, 374, 385, 396, 405, 451, 462, 473, 484, 495, 561, 572, 583, 594, 671, 682, 693

Offset

1,1

Comments

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

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.

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

Links

Marius A. Burtea, Table of n, a(n) for n = 1..95

Diophante, A333. Un chiffre à la trappe, Oct. 2011 (in French).

Example

264 is a term because 264/24 = 11.
34875 is a term because 34875/3875 = 9.

Mathematica

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 *)

Prog

(MATLAB) m=1;
for u=10:700 digit=dec2base(u, 10)-'0';
   if digit(length(digit))~=0 aa=str2num(strrep(num2str(digit), ' ', ''));
      digit(2)=[]; a=str2num(strrep(num2str(digit), ' ', ''));
if mod(aa, a)==0 sol(m)=u;  m=m+1;  end; end; end;
sol % Marius A. Burtea, May 16 2019
(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

Crossrefs

Cf. A034837, A178157.

Sequence in context: A267085 A284062 A261020 * A362023 A347471 A162672

Adjacent sequences: A308234 A308235 A308236 * A308238 A308239 A308240

Keyword

nonn,base,fini

Author

Bernard Schott, May 16 2019

Status

approved