A203565 Numbers that contain the product of any two adjacent digits as a substring.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 30, 31, 40, 41, 50, 51, 60, 61, 70, 71, 80, 81, 90, 91, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 126, 130, 131, 140, 141, 150, 151

Offset: 1

Author

M. F. Hasler, Jan 03 2012

Keywords

• nonn
• base

Comments

Inspired by the problem restricted to pandigital numbers suggested by E. Angelini (cf. link).

E. Angelini observes that up to a(86) this is the same as "Numbers that contain the product of (all) their digits as a substring" (cf. A227510 for the zeroless terms); then 212 is here but not there, and 236 is there and not here. - M. F. Hasler, Oct 14 2014

Examples

			Any number having no two adjacent digits larger than 1 is trivially in the sequence.
The smallest nontrivial example is the number 126, which is in the sequence since 1*2=2 and 2*6=12 are both substrings of "126".
		

Links

• Jayanta Basu, Table of n, a(n) for n = 1..1000
• E. Angelini, 10 different digits, 9 products, seqfan list, Jan 03 2012.

Crossrefs

Cf. A203569 (digits are permutations of 0...n).

Cf. also A203566, A198298, A236402, A236403, A236404.

Cf. A227510 (product of all digits is a substring and > 0).

Programs

Maple

filter:= proc(n)
local L,S,i;
S:= convert(n,string);
for i from 1 to length(S)-1 do
  if StringTools:-Search(convert(parse(cat(S[i],"*",S[i+1])),string),S) = 0 then
      return false
  fi
od:
true
end proc:
select(filter, [$0..1000]); # Robert Israel, Oct 15 2014

Mathematica

d[n_] := IntegerDigits[n]; Select[Range[0, 151], And @@ Table[MemberQ[FromDigits /@ Partition[d[#], IntegerLength[k], 1], k], {k, Times @@@ Partition[d[#], 2, 1]}] &] (* Jayanta Basu, Aug 10 2013 *)

PARI

has(n,m)={ my(p=10^#Str(m)); until( m>n\=10, n%p==m & return(1))}
is_A203565(n)={ my(d); for(i=2,#d=eval(Vec(Str(n))), has(n,d[i]*d[i-1]) | return);1 }
is_A203565(n)={ my(d=Vecsmall(Str(n))); for(i=2,#d, d[i]<50 & i++ & next; has(n,d[i-1]%48*(d[i]-48)) | return);1 } /* twice as fast */
for( n=0,999, is_A203565(n) & print1(n","))