A244471 Lexicographically earliest sequence of integers with property that if a vertical line is drawn between any pair of adjacent digits, the number Z formed by the digits to the left of the line is divisible by the digit to the right of the line.

1, 11, 3, 7, 71, 31, 111, 113, 33, 117, 77, 13, 37, 711, 1111, 19, 9, 91, 1117, 73, 311, 131, 1131, 1133, 93, 331, 11111, 39, 99, 97, 119, 333, 911, 133, 931, 1139, 771, 337, 713, 339, 933, 391, 1137, 773, 1113, 991, 11171, 3111, 777, 3311, 79, 17, 191, 171, 11311, 137, 719, 993

Offset: 1

Author

N. J. A. Sloane, Jul 02 2014

Keywords

• nonn
• base

Comments

"Lexicographically earliest" means in the sense of a sequence of integers, not digits.

No digit can be even or five. - Hans Havermann, Jul 02 2014 [Proof: if not, let d be the first digit in the sequence that is even or 5, and let Z be the concatenation of all earlier digits. But then Z is odd and does not end in 5, so is not divisible by d. Contradiction. - N. J. A. Sloane, Jul 03 2014] So any term must have only the odd digits {1, 3, 7, 9} (see A136333). - Robert G. Wilson v, Jul 02 2014

We choose the next term, a(n), to be the minimal number not already in the sequence such that the property "if a vertical line is drawn between any pair of adjacent digits, the number Z formed by the digits to the left of the line is divisible by the first digit following Z" holds.

So even if Z is prime, the next term can start with a 1.

So if Z is divisible by any d in {2,3,...,9} the next term can start with 1 or d, otherwise it must start with 1.

This sequence is missing A136333 terms 313, 319, 373, 379, 717, 737, 797, 913, 919, 939, 973, 979, 1313, ... The earliest occurrences of n-digit numbers are the repunits at indices 1, 2, 7, 15, 27, 97, 372, 939, 2164, 4781, 10851, 22779, 47056, ... The latest n-digit numbers and their indices are: (9,17), (17,52), (397,290), (1917,867), (19317,2003), (199117,7241), (1999117,17953), (19999997,44173), ... - Hans Havermann, Jul 04 2014, Jul 07 2014, Jul 15 2014

Examples

			After 1,11,3,7, let a(5) = x be the next term. Now 11137 = 7*37*43, so x must begin with 1 or 7. The candidates for x are therefore 12,13,...,19,71,72,,...,79,111,...
If x=12, we would get 1 11 3 7 12 ... but Z = 11371 is prime and is not divisible by 2, ..., 9. So x is not 12, ...,19. The next candidate is x=71, and this works. So a(5)=71.
		

References

• Eric Angelini, Posting to Sequence Fans Mailing List, Jun 26 2014

Links

• Robert G. Wilson v and Hans Havermann (Robert G. Wilson v to 1000), Table of n, a(n) for n = 1..10850

Crossrefs

A sister sequence to A243357 and A244496. A subsequence of A136333.

Programs

Mathematica

r=f=e={1,3,7,9};Do[e=10*e;f=Flatten[Table[e[[i]]+f,{i,4}]];r=Join[r,f],{9}];r=Select[r,Intersection[Partition[IntegerDigits[#],3,1],IntegerDigits[{313,319,373,379,717,737,797,913,919,939,973,979}]]=={}&];t=0;Do[c=1;While[d=IntegerDigits[r[[c]]];Union[Table[IntegerQ[(10^i*t+FromDigits[Take[d,i]])/d[[i+1]]],{i,0,Length[d]-1}]]!={True},c++];Print[r[[c]]];t=10^Length[d]*t+r[[c]];r=Delete[r,c],{10850}] (* Hans Havermann, Jul 04 2014 *)

Extensions

Corrected and extended by Hans Havermann, Jul 02 2014