A342304 k-digit positive numbers exactly one of whose substrings is divisible by k.

1, 2, 3, 4, 5, 6, 7, 8, 9, 21, 23, 25, 27, 29, 41, 43, 45, 47, 49, 61, 63, 65, 67, 69, 81, 83, 85, 87, 89, 101, 104, 107, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 122, 125, 128, 131, 134, 137, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 152, 155, 158, 161

Offset

1,2

Comments

Inspired by the 413th problem of Project Euler (see link) where such a number is called "one-child number".

There are k*(k+1)/2 substrings. All are considered, even when some are duplicates as strings or as numbers (see the Example section). 0 is always divisible by k so any number with two or more 0 digits is not a term. - Kevin Ryde, Mar 08 2021

The 2-digit terms are odd.

The number of k-digit terms for k = 1, 2, 3 is respectively 9, 20, 360.

From Robert Israel, Mar 11 2021: (Start)

5-digit terms are numbers starting with 5, and with no other digits 5 or 0.

There are no 10-digit terms. (End)

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Project Euler, Problem 413: One-child Numbers.

Example

107 is a 3-digit one-child number since among its substrings 1, 0, 7, 10, 07, 107 only 0 is divisible by 3.
222 is a 3-digit one-child number since among its substrings 2, 2, 2, 22, 22, 222 only 222 is divisible by 3.
572 is not a 3-digit one-child number, since among its substrings 5, 7, 1, 57, 72, 572 both 57 and 72 are divisible by 3.
616 is not a 3-digit one-child number, since among its substrings 6, 1, 6, 61, 16, 616 the two 6's are both divisible by 3.

Maple

filter:= proc(n) local L, d, i, j, k, ct, x;
  L:= convert(n, base, 10);
  d:= nops(L);
  ct:= 0:
  for i from 1 to d do
    for j from i to d do
      x:= add(L[k]*10^(k-i), k=i..j);
      if x mod d = 0 then ct:= ct+1; if ct = 2 then return false fi fi;
  od od;
  evalb(ct = 1)
end proc:
select(filter, [$1..200]); # Robert Israel, Mar 11 2021

Prog

(Python)
def ok(n):
    s, c = str(n), 0
    ss = (s[i:j] for i in range(len(s)) for j in range(i+1, len(s)+1))
    for w in ss:
        if int(w)%len(s) == 0: c += 1
        if c == 2: return False
    return n > 0 and c == 1
print([k for k in range(162) if ok(k)]) # Michael S. Branicky, Aug 15 2022

Crossrefs

Cf. A063527.

Sequence in context: A069571 A039173 A103951 * A098951 A097962 A247813

Adjacent sequences: A342301 A342302 A342303 * A342305 A342306 A342307

Keyword

nonn,base

Author

Bernard Schott, Mar 08 2021

Status

approved