A358610 Numbers k such that the concatenation 1,2,3,... up to (k-1) is one less than a multiple of k.

1, 2, 4, 5, 8, 10, 13, 20, 25, 40, 50, 52, 100, 125, 200, 250, 400, 475, 500, 601, 848, 908, 1000, 1120, 1250, 1750, 2000, 2500, 2800, 2900, 3670, 4000, 4375, 4685, 5000, 6085, 7000, 7640, 7924, 8375, 10000, 10900, 12500, 13346, 14000, 17800, 20000, 21568, 25000

Offset: 1

Author

Martin Renner, Nov 23 2022

Keywords

• nonn
• base

Comments

For a >= 0, the infinite subsequence of numbers 10^a, 2^b*10^a (for 1 <= b <= 2) and 5^c*10^a (for 1 <= c <= 3), i.e., 1, 2, 4, 5, 10, 20, 25, 40, 50, 100, 125, 200, 250, 400, 500, 1000, 1250, 2000, 2500, 4000, 5000, ... are terms in the sequence because first, the concatenation 1, 2, 3, ... up to (10^a - 1) mod 10^a is equal to 10^a times the concatenation 1, 2, 3, ... up to (10^a - 2) + (10^a - 1) mod 10^a, which results in 10^a - 1 and second, the concatenation 1, 2, 3, ... up to (2^b*10^a - 1) mod 2^b*10^a is equal to 10^(a+1) times the concatenation 1, 2, 3, ... up to (2^b*10^a - 2) + (2^b*10^a - 1) mod 2^b*10^a, which results in 2^b*10^a - 1 and third, the concatenation 1, 2, 3, ... up to (5^c*10^a - 1) mod 5^c*10^a is equal to 10^(a+c) times the concatenation 1, 2, 3, ... up to (5^c*10^a - 2) + (5^c*10^a - 1) mod 5^c*10^a, which results in 5^c*10^a - 1.

Examples

			13 is a term because 123456789101112 mod 13 = 12.
20 is a term because 12345678910111213141516171819 mod 20 = 19.
		

Links

• Martin Renner, Table of n, a(n) for n = 1..92

Crossrefs

Cf. A094151, A110740.

Programs

Maple

a:=proc(m)
  local A, str, i;
  if m = 1 then return([1]);
  else
    if m = 2 then return([1, 2]);
    else
      A := [1, 2];
      str := 1;
      for i from 2 to m do
        str := str*10^length(i) + i;
        if str mod (i+1) = i then A := [op(A), i+1]; fi;
      od;
    fi;
  fi;
  return(A);
end:

Python

from itertools import count, islice
def agen():
    s = "0"
    for n in count(1):
        if int(s)%n == n - 1: yield n
        s += str(n)
print(list(islice(agen(), 30))) # Michael S. Branicky, Nov 23 2022