A375590 Numbers m such that there exists a nonnegative integer k for which the concatenation of m, m-1, ..., m-k is an m-digit number.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 150, 153, 156, 159, 162, 165, 168

Offset: 1

Author

Nicholas M. R. Frieler, Aug 19 2024

Keywords

• easy
• nonn
• base

Comments

Let d be the number of digits of m (in base 10). One can show that the number m is a member of this sequence if and only if:

(a) m is divisible by d and m >= 10^(d-1) * d/(d+1)

or

(b) 10^(d-1) == 1 (mod d-1) (i.e., d-1 is a term of A014950) and m < 10^d * d/(d+1).

Examples

			13 is a term because the concatenation of 13, 12, ..., 5 is a 13-digit number.
100 is not a term because the concatenation of 100, 99, ..., 52 is a 99-digit number and concatenating this number with 51 yields a 101-digit number.
		

Crossrefs

Cf. A375461 (self-consecutive).

Programs

Mathematica

SelfDownwardConsecutiveQ[n_] :=
 Module[{len = Length@IntegerDigits[n], num, c = 1, numDigits = 0},
  numDigits = len*Ceiling[n + 1 - 10^(len - 1)];
If[numDigits >= n, Return[Mod[n, len] == 0]];
num = n - Ceiling[n + 1 - 10^(len - 1)];
While[numDigits < n + 1,
  If[(len - c)*Ceiling[num + 1 - 10^(len - c - 1)] >= n - numDigits,
   Return[Mod[n - numDigits, len - c] == 0],
   numDigits += (len - c)*Ceiling[num + 1 - 10^(len - c - 1)]
    ];
  num -= Ceiling[num + 1 - 10^(len - c - 1)];
  c++;
   ]
 ]
Select[Range[1000], SelfDownwardConsecutiveQ]