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

1,2

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).

Links

Table of n, a(n) for n=1..65.

Example

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.

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]

Crossrefs

Cf. A375461 (self-consecutive).

Sequence in context: A366186 A272322 A356800 * A246079 A246086 A194417

Adjacent sequences: A375587 A375588 A375589 * A375591 A375592 A375593

Keyword

easy,nonn,base

Author

Nicholas M. R. Frieler, Aug 19 2024

Status

approved