A375590 -id:A375590 - cp's OEIS frontend

A375461 Self-consecutive numbers: 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, 12, 14, 16, 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, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165, 168

Offset: 1

Nicholas M. R. Frieler, Aug 17 2024

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:
    1) m is divisible by d AND m < 10^d * d/(d+1)
     OR
    2) 10^d is divisible by d+1 AND m >= 10^d * d/(d+1).
Equivalently, integers m in the block 10^(d-1) <= m < 10^d are included in this sequence based on their relation to the value s = 10^d * d/(d+1). If m < s, then m is in this sequence iff m is divisible by d. If m >= s, m is in this sequence iff s is an integer. Note that for d = 2, s = 66.666... so no integers m in the range 66.666... <= m < 100 are in this sequence. For d = 3, s = 750 is an integer so all integers m in the range 750 <= m < 1000 are in this sequence. - Dominic McCarty, Sep 09 2024

			7 is a term because if we concatenate 7, 8, 9, 10, and 11, we get 7891011, which has 7 digits.
102 is a term because we can concatenate 102 with the next 33 consecutive integers to get a number with 34*3 = 102 digits.
68 is not a term because if we concatenate 68,69,...,100 we get a 67-digit number and concatenating 101 with this yields a 70-digit number, so it is not possible to achieve a 68-digit number.

Dominic McCarty, Table of n, a(n) for n = 1..10000

Cf. A035333 (concatenations), A375590.

function next(m){let d=(""+m).length;if(m*(d+1)d*(10**d)){return next(10**d)}else{return m+d}}else{if((10**d)%(d+1)==0){return m+1}else{return next(10**d)}}}let m=0;a="";while(m<168){m=next(m);a+=m+", "}console.log(a) // Dominic McCarty, Sep 09 2024

SelfConsecutiveQ[n_] :=
 Module[{len = Length@IntegerDigits[n], counter = 1, numDigits = 0},
  If[len*(Power[10, len] - n) >= n, Return[Mod[n, len] == 0],
   numDigits = len*(Power[10, len] - n)];
  Mod[n - numDigits, len + 1] == 0
 ]
Select[Range[1000], SelfConsecutiveQ[#] &]

cp's OEIS Frontend

A375461 Self-consecutive numbers: 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.

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