A110740 -id:A110740 - cp's OEIS frontend

A094151 Remainder when concatenation 1,2,3,...up to (n-1) is divided by n.

0, 1, 0, 3, 4, 3, 4, 7, 0, 9, 4, 3, 12, 9, 9, 7, 1, 9, 7, 19, 12, 15, 18, 15, 24, 9, 0, 11, 27, 9, 7, 7, 15, 9, 9, 27, 29, 21, 21, 39, 22, 33, 5, 15, 9, 39, 45, 39, 39, 49, 27, 51, 33, 27, 4, 23, 15, 49, 49, 39, 32, 13, 54, 7, 9, 15, 41, 59, 0, 39, 47, 63, 41, 17, 24, 23, 37, 21, 75, 39

Offset: 1

Amarnath Murthy, Apr 29 2004

			a(5) = 15 hence a(6) = least integer multiple of 15/6 = 5.

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Cf. A007908, A110740 (indices of 0's).

Cf. A095221.

Table[st = ""; For[i = 0, i <= n - 1, i++, st = st <> ToString[i]]; Mod[ ToExpression[st], n], {n, 1, 100}] (* Sam Handler (sam_5_5_5_0(AT)yahoo.com), Nov 12 2004 *)
Table[Mod[FromDigits[Flatten[IntegerDigits/@Range[n-1]]],n],{n,100}] (* Harvey P. Dale, May 21 2024 *)

a(n)=if(n<=9, return(1719%n)); my(m=Mod(123456789,n)); for(d=2,#Str(n-1), my(D=10^d); for(k=D/10,min(D,n)-1, m=D*m+k)); lift(m) \\ Charles R Greathouse IV, Nov 21 2022

from itertools import count, islice
def agen():
    s = "0"
    for n in count(1): yield int(s)%n; s += str(n)
print(list(islice(agen(), 80))) # Michael S. Branicky, Nov 21 2022

a(n) = A007908(n-1) mod n for n>1. - Michel Marcus, Nov 21 2022

More terms from Sam Handler (sam_5_5_5_0(AT)yahoo.com), Nov 12 2004

A316312 Numbers k such that the sum of the digits of the numbers 1, 2, 3, ... up to (k - 1) is divisible by k.

Original entry on oeis.org

1, 3, 5, 7, 9, 12, 15, 20, 27, 40, 45, 60, 63, 80, 81, 100, 180, 181, 300, 360, 363, 500, 540, 545, 700, 720, 727, 900, 909, 912, 915, 1137, 1140, 1200, 1500, 1560, 1563, 2000, 2700, 2720, 2727, 4000, 4500, 4540, 4545, 6000, 6300, 6360, 6363, 8000, 8100, 8180

Offset: 1

Debapriyay Mukhopadhyay, Jun 29 2018

Numbers k such that A007953(A007908(k - 1)) is divisible by k. - Felix Fröhlich, Jun 29 2018
From Robert Israel, Jun 29 2018: (Start)
Numbers k such that A037123(k - 1) is divisible by k.
If m is even, then 10^m, 3 * 10^m, 5 * 10^m, 7 * 10^m and 9 * 10^m are included.
If m is odd, then 2 * 10^m, 4 * 10^m, 6 * 10^m, and 8 * 10^m are included. (End)
Is it true that if k is a term then 100 * k is a term?

			For n = 7, sum of the digits of the numbers 1 to 6 is 21, which is divisible by 7.
For n = 12, sum of the digits of the numbers 1 to 11 is 48, which is divisible by 12.
For n = 15, sum of the digits of the numbers 1 to 14 is 60, which is divisible by 15.
16 is not in the sequence because the sum of the digits of the numbers 1 to 15 is 66, which is not divisible by 16.

Henry Bottomley, Table of n, a(n) for n = 1..118

Cf. A007953, A007908, A037123, A110740, A114136.

t:= 0: Res:= NULL:
for n from 1 to 10000 do
  t:= t + convert(convert(n-1,base,10),`+`);
  if (t/n)::integer then Res:= Res, n fi
od:
Res; # Robert Israel, Jun 29 2018

s = 0; Reap[Do[If[Mod[s, n] == 0, Sow[n]]; s += Plus @@ IntegerDigits@n, {n, 10000}]][[2, 1]] (* Giovanni Resta, Jun 29 2018 *)

sumsod(n) = sum(i=1, n, sumdigits(i))
is(n) = sumsod(n-1)%n==0 \\ Felix Fröhlich, Jun 29 2018

upto(n) = my(s=0,res=List()); for(i=0, n, s += vecsum(digits(i)); if(s%(i+1)==0, listput(res, i+1))); res \\ David A. Corneth, Jun 29 2018

More terms from Felix Fröhlich, Jun 29 2018

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

Original entry on oeis.org

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

Martin Renner, Nov 23 2022

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.

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

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

Cf. A094151, A110740.

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:

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

A362966 Numbers k such that A007908(k) == 1 (mod k).

Original entry on oeis.org

1, 121487, 293957, 13449179, 549999887

Offset: 1

Max Alekseyev, Jun 06 2023

a(6) > 10^11. - Jason Yuen, Oct 12 2024

Cf. A007908, A029455, A029479, A110740.

# See A029455 for concat_mod
def isok(k): return concat_mod(10, k, k)==1%k # Jason Yuen, Oct 12 2024

A302687 a(1) = 1; a(2) = 2; then a(n) is the smallest number > a(n-1) such that a(n) divides concat(a(1), a(2), ..., a(n-1)).

Original entry on oeis.org

1, 2, 3, 41, 43, 129, 9567001, 21147541, 22662659, 23817877, 24837187, 28850377, 28872229, 37916473, 48749751, 70416307, 439229167, 834385607, 2270365163, 2278377431, 3751789547, 4433933101, 4810754611, 14432263833, 15632412757, 30530543651, 42441819717, 65591903199, 65857498407

Offset: 1

Daniel Sterman, Apr 11 2018

			a(3) = 3, which makes the concatenation of the first three terms: 123. After 3, the next-highest factor of 123 is 41, so a(4) = 41. The concatenation of the first four terms is then 12341. After 41, the next-highest factor of 12341 is 43, so a(5) = 43.

Compare A240588, in which each term does not need to strictly increase as long as it has not yet appeared in the sequence.
Compare also A171785, in which each term must divide the concatenation of all terms in the sequence including itself.
In A029455, each term divides the concatenation of all smaller positive integers.
In A110740, each term divides the concatenation of all strictly smaller positive integers.

A[1]:= 1: A[2]:= 2: C:= 1:
for n from 3 to 20 do
  C:= A[n-1]+C*10^(ilog10(A[n-1])+1);
  A[n]:= min(select(`>`,numtheory:-divisors(C),A[n-1]))
od:
seq(A[i],i=1..20); # Robert Israel, Apr 12 2018

a(16)-a(20) from Robert Israel, Apr 12 2018

a(21)-a(29) from Daniel Suteu, Apr 12 2018

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A094151 Remainder when concatenation 1,2,3,...up to (n-1) is divided by n.

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A316312 Numbers k such that the sum of the digits of the numbers 1, 2, 3, ... up to (k - 1) is divisible by k.

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

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A362966 Numbers k such that A007908(k) == 1 (mod k).

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A302687 a(1) = 1; a(2) = 2; then a(n) is the smallest number > a(n-1) such that a(n) divides concat(a(1), a(2), ..., a(n-1)).

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