A171785 -id:A171785 - cp's OEIS frontend

A029455 Numbers k that divide the (right) concatenation of all numbers <= k written in base 10 (most significant digit on left).

1, 2, 3, 5, 6, 9, 10, 12, 15, 18, 20, 25, 27, 30, 36, 45, 50, 54, 60, 69, 75, 90, 100, 108, 120, 125, 135, 150, 162, 180, 200, 216, 225, 248, 250, 270, 300, 324, 360, 375, 405, 450, 470, 500, 540, 558, 600, 648, 675, 710, 750, 810, 900, 1000, 1053, 1080, 1116

Offset: 1

Olivier Gérard

Numbers k such that k divides A007908(k).

			k = 13 is not a term since 12345678910111213 is not divisible by 13.

Jason Yuen, Table of n, a(n) for n = 1..1195 (terms 1..236 from M. F. Hasler, terms 237..637 from Chai Wah Wu)

Cf. A007908.
Cf. A029447-A029470, A029471-A029494, A029495-A029518, A029519-A029542, A061931-A061954, A061955-A061978, A332580.
See A171785 for numbers that divide the concatenation of a(1) through a(n).

b = 10; c = {}; Select[Range[10^5], Divisible[FromDigits[c = Join[c, IntegerDigits[#, b]], b], #] &] (* Robert Price, Mar 11 2020 *)
Select[Range[1200],Divisible[FromDigits[Flatten[IntegerDigits/@Range[#]]],#]&] (* Harvey P. Dale, Dec 31 2020 *)
nxt[{rc_,n_}]:={rc*10^IntegerLength[n+1]+n+1,n+1}; Select[NestList[nxt,{1,1},1200],Mod[#[[1]],#[[2]]]==0&][[;;,2]] (* Harvey P. Dale, Sep 26 2023 *)

c=0;for(d=1,1e9,for(n=d,-1+d*=10,(c=c*d+n)%n || print1(n","));d--) \\ M. F. Hasler, Sep 11 2011

A029455_list, r = [], 0
for n in range(1,10**4+1):
    r = r*10**len(str(n))+n
    if not (r % n):
        A029455_list.append(n) # Chai Wah Wu, Nov 05 2014

def concat_mod(base, k, mod):
  total, digits, n1 = 0, 1, 1
  while n1 <= k:
    n2, p = min(n1*base-1, k), n1*base
    # Compute ((p-1)*n1+1)*p**(n2-n1+1)-(n2+1)*p+n2 divided by (p-1)**2.
    # Since (a//b)%mod == (a%(b*mod))//b, compute the numerator mod (p-1)**2*mod.
    tmp = pow(p, n2-n1+1, (p-1)**2*mod)
    tmp = ((p-1)*n1+1)*tmp-(n2+1)*p+n2
    tmp = (tmp%((p-1)**2*mod))//(p-1)**2
    total = (total*pow(p, n2-n1+1, mod)+tmp)%mod
    digits, n1 = digits+1, p
  return total
for k in range(1, 10**10+1):
  if concat_mod(10, k, k) == 0: print(k) # Jason Yuen, Jan 27 2024

A110740 Numbers k such that the concatenation 1,2,3,...,(k-1) is divisible by k.

Original entry on oeis.org

1, 3, 9, 27, 69, 1053, 1599, 2511, 8167, 21371, 73323, 225681, 313401, 362703, 371321, 1896939, 2735667, 3426273, 3795093, 5433153, 302278903, 1371292077, 19755637749, 23560349643, 33184178631

Offset: 1

Amarnath Murthy, Aug 10 2005

Subsequence of A029455 composed of the terms coprime to 10. - Max Alekseyev, Jun 07 2023

a(26) > 10^11. - Jason Yuen, Oct 12 2024

			3 divides 12, 9 divides 12345678.

Cf. A007908, A029455, A094151, A171785, A332580, A362966.

s = ""; Do[s = s <> ToString[n]; If[Mod[ToExpression[s], n + 1] == 0, Print[n + 1]], {n, 0, 5*10^6}] (* Ryan Propper, Aug 28 2005 *)
Select[Range[55*10^5],Divisible[FromDigits[Flatten[IntegerDigits/@Range[ #-1]]],#]&] (* Harvey P. Dale, Mar 28 2020 *)

# See A029455 for concat_mod
def isok(k): return concat_mod(10, k-1, k)==0 # Jason Yuen, Oct 06 2024

More terms from Ryan Propper, Aug 28 2005
a(20) from Giovanni Resta, Apr 10 2018
a(21)-a(24) from Scott R. Shannon, using a modified version of an algorithm by Joseph Myers, Apr 10 2020
a(25) from Jason Yuen, Oct 06 2024

A249398 Start with a(1) = 1; then a(n) = smallest number > a(n-1) such that a(n) divides concat(a(n-1),a(n)).

Original entry on oeis.org

1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 200, 250, 400, 500, 625, 1000, 1250, 2000, 2500, 3125, 5000, 6250, 10000, 12500, 15625, 20000, 25000, 31250, 40000, 50000, 62500, 78125, 100000, 125000, 156250, 200000, 250000, 312500, 390625, 500000, 625000, 781250, 1000000

Offset: 1

Paolo P. Lava, Dec 01 2014

			a(1) = 1;
a(2) = 2 -> 12 /2 = 6;
Now we cannot use 3 as the next term because it does not divide 23.
a(3) = 4 -> 24 / 4 = 6;
a(4) = 5 -> 45 / 5 = 9;
Again, 6, 7, 8 and 9 cannot be used as the next term.
a(5) = 10 -> 510 / 10 = 51;
a(6) = 20 -> 1020 / 20 = 51;
a(7) = 25 -> 2025 / 25 = 81; etc.

Paolo P. Lava, Table of n, a(n) for n = 1..100

Cf. A171785, A249399, A250745, A250746, A250747.

with(numtheory); P:=proc(q) local a,k,n; print(1); a:=1;
for n from 2 to q do if type((a*10^(1+ilog10(n))+n)/n,integer)
then a:=n; print(n); fi; od; end: P(10^12);

A249399 Start with a(1) = 1 then a(n) = smallest number, not already in the sequence, such that a(n) divides concat(a(n-1),a(n)).

Original entry on oeis.org

1, 2, 4, 5, 10, 20, 8, 16, 25, 50, 40, 32, 64, 80, 100, 125, 200, 160, 128, 250, 400, 320, 256, 500, 625, 1000, 800, 640, 512, 1024, 1250, 2000, 1280, 1600, 2500, 3125, 5000, 3200, 2048, 2560, 4000, 6250, 10000, 6400, 4096, 5120, 8000, 10240, 8192, 12500, 15625

Offset: 1

Paolo P. Lava, Dec 01 2014

Like A249398, but without the constraint a(n) > a(n-1).

			a(1) = 1;
a(2) = 2 -> 12 /2 = 6;
Now we cannot use 3 as the next term because it does not divide 23.
a(3) = 4 -> 24 / 4 = 6;
a(4) = 5 -> 45 / 5 = 9;
Again, 3, 6, 7, 8 and 9 cannot be used as the next term.
a(5) = 10 -> 510 / 10 = 51;
a(6) = 20 -> 1020 / 20 = 51;
a(7) = 8 -> 208 / 8 = 26; etc.

Paolo P. Lava, Table of n, a(n) for n = 1..100

Cf. A171785, A249398, A250745, A250746, A250747.

with(numtheory); P:=proc(q) local a,b,k,n; print(1); a:=1; b:={1};
for k from 1 to q do for n from 1 to q do if nops({n} intersect b)<1
then if type((a*10^(1+ilog10(n))+n)/n,integer)
then a:=n; b:=b union {n}; print(n); break;
fi; fi; od; od; end: P(10^12);

A250745 Start with a(1) = 1; then a(n) = smallest number, not already in the sequence, such that a(n) divides concat(a(1), a(2), ..., a(n)).

Original entry on oeis.org

1, 2, 3, 5, 10, 4, 8, 6, 11, 20, 13, 7, 9, 12, 15, 18, 14, 25, 30, 24, 16, 32, 40, 29, 50, 100, 26, 52, 39, 21, 28, 35, 42, 17, 34, 51, 23, 46, 27, 36, 45, 43, 19, 38, 68, 48, 60, 75, 90, 54, 56, 58, 22, 44, 33, 55, 97, 125, 200, 64, 80, 69, 66, 88, 70, 41, 82

Offset: 1

Paolo P. Lava, Nov 27 2014

Like A171785 but without the constraint a(n) > a(n-1).
Among the first 1000 terms, a(n) = n for n = 1, 2, 3, 15, 170, 577, 759, and the numbers not yet found are 149, 298, 347, 401, 447, 454, 457, 467, 487, 509, etc.
Is this sequence a rearrangement of the natural numbers?

			a(1) = 1;
a(2) = 2 -> 12 /2 = 6;
a(3) = 3 -> 123 / 3 = 41;
Then we cannot use 4 as the next term because 1234 / 4 = 617 / 2.
a(4) = 5 -> 1235 / 5 = 247;
Again, 4, 6, 7, 8 and 9 cannot be used as the next term.
a(5) = 10 -> 123510 / 10 = 12351;
a(6) = 4 -> 1235104 / 4 = 308776;
a(7) = 8 -> 12351048 / 8 = 1543881; etc.

Paolo P. Lava, Table of n, a(n) for n = 1..1000

Cf. A171785, A240588, A241811, A249398, A249399, A250746, A250747.

with(numtheory); P:=proc(q) local a,b,k,n; a:=0; b:={};
for k from 1 to q do for n from 1 to q do if nops({n} intersect b)<1
then if type((a*10^(1+ilog10(n))+n)/n,integer)
then a:=a*10^(1+ilog10(n))+n; b:= b union {n}; print(n); break;
fi; fi; od; od; end: P(10^5);

A250746 Start with a(0) = 0; then a(n) = smallest number > a(n-1) such that a(n) divides concat(a(n), a(n-1), ..., a(0)).

Original entry on oeis.org

0, 1, 2, 3, 5, 10, 15, 18, 19, 35, 42, 51, 55, 70, 85, 93, 95, 106, 155, 217, 310, 745, 1210, 1342, 3355, 5185, 6222, 6330, 9495, 10413, 11115, 12070, 13774, 34435, 41322, 61983, 68870, 1601065116264571, 2217993924228622, 2324778503347862, 2325380783693255

Offset: 0

Paolo P. Lava, Nov 27 2014

This sequence is infinite. - Robert G. Wilson v, Dec 09 2014

			a(0) = 0;
a(1) = 1 -> 10 / 1 = 10;
a(2) = 2 -> 210 / 2 = 105;
a(3) = 3 -> 3210 / 3 = 1070;
Now we cannot use 4 as the next term because 43210 / 4 = 21605 / 2.
a(4) = 5 -> 32105 / 5 = 6421; etc.

Robert G. Wilson v, Table of n, a(n) for n = 0..46

Cf. A171785, A240588, A241811, A249398, A249399, A250745, A250747.

with(numtheory); P:=proc(q) local a,k,n; print(0); print(1); a:=10;
for n from 2 to q do if type((n*10^(1+ilog10(a))+a)/n,integer)
then a:=n*10^(1+ilog10(a))+a; print(n);
fi; od; end: P(10^9);

f[lst_List] := Block[{k = lst[[-1]] + 1, id = FromDigits@ Flatten@ IntegerDigits@ Reverse@ lst}, While[ Mod[ id, k] > 0, k++]; Append[lst, k]]; Nest[f, {0}, 36] (* or *)
f[lst_List] := Block[{mn = lst[[-1]], id = FromDigits@ Flatten@ IntegerDigits@ Reverse@ lst}, d = Divisors@ id; Append[lst, Min@ Select[d, # > mn &]]]; Nest[f, {0, 1}, 36] (* Robert G. Wilson v, Dec 08 2014 *)

a(37)-a(40) from Robert G. Wilson v, Dec 08 2014

A250747 Start with a(0) = 0; then a(n) = smallest number not already in the sequence such that a(n) divides concat(a(n), a(n-1), ..., a(0)).

Original entry on oeis.org

0, 1, 2, 3, 5, 10, 6, 9, 13, 26, 15, 18, 30, 431, 73, 67, 134, 7, 14, 21, 35, 29, 58, 127, 27, 39, 43, 70, 11, 22, 19, 38, 95, 190, 2748070932534311, 2768821759897, 5537643519794, 787, 191, 382, 955, 17, 31, 45, 54, 90, 101, 202, 303, 57, 114, 47, 55, 33, 66

Offset: 0

Paolo P. Lava, Nov 28 2014

Like A250746, but without the constraint a(n) > a(n-1).

			a(0) = 0;
a(1) = 1 -> 10 / 1 = 10;
a(2) = 2 -> 210 / 2 = 105;
a(3) = 3 -> 3210 / 3 = 1070;
Now we cannot use 4 as the next term because 43210 / 4 = 21605 / 2.
a(4) = 5 -> 32105 / 5 = 6421;
Again, we cannot use 4, 6, 7, 8 or 9.
a(5) = 10 -> 1053210 / 10 = 105321.
We still cannot use 4, but 6 is ok.
a(6) = 6 -> 61053210 / 6 = 10175535. Etc.

Jon E. Schoenfield, Table of n, a(n) for n = 0..165

Cf. A171785, A240588, A241811, A249398, A249399, A250745, A250746.

with(numtheory); P:=proc(q) local a,b,k,n; print(0); print(1); a:=10; b:={0,1};
for k from 1 to q do for n from 1 to q do if nops({n} intersect b)<1
then if type((n*10^(1+ilog10(a))+a)/n,integer)
then a:=n*10^(1+ilog10(a))+a; b:= b union {n}; print(n); break; fi; fi;
od; od; end: P(10^5);

More terms from Jon E. Schoenfield, Nov 29 2014

A228806 Smallest odd number greater than any previous term such that it divides the concatenation of all the previous terms and itself; begin with 1.

Original entry on oeis.org

1, 5, 15, 25, 29, 35, 125, 625, 2125, 2675, 3125, 15625, 20125, 21875, 23975, 24797, 25125, 36875, 47495, 47725, 51875, 53125, 78125, 135475, 390625, 1171875, 1903875, 1928595, 2142375, 2265625, 6617125, 8385625, 8790525, 8807085, 8818575, 10504785

Offset: 1

Robert G. Wilson v, Sep 04 2013

Terms not congruent to 0 (mod 5) are: 1, 29, 24797, 24848081, 91381387, 274144161, ..., .

Terms not congruent to 0 (mod 25) are: 1, 5, 15, 29, 35, 24797, 47495, 1928595, 8807085, 10504785, 24848081, 91381387, 274144161, ..., .

			a(5) equals 29 because 15152527 (mod 27) == 19, but 15152529 (mod 29) == 0.

Robert G. Wilson v, Table of n, a(n) for n = 1..55

Cf. A171785.

f[s_List] := Block[{k = s[[-1]] + 2, conc = FromDigits[ Flatten@ IntegerDigits@ s]}, While[ Mod[ conc*10^Floor[ Log[10, k] + 1] + k, k] != 0, k += 2]; Append[s, k]]; Nest[f, {1}, 25]

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

cp's OEIS Frontend

A029455 Numbers k that divide the (right) concatenation of all numbers <= k written in base 10 (most significant digit on left).

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A110740 Numbers k such that the concatenation 1,2,3,...,(k-1) is divisible by k.

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A249398 Start with a(1) = 1; then a(n) = smallest number > a(n-1) such that a(n) divides concat(a(n-1),a(n)).

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A249399 Start with a(1) = 1 then a(n) = smallest number, not already in the sequence, such that a(n) divides concat(a(n-1),a(n)).

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A250745 Start with a(1) = 1; then a(n) = smallest number, not already in the sequence, such that a(n) divides concat(a(1), a(2), ..., a(n)).

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A250746 Start with a(0) = 0; then a(n) = smallest number > a(n-1) such that a(n) divides concat(a(n), a(n-1), ..., a(0)).

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A250747 Start with a(0) = 0; then a(n) = smallest number not already in the sequence such that a(n) divides concat(a(n), a(n-1), ..., a(0)).

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A228806 Smallest odd number greater than any previous term such that it divides the concatenation of all the previous terms and itself; begin with 1.

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