A029455 -id:A029455 - cp's OEIS frontend

A332580 a(n) = minimal positive k such that the concatenation of the decimal digits of n,n+1,...,n+k is divisible by n+k+1, or -1 if no such k exists.

1, 80, 1885, 6838, 1, 44, 13, 2, 1311, 18, 197, 20, 53, 134, 993, 44, 175, 124518, 263, 26, 107, 10, 5, 62, 15, 33172, 9, 14, 317, 708, 1501, 214, 37, 34, 67, 270, 19, 20188, 78277, 10738, 287, 2390, 695, 2783191412912, 3, 700, 8303, 350, 21, 100, 2249, 21326

Offset: 1

Scott R. Shannon and N. J. A. Sloane, Feb 16 2020

Certainly n+k must be even, since no odd number can be divisible by an even number.
The values of n+k = n+a(n) are given in the companion sequence A332584.
A heuristic argument suggests that k should always exist.
As of Jul 10 2020, up to n = 1000 there are just two unknown values, a(158) and a(539).
The following remarks summarize program made during the first half of 2020.
On Feb 19 2020 Joseph Myers discovered that a(98) = 259110640. On Feb 20 2020 he reported that a(44) > 10^11 if it exists; a(92), a(158) and a(170) are all > 10^10 if they exist; a(494), a(539), a(563), a(761), a(854), a(944) and a(956) are all > 2*10^9 if they exist; and that he has found all the other values up to a(1000). - N. J. A. Sloane, Feb 23 2020.
Added Feb 26 2020: Joseph Myers has now checked all the numbers up to 1000 out to a limit of 10^11 (see link).
Update from Paul Zimmermann, Mar 17 2020: (Start)
I started a parallel program using the same algorithm as in Joseph Myers's "grow.c" program on the few sequences with unknown status in http://oeis.org/A332580/a332580_2.txt.
This program just found:
pzimmermann@wurst:~/A332580$ tail 956.out
n=956 kmax=200000000000
found k=162236437060
It thus seems that a(956) = 162236437060, i.e., the term of index n+k+1 is divisible by 162236438017 = 43 * 5051 * 746969. (End)
Partial confirmation from Scott R. Shannon, Mar 17 2020:  I set n = 956 and a k value a few less than 162236437060 in my Java version of Joseph Myer's program, and it found the results Paul Zimmermann gave. But that’s not much of a confirmation as it uses the same algorithm, just implemented in a different language.
Partial confirmation from Pierrick Gaudry, Mar 18 2020:  (Start)
I ran the attached small C program in order to check that a(956) = 162236437060. More precisely, I check only that the 162236437060-th integer obtained starting with 956 is indeed 0 modulo 162236438017.
For this there is no need to rely on multi-precision arithmetic. However, since 162236438017 > 2^32, it is not possible to use 64-bit arithmetic; or at least, it was easier to use the 128-bit arithmetic provided by the compiler.
The algorithm is then fairly simple: just compute iteratively the big number obtained by concatenating 956, 957, 958, ... and so on, and reduce all along the way modulo 162236438017. The result should be zero. This was tested on a few other known example.
After a bit more than 1 hour on my laptop, this indeed prints 0, thus confirming that a(956) <= 162236437060 (this simple method does not check if there is a smaller value). (End)
Full confirmation for a(956) from Joseph Myers, Mar 18 2020: I restarted computations for 956 where I had stopped them before (at 101 * 10^9) and ran them up to 163 * 10^9; I also get 162236437060.
Update from Paul Zimmermann, Mar 22 2020: (Start)
Here are four more values to check, confirmed independently by Pierrick Gaudry:
a(44) <= 2783191412912
a(92) <= 218128159460
a(494) <= 2314160375788
a(854) <= 440578095296 (also k=587470935254 divides)
All four values were found with the "sieving" algorithm I described in an earlier email (see the Alekseyev et al. paper), sieving all primes up to 5000000000. Thus it is possible that smaller solutions exist.
Up to n=1000, the remaining cases where we have no bound at present are 158, 539, 761, 944. (End)
a(761) <= 111508066823971. Now only 3 values remain up to n=1000 (158, 539, 944). Paul Zimmermann, Mar 23 2020
I restarted my exhaustive search for 92 where I had previously stopped it, and can confirm a(92) = 218128159460.  - Joseph Myers, Mar 23 2020
The remaining values to check are:
a(44)  <= 2783191412912, a(494) <= 2314160375788, a(761) <= 111508066823971, a(854) <= 440578095296. - Paul Zimmermann, Mar 24 2020
a(854) = 440578095296 confirmed by Joseph Myers on Mar 26 2020.
Summary: As of Apr 15 2020, a(n) is known for all n <= 1000 except for four values where we have only an upper bound (44, 494, 539, and 761), and two values (158, 944) where all we know is that if k exists then it is greater than 10^15. See the table in the Links section. - Joseph Myers and Paul Zimmermann.
From Paul Zimmermann, Apr 17 2020: I have completed the full check for n=494 up to n+k=10^12. Thus a(494) >= 10^12-494. It took about 4 hours. The final check from 10^12 to 2314160375788+494+1 should take another 4-5 hours.  (I don't want this comment to be lost, even though it will probably be replaced by something stronger very soon. - N. J. A. Sloane, Apr 17 2020)
From Paul Zimmermann, Apr 18 2020: (Start)
I confirm that a(44) = 2783191412912 and a(494) = 2314160375788. These were checked with a parallel version of Joseph's program (attached). For n=44 I ran the following script which submits 28 jobs checking each a range of 10^11 values:
for i in `seq 0 27`; do
   kmin=`expr 1 + $i \* 100000000000`
   kmax=`expr $kmin + 100000000000 - 1`
   oarsub -p "cluster='grvingt'" -q production -l walltime=5 "./A332580 -kmin $kmin 44 $kmax"
done
The last job took a little less than 4 hours (wall clock time) on a 32-core cpu (64 virtual cores), thus it took a total of about 300 cpu days. (End)
a(944) <= 1032422879252. - Paul Zimmermann, Apr 19 2020

			a(1) = 1 as '1' || '2' = '12', which is divisible by 3 (where || denotes decimal concatenation).
a(2) = 80: the concatenation 2 || 3 || ... || 82 is
  23456789101112131415161718192021222324252627282930313233343536373839\
  40414243444546474849505152535455565758596061626364656667686970717273747\
  576777879808182, which is divisible by 83.
a(7) = 13 as '7' || '8' || '9' || '10' || '11' || '12' ||  ... || '20' = 7891011121314151617181920, which is divisible by 21.
a(8) = 2 as '8' || '9' || '10' = 8910, which is divisible by 11.

Joseph Myers and Paul Zimmermann, Table of n, a(n) for n = 1..157
M. A. Alekseyev, J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, Three Cousins of Recaman's Sequence, Fibonacci Quart. 60:3 (2022), 201-219 (preprint arXiv:2004:14000 [math.NT], 2021).
Pierrick Gaudry, C program for partial check
Joseph Myers and Paul Zimmermann, Table of n, a(n) for n <= 1000. The single remaining UNKNOWN entry at n = 158 either -1 or greater than 10^14.
N. J. A. Sloane, Conant's Gasket, Recamán Variations, the Enots Wolley Sequence, and Stained Glass Windows, Experimental Math Seminar, Rutgers University, Sep 10 2020 (video of Zoom talk).
Paul Zimmermann, A parallel version of Joseph's C program

Cf. A061836 (multiplication instead of concatenation), A281232, A332584, A332585 (length of the final concatenation). See A058183 for finding the length of a concatenation.
For records see A333546, A333547.
For n=44, see A332562.
See A332563, A332586 for a base 2 version.
See A281232 for the positions of the 1's.
A029455 is an older sequence in the same spirit.

grow := proc(n,M) # searches out to a limit of M, returns [n,n+k] or [n,-1] if no k was found
  local R,i;
  R:=n;
  for i from n+1 to M do
    R:=R*10^length(i)+i;
    if (i mod 2) = 0 then
      if (R mod (i+1)) = 0 then return([n, i]); fi;
    fi;
  od:
  [n, -1];
end;
for n from 1 to 100 do lprint(grow(n,20000)); od;

apply( {a(n,L=10^logint(n*10,10),c=n)= n%2||c=c*L+n+1; for(k=n+++n%2,oo, kM. F. Hasler, Feb 20 2020

Edited by Max Alekseyev, Dec 26 2024

A171785 Start with a(1) = 1; then a(n) = smallest number > a(n-1) such that a(n) divides concat(a(1), a(2), ..., a(n)).

Original entry on oeis.org

1, 2, 3, 5, 10, 12, 15, 20, 25, 30, 39, 44, 50, 100, 101, 125, 150, 188, 200, 220, 230, 250, 272, 304, 320, 370, 376, 400, 500, 525, 600, 615, 625, 1000, 1250, 1487, 1500, 1590, 1696, 1750, 2000, 2245, 2500, 3000, 3090, 3125, 3800, 4000, 5000, 5725, 6122, 7025

Offset: 1

David Scambler, Sep 30 2010

Primes appearing so far are 2, 3, 5, 101, 1487.

			1: 1 divides 1
1,2: 2 divides 12
1,2,3: 3 divides 123
1,2,3,4: 4 does NOT divide 1234, so
1,2,3,5: 5 divides 1235
etc.

Robert G. Wilson v, Table of n, a(n) for n = 1..584

See A029455 for numbers that divide the concatenation of all numbers <= n.

f[s_List] := Block[{k = s[[ -1]] + 1, conc = FromDigits[Flatten@ IntegerDigits@s]}, While[ Mod[conc*10^Floor[ Log[10, k] + 1] + k, k] != 0, k++ ]; Append[s, k]]; Nest[f, {1}, 51] (* Robert G. Wilson v, Oct 14 2010 *)
nxt[{a_,c_}]:=Module[{k=a+1},While[!Divisible[c*10^IntegerLength[k]+ k, k], k++];{k,c*10^IntegerLength[k]+k}]; Transpose[NestList[nxt,{1,1},60]][[1]] (* Harvey P. Dale, Mar 08 2015 *)

More terms from Robert G. Wilson v, Oct 14 2010

A075001 Smallest k such that the concatenation of n consecutive numbers starting with k (from k to n+k-1) is a multiple of n; or 0 if no such number exists.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 9, 1, 4, 7, 1, 5, 23, 1, 14, 1, 9, 9, 13, 5, 1, 21, 1, 13, 12, 1, 36, 21, 9, 3, 41, 1, 34, 33, 9, 21, 12, 9, 33, 9, 1, 13, 28, 5, 48, 1, 23, 21, 3, 1, 11, 13, 14, 41, 28, 1, 114, 115, 9, 41, 21, 9, 23, 69, 1, 61, 73, 5, 14, 43, 1, 145, 13, 9, 127, 41, 9, 95

Offset: 1

Amarnath Murthy, Aug 31 2002

Conjecture: For every n there exists a k.
First occurrence of k where a(n)=k: 1, 103, 4, 13, 8, 105, 14, 87, 11, 699, 55, 29, 23, 19, 114, 261, 102, 97, 178, 219, 26, 121, 17, 151, 92, ..., . - Robert G. Wilson v
a(n)=1 iff n is in A029455. - Robert G. Wilson v
Increasing a(n)'s: 1, 3, 5, 9, 23, 36, 41, 48, 114, 115, 145, 166, 175, 221, 251, ..., at n = 1, 4, 8, 11, 17, 31, 35, 49, 61, 62, 76, 85, 122, 133, 170, 179, 217, 229, ..., . - Robert G. Wilson v

			a(11) = 9 as 910111213141516171819 the concatenation of 11 numbers from 9 to 19 is divisible by 11 (11*82737383012865106529).

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 2500 terms from Robert G. Wilson v)

Cf. A058183, A074991, A074992, A074993, A074994, A074995, A074996, A036377, A073086, A074999, A075000, A077306, A075000.

Cf. A029455 (indices of 1's).

f[n_] := Block[{c = 1, id = Range@n}, While[k = FromDigits@Flatten@IntegerDigits@id/n; ! IntegerQ@k, id++; c++ ]; c]; Array[f, 82] (* Robert G. Wilson v, Oct 20 2007 *)

/* The following program assumes the conjecture is true. */ /* It has found nonzero a(n) for n up to 500. */ {for(n=1,500, k=0; until(s%n==0,k++; s=0; for(m=k,k+n-1, s=s*(10^length(Str(m)))+m)); print1(k,","))}

a(n) = {my(ld = 1, hd = n, qd, m = Mod(1, n), pow10, qdn = #digits(n), t=log(10*n+.5)\log(10)); qd = n*t+t-10^t\9; pow10 = Mod(10, n)^(qd-1); for(i = 2, n, m = m * Mod(10, n)^#digits(i) + i; ); while(1, if(lift(m) == 0, return(ld)); m -= ld * pow10; hd++; m = m * Mod(10, n)^#digits(hd) + hd; ld++; pow10*=10^(#digits(hd) - #digits(ld)); ) } \\ David A. Corneth, Aug 23 2020

More terms from Rick L. Shepherd, Sep 03 2002

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

A263623 a(1)=1; thereafter, a(n) = smallest k such that the decimal concatenation [a(n-2)+1 a(n-2)+2, ... a(n-1)] divides the decimal concatenation [a(n-1)+1 a(n-1)+2 ... k].

Original entry on oeis.org

1, 2, 4, 8, 36

Offset: 1

N. J. A. Sloane, Oct 23 2015

a(6), if it exists, is > 10^6. - Lars Blomberg, Dec 01 2016

a(6) <= 86794654347484748866500883685475458354620023089553379437308257589024531796179370608623026912768. - Max Alekseyev, Dec 25 2024

			n=3: a(3) = 4 because k=4 is the smallest number such that 2 divides the concatenation 345...k.
n=4: a(4) = 8 because k=8 is the smallest number such that 34 divides the concatenation 567...k. See A002782 for the relevant concatenations.

Cf. A002782, A029455, A332580.

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

A068090 Numbers n which divide the right concatenation of the first n even numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 18, 20, 24, 25, 30, 35, 36, 39, 40, 45, 50, 60, 72, 75, 80, 90, 100, 120, 125, 144, 150, 180, 200, 204, 225, 240, 250, 300, 360, 375, 400, 450, 500, 507, 540, 585, 600, 624, 625, 675, 702, 720, 750, 780, 800, 804, 864, 900

Offset: 1

Amarnath Murthy, Feb 19 2002

			The right concatenation of the first six even numbers is 24681012 and 6 divides this number. So 6 is a member of the sequence.

Cf. A029455.

a := proc(n): if n=1 then RETURN(2) fi: 2*n+a(n-1)*10^(ceil(log[10](2*n+.01))) end: for n from 1 to 1000 do if a(n) mod n = 0 then printf(`%d,`,n) fi; od:

concat[n_]:=FromDigits[Flatten[IntegerDigits/@Range[2,2n,2]]]; Select[ Range[700],Divisible[concat[#],#]&] (* Harvey P. Dale, Nov 02 2011 *)

Edited and extended by James Sellers, Feb 20 2002

Offset corrected by Sean A. Irvine, Jan 25 2024

A072712 Primes p such that p divides the (right) concatenation of all numbers from 1 to p written in base 10 (most significant digit on left).

Original entry on oeis.org

2, 3, 5, 8167, 371321

Offset: 1

Jeff Heleen, Aug 07 2002

Right concatenation, normal order.

Primes in A029455. - Derek Orr, Oct 04 2014

			p=17 is not a term since 1234567891011121314151617 is not divisible by 17.

Carlos Rivera, Puzzle 190. A follow up for the puzzle 188, The Prime Puzzles & Problems Connection.

Cf. A029447-A029542, A061931-A061978.

p=""; n=1; while(n<10^4, p=concat(p,Str(n)); if(eval(p)%n==0&&isprime(n), print1(n,", ")); n++) \\ Derek Orr, Oct 04 2014

Another term from Jeff Heleen, Oct 11 2009

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

A332580 a(n) = minimal positive k such that the concatenation of the decimal digits of n,n+1,...,n+k is divisible by n+k+1, or -1 if no such k exists.

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A171785 Start with a(1) = 1; then a(n) = smallest number > a(n-1) such that a(n) divides concat(a(1), a(2), ..., a(n)).

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A075001 Smallest k such that the concatenation of n consecutive numbers starting with k (from k to n+k-1) is a multiple of n; or 0 if no such number exists.

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

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A263623 a(1)=1; thereafter, a(n) = smallest k such that the decimal concatenation [a(n-2)+1 a(n-2)+2, ... a(n-1)] divides the decimal concatenation [a(n-1)+1 a(n-1)+2 ... k].

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

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Python

A068090 Numbers n which divide the right concatenation of the first n even numbers.

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Maple

Mathematica

Extensions

A072712 Primes p such that p divides the (right) concatenation of all numbers from 1 to p written in base 10 (most significant digit on left).

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