A276348 -id:A276348 - cp's OEIS frontend

A052983 Least multiple of n consisting of a succession of 1's followed by a succession of 0's.

10, 10, 1110, 100, 10, 1110, 1111110, 1000, 1111111110, 10, 110, 11100, 1111110, 1111110, 1110, 10000, 11111111111111110, 1111111110, 1111111111111111110, 100, 1111110, 110, 11111111111111111111110, 111000, 100, 1111110, 1111111111111111111111111110

Offset: 1

Lekraj Beedassy, Jun 26 2003

All entries are differences of two terms of A000042. Since the pigeonhole principle guarantees that, for any m, two among the first m+1 entries of A000042 are congruent modulo m, their difference (i.e. belonging to this sequence) is therefore divisible by m, so that such numbers exist for all m. This sequence is thus infinite.

For n>1, a(n) consists of s 1's and t 0's, where s=A084681(X) and t is the greater of p or q (s=1 for X=1, t=1 for p=q=0), when we write n=X*Y with (X,Y)=1 and Y=2^p*5^q.

			We have a(6)=1110 because 6 divides 1110=6*185, the smallest such one with a string of 1's followed by that of 0's

Cf. A000042, A002371, A007732.

Cf. A084681.

f[n_] := Select[ Map[ FromDigits, IntegerDigits[ Table[ Sum[2^i, {i, k, j, -1}], {j, k, 1, -1}], 2]]/n, IntegerQ[ # ] & ]; g[n_] := Block[{k = 1}, While[ f[n] == {}, k++ ]; n*Min[ f[n]]]; Table[ g[n], {n, 1, 27}]
nn=30;With[{nos=Sort[Flatten[Table[FromDigits[Join[Table[1,{n}], Table[ 0,{i}]]],{n,nn},{i,5}]]]},Flatten[Table[Select[nos,Divisible[#,n]&,1],{n,nn}]]] (* Harvey P. Dale, Mar 09 2014 *)

a(n) = A276348(n) * n; A227362(a(n)) = 10. - Jaroslav Krizek, Aug 30 2016

Edited, corrected and extended by Robert G. Wilson v, Jun 26 2003

A276349 Numbers consisting of a nonempty string of 1's followed by a nonempty string of 0's.

Original entry on oeis.org

10, 100, 110, 1000, 1100, 1110, 10000, 11000, 11100, 11110, 100000, 110000, 111000, 111100, 111110, 1000000, 1100000, 1110000, 1111000, 1111100, 1111110, 10000000, 11000000, 11100000, 11110000, 11111000, 11111100, 11111110, 100000000, 110000000, 111000000

Offset: 1

Jaroslav Krizek, Aug 30 2016

Intersection of A037415 and A009996 except for 1 [Corrected by David A. Corneth, Aug 30 2016].
Set of terms from sequence A052983.
a(n) is the binary expansion of A043569(n). - Michel Marcus, Sep 04 2016

			60 is of the form binomial(a, 2) + b where 0 < b <= a and a = 11, b = 5. So a(60) has (11 + 1) digits and 5 leading ones. The other digits are 0. Giving a(60) = 111110000000. It has 7 (more than 1) trailing zeros so the next one, a(61) is a(60) + 10^(7 - 1). - _David A. Corneth_, Aug 30 2016

Robert Israel, Table of n, a(n) for n = 1..10000
Luboš Pick, Dirichletovy šuplíčky, Pokroky matematiky, fyziky a astronomie, Vol. 61, No. 2 (2016), pp. 106-118. (In Czech; The Dirichlet pigeonhole principle)

Cf. A002024, A009996, A037415, A043569, A052983, A073668, A227362, A276348, A309761.

[n: n in [1..10^7] | Seqint(Setseq(Set(Sort(Intseq(n))))) eq 10 and Seqint(Sort((Intseq(n)))) eq n];

seq(seq(10^(m+1)*(1-10^(-j))/9,j=1..m),m=1..20); # Robert Israel, Sep 02 2016

Table[FromDigits@ Join[ConstantArray[1, #1], ConstantArray[0, #2]] & @@@ Transpose@ {#, n - #} &@ Range[n - 1], {n, 2, 9}] // Flatten (* Michael De Vlieger, Aug 30 2016 *)
Flatten[Table[FromDigits[Join[PadRight[{},n,1],PadRight[{},k,0]]],{n,8},{k,8}]]//Sort (* Harvey P. Dale, Jan 09 2019 *)

is(n) = vecmin(digits(n))==0 && vecmax(digits(n))==1 && digits(n)==vecsort(digits(n), , 4) \\ Felix Fröhlich, Aug 30 2016

a(n) = my(r =  ceil((sqrt(1+8*n)+1)/2), k = n - binomial(r-1, 2));10^(r-k)*(10^(k)-1)/9
\\ given an element n, computes the next element of the sequence.
nxt(n) = my(d = digits(n), qd=#d, s = vecsum(d)); if(qd-s>1, n+10^(qd-s-1), 10^qd)
\\ given an element n of the sequence, computes its place in the sequence.
inv(n) = my(d = digits(n)); binomial(#d-1,2) + vecsum(d) \\ David A. Corneth, Aug 31 2016

from math import isqrt, comb
def A276349(n): return 10*(10**(m:=isqrt(n<<3)+1>>1)-10**(comb(m+1,2)-n))//9 # Chai Wah Wu, Jun 16 2025

A227362(a(n)) = 10.
From Robert Israel, Sep 02 2016: (Start)
a((m^2-m)/2+j) = 10^(m+1)*(1-10^(-j))/9 for m>=1, 1<=j<=m.
a(n) = 10*(10^m - 10^(-n+m*(m+1)/2))/9 where m = A002024(n). (End)
A002275(A002260(n)) * 10^A004736(n) - Peter Kagey, Sep 02 2016
Sum_{n>=1} 1/a(n) = A073668. - Amiram Eldar, Feb 20 2022
a(n) = 10*A309761(n). - Chai Wah Wu, Jun 16 2025

A385290 Indices of records in A378865.

Original entry on oeis.org

1, 12, 24, 49, 101, 102, 497, 498, 499, 501, 1001, 1002, 2864, 4999, 5001, 10001, 10002, 10624, 12864, 18624, 27648, 123648, 249856, 442368, 786432, 1259874, 2159784, 8249175, 8759124, 10236587, 10236758, 10237649, 10239674, 10239786, 10268473, 10427539, 10476523

Offset: 1

Jinyuan Wang, Jun 24 2025

Cf. A276348, A378865, A379673.

A306544 Any positive integer n has a smallest multiple consisting of a succession of 1's followed by a succession of 0's (A052983). This multiple is regarded as a binary number and a(n) is its conversion to base 10.

Original entry on oeis.org

2, 2, 14, 4, 2, 14, 126, 8, 1022, 2, 6, 28, 126, 126, 14, 16, 131070, 1022, 524286, 4, 126, 6, 8388606, 56, 4, 126, 268435454, 252, 536870910, 14, 65534, 32, 126, 131070, 126, 2044, 14, 524286, 126, 8, 62, 126, 4194302, 12, 1022, 8388606, 140737488355326, 112, 8796093022206

Offset: 1

Bernard Schott, Feb 22 2019

For any odd number m not divisible by 5 (A045572), Euler's theorem (lcm(9*m,10) = 1, so 10^phi(9*m) == 1 (mod 9*m); i.e., 9*m | 10^d - 1 = 9*R_d with d = phi(9*m)) guarantees that the repunit R_d is always some multiple of m.
The numbers of the form 2^i*5^j with i, j >= 0 (A003592) clearly have a multiple equal to 10^r, for r = max(i,j).
These multiples of n end in a string of one or more 0's, so all the terms of this sequence are even.
The powers 2^k are fixed points of this sequence: the smallest multiple of 2^k consisting of a succession of 1's followed by a succession of 0's is 10^k, and 10^k in base 2 is 2^k in base 10.

			The smallest multiple of the integer 7 consisting of a succession of 1's followed by a succession of 0's is 1111110, and 1111110_2 = 126_10, so a(7) = 126. This is also the case for n=13, 14, 21, 26, 33, 35, 37, ...

Makoto Kamada, Factorization of 11...11 (Repunit).

Cf. A000079, A052983, A045572, A099679, A002275, A007088, A276349, A003592, A011557, A276348.

More terms from Michel Marcus, Feb 28 2019

cp's OEIS Frontend

A052983 Least multiple of n consisting of a succession of 1's followed by a succession of 0's.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

Extensions

A276349 Numbers consisting of a nonempty string of 1's followed by a nonempty string of 0's.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Python

Formula

A385290 Indices of records in A378865.

Views

Author

Keywords

Crossrefs

A306544 Any positive integer n has a smallest multiple consisting of a succession of 1's followed by a succession of 0's (A052983). This multiple is regarded as a binary number and a(n) is its conversion to base 10.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Extensions