A052983 -id:A052983 - cp's OEIS frontend

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.

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

A276348 a(n) = the smallest number k such that k*n is a number with a string of 1's followed by a string of 0's.

Original entry on oeis.org

10, 5, 370, 25, 2, 185, 158730, 125, 123456790, 1, 10, 925, 85470, 79365, 74, 625, 653594771241830, 61728395, 58479532163742690, 5, 52910, 5, 483091787439613526570, 4625, 4, 42735, 41152263374485596707818930, 396825, 383141762452107279693486590, 37

Offset: 1

Jaroslav Krizek, Aug 30 2016

a(n) = the smallest number k such that k*n is a number from A276349.

a(n) > 0 for all n.

			For n=3; 3*370 = 1110 (term of A276349).

L. Pick, Dirichletovy šuplíčky. Pokroky matematiky, fyziky & astronomie; 2 (2016), 106-118. (In Czech; The Dirichlet pigeonhole principle)

Robert Israel, Table of n, a(n) for n = 1..1016

Cf. A052983, A084680, A276349.

a:=10; S:=[a]; for n in [2..6] do k:=0; flag:= true; while flag do k+:=1; if [k*n] subset [n: n in [1..10000] | Seqint(Setseq(Set(Sort(Intseq(n))))) eq 10 and Seqint(Sort((Intseq(n)))) eq n] then Append(~S, k); a:=k; flag:=false; end if; end while; end for; S;

f:= proc(n) local b,c,d,m,q;
    b:= padic:-ordp(n,2); c:= padic:-ordp(n,5); if b+c=0 then d:= 1 else d:= max(b,c) fi; m:= n/2^b/5^c; q:= numtheory:-order(10,9*m);
     2^(d-b)*5^(d-c)*(10^q-1)/(9*m)
end proc:
map(f, [$1..100]); # Robert Israel, Aug 30 2016

Table[k = 1; While[! If[Length@ # == 2, Flatten@ Map[Union, #] == {1, 0}, False] &@ Split@ IntegerDigits[k n], k++]; k, {n, 8}] (* Michael De Vlieger, Aug 30 2016 *)

a(n) = A052983(n)/n.
From Robert Israel, Aug 30 2016: (Start)
Let n = 2^b*5^c*m where GCD(m,10)=1, and q = A084680(9*m).
If b=c=0 let d=1, otherwise d=max(b,c).
Then a(n) = 2^(d-a)*5^(d-b)*(10^q-1)/(9*m). (End)

A074157 Smallest multiple of n which uses no digits of n, or 0 if no such number exists.

Original entry on oeis.org

2, 4, 6, 8, 10, 12, 14, 16, 18, 0, 22, 36, 26, 28, 30, 32, 34, 36, 38, 0, 63, 44, 46, 96, 100, 78, 54, 56, 58, 0, 62, 64, 66, 68, 70, 72, 111, 76, 78, 0, 82, 168, 86, 88, 90, 92, 188, 96, 588, 0, 204, 104, 106, 108, 110, 112, 114, 116, 118, 0, 244, 310, 189, 128

Offset: 1

Amarnath Murthy, Aug 29 2002

For numbers n of the form 10k, of the form 10k+5 with a zero somewhere, and those using all 10 digits, a(n) = 0.
Are there numbers using 8 of the 9 nonzero digits for which a(n) is nonzero? If so, what is the smallest such number?
All such numbers have a(n) nonzero since a candidate value for a(n) is d*A052983(n) where d is the nonzero digit not in n. - Gonzalo Martínez, Dec 12 2024

Michel Marcus, Table of n, a(n) for n = 1..1000

Cf. A052983, A076924, A378865.

a(n) = if(!(n%10), return(0)); my(dn = Set(digits(n))); if(#dn==10 || (#dn==9 && dn[1]==1), return(0)); if(!(n%5) && !dn[1], return(0)); if(!(n%2), my(v=setminus([0, 2, 4, 6, 8], dn)); if(!#v || (#v==1 && #dn==9 && valuation(n, 2)>valuation(v[1], 2)) || (v==[2, 6] && #dn==8 && valuation(n, 2)>1), return(0))); forstep(m=2*n, oo, n, if(!#setintersect(Set(digits(m)), dn), return(m))); \\ Michel Marcus, Aug 04 2017 (Corrected by Jinyuan Wang, Jun 15 2025)

a(24) and a(49) corrected by Michel Marcus, Aug 04 2017

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

A343447 Smallest m such that alternating integer 101...101 = A094028(m) is a multiple of A045572(n), (i.e., integers coprime with 10).

Original entry on oeis.org

0, 2, 2, 8, 10, 2, 7, 8, 2, 10, 26, 13, 14, 32, 2, 2, 4, 20, 22, 20, 23, 12, 8, 28, 29, 8, 32, 32, 34, 3, 32, 12, 80, 40, 41, 21, 2, 14, 47, 98, 1, 16, 52, 53, 2, 55, 8, 23, 120, 14, 20, 20, 64, 8, 3, 22, 68, 32, 20, 73, 74, 71, 38, 38, 32, 80, 82, 38, 8, 42

Offset: 1

Bernard Schott, Apr 15 2021

Every number coprime with 10 has a smallest multiple that is repunit (A099679).
Every positive number has a smallest multiple consisting of a succession of 1's followed by a succession of 0's (A052983).
Every number coprime with 10 has a smallest multiple that is alternating of the form 1010...0101 (this sequence).

			A045572(3) = 7, the smallest alternating multiple of 7 in A094028 is A094028(2) = 10101 because 1443*7 = 10101, as 1 and 101 are not divisible by 7, so a(3) = 2.

Cf. A030141, A045572, A052983, A094028, A099679.

a[n_] := Module[{k = (5*n + (Mod[3*n + 2, 4] - 4))/2, m = 0}, While[! Divisible[1 + 100*(100^m - 1)/99, k], m++]; m]; Array[a, 100] (* Amiram Eldar, Apr 15 2021 *)

a045572(n)=10*(n>>2)+[-1,1,3,7][n%4+1] \\ after Charles R Greathouse IV in A045572
a094028(n) = 1+100*(100^n-1)/99
a(n) = for(m=0, oo, if(a094028(m)%a045572(n)==0, return(m))) \\ Felix Fröhlich, Apr 15 2021

More terms from Felix Fröhlich, Apr 15 2021

cp's OEIS Frontend

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

A276348 a(n) = the smallest number k such that k*n is a number with a string of 1's followed by a string of 0's.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A074157 Smallest multiple of n which uses no digits of n, or 0 if no such number exists.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

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

A343447 Smallest m such that alternating integer 101...101 = A094028(m) is a multiple of A045572(n), (i.e., integers coprime with 10).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Extensions