A276349 -id:A276349 - cp's OEIS frontend

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.

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)

A309761 Numbers that are sums of consecutive powers of 10.

Original entry on oeis.org

1, 10, 11, 100, 110, 111, 1000, 1100, 1110, 1111, 10000, 11000, 11100, 11110, 11111, 100000, 110000, 111000, 111100, 111110, 111111, 1000000, 1100000, 1110000, 1111000, 1111100, 1111110, 1111111, 10000000, 11000000, 11100000, 11110000, 11111000, 11111100

Offset: 1

Ilya Gutkovskiy, Aug 15 2019

nonn

Numbers of the form (10^i - 10^j)/9 with i > j.

			11100 = 10^2 + 10^3 + 10^4, so 11100 is in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10011

Cf. A000042, A007088, A011557, A023758, A038444, A276349, A309758, A309759.

seq(seq((10^i-10^(i-j))/9, j=1..i),i=1..10); # Robert Israel, Aug 16 2019

from math import isqrt
def A309761(n): return (10**(m:=isqrt(n<<3)+1>>1)-10**(m*(m+1)-(n<<1)>>1))//9 # Chai Wah Wu, Apr 04 2025

a(n) = A007088(A023758(n+1)).
a(i*(i-1)/2 + j) = (10^i - 10^(i-j))/9 for 1<=j<=i. - Robert Israel, Aug 16 2019
a(n) = A276349(n)/10. - Chai Wah Wu, Jun 16 2025

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

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.

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A309761 Numbers that are sums of consecutive powers of 10.

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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.

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