A128857 a(n) = least number m beginning with 1 such that the quotient m/n is obtained merely by shifting the leftmost digit 1 of m to the right end.

1, 105263157894736842, 1034482758620689655172413793, 102564, 102040816326530612244897959183673469387755, 1016949152542372881355932203389830508474576271186440677966

Offset: 1

Author

Anton V. Chupin (chupin(X)icmm.ru), Apr 12 2007

Keywords

• nonn
• base

Comments

a(n) is simply the decimal period of the fraction n/(10n-1). Thus, we have: n/(10n-1) = a(n)/(10^A128858(n)-1). With the usual convention that the decimal period of 0 is zero, that definition would allow the extension a(0)=0. a(n) is also the period of the decadic integer -n/(10n-1). - Gerard P. Michon, Oct 31 2012

Examples

			a(4) = 102564 since this is the smallest number that begins with 1 and which is divided by 4 when the first digit 1 is made the last digit (102564/4 = 25641).
		

Links

• A. V. Chupin, Table of n, a(n) for n=1..101
• G. P. Michon, Integers that are divided by k if their digits are rotated left.

Crossrefs

Minimal numbers for shifting any digit from the left to the right (not only 1) are in A097717.

By accident, the nine terms of A092697 coincide with the first nine terms of the present sequence. - N. J. A. Sloane, Apr 13 2009

Programs

Mathematica

(*Moving digits a:*) Give[a_,n_]:=Block[{d=Ceiling[Log[10,n]],m=(10n-1)/GCD[10n-1, a]}, If[m!=1,While[PowerMod[10,d,m]!=n,d++ ],d=1]; ((10^(d+1)-1) a n)/(10n-1)]; Table[Give[1,n],{n,101}]

Python

from sympy import n_order
def A128857(n): return n*(10**n_order(10,(m:=10*n-1))-1)//m # Chai Wah Wu, Apr 09 2024

Extensions

Edited by N. J. A. Sloane, Apr 13 2009

Code and b-file corrected by Ray Chandler, Apr 29 2009