A178335 Integers for which the decimal expansion of the reciprocal contains the repeating digits 1,4,2,8,5,7 (corresponding to the decimal expansion of 1/7).

7, 14, 28, 35, 56, 70, 112, 140, 175, 224, 280, 350, 448, 560, 700, 875, 896, 1120, 1400, 1750, 1792, 2240, 2800, 3500, 3584, 4375, 4480, 5600, 7000, 7168, 8750, 8960, 11200, 14000, 14336, 17500, 17920, 21875, 22400, 28000, 28672, 35000, 35840, 43750

Offset: 1

Author

Joost de Winter, May 25 2010

Keywords

• nonn
• base

Comments

All terms are == 0 (mod 7).

If m appears, so does 10*m. Therefore the primitive terms (they will not end in 0) are: 7, 14, 28, 35, 56, 112, 175, 224, 448, 875, 896, 1792, 3584, 4375, 7168, 14336, 21875, 28672,... (see A158204).

From R. J. Mathar, Jul 13 2010: (Start)

a(n) = 7*A003592(n). [Proof: the a(n) are defined demanding that 1/a(n) = t/10^b+1/(7*10^c) for a transient integer t>=0 and a periodic part 1/(7*10^c) for some b>=0 and c>=0.

Note this splits the chain of decimal digits right in front of the period 142857, which means the least significant digits of t may be some of the least significant digits of 142857. We may assume that 1/(7*10^c) < 1/10^b, so c>=b.

Multiply by a(n)*7*10^c to get 7*10^c = a(n)*(7*t*10^(c-b)+1). Reduction modulo 7 shows that a(n)=7*k, so 10^c = k*(7*t*10^(c-b)+1).

Decomposition of both sides into prime factors shows that k must be of the form 2^i*5^j, which shows that the a(n) are of the form 7*A003592(.)

To demonstrate that none of the A003592 are missed it remains to show that the other factor, 7*t*10^(c-b)+1, can always be chosen of the form 2^(i')*5^(i'+i-j) to cancel the excess of the two exponents that the prime factorization of k may have: 10^c =2^c*5^c demands equal exponents.

Because t and 10^(c-b) can chosen freely, this is equivalent to showing that there is always a t, a c-b and an i' such that 7*t*10^(c-b)+1 = 10^i'*(excess power of 2 or 5).

On the right hand side, the (power of 2 or 5) mod 7 is a fixed number between 1 and 6.

As i' runs through 7 consecutive numbers, 10^i' mod 7 attains all numbers between 1 and 6; the product 10^i'*(power of 2 or 5) can always be tuned to == 1 (mod 7) by selection of i', and t*10^(c-b) follows by division. This shows that all k of the form 2^i*5^j contribute to the sequence.] (End)

Examples

			1/7 = 0.142857142857143..., 1/14 = 0.0714285714285714...
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..225 . [From _Robert G. Wilson v_, Jun 08 2010] [Corrected by Zak Seidov, Jul 12 2010]
• Eric Weisstein's World of Mathematics, Repeating Decimal

Crossrefs

Cf. A051626, A085837, A158204.

Programs

Mathematica

digitCycleLength[ r_Rational, b_Integer?Positive ] := MultiplicativeOrder[ b, FixedPoint[ Quotient[ #, GCD[ #, b ] ] &, Denominator[ r ] ] ] (* from Wolfram Library, Help Menu for MultiplicativeOrder *); fQ[ n_ ] := MemberQ[ {{1, 4, 2, 8, 5, 7}, {4, 2, 8, 5, 7, 1}, {2, 8, 5, 7, 1, 4}, {8, 5, 7, 1, 4, 2}, {5, 7, 1, 4, 2, 8}, {7, 1, 4, 2, 8, 5}}, RealDigits[ 1/n ][ [ 1, -1 ] ] ]; k = 0; lst = {}; While[ k < 10^9, If[ digitCycleLength[ 1/k, 10 ] == 6 && fQ[ k ], Print@k; AppendTo[ lst, k ] ]; k += 7 ]; lst

Extensions

Edited, corrected and extended by Robert G. Wilson v, May 31 2010