cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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).

Original entry on oeis.org

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

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Author

Joost de Winter, May 25 2010

Keywords

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...
		

Crossrefs

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
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