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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).

%I A178335 #27 Jun 02 2025 02:58:21
%S A178335 7,14,28,35,56,70,112,140,175,224,280,350,448,560,700,875,896,1120,
%T A178335 1400,1750,1792,2240,2800,3500,3584,4375,4480,5600,7000,7168,8750,
%U A178335 8960,11200,14000,14336,17500,17920,21875,22400,28000,28672,35000,35840,43750
%N 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).
%C A178335 All terms are == 0 (mod 7).
%C A178335 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).
%C A178335 From _R. J. Mathar_, Jul 13 2010: (Start)
%C A178335 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.
%C A178335 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.
%C A178335 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).
%C A178335 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(.)
%C A178335 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.
%C A178335 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).
%C A178335 On the right hand side, the (power of 2 or 5) mod 7 is a fixed number between 1 and 6.
%C A178335 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)
%H A178335 Robert G. Wilson v, <a href="/A178335/b178335.txt">Table of n, a(n) for n = 1..225 </a>. [From _Robert G. Wilson v_, Jun 08 2010] [Corrected by Zak Seidov, Jul 12 2010]
%H A178335 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RepeatingDecimal.html">Repeating Decimal</a>
%e A178335 1/7 = 0.142857142857143..., 1/14 = 0.0714285714285714...
%t A178335 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
%Y A178335 Cf. A051626, A085837, A158204.
%K A178335 nonn,base
%O A178335 1,1
%A A178335 _Joost de Winter_, May 25 2010
%E A178335 Edited, corrected and extended by _Robert G. Wilson v_, May 31 2010