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Integers x with h+1 digits that have the property that there exists an integer k, with x <= k < 2*x, such that k/x = 1 + (x-10^h)/(10^h-1), i.e., the same digits appear in the denominator and in the recurring decimal.
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%I #16 Jul 02 2017 03:57:33

%S 10,18,100,144,154,198,1000,1296,1702,1998,10000,12222,12727,14949,

%T 15049,17271,17776,19998,100000,104878,117343,122221,177777,182655,

%U 195120,199998,1000000,1005291,1038961,1142856,1148148,1181818,1187109,1208494,1318681

%N Integers x with h+1 digits that have the property that there exists an integer k, with x <= k < 2*x, such that k/x = 1 + (x-10^h)/(10^h-1), i.e., the same digits appear in the denominator and in the recurring decimal.

%C The numbers appear to be in pairs that add up to 299...998; e.g., 144 + 154 = 298, 12222 + 17776 = 29998.

%H Giovanni Resta, <a href="/A288781/b288781.txt">Table of n, a(n) for n = 1..10000</a>

%t Union @@ Reap[Do[Sow[x /. List@ ToRules@ Reduce[k/x == 1 + (x - 10^n)/(10^n - 1) && 10^n <= x < 10^(n + 1) && x <= k < 2 x, {k, x}, Integers]], {n, 6}]][[2, 1]] (* _Giovanni Resta_, Jun 30 2017 *)

%o (Python 3)

%o from math import sqrt

%o def is_square(n):

%o root = int(sqrt(n))

%o return root*root == n

%o def find_sols(length):

%o count = 0

%o k=10**length

%o for n in range(k,4*k-2):

%o discr= (2*k-1)*(2*k-1) - 4*(k*(k-1)-(k-1)*n)

%o if is_square(discr):

%o count+=1

%o b=(-(2*k-1)+sqrt(discr))/2

%o print(n, k+b, n/(k+b))

%o return count

%o for i in range(8):

%o print(find_sols(i))

%Y Cf. A285273, A288782 (numerators).

%K nonn,base

%O 1,1

%A _James Kilfiger_, Jun 15 2017

%E Definition corrected by and more terms from _Giovanni Resta_, Jun 30 2017