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A288782
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Integers k that have the property that there exists an integer x with n+1 digits, such that 1 <= k/x < 2 and k/x = 1 + (x-10^n)/(10^n-1), i.e., the same digits appear in the denominator and in the recurring decimal.
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3
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10, 34, 100, 208, 238, 394, 1000, 1680, 2898, 3994, 10000, 14938, 16198, 22348, 22648, 29830, 31600, 39994, 100000, 109994, 137694, 149380, 316048, 333630, 380720, 399994, 1000000, 1010610, 1079440, 1306120, 1318244, 1396694, 1409228, 1460458, 1738920, 1768810, 1826150
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OFFSET
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1,1
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COMMENTS
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The values 399..994 all seem to appear.
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LINKS
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MATHEMATICA
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Union @@ Reap[ Do[Sow[k /. 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 *)
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PROG
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(Python 3)
from math import sqrt
def is_square(n):
root = int(sqrt(n))
return root*root == n
def find_sols(length):
count = 0
k=10**length
for n in range(k, 4*k-2):
discr= (2*k-1)*(2*k-1) - 4*(k*(k-1)-(k-1)*n)
if is_square(discr):
count+=1
b=(-(2*k-1)+sqrt(discr))/2
print(n, k+b, n/(k+b))
return count
for i in range(8):
print(find_sols(i))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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