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A285273
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Number of integers, x, with n+1 digits, that have the property that there exists an integer k, with x <= k < 2*x, such that 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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2
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2, 4, 4, 8, 8, 32, 8, 32, 8, 32, 8, 128, 16, 32, 64, 128, 8, 256, 4, 256, 128, 128, 4, 1024, 64, 128, 32, 512, 64, 8192, 16, 4096, 64, 128, 256, 2048, 16, 16, 64, 4096, 32, 16384, 32, 2048, 512, 128, 8, 8192, 32, 2048, 256, 1024, 32, 4096, 512, 8192, 64, 512, 8
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OFFSET
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1,1
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COMMENTS
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This was suggested by generalizing an exam question which asked "Jack typed a whole number into his calculator and divided by 154. The result was 1.545454545. What was his number?"
It appears that a(n) is always a power of 2.
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LINKS
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EXAMPLE
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The number 154 has the property that there exists an integer, 238, for which
238/154 = 1 + 54/99 = 1.545454545...
There are 4 three-digit values that give rise to a 2-digit recurring decimal:
100/100.0 = 1.0000000000000000
208/144.0 = 1.4444444444444444...
238/154.0 = 1.5454545454545454...
394/198.0 = 1.9898989898989898...
thus a(2) = 4.
For n=3, a(3) = 8:
10000/10000.0 = 1.0000000000000000
14938/12222.0 = 1.2222222222222222...
16198/12727.0 = 1.2727272727272727...
22348/14949.0 = 1.4949494949494949...
22648/15049.0 = 1.5049504950495049...
29830/17271.0 = 1.7271727172717271...
31600/17776.0 = 1.7776777677767776...
39994/19998.0 = 1.9998999899989998...
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MATHEMATICA
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a[n_] := Length@ 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]; Array[a, 20] (* for n<60, 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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