A285273 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.

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

Offset: 1

Author

James Kilfiger, Jun 14 2017

Keywords

• nonn
• base

Comments

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.

Examples

			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...
		

Crossrefs

Cf. A288781, A288782.

Programs

Mathematica

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 *)

Python

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))

Extensions

Definition corrected and a(11)-a(59) from Giovanni Resta, Jun 30 2017