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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
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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.

LINKS

Table of n, a(n) for n=1..59.

EXAMPLE

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...

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 *)

PROG

(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))

CROSSREFS

Cf. A288781, A288782.

Sequence in context: A248692 A048656 A107848 * A188824 A181212 A233394

Adjacent sequences:  A285270 A285271 A285272 * A285274 A285275 A285276

KEYWORD

nonn,base

AUTHOR

James Kilfiger, Jun 14 2017

EXTENSIONS

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

STATUS

approved

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Last modified February 25 22:37 EST 2021. Contains 341618 sequences. (Running on oeis4.)