cp's OEIS Frontend

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.

%I A285273 #37 Nov 09 2024 19:30:57
%S A285273 2,4,4,8,8,32,8,32,8,32,8,128,16,32,64,128,8,256,4,256,128,128,4,1024,
%T A285273 64,128,32,512,64,8192,16,4096,64,128,256,2048,16,16,64,4096,32,16384,
%U A285273 32,2048,512,128,8,8192,32,2048,256,1024,32,4096,512,8192,64,512,8
%N 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.
%C A285273 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?"
%C A285273 It appears that a(n) is always a power of 2.
%e A285273 The number 154 has the property that there exists an integer, 238, for which
%e A285273      238/154 = 1 + 54/99 = 1.545454545...
%e A285273 There are 4 three-digit values that give rise to a 2-digit recurring decimal:
%e A285273   100/100.0 = 1.0000000000000000
%e A285273   208/144.0 = 1.4444444444444444...
%e A285273   238/154.0 = 1.5454545454545454...
%e A285273   394/198.0 = 1.9898989898989898...
%e A285273 thus a(2) = 4.
%e A285273 For n=3, a(3) = 8:
%e A285273   10000/10000.0 = 1.0000000000000000
%e A285273   14938/12222.0 = 1.2222222222222222...
%e A285273   16198/12727.0 = 1.2727272727272727...
%e A285273   22348/14949.0 = 1.4949494949494949...
%e A285273   22648/15049.0 = 1.5049504950495049...
%e A285273   29830/17271.0 = 1.7271727172717271...
%e A285273   31600/17776.0 = 1.7776777677767776...
%e A285273   39994/19998.0 = 1.9998999899989998...
%t A285273 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 *)
%o A285273 (Python)
%o A285273 from math import sqrt
%o A285273 def is_square(n):
%o A285273   root = int(sqrt(n))
%o A285273   return root*root == n
%o A285273 def find_sols(length):
%o A285273     count = 0
%o A285273     k=10**length
%o A285273     for n in range(k,4*k-2):
%o A285273         discr= (2*k-1)*(2*k-1) - 4*(k*(k-1)-(k-1)*n)
%o A285273         if is_square(discr):
%o A285273             count+=1
%o A285273             b=(-(2*k-1)+sqrt(discr))/2
%o A285273             print(n, k+b, n/(k+b))
%o A285273     return count
%o A285273 for i in range(8):
%o A285273     print(find_sols(i))
%Y A285273 Cf. A288781, A288782.
%K A285273 nonn,base
%O A285273 1,1
%A A285273 _James Kilfiger_, Jun 14 2017
%E A285273 Definition corrected and a(11)-a(59) from _Giovanni Resta_, Jun 30 2017