perplexus.info :: Numbers : Aleph<sub>0</sub>

Aleph₀ (Posted on 2024-05-20)

All rational numbers can be put into a 1-to-1 matching with the integers. For the positive of each, the below table shows a 1-to-1 matching where the rationals are laid out in order of the total of the numerator and denominator of the number expressed as a fraction in simplest (reduced) form. Within one total of numerator and denominator, they are in order of the numerator.

    1     1              1/14    64
  1/2     2              2/13    65
    2     3              4/11    66
  1/3     4               7/8    67
    3     5               8/7    68
  1/4     6              11/4    69
  2/3     7              13/2    70
  3/2     8                14    71
    4     9              1/15    72
  1/5    10              3/13    73
    5    11              5/11    74
  1/6    12               7/9    75
  2/5    13               9/7    76
  3/4    14              11/5    77
  4/3    15              13/3    78
  5/2    16                15    79
    6    17              1/16    80
  1/7    18              2/15    81
  3/5    19              3/14    82
  5/3    20              4/13    83
    7    21              5/12    84
  1/8    22              6/11    85
  2/7    23              7/10    86
  4/5    24               8/9    87
  5/4    25               9/8    88
  7/2    26              10/7    89
    8    27              11/6    90
  1/9    28              12/5    91
  3/7    29              13/4    92
  7/3    30              14/3    93
    9    31              15/2    94
 1/10    32                16    95
  2/9    33              1/17    96
  3/8    34              5/13    97
  4/7    35              7/11    98
  5/6    36              11/7    99
  6/5    37              13/5   100
  7/4    38                17   101
  8/3    39              1/18   102
  9/2    40              2/17   103
   10    41              3/16   104
 1/11    42              4/15   105
  5/7    43              5/14   106
  7/5    44              6/13   107
   11    45              7/12   108
 1/12    46              8/11   109
 2/11    47              9/10   110
 3/10    48              10/9   111
  4/9    49              11/8   112
  5/8    50              12/7   113
  6/7    51              13/6   114
  7/6    52              14/5   115
  8/5    53              15/4   116
  9/4    54              16/3   117
 10/3    55              17/2   118
 11/2    56                18   119
   12    57              1/19   120
 1/13    58              3/17   121
 3/11    59              7/13   122
  5/9    60              9/11   123
  9/5    61              11/9   124
 11/3    62              13/7   125
   13    63              17/3   126
                           19   127

Using this scheme, what rational number is matched to the integer 1 million?

When 1 million is considered as a rational number, to what integer is it matched?

See The Solution Submitted by Charlie

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Part 1

Comment 4 of 4 |

Part 1:

I wrote a program to match the list in the statement of the problem. Since Python lists are zero-based, the "zero-th" element is a blank.

Here are first 10 elements, 0 through 9:

alist[:10]

['', '1/1', '1/2', '2/1', '1/3', '3/1', '1/4', '2/3', '3/2', '4/1']

What rational number is matched to the integer 1 million?

alist[1000000] = '113/1701' (whose n + d sum is 1814)

------ ------

import math

count = 0

alist = ['']

finished = False

for asum in range(2,1000003):

for p in range(1,asum):

q = asum - p

if math.gcd(p,q) != 1:

continue

alist.append(str(p) + '/' + str(q))

if len(alist) == 1000001:

print('alist[1000000] = ', alist[1000000])

finished = True

break

if finished:

break

Part 2: I rewrote the code in order to find the answer to part 2.

All I had to do was run the program a lot farther looking for:

alist.index('1000000/1') (whose n + d sum is 1000001)

But my computer ran out of memory and choked.

But I have some intermediate results. First the modified code:

count = 0

a_dict = {}

asum = 2

while asum < 1000005:

for p in range(1,asum):

q = asum - p

if math.gcd(p,q) != 1:

continue

count += 1

if q == 1 and str(p)[-4:] == '0000':

a_dict[str(p) + '/' + str(q)] = count

print(p,count)

asum += 1

Intermediate results where p is the numerator (and the denominator is 1):

p integer

10000 30407277

20000 121603387

30000 273600177

40000 486380275

50000 759952823

60000 1094335409

70000 1489495819

80000 1945453481

90000 2462193829

100000 3039741653

110000 3678059147

120000 4377196945

130000 5137116825

140000 5957809263

150000 6839354027

Edited on May 20, 2024, 8:25 pm

Posted by Larry on 2024-05-20 20:09:47

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