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| Page under construction | <<TableOfContents>> '''Page under construction ''' Here's one solution to the Taxi Cab function: {{{ ∇R←TaxiCab N;⎕IO;vals;sumCubes;cubes [1] ⎕IO←1 [2] vals←⍳N [3] sumCubes←(vals∘.<vals)×cubes∘.+cubes←vals*3 [4] R←(R=1⌽R)/R←R[⍋R←(,sumCubes)~0] [5] R←⍉R,[0.5](,¨(R⍷¨⊂sumCubes)×⊂vals∘.,vals)~¨⊂⊂0 0 ∇ }}} For example: {{{ TaxiCab 30 1729 1 12 9 10 4104 2 16 9 15 13832 2 24 18 20 20683 10 27 19 24 }}} The output shows the taxi cab number followed by the pairs of terms. For example 1729 = 1^3^ + 12^3^ = 9^3^ + 10^3^ = An explanation of the APL code = Blah : Under construction = Other investigations concerning 1729 = First we define a handy function to break a number up into its digits: {{{ ∇R←base Digits val [1] ⍝ Returns individual digits of 'val' when expressed in base 'base', [2] ⍝ e.g. 10 Digits 1729 gives 1 7 2 9 [3] R←((1+⌊base⍟val+0=val)⍴base)⊤val ∇ }}} (a) Now to find out which numbers are evenly divisible by the sum of their digits. Here's a test of the range 1720-1770 {{{ (0=(+/¨10 Digits¨ X)|X)/X←1719+⍳50 1725 1728 1729 1740 1744 1746 1764 1770 }}} (b) Now we investigate which of the first 100000 numbers behave in the same way as 1729, i.e. 1 + 7 + 2 + 9 = 19 ; 19 × 91 = 1729 {{{ (vals=sums×10⊥¨⌽¨10 Digits¨sums←+/¨10 Digits¨ vals)/vals←⍳100000 1 81 1458 1729 }}} In fact these are the only four natural numbers which have this property. |
Ramanujan's Taxi Cab Numbers : A Solution
Contents
Page under construction
Here's one solution to the Taxi Cab function:
∇R←TaxiCab N;⎕IO;vals;sumCubes;cubes
[1] ⎕IO←1
[2] vals←⍳N
[3] sumCubes←(vals∘.<vals)×cubes∘.+cubes←vals*3
[4] R←(R=1⌽R)/R←R[⍋R←(,sumCubes)~0]
[5] R←⍉R,[0.5](,¨(R⍷¨⊂sumCubes)×⊂vals∘.,vals)~¨⊂⊂0 0
∇ For example:
TaxiCab 30 1729 1 12 9 10 4104 2 16 9 15 13832 2 24 18 20 20683 10 27 19 24
The output shows the taxi cab number followed by the pairs of terms. For example 1729 = 13 + 123 = 93 + 103
An explanation of the APL code
Blah : Under construction
Other investigations concerning 1729
First we define a handy function to break a number up into its digits:
∇R←base Digits val
[1] ⍝ Returns individual digits of 'val' when expressed in base 'base',
[2] ⍝ e.g. 10 Digits 1729 gives 1 7 2 9
[3] R←((1+⌊base⍟val+0=val)⍴base)⊤val
∇ (a) Now to find out which numbers are evenly divisible by the sum of their digits. Here's a test of the range 1720-1770
(0=(+/¨10 Digits¨ X)|X)/X←1719+⍳50 1725 1728 1729 1740 1744 1746 1764 1770
(b) Now we investigate which of the first 100000 numbers behave in the same way as 1729, i.e.
1 + 7 + 2 + 9 = 19 ; 19 × 91 = 1729
(vals=sums×10⊥¨⌽¨10 Digits¨sums←+/¨10 Digits¨ vals)/vals←⍳100000 1 81 1458 1729
In fact these are the only four natural numbers which have this property.
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