proof
| - Let the digits of be , , and . Then
:
Thus is a perfect digital invariant for for all . (en)
- Let
:
be a natural number with digits, where , and , where is a natural number greater than 1.
According to the divisibility rules of base , if , then if , then the digit sum
:
If a digit , then . According to Euler's theorem, if , . Thus, if the digit sum , then .
Therefore, for any natural number , if , and , then for every natural number , if , then . (en)
- Let the digits of be , , and . Then
:
Thus is a perfect digital invariant for for all . (en)
- Let the digits of be , , and . Then
:
::
::
::
::
::
::
::
::
::
::
::
::
Thus is a perfect digital invariant for for all . (en)
- Let the digits of be , , and . Then
:
::
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::
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::
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::
Thus is a perfect digital invariant for for all . (en)
- If , then the bound can be reduced.
Let be the number for which the sum of squares of digits is largest among the numbers less than .
:
: because
Let be the number for which the sum of squares of digits is largest among the numbers less than .
:
: because
Let be the number for which the sum of squares of digits is largest among the numbers less than .
:
:
Let be the number for which the sum of squares of digits is largest among the numbers less than .
:
:
. Thus, numbers in base lead to cycles or fixed points of numbers . (en)
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