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Happy New Palindromic Year 11111011111 !

Very nice palindromic binary year for everybody!

IntegerDigits[2015, 2]

{1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1}

4 Replies

It is also a Lucas-Carmichel number.

n = 2015;
factors = FactorInteger[n][[All, 1]];
AllTrue[factors + 1, Divisible[n + 1, #] &]

True

See this link: Lucas–Carmichael number

-adk-

POSTED BY: Abdul Dakkak

Palindromes in base 2 are rare but for a number to be a palindrome in some base is not so rare. For instance there are a few other bases that make 2015 into a palindrome:

Select[Range[2, 2013], # == Reverse[#] &[IntegerDigits[2015, #]] &]

{2, 38, 64, 154, 402}

For {1,1} is always b+1 palindrome, so any nontrivial palindrome base is between 2 and n-2. It is rare to not be a palindrome in any such base. The last one was 2011 and the next one is 2063.

Select[Range[1900, 2200], Function[n, Select[Range[2, n - 2], # == Reverse[#] &[IntegerDigits[n, #]] &] === {}]]

{1907, 1949, 1993, 1997, 2011, 2063, 2087, 2099, 2111, 2137, 2179}

POSTED BY: Todd Rowland

It's (trivially) a palindrome in any base >= 2016..

POSTED BY: Daniel Lichtblau

I noticed another interesting coincidence (?) about "2015":

which are:

  • 2015 (base 11) converted to base 10 is 2678.
  • 2015 (base 10) converted to base 9 is 2678.

Well, these have actually been in alignment since the year 2010, when both computations yielded 2673, but did anyone notice before?

POSTED BY: Andrew Walters
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