A177062 Length of the longest common substring in binary representations of n and n^2.

1, 2, 1, 3, 2, 2, 2, 4, 3, 3, 2, 3, 3, 3, 3, 5, 4, 4, 4, 4, 3, 3, 2, 4, 4, 4, 5, 4, 4, 4, 4, 6, 5, 5, 4, 5, 4, 4, 4, 5, 6, 3, 3, 4, 4, 2, 3, 5, 5, 4, 4, 5, 5, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7, 6, 6, 5, 6, 5, 5, 5, 6, 5, 5, 5, 4, 4, 4, 4

Offset

1,2

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Maple

f:= n -> length(StringTools:-LongestCommonSubString(convert(convert(n, binary), string), convert(convert(n^2, binary), string))):
map(f, [$ 1..100]); # Robert Israel, Sep 13 2016

Mathematica

longestRun[pairs_] := Split[pairs, Equal @@ #1 && Equal @@ #2 &] // Sort[#, Length[#1] < Length[#2] &] & // Last; a[n_] := (bits1 = IntegerDigits[n, 2]; bits2 = IntegerDigits[n^2, 2]; longestRun /@ ListConvolve[ Reverse[bits1], bits2, {1, -1}, -1, List, List] // Sort[#, Length[#1] < Length[#2] &] & // Last // Length); a /@ Range[79] (* Jean-François Alcover, Jun 05 2013 *)
Table[Length[LongestCommonSubsequence[IntegerDigits[n^2, 2], IntegerDigits[ n, 2]]], {n, 1, 100}] (* Vladimir Reshetnikov, Apr 26 2016 *)

Prog

(Haskell)
import Data.List
toBinary 0 = []
toBinary n = toBinary (n `div` 2) ++ [odd n]
lcstr xs ys = maximum . concat $ [f xs' ys | xs' <- tails xs] ++ [f xs ys' | ys' <- drop 1 $ tails ys] where f xs ys = scanl g 0 $ zip xs ys; g z (x, y) = if x == y then z + 1 else 0
a = [lcstr (toBinary $ n) (toBinary $ n^2) | n <- [1..]]

Crossrefs

Cf. A276692

Sequence in context: A325224 A303389 A276166 * A133924 A023135 A345058

Adjacent sequences: A177059 A177060 A177061 * A177063 A177064 A177065

Keyword

base,easy,nonn

Author

Vladimir Reshetnikov, May 02 2010

Status

approved