A050430 - OEIS

A050430

Length of longest palindromic subword of (n base 2).

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

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OFFSET

1,3

COMMENTS

a(A083318(n-1)) = n; a(A193159(k)) = 3, 1 <= k <= 26. [Reinhard Zumkeller, Jul 17 2011]

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..16385 = 2^14 + 1

FORMULA

a(n) <= min(a(2*n), a(2*n+1)). [Reinhard Zumkeller, Jul 31 2011]

EXAMPLE

(11 base 2) = 1011, containing the palindrome 101, therefore a(11) = 3.

MAPLE

# A050430 Length of longest palindromic factor of n for n in [M1..M2] - from N. J. A. Sloane, Aug 07 2012, revised Aug 11 2012

isPal := proc(L)

local d ;

for d from 1 to nops(L)/2 do

if op(d, L) <> op(-d, L) then

return false;

end if;

end do:

return true;

end proc:

# start of main program

ans:=[];

M1:=0; M2:=64;

for n from M1 to M2 do

t1:=convert(n, base, 2);

rec:=0:

l1:=nops(t1);

for j1 from 0 to l1-1 do

for j2 from j1+1 to l1 do

F1 := [op(j1+1..j2, t1)];

if (isPal(F1) and j2-j1>rec) then rec:=j2-j1; fi;

od:

ans:=[op(ans), rec]:

od:

ans;

MATHEMATICA

f[n_] := Block[{id = IntegerDigits[n, 2]}, k = Length@ id; While[ Union[# == Reverse@# & /@ Partition[id, k, 1]][[-1]] != True, k--]; k]; Array[f, 105] (* Robert G. Wilson v, Jul 16 2011 *)

PROG

(Haskell)

import Data.Char (intToDigit, digitToInt)

import Numeric (showIntAtBase)

a050430 n = a050430_list !! (n-1)

a050430_list = f 1 where

f n = g (showIntAtBase 2 intToDigit n "") : f (n+1)

g zs | zs == reverse zs = length zs

| otherwise = max (h $ init zs) (h $ tail zs)

h zs@('0':_) = g zs

h zs@('1':_) = a050430 $ foldl (\v d -> digitToInt d + 2*v) 0 zs

-- Reinhard Zumkeller, Jul 16 2011