A050431 -id:A050431 - cp's OEIS frontend

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

Offset: 1

Clark Kimberling

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

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

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

Cf. A007088; A050431 (base 3), A050432 (base 4), A050433 (base 5).

Cf. A215244, A215467, A215256.

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

# 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:
od:
ans:=[op(ans),rec]:
od:
ans;

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 *)

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

Extended by Ray Chandler, Mar 11 2010

A050432 Length of longest palindromic subword of (n base 4).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 3, 1, 1, 2, 3, 2, 2, 1, 3, 2, 1, 1, 3, 1, 2, 2, 1, 3, 1, 1, 2, 3, 1, 2, 2, 3, 2, 1, 1, 3, 2, 2, 1, 1, 3, 1, 2, 1, 3, 1, 1, 2, 3, 2, 2, 2, 3, 3, 4, 2, 2, 3, 3, 3, 3, 3, 1, 2, 1, 3, 1, 1, 2, 2, 3, 2, 2, 3, 4, 3, 3, 2, 3, 2

Offset: 1

Clark Kimberling

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

Cf. A050430, A050431, A050433.

LPS:= proc(L) local nL,n,i;
  nL:= nops(L);
  for n from nL to 1 by -1 do
    for i from 1 to nL-n+1 do
      if L[i..i+n-1] = ListTools:-Reverse(L[i..i+n-1]) then return n fi
  od od:
end proc:
seq(LPS(convert(n,base,4)),n=1..100); # Robert Israel, Dec 17 2020

a(n) <= min(a(4*n+k): 0 <= k < 4). [Reinhard Zumkeller, Jul 31 2011]

A050433 Length of longest palindromic subword of (n base 5).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 3, 1, 1, 1, 2, 3, 2, 2, 2, 1, 3, 2, 1, 1, 1, 3, 1, 2, 1, 1, 3, 1, 1, 2, 2, 1, 3, 1, 1, 1, 2, 3, 1, 1, 2, 2, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 3, 1, 2, 2, 1, 1, 3, 1, 1, 2, 1, 3, 1, 1, 1, 2, 3, 1, 2

Offset: 1

Clark Kimberling

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

Cf. A050430, A050431, A050432

LPS:= proc(L) local nL,n,i;
  nL:= nops(L);
  for n from nL to 1 by -1 do
    for i from 1 to nL-n+1 do
      if L[i..i+n-1] = ListTools:-Reverse(L[i..i+n-1]) then return n fi
  od od:
end proc:
seq(LPS(convert(n,base,5)),n=1..100); # Robert Israel, Dec 17 2020

a(n) <= min(a(5*n+k): 0 <= k < 5). [Reinhard Zumkeller, Jul 31 2011]

cp's OEIS Frontend

A050430 Length of longest palindromic subword of (n base 2).

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A050432 Length of longest palindromic subword of (n base 4).

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A050433 Length of longest palindromic subword of (n base 5).

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